mm-3509 === Subject: Re: New Address for the Missouri State Problem Corner > We have a new URL: http://faculty.missouristate.edu/l/lesreid/POTW.html > (and a new set of problems). And you have no link to BrainBashers! === Subject: Re: Capacitors > Sorry if this is the wrong place to post this, but it doesn't quite > make sense to me (sadly enough) How is it that a capacitor rated at 300 V, such as that in a > disposable camera, gets charged to that voltage even though it only > has like a 1.5 V AA battery charging it? Does it just take longer? o.O > there is a small oscillator and small transformer that takes the 1.5 volt up to about 150 volts needed for the flash unit. done all the time in Flash units. Flash unit has a high trigger voltage too. === Subject: Re: Logical Formedness Axioms > http://mymath.blogspot.com/2005_05_01_archive.html > Logical Formedness Axioms 1. Identical sets are identical. All members of one set are included in the other set and no more. 2. Different sets are different. Different set means it is different than the one that you have. Therefore #2 only applies to two sets. you could have one set that is different than three others, but two of the others can exactly match. 3. Statements contradicting axioms 1 or 2 are false or malformed. I have Idential set A and A is different to set B, but B is also Identical to B, therefore this statement is false OR malformed. 4. A malformed statement is one for which a conclusion does not follow > given its structure. so the above#3 responce is structurly OK, and not malformed. 5. A false statement is one that while structurally correct is not true. so the above responce to #3 is not true, however it is true. So your axiom system brakes down, as it has proven NOT TRUE = TRUE The structure I mean has to do with how the sentence is put together. For instance, the following is a badly structured syllogism: If x=1, and y=2, then x=y. The basic structure for the syllogism--the well-formed structure--is, if a=b, and b=c, then a=c, and variations on that structure are to be considered malformed. Notice that with the first given that x does not equal y, the conclusion > is false, but the entire statement is malformed so that is secondary. With the basic axioms established--and notice how simple they are--it's > trivial to handle supposed paradoxes which reduce to attacking one of the > first two axioms. For instance, the so-called Russell Paradox reduces to the assertion that > a set includes and excludes itself. Let A be a set that includes itself, and let B be a set that excludes > itself. B is different from A. Therefore, by axiom 2 any statement that B is A is malformed or false. Stating that B is A and B is different from A is structurally wrong, so > the full statement is malformed. Notice also that the resolution to the supposed paradox is the well-formed > statement: Consider a set A that includes all and only sets, except itself, that > exclude themselves. Notice that the axioms prevent that set from including itself, as then you > reduce to a set that both includes and excludes itself, against axiom 2 as > I explained. > > === Subject: Regression/covariance I have some trouble understanding how the covariance matrix is computed, i.e. where the numbers really come from. To compute covariance between say a and b coeffs in a linear regression, you'd need a list of numbers, a[1],a[2],b[1], b[2] and so on. How do I compute these values so that I can numerically calculate the covariance? Any help would be appreciated. -Thomas === Subject: Re: How precise is JordanForm? > Does this imply that other programs like Mathematica, Matlab, > MuPAD,... have no reliable way to calculate Jordan normal forms? MuPAD gives the following result array(1..4, 1..4, (1, 1) = 2.019218126e-14 + 3.50531236e-14 I, (1, 2) = 5.181029564e-11 + 1.265992693e-11 I, (1, 3) = 2.978268673e-11 + 7.299471791e-12 I, (1, 4) = - 7.674763602e-14 - 3.391254291e-14 I, (2, 1) = - 5.967448757e-16 - 2.181761716e-14 I, (2, 2) = 4.393933134e-13 + 2.02361114e-11 I, (2, 3) = 2.403910404e-13 + 1.161504139e-11 I, (2, 4) = - 4.069661275e-15 - 1.414797099e-14 I, (3, 1) = - 4.7808979e-15 + 2.551778233e-14 I, (3, 2) = - 8.204201207e-13 - 4.161151979e-11 I, (3, 3) = - 4.527003772e-13 - 2.389237853e-11 I, (3, 4) = 9.388323452e-15 + 3.799825038e-14 I, (4, 1) = 4.898859096e-14 + 9.582439009e-14 I, (4, 2) = 1.437301667e-12 - 2.439451419e-11 I, (4, 3) = 8.568146193e-13 - 1.394149379e-11 I, (4, 4) = - 5.100087019e-15 - 1.686151219e-14 I ) when using the dotted entries. Is that close enough to zero? (I did not change the precision). It has been working for more than 10 minutes so far on the undotted version. Ralf === Subject: cancel: Elvis Presley on a magic square Control: cancel <45c1de42$0$2019$ba620dc5@text.nova.planet.nl> Excessive crossposting. (<45c1de42$0$2019$ba620dc5@text.nova.planet.nl>) === Subject: Converge or Diverge sigma[(tan n)/(2^n)] (infinite sum) HELP!! === Subject: Re: Converge or Diverge <20070201110030.323$bZ@newsreader.com sigma[(tan n)/(2^n)] > (infinite sum) I merely conjecture that the sum (starting with n = 0) converges > to 0.0468... Perhaps Robert Israel or someone else could give us an idea if a proof > seems feasible. By a result of Mahler, we know there is a positive constant p such that |pi - m/n| > 1/n^p for all positive integers m,n. The current best p, |tan n| < c n^(p-1) for some constant c, and therefore the sum does converge. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Converge or Diverge > sigma[(tan n)/(2^n)] > (infinite sum) HELP!! -1/2^n <= tan n/2^n <= 1/2^n === Subject: Proving a^2 + b^2 is not a perfect square Hi I saw a somewhat complicated proof of the fact that, if a and b are positive odd numbers, then a^2 + b^2 is not a perfect square. Maybe I'm making a mistake, but I think the proof is simple: If a^2 + b^2 is a perfect square, then there's a positive integer m such that a^2 + b^2 = m^2. Since a and b are odd, so are a^2 and b^2, and, therefore, m^2, and m, must be even. So, m = 2n for some positive integer n, and this gives a^2 + b^2 = 4 n^2 Since a and b are odd, we have a^2 = 1 (mod 8) and b^2 = 1(mod 8), so that a^2 + b^2 = 2 (mod 8). This implies that a^2 + b^2 = 8k + 2 for some postive integer k. Therefore, 8k + 2 = 4 n^2 => 4k +1 = 2 n^2. Since 4k + 1 is odd and 2 n^2 is even, we have reached a contradiction, which shows a^2 + b^2 is not a perfect square. Is there any flaw in this proof? Sharon === Subject: Re: Proving a^2 + b^2 is not a perfect square > Hi I saw a somewhat complicated proof of the fact that, if a and b are > positive odd numbers, then a^2 + b^2 is not a perfect square. Maybe > I'm making a mistake, but I think the proof is simple: If a^2 + b^2 is a perfect square, then there's a positive integer m > such that a^2 + b^2 = m^2. Since a and b are odd, so are a^2 and b^2, > and, therefore, m^2, and m, must be even. So, m = 2n for some positive > integer n, and this gives a^2 + b^2 = 4 n^2 Since a and b are odd, we have a^2 = 1 (mod 8) and b^2 = 1(mod 8), so > that a^2 + b^2 = 2 (mod 8). This implies that a^2 + b^2 = 8k + 2 for > some postive integer k. Therefore, 8k + 2 = 4 n^2 => 4k +1 = 2 n^2. Since 4k + 1 is odd and 2 n^2 is > even, we have reached a contradiction, which shows a^2 + b^2 is not a > perfect square. Is there any flaw in this proof? Sharon As William showed, there is a simpler proof, but I don't see anything wrong with your original one either. If you are doing this for a class, your instructor may want you to show that for odd a, a^2 = 1 (mod 8), but this is not difficult (write a=2a'+1 ...) Also, your last statement can be shortened: 8k+2 is not a multiple of 4, but 4n^2 is. (same arguement as you did, just without dividing by 2 first) === Subject: Re: Proving a^2 + b^2 is not a perfect square I saw a somewhat complicated proof of the fact that, if a and b are > positive odd numbers, then a^2 + b^2 is not a perfect square. Maybe > I'm making a mistake, but I think the proof is simple: Complicated? Nay, it is simple. > If c is even, then c^2 = 0 (mod 4) > If c is odd, then c^2 = 1 (mod 4) Since a and b are odd, a^2 + b^2 = 2 (mod 4) > which from the above cannot be a square. If a^2 + b^2 is a perfect square, then there's a positive integer m > such that a^2 + b^2 = m^2. Since a and b are odd, so are a^2 and b^2, > and, therefore, m^2, and m, must be even. So, m = 2n for some positive > integer n, and this gives a^2 + b^2 = 4 n^2 This IS complicated. Since a and b are odd, we have a^2 = 1 (mod 8) and b^2 = 1(mod 8), so > that a^2 + b^2 = 2 (mod 8). This implies that a^2 + b^2 = 8k + 2 for > some postive integer k. Therefore, 8k + 2 = 4 n^2 => 4k +1 = 2 n^2. Since 4k + 1 is odd and 2 n^2 is > even, we have reached a contradiction, which shows a^2 + b^2 is not a > perfect square. Is there any flaw in this proof? Too complicated to read through. Yes, your proof is kinda simple. Sharon === Subject: Re: frequency analysis of long transient signal your any possible help. I have a long time transient signal, i.e. it takes a quite a long time > to decay to zero. My aim is to get the frequency spectrum of this > signal as accurately as possible. But now I can only obtain a limited > period of this signal from the start time. And all the material is > only this limited period of signal. So is there any advanced technique which can help me to work out the > frequency spectrum of the whole signal to a relatively high precision > by only analyzing this short period signal? I totally have no idea, so any help or any tip is appreciated. No. By definition a transient is a signal that is not part of the long term > behavor(which is usually periodic or quasi-periodic). If you sample only the first 10 seconds then you have no idea what happens > in the next 10 and the frequency spectrum could be totally different. I think this sorta might have to do with causality too if I understand your > problem. Suppose your tansient is a sinc function. Then your frequency > spectrum is infinite and no matter how long you sample you'll never get it > all. So what you really need to do is look at your problem more and find some > criteria that can help you reduce your problem to something thats workable. I might not be understand what you want though. But suppose we have a a > signal we are trying to measure such as an impulse that might happen at any > time between 0 and 10 seconds. If we just look at the first 5 seconds then > there is no way we can know when the impulse occurs if it doesn't occur in > the first 5 seconds. In this case though we do know the spectrum because we > know the signal. In general though its impossible if you don't have some > idea what is going to happen. By your logic we could just sample for 1 second and be able to know the > spectrum of all future events no matter what happens.- Hide quoted text - - Show quoted text - But why wavelet can use a period of signal to calculate the spectrum? === Subject: Integral Involving Bessel Function and Sqrt I was wondering if any one know an analytic solution to the following integral. I'm using Mathematica notation. G[alph_,bet_]=Integrate[BesselK[0,Sqrt[bet^2+x^2]]*Exp[-alph*x],{x, 0,Infinity}] alph and bet >0 and real. Any ideas on how to get an explicit solution. === Subject: Re: Integral Involving Bessel Function and Sqrt I was wondering if any one know an analytic solution to the following > integral. I'm using Mathematica notation. G[alph_,bet_]=Integrate[BesselK[0,Sqrt[bet^2+x^2]]*Exp[-alph*x],{x, > 0,Infinity}] alph and bet >0 and real. Any ideas on how to get an explicit solution. Well, I quickly tried using equation (11) from http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html, and then interchanging the order of integration. The inner integral becomes: Integrate[Cos[Sqrt[bet^2 + x^2]*t] * Exp[-alph*x], {x, 0, Infinity}] So, essentially the bessel function is replaced by a trigonometric function, but it doesn't really seem to help. Maybe others can proceed from here. -Michael. === Subject: Re: Integral Involving Bessel Function and Sqrt <45c20eb2$0$61506$edfadb0f@dread12.news.tele.dk Integrate[Cos[Sqrt[bet^2 + x^2]*t] * Exp[-alph*x], {x, 0, Infinity}] What is t? === Subject: Re: Integral Involving Bessel Function and Sqrt > Integrate[Cos[Sqrt[bet^2 + x^2]*t] * Exp[-alph*x], {x, 0, Infinity}] What is t? 't' is the integration variable in the outer integral. It comes from the integral representation of the Bessel function. Look at the link I gave in the previous post. -Michael. === Subject: Integrating Show that integral{0~inf} [sin(x^2) dx] = sqrt(pi/8) HELP!!!! === Subject: Re: Integrating >> Show that >> integral{0~inf} [sin(x^2) dx] = sqrt(pi/8) > hmm.. do you really think this is true? What must be a necessary condition > for int(f(x),x=0..oo) to exist? > Are you thinking of something like int(|f(x)|,x=a..b)->0 for a,b > k->oo ? I don't think that is violated here, since x^2 has a squeezing effect for the sine which results in decreasing areas under the curvefor large x. === Subject: Re: Integrating What have you tried so far? > Are there any hints, are you eg. assumed to use Gauss' residue > theorem? > Alois This is not homework, but a friend asked me. I've tried just expand the sin function... and I can't find the Gauss' residue thm by googling. Can you link it? === Subject: Re: Odds I was originally thinking that A/B, C/D and E/F would each be likely per pair. However, you have piqued my interest here. How would one calculate the likelihood of F being true if... A or B (even) if B then C or D (even) if D then E or F (100:1) or A or B (even) if B then C or D (60:40) if D then E or F (100:1) Simon > How do I calculate and express the odds of F being true given: A or B > if B then > C or D > if D then > E or F If I take F as 1 out of 6 I get .16 But if I take B as .5 chance and then D as .5 of the .5 that is .25 > and then F as .5 of the .25 that is .125. So, is it .16 or .125 or something else again? Are A and B equally likely? Given B, are C and D equally likely? Given > D, are E and F equally likely? If the answers are all yes, then the > probability of F is 0.125. If the answers are not all yes, then you need > to specify the probabilities. -- > === Subject: Re: Odds per pair. However, you have piqued my interest here. How would one calculate the likelihood of F being true if... A or B (even) > if B then > C or D (even) > if D then > E or F (100:1) Let P(X|Y) mean the probability of X given Y. Write ^ for and. Then P(X|Y)P(Y) = P(X^Y). So, P(F^D^B) = P(F|D^B)P(D^B) > = P(F|D^B)P(D|B)P(B) In your case P(B) = 0.5. P(D|B) = 0.5. I'm not exactly sure what you > mean by 100:1, but perhaps you mean P(F|D^B) = 1/101. A or B (even) > if B then > C or D (60:40) > if D then > E or F (100:1) -- > By 100:1 I meant that e was 100 times more likely than F (odds against F being true when either it or E would be 100 to 1). Sorry, I have no familiarity with the correct terms to use. === Subject: Re: writing a proof > No. The null string is not an implicit member of the sets. It's > neither in S nor in T, which are sets of two strings each, none of > which is e. The correct answer is that ab is the concatenation of a string in S > (namely, a) and a string in T (namely, b). > I'm sorry I overlooked that. I missed seeing the second element of S. :-p > Now let U = { b, e }, where e stands for the empty string. Why is > ab in S U? There are two correct answers. >> Similar explanation is what I can think of here as well. But, I >> can't guess what the other answer could be, since you say that there >> exists two solutions! This time your answer works, because ab is the concatenation of ab and > e, and ab in S and e in U. The other answer is the same as for S and T. We can split ab in two > different ways that work. (A third way doesn't work.) > That's clear now! > Can you now state when an arbitrary string w is a member of the > concatenated set of strings, say Q R? IMHO, I now look at defining w as say a set of characters such that each of the character is either a member of Q or R. -- _ _ _]{5pitph!r3}[_ _ _ __________________________________________________ ñI'm smart enough to know that I'm dumb.î - Richard P Feynman === Subject: A help with a sequence please Hi I'd like some hints about the following. I'm really stuck. Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) - x(n)) = 0. Then, does (x_n) need to be convergent? Sharon === Subject: Re: A help with a sequence please > Hi I'd like some hints about the following. I'm really stuck. Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x_n) need to be convergent? No. You know that the harmonic series diverges, yes? (sum_n 1/n = infty). So you can do something like this: x(1) = 1 x(2) = 1 + 1/2 x(3) = 1 + 1/2 + 1/3 ... Keep doing that until you reach an x(n) which is > 2 (that happens when n = 4). Now proceed to keep SUBTRACTING until the sum is < -2. That is, x(5) = 1 + 1/2 + 1/3 + 1/4 - 1/5, x(6) = 1 + 1/2 + 1/3 + 1/4 - 1/5 - 1/6, etc. Then start ADDING, etc., etc. Clearly the sequence {x(n)} is bounded (from just a little less than -2 to a little more than 2), and x(n) - x(n+1) --> 0, but the set of subsequential limit points is [-2,2]. You might try proving the following theorem: if {x(n)} is a sequence in a COMPACT metric space such that d(x(n), x(n+1)) --> 0, then the set of subsequential limits is connected. This is useful when studying solving equations, for example. Suppose you can prove that the solutions of f(x) = 0 are discrete, i.e. isolated. And suppose you have a solution method which generates a sequence {x(n)} whose subsequential limits are known to be solutions of f(x) = 0. Then {x(n)} must converge to a solution of f(x) = 0. -- Ron Bruck === Subject: Re: A help with a sequence please <010220071626523372%bruck@math.usc.edu > You might try proving the following theorem: if {x(n)} is a sequence > in a COMPACT metric space such that d(x(n), x(n+1)) --> 0, then the set > of subsequential limits is connected. This is useful when studying > solving equations, for example. Suppose you can prove that the > solutions of f(x) = 0 are discrete, i.e. isolated. And suppose you > have a solution method which generates a sequence {x(n)} whose > subsequential limits are known to be solutions of f(x) = 0. Then > {x(n)} must converge to a solution of f(x) = 0. -- > Ron Bruck I tried to prove this, but couldn't finish my proof. Let X be a compact metric space and let (x(n)) be a sequence in X with the given property. Let S be the set of subsequential limits of X. We know S is closed, which implies (being a closed subset of a compact metric space) it is compact. Suppose A and B form a disconnection of S. Then A and B are disjoint nonempty sets, closed with respect to S, whose union is S. This implies A and B are compact subsets of X and, since they are also disjoint, the distance L = d(A,B) = infimum {d(a,b) | a is in A , b is in B} >0. But since x(n) doesn't need to be Cauchy, this is no contradiction. Could you give a hint, please? Artur Artur === Subject: Re: A help with a sequence please You might try proving the following theorem: if {x(n)} is a sequence > in a COMPACT metric space such that d(x(n), x(n+1)) --> 0, then the set > of subsequential limits is connected. This is useful when studying > solving equations, for example. Suppose you can prove that the > solutions of f(x) = 0 are discrete, i.e. isolated. And suppose you > have a solution method which generates a sequence {x(n)} whose > subsequential limits are known to be solutions of f(x) = 0. Then > {x(n)} must converge to a solution of f(x) = 0. -- > Ron Bruck I tried to prove this, but couldn't finish my proof. Let X be a > compact metric space and let (x(n)) be a sequence in X with the given > property. Let S be the set of subsequential limits of X. We know S is > closed, which implies (being a closed subset of a compact metric > space) it is compact. Suppose A and B form a disconnection of S. Then > A and B are disjoint nonempty sets, closed with respect to S, whose > union is S. This implies A and B are compact subsets of X and, since > they are also disjoint, the distance L = d(A,B) = infimum {d(a,b) | > a is in A , b is in B} >0. But since x(n) doesn't need to be Cauchy, > this is no contradiction. Could you give a hint, please? Because d(x(n), x(n+1)) -> 0, d(A,B) > 0 implies the tail end of the sequence lies in one of A, B. === Subject: Re: A help with a sequence please > I tried to prove this, but couldn't finish my proof. Let X be a > compact metric space and let (x(n)) be a sequence in X with the given > property. Let S be the set of subsequential limits of X. We know S is > closed, which implies (being a closed subset of a compact metric > space) it is compact. Suppose A and B form a disconnection of S. Then > A and B are disjoint nonempty sets, closed with respect to S, whose > union is S. This implies A and B are compact subsets of X and, since > they are also disjoint, the distance L = d(A,B) = infimum {d(a,b) | > a is in A , b is in B} >0. But since x(n) doesn't need to be Cauchy, > this is no contradiction. Could you give a hint, please? Because d(x(n), x(n+1)) -> 0, d(A,B) > 0 implies the tail end of > the sequence lies in one of A, B. I'm confused, the way Arthur put it, A and B contains only > subsequential limits, not actual terms of the sequence. > Sharon Good catch, actually I was the one confused, although it's sort of the right idea. To fix it, let's enlarge A and B a bit. We can choose open sets V and W in X containing A and B resp. such that d(V,W) > 0. Then K = X (V U W) is compact. Now x_n is in K only for finitely many n; otherwise, by the compactness of K, you'd have a subsequence of x_n converging to a point in K, contradiction (since all the subsequential limits of x_n are in A U B.) So now I think it's right to claim the tail end of the sequence lies in one of V, W, contradiction. === Subject: Re: A help with a sequence please > Hi I'd like some hints about the following. I'm really stuck. Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x_n) need to be convergent? Sharon x_n = sin(ln(n)) should be a counterexample === Subject: Re: A help with a sequence please Hi I'd like some hints about the following. I'm really stuck. Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x_n) need to be convergent? Sharon x_n = sin(ln(n)) should be a counterexample For every n, |sin(ln(n+1) - sin(ln(n)| = 2 |sin(ln(n+1) - ln(n)) cos(ln(n+1) + ln(n))| <= 2 | sin ln(1 + 1/n)| If n -> oo then the right hand side goes to 0, But though this seems intuitive, how can we prove this sequence diverges? Sharon === Subject: Re: A help with a sequence please On Feb 1, 7:55 pm, The World Wide Wade > Hi > I'd like some hints about the following. I'm really stuck. > Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x_n) need to be convergent? > Sharon > x_n = sin(ln(n)) should be a counterexample For every n, |sin(ln(n+1) - sin(ln(n)| = 2 |sin(ln(n+1) - ln(n)) > cos(ln(n+1) + ln(n))| <= 2 | sin ln(1 + 1/n)| If n -> oo then the > right hand side goes to 0, But though this seems intuitive, how can > we prove this sequence diverges? Imagine the points on the unit circle corresponding to ln(n) > radians, n = 1, 2, ... Think of an arc of length Pi/2 centered at > (0,1), and a similar one centered at (0,-1). This sequence must > visit each of these arcs infinitely many times. Can you see that? > If so, you'll see that sin(ln(n)) > 1/sqrt(2) for infinitely many > n, and sin(ln(n)) < -1/sqrt(2) for infinitely many n. Therefore > the sequence diverges. (In fact, it's much worse; this sequence > is dense in [-1,1].)- Hide quoted text - - Show quoted text - the sequence will also be dense if, instead of sine, we have a periodic continuous function, and, instead of ln, we have a function g that increases monotonically to oo and is such that lim (x -> oo) g(x +1) - g(x) =0. In this case, the sequence is dense in the range of g, which is a closed and bounded interval. Is that right? Sharon === Subject: Re: A help with a sequence please > On Feb 1, 7:55 pm, The World Wide Wade > Hi > I'd like some hints about the following. I'm really stuck. > Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x_n) need to be convergent? > Sharon > x_n = sin(ln(n)) should be a counterexample > For every n, |sin(ln(n+1) - sin(ln(n)| = 2 |sin(ln(n+1) - ln(n)) > cos(ln(n+1) + ln(n))| <= 2 | sin ln(1 + 1/n)| If n -> oo then the > right hand side goes to 0, But though this seems intuitive, how can > we prove this sequence diverges? Imagine the points on the unit circle corresponding to ln(n) > radians, n = 1, 2, ... Think of an arc of length Pi/2 centered at > (0,1), and a similar one centered at (0,-1). This sequence must > visit each of these arcs infinitely many times. Can you see that? > If so, you'll see that sin(ln(n)) > 1/sqrt(2) for infinitely many > n, and sin(ln(n)) < -1/sqrt(2) for infinitely many n. Therefore > the sequence diverges. (In fact, it's much worse; this sequence > is dense in [-1,1].)- Hide quoted text - - Show quoted text - the sequence will also be dense if, instead of sine, we have a > periodic continuous function, and, instead of ln, we have a function g > that increases monotonically to oo and is such that lim (x -> oo) g(x > +1) - g(x) =0. In this case, the sequence is dense in the range of g, > which is a closed and bounded interval. Is that right? I think you meant to say: If f is the periodic continuous function, then f(g(x)) will be dense in the range of f, which is a closed and bounded interval. === Subject: Re: A help with a sequence please Hi > I'd like some hints about the following. I'm really stuck. > Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x_n) need to be convergent? > Sharon > x_n = sin(ln(n)) should be a counterexample > For every n, |sin(ln(n+1) - sin(ln(n)| = 2 |sin(ln(n+1) - ln(n)) > cos(ln(n+1) + ln(n))| <= 2 | sin ln(1 + 1/n)| If n -> oo then the > right hand side goes to 0, But though this seems intuitive, how can > we prove this sequence diverges? Imagine the points on the unit circle corresponding to ln(n) > radians, n = 1, 2, ... Think of an arc of length Pi/2 centered at > (0,1), and a similar one centered at (0,-1). This sequence must > visit each of these arcs infinitely many times. Can you see that? I think the issue was that she *can* see it but didn't know how to write up an acceptable proof of it. Would it be just too anal to insist on one? Question of style. Clearly as you go round the circle, the unboundedness of ln means that you will eventually get to each of the arcs, each time you go around, and furthermore that you never stop going around. Also, convergence of the differences to zero means that you cannot overshoot an arc without hitting it at least once. But, writing that up more formally, in a clever enough way that it doesn't overwhelm the readability of the whole, seems like more of a chore than I'm prepared to do.... > If so, you'll see that sin(ln(n)) > 1/sqrt(2) for infinitely many > n, and sin(ln(n)) < -1/sqrt(2) for infinitely many n. Therefore > the sequence diverges. (In fact, it's much worse; this sequence > is dense in [-1,1].) === Subject: Re: A help with a sequence please > Hi > I'd like some hints about the following. I'm really stuck. > Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x_n) need to be convergent? > Sharon x_n = sin(ln(n)) should be a counterexample > For every n, |sin(ln(n+1) - sin(ln(n)| = 2 |sin(ln(n+1) - ln(n)) > cos(ln(n+1) + ln(n))| <= 2 | sin ln(1 + 1/n)| If n -> oo then the > right hand side goes to 0, But though this seems intuitive, how can > we prove this sequence diverges? Hmm, probably there is a more elegant way, but here's something at least: Let e = 1/10, and let m be any natural number. We know that pi/3 < ln(3) < pi/2 and from this we can show that either |a_n - a_3n| > e, or |a_n - a_6n| > e. This shows that there does *not* exist M such that for all n>M, |a_n - a_M| < e. Actually writing out a proof of the key detail (which I didn't) could be messy, but it's self evident if you look at the graph of sine. Perhaps the messiness would be reduced if I varied the approach slightly, e.g. picked a constant other than 3. === Subject: Re: A help with a sequence please Hi I'd like some hints about the following. I'm really stuck. Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) >- x(n)) = 0. Then, does (x_n) need to be convergent? I happen to know that you know of an infinite _series_ > such that the sequence of partial sums is a counterexample. Sharon ************************ David C. Ullrich Do you mean Sum (1/n)? But the sequence of partial sums goes to infinity, right? Sharon === Subject: Re: A help with a sequence please >>Hi >>I'd like some hints about the following. I'm really stuck. >>Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) >>- x(n)) = 0. Then, does (x_n) need to be convergent? >> I happen to know that you know of an infinite _series_ >> such that the sequence of partial sums is a counterexample. >>Sharon >> ************************ >> David C. Ullrich Do you mean Sum (1/n)? But the sequence of partial sums goes to >infinity, right? Yes - I missed the word bounded in your question, sorry. >Sharon ************************ David C. Ullrich === Subject: Re: A help with a sequence please On Feb 1, 4:23 pm, The World Wide Wade >Hi >I'd like some hints about the following. I'm really stuck. >Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) >- x(n)) = 0. Then, does (x_n) need to be convergent? > I happen to know that you know of an infinite _series_ > such that the sequence of partial sums is a counterexample. >Sharon > ************************ > David C. Ullrich Do you mean Sum (1/n)? But the sequence of partial sums goes to > infinity, right? Yes, which is why he mentioned it. What is x(n+1) - x(n)) for > this sequence?- Hide quoted text - - Show quoted text - Ok, for the sequence of partial sums, (x(n+1) - x(n) = 1/(n+1) -> 0 as n-> oo. But this is not a counter example, because x(n) -> oo and the statement asked for a bounded sequence. So, like Jose Carlos did, we had to find an example of a bounded and divergent sequence such that x(n+1) - x(n) -> 0. Maybe, like that other example, x(n) = sin(ln(n)) satisfies such conditions. We have lim x(n+1) - x(n) = 0, but though it seems intuitive, I don't how to prove this sequence diverges. Sharon === Subject: Re: A help with a sequence please > I'd like some hints about the following. I'm really stuck. Let (x_n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x_n) need to be convergent? No. Consider the sequence: 0, 1, 1/2, 0, 1/3, 2/3, 1, 3/4, 1/2, 1/4, 0, 1/5, ... Jose Carlos Santos === Subject: Re: A help with a sequence please <52edblF1nquq3U1@mid.individual.net > I'd like some hints about the following. I'm really stuck. Let (x n) be a bounded sequence of real numbers such that lim (x(n+1) > - x(n)) = 0. Then, does (x n) need to be convergent? No. Consider the sequence: 0, 1, 1/2, 0, 1/3, 2/3, 1, 3/4, 1/2, 1/4, 0, 1/5, ... > Jose Carlos Santos But I'm a bit confused, I didn't get the law of this sequence, how it's formed Sharon === Subject: Re: More primality testing (was: Elementary group theory: Proof of Fermat-Maas ...) Since your post is long enough and since Tim Peters already answered most parts (better than I could have done), I only answered your questions regarding ECPP. robert maas, see http://tinyurl.com/uh3t a .8ecrit : > An addendum to my earlier followup. You cited several large numbers > you believe to be prime, but which you can't prove prime, and > challeged me to try to prove them prime. I counter-challenge. Below > are several large numbers I've already proven to be prime. Can you > also prove them prime? Some are really easy to prove prime if you > know the trick, while others are much more difficult. >> Difficult with a N-1 method, not with ECPP (Elliptic Curve Primality >> Proving). Did you program the ECPP yourself, Yes. > or take somebody else's program > source and study it carefully to make sure you understand how it > works and trust it based on that understanding, The only published ECPP source I know is in the LiDIA library. But it is still too new to be fast (optimizing an ECPP implementation takes time). > or did you just > trust some canned program without actually checking what it does > inside? With ECPP, there is no need to check the program itself. ECPP has a redhibitory advantage over all of its competitors [*], it produces primality certificates. When running ECPP, if there is a hardware failure or a bug in the program, either the program stops because of the error or it produces an invalid certificate. In any case, we know that a problem occurred. One could almost say that ECPP doesn't answer the question 'Is this number prime?' but replaces it by a more easy one: 'Is this certificate valid?' [*] i.e., over APR-CL. No, it has no other one. AKS is a theoretical revolution but, practically, to prove primality it is about as useful as the Wilson theorem. (I am making a lot of friends there :-) ) >[...] > Is the Atkin-Goldwasser-kilian-morain certificate really faster > than the Pratt/Maas certificate if factorization of p-1 is > recursively known a priori due to direct construction of (p-1)/2 > as a product already, for very large primes? No. And, by using the Pocklington theorem, the difference would still be greater. An ECPP certificate contains steps with N-1 and N+1 (Lucas) tests. They are used when this is possible precisely because they are faster than a EC test. > I did a little experiment here just now. The following rather large > numbers are proven primes (by Pratt/Maas algorithm, formerly called > Fermat-Maas algorithm): p1=9546221476226082289729570071623288138324497060784770277001712162915439737 6 1033616918333544268480756481510147656340593480746164697244619229336689663500 8 03 p2=2384642503849816800941343711487142824042270280685640271565002482800548082 3 9211936121703390658836551417208081166489028260466711956419068655903034092225 3 7205536939245274685830819751464175827705089640690181338163281019281134304663 6 4564150323928621121292022971736935858447273590923968941233041233067058563045 8 6616598339768773337899176643 Constructing a certificate for each requires three steps: Run the > algorithm to prove it's really prime (which doesn't take long). > Convert the proof to a single base that is a primitive element > (which doesn't take long either). Issue a certificate stating the > base (virtually instant, limited by I/O rate rather than any > computing that is needed). (And do same recursively for factors of > p-1 if they are large enough to not be apparently prime already.) > Let me time my method: (setq p1 9546221476226082289729570071623288138324497060784770277001712162915439737610 3 3616918333544268480756481510147656340593480746164697244619229336689663500803 ) > (setq p2 2384642503849816800941343711487142824042270280685640271565002482800548082392 1 1936121703390658836551417208081166489028260466711956419068655903034092225372 0 5536939245274685830819751464175827705089640690181338163281019281134304663645 6 4150323928621121292022971736935858447273590923968941233041233067058563045866 1 6598339768773337899176643) > [...] > (72 517) > That's just over 0.7 second to prove&make the certificate for the > 153-digit prime, and just over 5 seconds to prove&make the > certificate for the 333-digit prime. > (list (- 13310 13283) (- 13924 13827)) > (27 97) > That's just over a quarter-second to verify the certificate for the > 153-digit prime, and just under 1 second to verify the certificate > for the 333-digit prime. And my code isn't even compiled (except by incremental compiler > built into CMUCL interpretor). How long does it take to generate, and to verify, an > Atkin-Goldwasser-kilian-morain certificate for those same primes? Certifying p1 -> 0.32s Verifying the certificate -> 0.10s Certifying p2 -> 10.24s Verifying the certificate -> 1.32s The 0.10s is not meaningful. Any running time below 0.10s is rounded to 0.10s. > Here's another example: > p=74098166510356888853578081205621951902040062914242490543288006521731051445 9 2546051523427394374806971810900275606483718086800588674400571045916946512639 4 8292601070436548121712978624275175836999294563535821143089523835255046905063 0 661 is prime. > [...] > (203 91) > That's just for the top level of the certificate, 2 seconds to > prove and issue, 1 second to verify, for a 232-digit prime. > It'd take too long to manually count up all the times for recursive > certification, but maybe I'll write the code someday soon. > So what's time to prove&issue, and to just verify, an > Atkin-Goldwasser-kilian-morain certificate for that prime? Certifying p -> 1.73s Verifying the certificate -> 0.46s > Here's another example: > p=19672623156857094972133172659022419078939060026617767829942707264261372108 0 2339504799028792336893901860665434931426221769816416700666235981397145023810 6 0049021799452707325155436356897242866104261954226258240316459896534018619093 7 9797410744021474238350587924893282639226806866312333991764376612244405086003 1 6102692803798649168692368815534817924585809581368627097601903741841699429716 2 4341851844918266906364398228743265911416042869402727140024390711019593139839 9 3348023153 is prime. > [...] > (1484 268) > That's nearly 15 seconds to prove primality and generate primitive > element, and just over 2.5 seconds to verify the certificate, > toplevel only again, for 470-digit prime. The next level down would > be each similar to the 232-digit prime earlier, 2 of them, i.e. 2 * > (2secgen 1secver), total for top two levels 19 sec generate and 4.5 > second verify. Third level would be 4 copies of the timing for a > 100-digit prime, which is very little time. Grand total probably > about 20 seconds to prove primality and generate recursive > certificate, 5 seconds to verify all levels. How does timing of > Atkin-Goldwasser-kilian-morain certificate for that 470-digit prime > compare, for prove&generate, and for verify, separately? Certifying p -> 27.34s Verifying the certificate -> 3.87s mm === Subject: Re: More primality testing (was: Elementary group theory: Proof of Fermat-Maas ...) > We learned in grade school that anulling the sale of a slave > because he was sickly was redhibition. Well, that explains my problem - by the time I got to grade school, slavery had already been outlawed. -- === Subject: Re: More primality testing (was: Elementary group theory: Proof of Fermat-Maas ...) >> We learned in grade school that anulling the sale of a slave >> because he was sickly was redhibition. Well, that explains my problem - by the time I got to grade school, > slavery had already been outlawed. They didn't have Taco Bell where you lived?? -- The man without a .sig === Subject: Re: More primality testing (was: Elementary group theory: Proof of Fermat-Maas ...) Gerry Myerson a .8ecrit : > >> We learned in grade school that anulling the sale of a slave >> because he was sickly was redhibition. Well, that explains my problem - by the time I got to grade school, > slavery had already been outlawed. > What's all that noise around the word redhibitory? I am French and I know the expression redhibitory defect [*] but you, who are English speaking, you never saw it? [*] And my English is far from being good. I go on to learn every day. For instance, today I learned that we learned in grade school is synonymous with I just saw on a wiki page. mm === Subject: Re: More primality testing (was: Elementary group theory: Proof of Fermat-Maas ...) > What's all that noise around the word redhibitory? I am French and > I know the expression redhibitory defect [*] but you, who are English > speaking, you never saw it? Such things are not uncommon. When I was learning German in college we had to memorize all these $50 words so we could discuss relevant and timely issues. So I knew the word for union political stuff (6 syllables, I think) but not the word for raisin. And in a related, but exactly opposite sense: My French office mate in grad school was surprised when I made a pun Au reservoir. He wanted to know how I (knowing no French) knew that reservoir was a French word. He was surprised to find out that it was also an English word. No doubt there is some French concept for which we don't have a decent English word. So you, being French, had a French thought and needed a word that we never use, because we never have such thoughts....? Bart -- The man without a .sig === Subject: Re: More primality testing (was: Elementary group theory: Proof of Fermat-Maas ...) Gerry Myerson a .8ecrit : > >> ECPP has a redhibitory advantage over all of its competitors I can usually figure out what people really meant to type, > but redhibitory has me stumped. And what about oxymoron? mm P.S. I could have written Compared to ECPP, its competitors have a redhibitory defect but I thought that, expressed this way, someone could have regarded it as a declaration of war :-) === Subject: Re: More primality testing (was: Elementary group theory: Proof of Fermat-Maas ...) Phil Carmody a .8ecrit : >> The only published ECPP source I know is in the LiDIA library. But it >> is still too new to be fast (optimizing an ECPP implementation takes >> time). Francois' ECPP (not the recent distributable FastECPP) was released > as source code. And (I nearly missed it) no, the current version he is distributing is not FastECPP but always a version published in 2001. mm === Subject: Re: More primality testing (was: Elementary group theory: Proof of Fermat-Maas ...) Phil Carmody a .8ecrit : >> The only published ECPP source I know is in the LiDIA library. But it >> is still too new to be fast (optimizing an ECPP implementation takes >> time). Francois' ECPP (not the recent distributable FastECPP) was released > as source code. Yes, but I believe he stopped the source distribution in 1993-1994 (?) mm === Subject: topological proper closed Let X, Y be two topological spaces and f: X --> Y be a continuous map. Suppose that the subset f(X) of Y intersects every compact set in a compact subset. May you help me to show that f(X) is closed, please? === Subject: Re: topological proper closed Let X, Y be two topological spaces and f: X --> Y be a continuous map. Suppose that the subset f(X) of Y intersects every compact set in a >compact subset. May you help me to show that f(X) is closed, please? Are you sure you've stated the question correctly? The question you're asking really has nothing > at all to do with f - it's the same as this > question: Suppose that Y is a topological space, A is > a subset of Y, and A intersects every compact > subset of Y in a compact set. Show that A is closed. ************************ David C. Ullrich I was trying to understand the following: http://i7.tinypic.com/2eq5n5j.jpg in which I do not understand the following two things: (i) What is the reason for the intersection is compact? I know that -a priori- the intersection of two compact sets need not to be compact itself. (ii) What is the reason for f(X) is closed? === Subject: Re: topological proper closed Let X, Y be two topological spaces and f: X --> Y be a continuous map. >Suppose that the subset f(X) of Y intersects every compact set in a >compact subset. >May you help me to show that f(X) is closed, please? Are you sure you've stated the question correctly? The question you're asking really has nothing > at all to do with f - it's the same as this > question: Suppose that Y is a topological space, A is > a subset of Y, and A intersects every compact > subset of Y in a compact set. Show that A is closed. Right, so now, getting back to the main point, how do you prove this assuming Y is Hausdorff? Here's my try; I confess it took me awhile to arrive at it so I was wondering if there's a tidier or more obvious approach: Suppose p is a limit point of A that is not in A. Then (using the Hausdorff condition) there is an infinite sequence of distinct points in A converging to p. The union of this sequence with {p} is a compact subset of Y, since any neighborhood of p is disjoint from at most a finite number of sequence points. And yet, the sequence itself is not compact. (We can enclose each point in a neighborhood distinct from all the other neighborhoods, and no finite subset of these covers the sequence.) Thus, the intersection of A with a compact set is noncompact, a contradiction. It follows that A contains all of its limit points and is therefore closed, QED. It's easy to see the proposition is false for non-T1 spaces -- just pick A to be some point that is not closed. Not sure how to find a good T1, non-T2 counterexample. ************************ David C. Ullrich I was trying to understand the following: http://i7.tinypic.com/2eq5n5j.jpg The other link that you posted has (on p.5) a statement that the author will henceforth assume all spaces to be Hausdorff unless explicitly stated otherwise. The above link is from p. 28. in which I do not understand the following two things: > (i) What is the reason for the intersection is compact? I know that -a > priori- the intersection of two compact sets need not to be compact > itself. But in a Hausdorff space it is. This follows directly from the basic theorems that any compact subset of a Hausdorff space is closed, and any closed subset of a compact space is compact. (ii) What is the reason for f(X) is closed? See above. > - Show quoted text - === Subject: topological proper closed === Subject: Re: topological proper closed Suppose that Y is a topological space, A is > a subset of Y, and A intersects every compact > subset of Y in a compact set. Show that A is closed. > Right, so now, getting back to the main point, how do you prove > this assuming Y is Hausdorff? > Here's my try; I confess it took me awhile to arrive at it so > I was wondering if there's a tidier or more obvious approach: > Suppose p is a limit point of A that is not in A. Then (using the > Hausdorff condition) there is an infinite sequence of distinct > points in A converging to p. Hm, you're assuming the space is first countable. Let Y = omega_1 + 1 and A = Yomega_1. omega_1 is a limit point of A and no sequence of points within A will converge to omega_1. ---- === Subject: Re: topological proper closed Suppose that Y is a topological space, A is > a subset of Y, and A intersects every compact > subset of Y in a compact set. Show that A is closed. Right, so now, getting back to the main point, how do you prove > this assuming Y is Hausdorff? Here's my try; I confess it took me awhile to arrive at it so > I was wondering if there's a tidier or more obvious approach: Suppose p is a limit point of A that is not in A. Then (using the > Hausdorff condition) there is an infinite sequence of distinct > points in A converging to p. Hm, you're assuming the space is first countable. > Let Y = omega_1 + 1 and A = Yomega_1. > omega_1 is a limit point of A and no sequence of points within A > will converge to omega_1. > You're approach will work to show if space is 1st countable and for all compact K, A / K is closed, then A is closed. and for all compact K, A / K is compact, then A is closed. === Subject: Re: topological proper closed Suppose that Y is a topological space, A is > a subset of Y, and A intersects every compact > subset of Y in a compact set. Show that A is closed. > Right, so now, getting back to the main point, how do you prove > this assuming Y is Hausdorff? > Suppose p is a limit point of A that is not in A. Then (using the > Hausdorff condition) there is an infinite sequence of distinct > points in A converging to p. Hm, you're assuming the space is first countable. > Let Y = omega_1 + 1 and A = Yomega_1. > omega_1 is a limit point of A and no sequence of points within A > will converge to omega_1. You're approach will work to show if space is 1st countable > and for all compact K, A / K is closed, then A is closed. > If the space is not 1st countable, then the same approach will work using nets instead of sequences, to show If for all compact K, A / K is closed, then A is closed. Hence for Hausdorff spaces if for all compact K, A / K is compact, then A is closed. The requirment for Hausdorff space may be slightly weaked to kc-space, ie a space for which all compact sets are closed. > and for all compact K, A / K is compact, then A is closed. > === Subject: Re: topological proper closed Retraction. The lines marked below from //** through ** may be vaporized, shredded or vanquished and vanished as faultly intelligence. > Suppose that Y is a topological space, A is > a subset of Y, and A intersects every compact > subset of Y in a compact set. Show that A is closed. > Right, so now, getting back to the main point, how do you prove > this assuming Y is Hausdorff? > Suppose p is a limit point of A that is not in A. Then (using the > Hausdorff condition) there is an infinite sequence of distinct > points in A converging to p. > Hm, you're assuming the space is first countable. > Let Y = omega_1 + 1 and A = Yomega_1. > omega_1 is a limit point of A and no sequence of points within A > will converge to omega_1. > You're approach will work to show if space is 1st countable > and for all compact K, A / K is closed, then A is closed. //** > If the space is not 1st countable, then the same approach will work using > nets instead of sequences, to show If for all compact K, A / K is closed, then A is closed. > Hence for Hausdorff spaces > if for all compact K, A / K is compact, then A is closed. The requirment for Hausdorff space may be slightly weaked to > kc-space, ie a space for which all compact sets are closed. > ** > and for all compact K, A / K is compact, then A is closed. === Subject: Re: topological proper closed >>Let X, Y be two topological spaces and f: X --> Y be a continuous map. >>Suppose that the subset f(X) of Y intersects every compact set in a >>compact subset. >>May you help me to show that f(X) is closed, please? >> Are you sure you've stated the question correctly? >> The question you're asking really has nothing >> at all to do with f - it's the same as this >> question: >> Suppose that Y is a topological space, A is >> a subset of Y, and A intersects every compact >> subset of Y in a compact set. Show that A is closed. >> ************************ >> David C. Ullrich I was trying to understand the following: http://i7.tinypic.com/2eq5n5j.jpg in which I do not understand the following two things: >(i) What is the reason for the intersection is compact? I know that -a >priori- the intersection of two compact sets need not to be compact >itself. What??? The intersection of two compact sets need not be compact??????? Again: You really need to learn a lot of background material much more thoroughly, then come back to whatever it is you're trying to study right now. >(ii) What is the reason for f(X) is closed? > ************************ David C. Ullrich === Subject: Re: topological proper closed Let X, Y be two topological spaces and f: X --> Y be a continuous map. >Suppose that the subset f(X) of Y intersects every compact set in a >>compact subset. >May you help me to show that f(X) is closed, please? > Are you sure you've stated the question correctly? > The question you're asking really has nothing >> at all to do with f - it's the same as this >> question: > Suppose that Y is a topological space, A is >> a subset of Y, and A intersects every compact >> subset of Y in a compact set. Show that A is closed. > ************************ > David C. Ullrich I was trying to understand the following: http://i7.tinypic.com/2eq5n5j.jpg in which I do not understand the following two things: >(i) What is the reason for the intersection is compact? I know that -a >priori- the intersection of two compact sets need not to be compact >itself. What??? The intersection of two compact sets need not be > compact??????? > Look at here, please: http://i5.tinypic.com/2jg9fex.jpg On the other hand, perhaps situation changes for metrizable spaces. (ii) What is the reason for f(X) is closed? > ************************ David C. Ullrich- Nascondi testo tra virgolette - - Mostra testo tra virgolette - === Subject: Re: topological proper closed >Let X, Y be two topological spaces and f: X --> Y be a continuous map. >Suppose that the subset f(X) of Y intersects every compact set in a >compact subset. >May you help me to show that f(X) is closed, please? > Are you sure you've stated the question correctly? > The question you're asking really has nothing > at all to do with f - it's the same as this > question: > Suppose that Y is a topological space, A is > a subset of Y, and A intersects every compact > subset of Y in a compact set. Show that A is closed. > ************************ > David C. Ullrich >>I was trying to understand the following: >>http://i7.tinypic.com/2eq5n5j.jpg >>in which I do not understand the following two things: >>(i) What is the reason for the intersection is compact? I know that -a >>priori- the intersection of two compact sets need not to be compact >>itself. >> What??? The intersection of two compact sets need not be >> compact??????? > >Look at here, please: http://i5.tinypic.com/2jg9fex.jpg On the other hand, perhaps situation changes for metrizable spaces. Again and again and again: You need to learn some preliminary material first, and also you need to learn to read more carefully. Right there in that jpg you posted it specifies that if we are to have two compact sets with non-compact intersection then the space must not be Hausdorff. All the spaces you've been dealing with are Hausdorff. (No, I'm not going to tell you what Hausdorff means. You can easily look this up in many books or even on the internet.) >>(ii) What is the reason for f(X) is closed? >> ************************ >> David C. Ullrich- Nascondi testo tra virgolette - >> - Mostra testo tra virgolette - > ************************ David C. Ullrich === Subject: Re: topological proper closed Philips a .8ecrit : > Let X, Y be two topological spaces and f: X --> Y be a continuous map. > Suppose that the subset f(X) of Y intersects every compact set in a > compact subset. > May you help me to show that f(X) is closed, please? >> Are you sure you've stated the question correctly? >> The question you're asking really has nothing >> at all to do with f - it's the same as this >> question: >> Suppose that Y is a topological space, A is >> a subset of Y, and A intersects every compact >> subset of Y in a compact set. Show that A is closed. >> ************************ >> David C. Ullrich I was trying to understand the following: http://i7.tinypic.com/2eq5n5j.jpg in which I do not understand the following two things: > (i) What is the reason for the intersection is compact? I know that -a > priori- the intersection of two compact sets need not to be compact > itself. What? Do you have a counter-example ? (ii) What is the reason for f(X) is closed? === Subject: Re: topological proper closed <45c2340e$0$21142$7a628cd7@news.club-internet.fr> On 1 Feb, 19:40, Denis Feldmann Philips a .8ecrit : > Let X, Y be two topological spaces and f: X --> Y be a continuous map. > Suppose that the subset f(X) of Y intersects every compact set in a > compact subset. > May you help me to show that f(X) is closed, please? >> Are you sure you've stated the question correctly? > The question you're asking really has nothing >> at all to do with f - it's the same as this >> question: > Suppose that Y is a topological space, A is >> a subset of Y, and A intersects every compact >> subset of Y in a compact set. Show that A is closed. > ************************ > David C. Ullrich I was trying to understand the following: http://i7.tinypic.com/2eq5n5j.jpg in which I do not understand the following two things: > (i) What is the reason for the intersection is compact? I know that -a > priori- the intersection of two compact sets need not to be compact > itself. What? Do you have a counter-example ? (ii) What is the reason for f(X) is closed? > - Mostra testo tra virgolette -- Nascondi testo tra virgolette - - Mostra testo tra virgolette - go to page 28 of: http://www.math.utah.edu/~bestvina/6510/notes.pdf you'll find http://i7.tinypic.com/2eq5n5j.jpg === Subject: Re: topological proper closed <45c2340e$0$21142$7a628cd7@news.club-internet.fr On 1 Feb, 19:40, Denis Feldmann Let X, Y be two topological spaces and f: X --> Y be a continuous map. > Suppose that the subset f(X) of Y intersects every compact set in a > compact subset. > May you help me to show that f(X) is closed, please? >> Are you sure you've stated the question correctly? >> The question you're asking really has nothing >> at all to do with f - it's the same as this >> question: >> Suppose that Y is a topological space, A is >> a subset of Y, and A intersects every compact >> subset of Y in a compact set. Show that A is closed. >> ************************ >> David C. Ullrich > I was trying to understand the following: >http://i7.tinypic.com/2eq5n5j.jpg > in which I do not understand the following two things: > (i) What is the reason for the intersection is compact? I know that -a > priori- the intersection of two compact sets need not to be compact > itself. What? Do you have a counter-example ? > (ii) What is the reason for f(X) is closed? > - Mostra testo tra virgolette -- Nascondi testo tra virgolette - - Mostra testo tra virgolette - go to page 28 of: http://www.math.utah.edu/~bestvina/6510/notes.pdf you'll find http://i7.tinypic.com/2eq5n5j.jpg that is not a counterexample to the intersection of two compact sets being compact in fact it uses the idea intersection preserves closedness boundedness ? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: topological proper closed <45c2340e$0$21142$7a628cd7@news.club-internet.fr On 1 Feb, 19:40, Denis Feldmann Philips a .8ecrit : > Let X, Y be two topological spaces and f: X --> Y be a continuous map. > Suppose that the subset f(X) of Y intersects every compact set in a > compact subset. > May you help me to show that f(X) is closed, please? >> Are you sure you've stated the question correctly? >> The question you're asking really has nothing >> at all to do with f - it's the same as this >> question: >> Suppose that Y is a topological space, A is >> a subset of Y, and A intersects every compact >> subset of Y in a compact set. Show that A is closed. >> ************************ >> David C. Ullrich > I was trying to understand the following: >http://i7.tinypic.com/2eq5n5j.jpg > in which I do not understand the following two things: > (i) What is the reason for the intersection is compact? I know that -a > priori- the intersection of two compact sets need not to be compact > itself. > What? Do you have a counter-example ? Does your definition of compact require the space to be Hausdorff? It would seem that the O.P. and his sources are not using the term that way. Elsewhere, in response to David C. Ullrich, he posted a link pointing out that counterexamples exist in R x {a,b} where the two-element set is in the indiscrete topology. For example, if we let A = [0,1] x {a} B = [0,1) x {a} union {1} x {b} then both A and B are compact in the less restricted sense of that term, yet their intersection is not. Of course this space is not Hausdorff (nor even T1). > (ii) What is the reason for f(X) is closed? > - Mostra testo tra virgolette -- Nascondi testo tra virgolette - > - Mostra testo tra virgolette - go to page 28 of: http://www.math.utah.edu/~bestvina/6510/notes.pdf you'll find http://i7.tinypic.com/2eq5n5j.jpg > that is not a counterexample to > the intersection of two compact sets being compact in fact it uses the idea intersection preserves > closedness > boundedness Boundedness can't be used if we are talking about general topological spaces. > ? === Subject: Re: topological proper closed <45c2340e$0$21142$7a628cd7@news.club-internet.fr > On 1 Feb, 19:40, Denis Feldmann Philips a .8ecrit : > Let X, Y be two topological spaces and f: X --> Y be a continuous map. > Suppose that the subset f(X) of Y intersects every compact set in a > compact subset. > May you help me to show that f(X) is closed, please? >> Are you sure you've stated the question correctly? >> The question you're asking really has nothing >> at all to do with f - it's the same as this >> question: >> Suppose that Y is a topological space, A is >> a subset of Y, and A intersects every compact >> subset of Y in a compact set. Show that A is closed. >> ************************ >> David C. Ullrich > I was trying to understand the following: >http://i7.tinypic.com/2eq5n5j.jpg > in which I do not understand the following two things: > (i) What is the reason for the intersection is compact? I know that -a > priori- the intersection of two compact sets need not to be compact > itself. > What? Do you have a counter-example ? Does your definition of compact require the space to be > Hausdorff? It would seem that the O.P. and his sources > are not using the term that way. Elsewhere, in response to David C. Ullrich, he posted a > link pointing out that counterexamples exist in R x {a,b} > where the two-element set is in the indiscrete topology. For example, if we let A = [0,1] x {a} > B = [0,1) x {a} union {1} x {b} then both A and B are compact in the less restricted > sense of that term, yet their intersection is not. Of course > this space is not Hausdorff (nor even T1). i see! another definition ambiguity... finite subcovers are not always available without some type of metrisability > (ii) What is the reason for f(X) is closed? > - Mostra testo tra virgolette -- Nascondi testo tra virgolette - > - Mostra testo tra virgolette - > go to page 28 of: >http://www.math.utah.edu/~bestvina/6510/notes.pdf > you'll find >http://i7.tinypic.com/2eq5n5j.jpg that is not a counterexample to > the intersection of two compact sets being compact in fact it uses the idea intersection preserves > closedness > boundedness Boundedness can't be used if we are talking about > general topological spaces. even my book on topological uniform structures makes the assumption (but proves, for its case) that compact is closed and bounded in the pseudometric but i understand this still isn't the most general and can see why the finite subcover definition is useful another day another assumption exposed the OP's quote appears to come from a section on proper maps of manifolds so the assumption appears valid here if that clears up anything... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: what's the condition for the product of first derivative to equate the second derivative? > I don't think this equation is always true, i.e. it's not a theorem. > Unless you mean to say: It obviously isn't (take f (x,y) = xy, then f_x * f_y = xy, but f_xy = 1). However, the OP asked for what condition(s) it IS true. I have no more to add than others have already, however. === Subject: Just a quick question on Euler's theorem. (Number Theory). hi what is the last 2 decimal digits of 7^77^777 i get it to be 43 im not sure if it is correct. === Subject: Re: Just a quick question on Euler's theorem. (Number Theory). , > hi what is the last 2 decimal digits of 7^77^777 i get it to be 43 im not sure if it is correct. 7^2 = 49 = 100/2 - 1 7^4 = 100 * 100 / 4 - 100 + 1 = 1 (mod 100). 77^777 = (19 * 4 + 1)^777 7^77^777 = 7^(19 * 4 + 1)^777 = 7^(4 * k + 1^777) = 7 (mod 100) -- Michael Press === Subject: Re: Just a quick question on Euler's theorem. (Number Theory). <010220071113249261%anniel@nym.alias.net.invalid> 777 77 i mean 7 what are the last two digits of this huge number. what am i doing? by Eulers theorem; 7 ^ (psi (100)) = 7^40 = 1(mod 100) by Eulers theorem again; 77 ^ (psi (40)) = 7^16 = 1(mod 40) then 777 = 16*48 + 9 then 77 ^ 777 = 77^(16*48) * 7^(9) = 7^9 (mod40) = 7(mod40) thus 77 ^ 777 = 7 + 40k, for some interger k. now 777 77 7 = 7^(7+40k) = (7^7) * 7^40k) = 7^7 (mod 100) = 43(mod 100). i.e. 43 is the last 2 digits of the huge number. === Subject: Re: Just a quick question on Euler's theorem. (Number Theory). > 777 > 77 > i mean 7 what are the last two digits of this huge number. what am i doing? by Eulers theorem; 7 ^ (psi (100)) = 7^40 = 1(mod 100) by Eulers theorem again; 77 ^ (psi (40)) = 7^16 = 1(mod 40) then 777 = 16*48 + 9 then 77 ^ 777 = 77^(16*48) * 7^(9) = 7^9 (mod40) = 7(mod40) > should be: 77 ^ 777 = 77^(16*48) * 77^(9) = (-3)^9 (mod40) = ((-3)^4)^2 * (-3) (mod 40) = -3 (mod 40) > thus 77 ^ 777 = 7 + 40k, for some interger k. now 777 > 77 > 7 = 7^(7+40k) = (7^7) * 7^40k) = 7^7 (mod 100) = 43(mod 100). i.e. 43 is the last 2 digits of the huge number. === Subject: Re: Just a quick question on Euler's theorem. (Number Theory). <010220071113249261%anniel@nym.alias.net.invalid> <87sldplsix.fsf@nonospaz.fatphil.org> some stupid ramblings by Phil Carmody Where's the 48 from? Phil > -- > Home taping is killing big business profits. We left this side blank > so you can help. -- Dead Kennedys, written upon the B-side of tapes of > /In God We Trust, Inc./. IT'S FROM UP YOUR ASS, ASSHOLE Hope the above helps, Cordially, Aktivator === Subject: Re: Just a quick question on Euler's theorem. (Number Theory). <010220071113249261%anniel@nym.alias.net.invalid> <87sldplsix.fsf@nonospaz.fatphil.org> <87k5z1lm0y.fsf@nonospaz.fatphil.org some stupid ramblings by Phil Carmody > Where's the 48 from? IT'S FROM UP YOUR ASS, ASSHOLE Hope the above helps, Helps me conclude you're an ignorant , devoid of any > mathematical faculties, and prepared to demonstrate that > lack in public, yes. Phil > -- > Home taping is killing big business profits. We left this side blank > so you can help. -- Dead Kennedys, written upon the B-side of tapes of > /In God We Trust, Inc./. BEGIN QUOTE Pulling numbers out of your arse? > by Eulers theorem; > 7 ^ (psi (100)) = 7^40 = 1(mod 100) > by Eulers theorem again; > 77 ^ (psi (40)) = 7^16 = 1(mod 40) > then > 777 = 16*48 + 9 Where's the 48 from? Phil END QUOTE PROVE IT ASS HOLE.. MAY I REMIND YOU THAT MATH IS DONE WITH ONES BRAIN, HOWEVER, YOU SEEM TO DO MATH WITH YOUR ASS HOLE Hope the above helps, Cordially, Aktivator === Subject: finite differences and diffusion equation I am trying to obtain a finite difference solution to the simple 1- dimensional diffusion equation Dt(psi) = D^2x(psi) with Neumann (derivative) boundary conditions. I am using relatively large step sizes for x and t (0.25 and 0.2 respectively). It is known that the solution to this equation with Dirichlet (non- derivative) boundary conditions can be achieved with the Crank- Nicolson scheme. When I apply this scheme to the above problem, the instability is not removed. I am using Milne finite differences for both derivatives. Dt(psi) = [psi(k,m+1) - psi(k,m-1)]/2ht D^2x(psi) = [psi(k+1,m) - 2psi(k,m) + psi(k-1,m)]/hx^2 Does anyone have a reference to a discussion about this problem? Any help would be greatly appreciated. Please respond to my email address. Harold Cohen twohco@yahoo.com === Subject: Re: Roots of a multivariate polynomial > The top coefficient is a polynomial in one degree less. > ... one *variable* less, of course ... > the first case, but not the second. Would you mind elaborating on the > following example? > Say we have (2x+2y+1)(2x-2y-1)=0 or equivalently 4x^2-4y^2-4y-1=0. The > top coefficient (4) is a polynomial like 4x^2-4y^2 and has two > variables, while you said it should be one *variable* less (in this > case losing one of the two variables would leave us with one). So it > seems I did not understand your suggestion for the second part. Actually, I am for sure in the second case. This is because my > equation is of form (f(x,y,,...z))^2=c where c is a positive constant > and f(x,y,...z) is a polynomial in variables x,y,...z. It's probably better to try to solve for f(x,y,..z) = +sqrt(c) or f(x,y,..z) = -sqrt(c) === Subject: Re: Roots of a multivariate polynomial > The top coefficient is a polynomial in one degree less. ... one *variable* less, of course ... > the first case, but not the second. Would you mind elaborating on the > following example? > Say we have (2x+2y+1)(2x-2y-1)=0 If you know this, it's simple: Solve 2x+2y+1 = 0 or 2x-2y-1=0. > or equivalently 4x^2-4y^2-4y-1=0. The > top coefficient (4) is a polynomial like 4x^2-4y^2 and has two > variables, while you said it should be one *variable* less (in this > case losing one of the two variables would leave us with one). So it > seems I did not understand your suggestion for the second part. The degree in x is 2 and the degree in y is two. The top coefficient of x is 4, but in general it could be a polynomial in y. Similar with x<->y reversed. Here, the top coefficient of x is 4 and thus has 2 variables less. The unfortunate thing is that this polynomial has no zeroes, thus my second trick is not applicable. === Subject: Re: Small Set Theory,Updated. > Just as a farewell, see the following small theory. > x is P_defined <-> Ay(yex<->(Py & ~y=x & Px)). > Ax1) Extensionality: As in ZFC > Ax2) Comprehension: Ex x is P_defined > Ax3) Infinity:As in ZFC > Ax4) Choice: As in ZFC.Once again, we obtain the empty set 0 as a (and in fact the) > (~y=y)_defined set. > But this time, 0 is P_defined for any P such that P(0) is false. > Put in another way, Ax.2 brings new sets into existence for a > predicate P only if P(0) is true. > And if P is any predicate such that P(0) is true, but P(y) is false > for any other set, > then Ax.2 is still fulfilled by x=0. In general, we have > 0ex <-> (P(0) & ~0=x & P(x)) > for any P_defined set as postulated by Ax.2. > Either this is true already for x=0 (and Ax.2 does not imply the > existence of any new set in this case), > or P(0) and P(x) and ~0=x holds, i.e. 0ex. > Ax.2 cannot guarantee the existence of a nonempty set that does not > have 0 as an element. > I claim that Ex(~x=0 & ~0ex) is NOT a theorem in this theory, unless the theory is inconsistent. Ok, let me see, > take the following P[y]<->((~y=0)&(AwAuAz~(weu&uez&zey))). Now by comprehension Ex x is P_defined. For such x we have Ay(yex<->(Py & ~y=x & Px)) i.e. Ay(yex<->(((~y=0)&(AwAuAz~(weu&uez&zey))) & ~y=x & ((~x=0)&(AwAuAz~(weu&uez&zex))))). As I claimed that this is already fulfilled by x=0, hence the existence of a set deserving the name {{0}} is not shown. > By the way, from Ax.5 we trivially have Ax(~x=0 -> Ey(yex & Az(~(zey & zex)))) i.e. the axiom of regularity is a THEOREM. > However, different from useful set theories, this does not immediately > imply that for example yex & xey is impossible. I think you might be mistaken, this theory cannot be well founded. A > simple example, would be V which is V is (y=y)_defined > and H which H is (|y|>2)_defined. where |y|>2 could be written as EaEbEc(aey & bey & cey & ~a=b & ~a=c & ~b=c) > You see that HeV and VeH. I don't see that. The possibility of two sets containing one another was exactly what I said could not be excluded. But OTOH, I doubt that VeH is necessarily true. Recall that my Ax.5 was Ax(x=0 v 0ex). With that we would find that H=0 (since |0|>2 is false) and then VeH is false. To see that Ax(~x=0 -> Ey(yex & Az(~(zey & zex)))) holds under the assumption of Ax.5: Let x be any set Assume ~x=0. Then 0ex by Ax5 and thus (using the fact that ze0 is always false) 0ex & Az(~(ze0 & zex)), hence Ey(yex & Az(~(zey & zex))). This shows that ~x=0 -> Ey(yex & Az(~(zey & zex))) Finally, we obtain Ax(~x=0 -> Ey(yex & Az(~(zey & zex)))) === Subject: Re: Small Set Theory,Updated. > Just as a farewell, see the following small theory. > x is P_defined <-> Ay(yex<->(Py & ~y=x & Px)). > Ax1) Extensionality: As in ZFC > Ax2) Comprehension: Ex x is P_defined > Ax3) Infinity:As in ZFC > Ax4) Choice: As in ZFC.Once again, we obtain the empty set 0 as a (and in fact the) > (~y=y)_defined set. > But this time, 0 is P_defined for any P such that P(0) is false. > Put in another way, Ax.2 brings new sets into existence for a > predicate P only if P(0) is true. > And if P is any predicate such that P(0) is true, but P(y) is false > for any other set, > then Ax.2 is still fulfilled by x=0. > In general, we have > 0ex <-> (P(0) & ~0=x & P(x)) > for any P_defined set as postulated by Ax.2. > Either this is true already for x=0 (and Ax.2 does not imply the > existence of any new set in this case), > or P(0) and P(x) and ~0=x holds, i.e. 0ex. > Ax.2 cannot guarantee the existence of a nonempty set that does not > have 0 as an element. > I claim that > Ex(~x=0 & ~0ex) > is NOT a theorem in this theory, unless the theory is inconsistent. Ok, let me see, > take the following P[y]<->((~y=0)&(AwAuAz~(weu&uez&zey))). Now by comprehension Ex x is P_defined. For such x we have > Ay(yex<->(Py & ~y=x & Px)) i.e. > Ay(yex<->(((~y=0)&(AwAuAz~(weu&uez&zey))) & ~y=x & > ((~x=0)&(AwAuAz~(weu&uez&zex))))). > As I claimed that this is already fulfilled by x=0, hence the > existence of a set deserving the name {{0}} is not shown. But this is silly. I can simply say the following. E(all) x ( x is P_defined) were all after E means all x such that x is P_defined is true. here we have two sets fulfilling P. one of them is 0 and therefore existing, and the other is {{0}} which therefore exist. Perhaps this is also symbolized by the following notation. E(many) x ( x is P_defined ). I don't need uniquenss here. I only need a source of existance for sets. > By the way, from Ax.5 we trivially have > Ax(~x=0 -> Ey(yex & Az(~(zey & zex)))) > i.e. the axiom of regularity is a THEOREM. > However, different from useful set theories, this does not immediately > imply that for example yex & xey is impossible. I think you might be mistaken, this theory cannot be well founded. A > simple example, would be V which is V is (y=y)_defined > and H which H is (|y|>2)_defined. where |y|>2 could be written as > EaEbEc(aey & bey & cey & ~a=b & ~a=c & ~b=c) You see that HeV and VeH. I don't see that. > The possibility of two sets containing one another was exactly what I > said could not be excluded. > But OTOH, I doubt that VeH is necessarily true. Recall that my Ax.5 was Ax(x=0 v 0ex). which is wrong. see the multiple type exitential quantifier I used. > With that we would find that H=0 (since |0|>2 is false) and then VeH > is false. To see that > Ax(~x=0 -> Ey(yex & Az(~(zey & zex)))) > holds under the assumption of Ax.5: Let x be any set > Assume ~x=0. > Then 0ex by Ax5 and thus (using the fact that ze0 is always false) > 0ex & Az(~(ze0 & zex)), > hence > Ey(yex & Az(~(zey & zex))). > This shows that > ~x=0 -> Ey(yex & Az(~(zey & zex))) > Finally, we obtain > Ax(~x=0 -> Ey(yex & Az(~(zey & zex))))- Hide quoted text - - Show quoted text - === Subject: Re: Small Set Theory,Updated. > Just as a farewell, see the following small theory. > x is P_defined <-> Ay(yex<->(Py & ~y=x & Px)). > Ax1) Extensionality: As in ZFC > Ax2) Comprehension: Ex x is P_defined > Ax3) Infinity:As in ZFC > Ax4) Choice: As in ZFC.Once again, we obtain the empty set 0 as a (and in fact the) > (~y=y)_defined set. > But this time, 0 is P_defined for any P such that P(0) is false. > Put in another way, Ax.2 brings new sets into existence for a > predicate P only if P(0) is true. > And if P is any predicate such that P(0) is true, but P(y) is false > for any other set, > then Ax.2 is still fulfilled by x=0. > In general, we have > 0ex <-> (P(0) & ~0=x & P(x)) > for any P_defined set as postulated by Ax.2. > Either this is true already for x=0 (and Ax.2 does not imply the > existence of any new set in this case), > or P(0) and P(x) and ~0=x holds, i.e. 0ex. > Ax.2 cannot guarantee the existence of a nonempty set that does not > have 0 as an element. > I claim that > Ex(~x=0 & ~0ex) > is NOT a theorem in this theory, unless the theory is inconsistent. > Ok, let me see, > take the following > P[y]<->((~y=0)&(AwAuAz~(weu&uez&zey))). > Now by comprehension Ex x is P_defined. For such x we have > Ay(yex<->(Py & ~y=x & Px)) i.e. > Ay(yex<->(((~y=0)&(AwAuAz~(weu&uez&zey))) & ~y=x & > ((~x=0)&(AwAuAz~(weu&uez&zex))))). > As I claimed that this is already fulfilled by x=0, hence the > existence of a set deserving the name {{0}} is not shown. But this is silly. It is not silly, at most by virtue of the garbage-in-garbage-out principle. I can simply say the following. E(all) x ( x is P_defined) were all after E means all x such that > x is P_defined is true. What?? here we have two sets fulfilling P. We do not have two such sets until their existence has been proved. > one of them is 0 and therefore > existing, and the other is {{0}} which therefore exist. Again: What?? Perhaps this is also symbolized by the following notation. E(many) x ( x is P_defined ). I don't need uniquenss here. I didn't require uniqueness either. BUT: ExQ is defined to mean that at least one (but possibly more) x exist such that Q holds. Therefore, once I have found a special x such that Q holds, I cannot be certain any more that another x with that property exists. So what should your new quantifier in E(many) x (x is (~y=y)_defined mean? That there are many empty sets?? I only need a source of existance for sets. Yes, definitely. > Recall that my Ax.5 was Ax(x=0 v 0ex). which is wrong. see the multiple type exitential quantifier I used. a) You entered that strange thing /after/ my post and proof that Ax.5 is relatively consistent. b) I am still wondering what that quantifier should mean. Can you reduce it to established logical concepts? === Subject: Re: Small Set Theory,Updated. > Just as a farewell, see the following small theory. > x is P_defined <-> Ay(yex<->(Py & ~y=x & Px)). > Ax1) Extensionality: As in ZFC > Ax2) Comprehension: Ex x is P_defined > Ax3) Infinity:As in ZFC > Ax4) Choice: As in ZFC.Once again, we obtain the empty set 0 as a (and in fact the) > (~y=y)_defined set. > But this time, 0 is P_defined for any P such that P(0) is false. > Put in another way, Ax.2 brings new sets into existence for a > predicate P only if P(0) is true. > And if P is any predicate such that P(0) is true, but P(y) is false > for any other set, > then Ax.2 is still fulfilled by x=0. > In general, we have > 0ex <-> (P(0) & ~0=x & P(x)) > for any P_defined set as postulated by Ax.2. > Either this is true already for x=0 (and Ax.2 does not imply the > existence of any new set in this case), > or P(0) and P(x) and ~0=x holds, i.e. 0ex. > Ax.2 cannot guarantee the existence of a nonempty set that does not > have 0 as an element. > I claim that > Ex(~x=0 & ~0ex) > is NOT a theorem in this theory, unless the theory is inconsistent. > Ok, let me see, > take the following > P[y]<->((~y=0)&(AwAuAz~(weu&uez&zey))). > Now by comprehension Ex x is P_defined. > For such x we have > Ay(yex<->(Py & ~y=x & Px)) i.e. > Ay(yex<->(((~y=0)&(AwAuAz~(weu&uez&zey))) & ~y=x & > ((~x=0)&(AwAuAz~(weu&uez&zex))))). > As I claimed that this is already fulfilled by x=0, hence the > existence of a set deserving the name {{0}} is not shown. But this is silly. It is not silly, at most by virtue of the garbage-in-garbage-out > principle. I can simply say the following. E(all) x ( x is P_defined) were all after E means all x such that > x is P_defined is true. What?? here we have two sets fulfilling P. We do not have two such sets until their existence has been proved. one of them is 0 and therefore > existing, and the other is {{0}} which therefore exist. Again: What?? Perhaps this is also symbolized by the following notation. E(many) x ( x is P_defined ). I don't need uniquenss here. I didn't require uniqueness either. > BUT: ExQ is defined to mean that at least one (but possibly more) x > exist such that Q holds. > Therefore, once I have found a special x such that Q holds, I cannot > be certain any more that another x with that property exists. > So what should your new quantifier in > E(many) x (x is (~y=y)_defined > mean? That there are many empty sets?? I only need a source of existance for sets. Yes, definitely. > Recall that my Ax.5 was Ax(x=0 v 0ex). which is wrong. see the multiple type exitential quantifier I used. a) You entered that strange thing /after/ my post and proof that Ax.5 > is relatively consistent. > b) I am still wondering what that quantifier should mean. Can you > reduce it to established logical concepts?- Hide quoted text - - Show quoted text - Forget about this quantifier. You win. let me try again. Aim: is for x to be not the empty set if there exist a x such that Px. and for x to be the empty set if there doesn't exist x such that Px. In other words I think comprehension should be changed to the following: Ex( (x=0 -> ~Ey(Py)) & Ay(yex<->(Py & ~y=x & Px)) ) . Now according to this axiom lets see if the following predicate works for x=0. Py<-> ( ~y=0 & ~y=2 & AwAuAz~(weu&uez&zey) ). Now P holds for y={{0}}. That's to say that Ey(Py) is a true sentence. therefore ~Ey(Py) is a false sentence. Now for if x=0 then we have ( x=0 -> ~Ey(Py) ) as a trivally false statement. Therefore x cannot be 0 in this case. And according to this axiom only the set {{0}} would exist. While for any P that is false for all values of y. i.e Ay(~Py). such a predicate would lead to the existance of 0. by the modefied comprehension above. I think that this is a better presentation of what is in my mind. What do you think? Zuhair === Subject: Re: Small Set Theory,Updated. > Just as a farewell, see the following small theory. > x is P_defined <-> Ay(yex<->(Py & ~y=x & Px)). > Ax1) Extensionality: As in ZFC > Ax2) Comprehension: Ex x is P_defined > Ax3) Infinity:As in ZFC > Ax4) Choice: As in ZFC.Once again, we obtain the empty set 0 as a (and in fact the) > (~y=y)_defined set. > But this time, 0 is P_defined for any P such that P(0) is false. > Put in another way, Ax.2 brings new sets into existence for a > predicate P only if P(0) is true. > And if P is any predicate such that P(0) is true, but P(y) is false > for any other set, > then Ax.2 is still fulfilled by x=0. > In general, we have > 0ex <-> (P(0) & ~0=x & P(x)) > for any P_defined set as postulated by Ax.2. > Either this is true already for x=0 (and Ax.2 does not imply the > existence of any new set in this case), > or P(0) and P(x) and ~0=x holds, i.e. 0ex. > Ax.2 cannot guarantee the existence of a nonempty set that does not > have 0 as an element. > I claim that > Ex(~x=0 & ~0ex) > is NOT a theorem in this theory, unless the theory is inconsistent. > Ok, let me see, > take the following > P[y]<->((~y=0)&(AwAuAz~(weu&uez&zey))). > Now by comprehension Ex x is P_defined. > For such x we have > Ay(yex<->(Py & ~y=x & Px)) i.e. > Ay(yex<->(((~y=0)&(AwAuAz~(weu&uez&zey))) & ~y=x & > ((~x=0)&(AwAuAz~(weu&uez&zex))))). > As I claimed that this is already fulfilled by x=0, hence the > existence of a set deserving the name {{0}} is not shown. > But this is silly. It is not silly, at most by virtue of the garbage-in-garbage-out > principle. > I can simply say the following. > E(all) x ( x is P_defined) > were all after E means all x such that > x is P_defined is true. What?? > here we have two sets fulfilling P. We do not have two such sets until their existence has been proved. > one of them is 0 and therefore > existing, and the other is {{0}} which therefore exist. Again: What?? > Perhaps this is also symbolized by the following notation. > E(many) x ( x is P_defined ). > I don't need uniquenss here. I didn't require uniqueness either. > BUT: ExQ is defined to mean that at least one (but possibly more) x > exist such that Q holds. > Therefore, once I have found a special x such that Q holds, I cannot > be certain any more that another x with that property exists. > So what should your new quantifier in > E(many) x (x is (~y=y)_defined > mean? That there are many empty sets?? > I only need a source of existance for sets. Yes, definitely. > Recall that my Ax.5 was Ax(x=0 v 0ex). > which is wrong. see the multiple type exitential quantifier I used. a) You entered that strange thing /after/ my post and proof that Ax.5 > is relatively consistent. > b) I am still wondering what that quantifier should mean. Can you > reduce it to established logical concepts?- Hide quoted text - - Show quoted text - Forget about this quantifier. > You win. > let me try again. Aim: is for x to be not the empty set if there exist a x such that Px. > and for x to be the empty set if there doesn't exist x such that Px. In other words I think comprehension should be changed to the > following: Ex( (x=0 -> ~Ey(Py)) & Ay(yex<->(Py & ~y=x & Px)) ) . Now according to this axiom lets see if the following predicate works > for x=0. Py<-> ( ~y=0 & ~y=2 & AwAuAz~(weu&uez&zey) ). Now P holds for y={{0}}. That's to say that > Ey(Py) is a true sentence. therefore ~Ey(Py) is a false sentence. Now for if x=0 then we have ( x=0 -> ~Ey(Py) ) as a trivally false > statement. Therefore x cannot be 0 in this case. And according to this axiom only the set {{0}} would exist. While for any P that is false for all values of y. > i.e Ay(~Py). such a predicate would lead to the existance of 0. by the > modefied comprehension above. I think that this is a better presentation of what is in my mind. What do you think? Zuhair- Hide quoted text - - Show quoted text - Correction: The axiom of comprehension should be modefied to the following: Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(Py & ~y=x & Px)) ) . Also x is P_defined should be modefied to x is P_defined <-> ( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(Py & ~y=x & Px)) ). were P is a formula in first order language in one free variable. that may or may not use the symbole x. Well I don't know if this would work, I think still I should impose further restriction on some types of P. anyhow. Zuhair === Subject: Re: Small Set Theory,Updated. Imagine the following theory. Primitives: e , = Axioms: Extensionality, Replacement, Infinity, Choice. All as in ZFC. + Axiom of comprehension as below: Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . I think this theory is inconsistent, but I don't know that for sure. According to this theory the set of all sets exist, and the set of all ordinals exist. Most sets can be constructed by replacement of an ordinal using the appropriate replacement predicate. There is no need for pairing,union,empty,separation and power since all of these can be theorums in this theory. Zuhair === Subject: Re: Small Set Theory,Updated. > Imagine the following theory. Primitives: e , = Axioms: Extensionality, Replacement, Infinity, Choice. All as in ZFC. + Axiom of comprehension as below: Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & > P(x))) ) . I think this should be further modefied to the following. Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . were y is P_embedded mean the following: y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). Zuhair I think this theory is inconsistent, but I don't know that for sure. According to this theory the set of all sets exist, and the set of all > ordinals exist. Most sets can be constructed by replacement of an > ordinal using the appropriate replacement predicate. There is no need for pairing,union,empty,separation and power since > all of these can be theorums in this theory. Zuhair === Subject: Re: Small Set Theory,Updated. > were y is P_embedded mean the following: y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). Suppose P(-z), P(y), z not in y, y is not P_embedded. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: Small Set Theory,Updated. <87abzv2ey8.fsf@laptop.at.pyenos were y is P_embedded mean the following: y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). Suppose P(-z), > P(y), > z not in y, > y is not P_embedded. I think what you mean is the following. suppose P(z)<-> z= -z Of course it is clear that we have ~P(y). Now what we are interested in is x as defined by comprehension. the result will be x=0 when x is (-z)_defined. which is true. Zuhair -- > ### Author Logan Lee ### > ### AuthorHomePagehttp://beam.to/pyenos### > ### AuthorQuote I just want to learn. ### === Subject: Re: Small Set Theory,Updated. <87abzv2ey8.fsf@laptop.at.pyenos > were y is P_embedded mean the following: > y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). Suppose P(-z), > P(y), > z not in y, > y is not P_embedded. I think what you mean is the following. suppose P(z)<-> z= -z What does -z mean? === Subject: Re: Small Set Theory,Updated. were y is P_embedded mean the following: y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). > Suppose P(-z), > P(y), Therefore, P(y) is not equal to Py. What is Py? > z not in y, > y is not P_embedded. -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: Small Set Theory,Updated. <87abzv2ey8.fsf@laptop.at.pyenos> <8764aj2ev7.fsf@laptop.at.pyenos > were y is P_embedded mean the following: > y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). > Suppose P(-z), > P(y), Therefore, > P(y) is not equal to Py. What is Py? z not in y, > y is not P_embedded. I am not understanding what you are talking about? -- > ### Author Logan Lee ### > ### AuthorHomePagehttp://beam.to/pyenos### > ### AuthorQuote I just want to learn. ### -- > ### Author Logan Lee ### > ### AuthorHomePagehttp://beam.to/pyenos### > ### AuthorQuote I just want to learn. ### === Subject: Re: Small Set Theory,Updated. > >> were y is P_embedded mean the following: >> >> y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). >> Suppose P(-z), >> P(y), > Therefore, > P(y) is not equal to Py. What is Py? > z not in y, >> y is not P_embedded. Wow. This development surely proves there is mathematical karma. -- Jesse F. Hughes. Me: It's very sad when one's husband or wife dies. Quincy (Age 4 1/2): Yeah. You might want to tell them something and you just can't. [Long pause] Like Take out the trash. === Subject: Re: Small Set Theory,Updated. > Imagine the following theory. Primitives: e , = Axioms: Extensionality, Replacement, Infinity, Choice. All as in ZFC. + Axiom of comprehension as below: Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & > P(x))) ) . I think this should be further modefied to the following. > Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) > <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . were y is P_embedded mean the following: y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). what is 'P' in Py? is it a logical quantifier? Zuhair I think this theory is inconsistent, but I don't know that for sure. According to this theory the set of all sets exist, and the set of all > ordinals exist. Most sets can be constructed by replacement of an > ordinal using the appropriate replacement predicate. There is no need for pairing,union,empty,separation and power since > all of these can be theorums in this theory. Zuhair -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: Small Set Theory,Updated. <87mz3v2gr6.fsf@laptop.at.pyenos > Imagine the following theory. > Primitives: e , = > Axioms: Extensionality, Replacement, Infinity, Choice. All as in ZFC. > + Axiom of comprehension as below: > Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & > P(x))) ) . I think this should be further modefied to the following. > Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) > <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . were y is P_embedded mean the following: y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). what is 'P' in Py? is it a logical quantifier? No P is not a logical quantifier. It is a formula in first order language in one free variable y. examples P(y)<-> Az(~zey) you see Az(~zey) is a first order language statement. Now z is quantified over therefore it is bound, Only y is free in this formula. Another example: P(y)<-> AwAz ~(wez & zey). here also you see that P(y) is a formula in first order language in one free variable y. Let me write everything again. Primitives e,= Definition schema 1:) x is P_embedded <->( P(x) &Ay(yex<->(P(y) &~y=x) ). were P is a formula in first order language in one free variable. so P(y) is a formula in first order language with only y as its free variable. also P(x) is a formula in first order language with only x as its free variable. Axioms: Extensionality,Replacement,Infinity,Power, Choice , Comprehension schema were Comprehesion schema is as below: Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . Of course 0 is the empty set,derived from Replacement. Example let P(y)<-> ~y=y. now we have (~Ey(y is P_embedded) since if not then E y (~y=y) violating extensionlity. So for such a P, you see that x=0. While if P(y)<-> AwAz~(wez&zey). here we have the set {0} is P_embedded. and so {0} is P_defined. while see the following formula. P(y)<-> (~y=0 & ~y=2 & AwAuAz ~(weu & uez & zey)). were y=2 is y is (AwAuAz ~(weu & uez & zey))_defined. Now x is P_defined would be x={{0}}. I am not sure of this theory. It need meticulous analysis. It is rather complex. anyhow. Zuhair Zuhair > I think this theory is inconsistent, but I don't know that for sure. > According to this theory the set of all sets exist, and the set of all > ordinals exist. Most sets can be constructed by replacement of an > ordinal using the appropriate replacement predicate. > There is no need for pairing,union,empty,separation and power since > all of these can be theorums in this theory. > Zuhair -- > ### Author Logan Lee ### > ### AuthorHomePagehttp://beam.to/pyenos### > ### AuthorQuote I just want to learn. ###- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Small Set Theory,Updated. <87mz3v2gr6.fsf@laptop.at.pyenos > Imagine the following theory. > Primitives: e , = > Axioms: Extensionality, Replacement, Infinity, Choice. All as in ZFC. > + Axiom of comprehension as below: > Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & > P(x))) ) . > I think this should be further modefied to the following. > Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) > <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . > were y is P_embedded mean the following: > y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). what is 'P' in Py? is it a logical quantifier? No P is not a logical quantifier. It is a formula in first order > language in one free variable y. examples P(y)<-> Az(~zey) you see Az(~zey) is a first order language statement. Now z is > quantified over therefore it is bound, Only y is free in this formula. Another example: P(y)<-> AwAz ~(wez & zey). here also you see that P(y) is a formula in first order language in > one free variable y. Let me write everything again. Primitives e,= > Definition schema 1:) x is P_embedded <->( P(x) &Ay(yex<->(P(y) &~y=x) ). were P is a > formula in first order language in one free variable. so P(y) is a formula in first order language with only y as its free > variable. also P(x) is a formula in first order language with only x as its free > variable. I should have mentioned also Definition schema 2:) x is P_defined<->( (x=0 <-> (~Ey(y is P_embedded) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . were P is a formula in first order language in one free variable. Axioms: Extensionality,Replacement,Infinity,Power, > Choice , Comprehension schema were Comprehesion schema is as below: Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) > <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . This also can be writtin as: Ex: x is P_defined. Of course 0 is the empty set,derived from Replacement. Example let P(y)<-> ~y=y. > now we have (~Ey(y is P_embedded) > since if not then E y (~y=y) violating extensionlity. > So for such a P, you see that x=0. While if P(y)<-> AwAz~(wez&zey). > here we have the set {0} is P_embedded. > and so {0} is P_defined. while see the following formula. P(y)<-> (~y=0 & ~y=2 & AwAuAz ~(weu & uez & zey)). were y=2 is y is (AwAuAz ~(weu & uez & zey))_defined. Now x is P_defined would be x={{0}}. I am not sure of this theory. It need meticulous analysis. It is > rather complex. anyhow. Zuhair > Zuhair > I think this theory is inconsistent, but I don't know that for sure. > According to this theory the set of all sets exist, and the set of all > ordinals exist. Most sets can be constructed by replacement of an > ordinal using the appropriate replacement predicate. > There is no need for pairing,union,empty,separation and power since > all of these can be theorums in this theory. > Zuhair -- > ### Author Logan Lee ### > ### AuthorHomePagehttp://beam.to/pyenos### > ### AuthorQuote I just want to learn. ###- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Small Set Theory,Updated. > Imagine the following theory. Primitives: e , = Axioms: Extensionality, Replacement, Infinity, Choice. All as in ZFC. + Axiom of comprehension as below: Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & > P(x))) ) . I think this should be further modefied to the following. > Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) > <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . were y is P_embedded mean the following: y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). what is 'P' in Py? is it a logical quantifier? Zuhair I think this theory is inconsistent, but I don't know that for sure. According to this theory the set of all sets exist, and the set of all > ordinals exist. Most sets can be constructed by replacement of an > ordinal using the appropriate replacement predicate. There is no need for pairing,union,empty,separation and power since > all of these can be theorums in this theory. Zuhair -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: Small Set Theory,Updated. > Imagine the following theory. > Primitives: e , = > Axioms: Extensionality, Replacement, Infinity, Choice. All as in ZFC. > + Axiom of comprehension as below: > Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & > P(x))) ) . I think this should be further modefied to the following. > Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) > <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . were y is P_embedded mean the following: > y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). > what is 'P' in Py? is it a logical quantifier? kind of think of it, no. 'P' in Py does not seem to be a logical quantifier. y is P_embedded iff Py and [ (forall z)(z in y iff (P(z) & ~z=y)) ] > Zuhair > I think this theory is inconsistent, but I don't know that for sure. > According to this theory the set of all sets exist, and the set of all > ordinals exist. Most sets can be constructed by replacement of an > ordinal using the appropriate replacement predicate. > There is no need for pairing,union,empty,separation and power since > all of these can be theorums in this theory. > Zuhair -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: Small Set Theory,Updated. > Imagine the following theory. > Primitives: e , = > Axioms: Extensionality, Replacement, Infinity, Choice. All as in ZFC. > + Axiom of comprehension as below: > Ex( (x=0 <-> (~Ey(Py) / (P(y)<->y=0)) & Ay(yex<->(P(y) & ~y=x & > P(x))) ) . > I think this should be further modefied to the following. > Ex( (x=0 <-> (~Ey(y is P_embedded) / (P(y) > <->y=0)) & Ay(yex<->(P(y) & ~y=x & P(x))) ) . > were y is P_embedded mean the following: > > y is P_embedded <->( Py & Az(zey<->(P(z) & ~z=y)) ). > what is 'P' in Py? is it a logical quantifier? > kind of think of it, no. 'P' in Py does not seem to be a logical quantifier. y is P_embedded iff > Py and [ (forall z)(z in y iff (P(z) & ~z=y)) ] > What is 'Py'? > Zuhair > I think this theory is inconsistent, but I don't know that for sure. > According to this theory the set of all sets exist, and the set of all > ordinals exist. Most sets can be constructed by replacement of an > ordinal using the appropriate replacement predicate. > There is no need for pairing,union,empty,separation and power since > all of these can be theorums in this theory. > Zuhair > > -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Vacuously true statements Sometimes I get confused with this so called vacuously statements. I think they mean that, if a statement is never true in an if ..then clause, than anything you say after then is true. For example, if x is a real number, than the statement if x^2 <0, than x = 12 is true.If we take the contrapositive, if x <>12 then x^2 >=0, then we see it's true. Is this the reason for the validity of vacuoulsy true statements? Does this mean that what doesn't exist satisfy any property? So, if I say, every dragon has a deep knowledge of Analysis, then is this a true statement? I have no kids, so if I say all my kids are great mathematicians, then, from the logical standpoint, I'm not lying,? But my kids can't solve any math problem at all, simply because they don't exist. Are those vacuoulsly true statements the reason why, for example, the empty set is subset of any set, including the empty set itself? Why is the following reasoning wrong? By definition, a set A is subset of B if every element of A is element of B. If A is empty, then A has no element and so no element of A is in B. According to the definition, it follows A is not subset of B. I know this is wrong, but apparently this makes sense. Sharon. ; === Subject: Re: Vacuously true statements So, if I > say, every dragon has a deep knowledge of Analysis, then is this a > true statement? I have no kids, so if I say all my kids are great > mathematicians, then, from the logical standpoint, I'm not lying,? Strictly speaking, there is no logical standpoint that gives a single answer on this. If we regiment every dragon has a deep knowledge of Analysis into a standard single-sorted first-order language, using a quantifier that applies to a single open wff, we'll come up with something like (forall x)(Dx --> Ax), where --> is the truth-functional conditional. If nothing in the doman is D, then (Dx --> Ax) always has a false antecedent, so is always true, whatever we assign to x. So (forall x) (Dx --> Ax) is true. So far so good, as others have said. However, if we regiment every dragon has a deep knowledge of Analysis into a language with quantifiers that take two open wffs, along the lines of (forall x)(Dx, Ax), there are natural ways of doing that where (forall x)(Dx, Ax) implies (exists x)(Dx, Ax) and is false if there are no Ds [just as, in standard single-sorted logic, (forall x)Dx implies (exists x)Dx]. In a way, you pays your money and you takes your choice. If you take the first route, you buy simplicity of formal system at the price (arguably) of a certain divorce from the logical form of ordinary language. For plausibly, All Ds are As and Most Ds are As should be treated together as far as logical form is concerned: but the latter can't be regimented using a quantifier that takes a single open wff (try it! (most x)(Dx --> Ax) has the wrong meaning!!). There's nothing sacrosanct about standard First-Order Logic and its particular regimentations. It involves a raft of choices, going for single-sorted rather than dyadic quantifiers, disallowing empty names, disallowing empty domains, not coping with plural reference, and so on and so forth. That's just fine of course. We happily pay those prices for certain good purposes: and the choices aren't just arbitrary, but are based on cost/benefit trade-offs, and standard FOL wins in many contexts. But in some contexts it is, e.g. important to work with a free logic, or a logic with dyadic quantifiers. So we mustn't think that, having made our choice for standard FOL, the fact that we then have to regiment every dragon has a deep knowledge of Analysis in a way that makes it come out true conclusively shows that it (the ordinary language statement) *is* really true. === Subject: Re: Vacuously true statements >Sometimes I get confused with this so called vacuously statements. I >think they mean that, if a statement is never true in an if ..then >clause, than anything you say after then is true. For example, if x >is a real number, than the statement if x^2 <0, than x = 12 is true.If >we take the contrapositive, if x <>12 then x^2 >=0, then we see it's >true. Is this the reason for the validity of vacuoulsy true >statements? Does this mean that what doesn't exist satisfy any property? So, if I >say, every dragon has a deep knowledge of Analysis, then is this a >true statement? I have no kids, so if I say all my kids are great >mathematicians, then, from the logical standpoint, I'm not lying,? Right. >But >my kids can't solve any math problem at all, simply because they don't >exist. No contradiction. It's true that all your kids are great mathematicians, and it's also true that all your kids are dumb as bricks. No contradiction simply because you don't have any kids. Seems strange at first, but it turns out that using the language this way _simplifies_ a lot (for example, as you point out, the equivalence of an implication and its contrapositive depends on this curiosity about the empty set.) >Are those vacuoulsly true statements the reason why, for example, the >empty set is subset of any set, including the empty set itself? Why >is the following reasoning wrong? By definition, a set A is subset of B if every element of A is element >of B. If A is empty, then A has no element and so no element of A is >in B. According to the definition, it follows A is not subset of B. No! Yes, it is true that no element of the empty set is an element of B. But it is _also_ true that _every_ element of the empty set is an element of B. Just like with your contradictory children. > I >know this is wrong, but apparently this makes sense. Sharon. ; ************************ David C. Ullrich === Subject: Re: Vacuously true statements >Are those vacuoulsly true statements the reason why, for example, the >empty set is subset of any set, including the empty set itself? Why >is the following reasoning wrong? It is not wrong. It is indeed true that the empty set is a subset of > any set. > >By definition, a set A is subset of B if every element of A is element >of B. If A is empty, then A has no element and so no element of A is >in B. Both the statements: For all x( if x is in A then x is in B) and For all x (if x is in A then x is not in B) are true when A is the emptyset. Since the FIRST statement above is the definition of A is a subset of > B, it follows that A is a subset of B, period. Remember that A is not a subset of B is ->not<- equivalent to the > second statement above; rather, A is not a subset of B means There exists x (x is in A and x is not in B). And this statement is False whenever A is the empty set. > >According to the definition, it follows A is not subset of B. I >know this is wrong, but apparently this makes sense. Why do you know this is wrong? It is not wrong. The empty set is > indeed a subset of every set. -- === Subject: Re: Vacuously true statements days. My association with the Department is that of an alumnus. [...] >>According to the definition, it follows A is not subset of B. I >>know this is wrong, but apparently this makes sense. >> >> Why do you know this is wrong? It is not wrong. The empty set is >> indeed a subset of every set. > Quite right. I missed that... It is false that the definition says that A (the empty set) is not a subset of B, since the definition of not a subset of would be essentially there exists x such that (x in A and x not in B) which is false when A is the empty set. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Cantor and Cardinality I think the following quotation states something obvious: like ''Cantor proved that there are just as many positive even integers as natural numbers.'' According to a cognitive account of our ordinary notion of ''As Many As'' Cantor proved no such thing. What Cantor did was simply to prove that the sets were pairable (...). http://www.cogsci.ucsd.edu/~nunez/web/TransfinitePrgmtcs.pdf p. 1735 Unless there are evident reasons to think that pairability implies equicardinality. Certainly it does for finite sets. But in the comparison of finite sets it happens too that if A is a proper subset of B, then A Certainly it does for finite sets. But in the comparison of finite > sets it happens too that if A is a proper subset of B, then A Applying this to N and Z we would have N > Why should the criterion of pairability be more evident? Because a comparison by subsetting of sets we get at most a lattice. That is that for most pairs of sets A, B it is undecidable whether |A| < |B| or not, because neither is a subset of the other. For finite sets you can alleviate that by counting the number of elements and comparing that. But that is nothing more, nor less, than pairing elements with initial subsets of the natural numbers, or, ulitmately, by pairing of elements of the sets. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and Cardinality > I think the following quotation states something obvious: like ''Cantor > proved that there are just as many positive even integers as natural > numbers.'' According to > a cognitive account of our ordinary notion of ''As Many As'' Cantor > proved no such thing. > What Cantor did was simply to prove that the sets were pairable > (...).http://www.cogsci.ucsd.edu/~nunez/web/TransfinitePrgmtcs.pdfp. 1735 Unless there are evident reasons to think that pairability implies > equicardinality. > If we have a bijection (pairability) between sets A and B we can use the bijection to change the names of the elements of A to the names of the elements of B. So if A and B have different cardinality, then we can change the cardinality of A by changing the names of the elements. Intuitively, we should not be able to change the cardinality of a set by changing the names of the elements. So we either use a non-intuitive definition of cardinality or we have pairability implies equicardinality. - William Hughes === Subject: Re: Cantor and Cardinality [...] Unless there are evident reasons to think that pairability > implies equicardinality. Certainly it does for finite sets. But in the comparison of finite > sets it happens too that if A is a proper subset of B, then |A| < |B|. > Applying this to N and Z we would have |N| < |Z|. Why should the criterion of pairability be more evident? > And what if for two sets A, B neither A c B nor B c A? Can't we compare them (in size)? Say A = { 1 , 2 } B = { 0.1, 0.01, 0.001, ... } Shouldn't we have |A| < |B| here? :-o Or consider the two sets: a = { 1 , 2 , 3 , ... } b = { 1/1 , 1/2 , 1/3 , ... } Isn't it quite obvious, that we _should_ have |a| = |b| here? How would YOU show that without the method of pairing the elements of the two sets? F. -- E-mail: infosimple-linede === Subject: Re: Cantor and Cardinality > I think the following quotation states something obvious: like ''Cantor > proved that there are just as many positive even integers as natural > numbers.'' According to > a cognitive account of our ordinary notion of ''As Many As'' Cantor > proved no such thing. > What Cantor did was simply to prove that the sets were pairable > (...).http://www.cogsci.ucsd.edu/~nunez/web/TransfinitePrgmtcs.pdfp. 1735 Unless there are evident reasons to think that pairability implies > equicardinality. Certainly it does for finite sets. But in the comparison of finite > sets it happens too that if A is a proper subset of B, then A Applying this to N and Z we would have N output node subNodeIds = select id from Tree where parentId = for each subNodeId in subNodeIds printSubTree(subNodeId) } Using O() notation, describe how many queries this algorithm makes to the database for a subtree with n nodes. === Subject: Re: Big O problem > how many nodes there are in a sub tree with n nodes? Is this a trick question, like Who's buried in Grant's tomb?? If not, the answer is n, and the corresponding big-O notation for it is O(n). === Subject: Re: Big O problem > Can someone help me understand Big O enough to determine the Big O for > how many nodes there are in a sub tree with n nodes? All of the > tutorials on the internet are just confusing me! Here is the original problem... The following recursive algorithm prints out all the tree items in the > specified subtree: printSubTree(id) > { > node = select * from Tree where id = printSubTree(subNodeId) } Using O() notation, describe how many queries this algorithm makes to > the database for a subtree with n nodes. Why don't you start with an estimate for the actual function of n that you think represents the number of queries? Then we can talk about the big-O of that function. - Randy === Subject: Re: Big O problem Can someone help me understand Big O enough to determine the Big O for > how many nodes there are in a sub tree with n nodes? All of the > tutorials on the internet are just confusing me! Here is the original problem... The following recursive algorithm prints out all the tree items in the > specified subtree: printSubTree(id) > { > node = select * from Tree where id = printSubTree(subNodeId) } Using O() notation, describe how many queries this algorithm makes to > the database for a subtree with n nodes. Why don't you start with an estimate for the actual function of n > that you think represents the number of queries? Then we > can talk about the big-O of that function. - Randy- Hide quoted text - - Show quoted text - Randy If someone knows of a VERY simple big O example that shows how to determine the number of nodes in a sub tree, I can probably figure out the solution myself. I just don't know anything about Big O. Bryan === Subject: Re: Big O problem > Can someone help me understand Big O enough to determine the Big O for > how many nodes there are in a sub tree with n nodes? All of the > tutorials on the internet are just confusing me! > Here is the original problem... > The following recursive algorithm prints out all the tree items in the > specified subtree: > printSubTree(id) > { > node = select * from Tree where id = output node > subNodeIds = select id from Tree where parentId = for each subNodeId in subNodeIds > printSubTree(subNodeId) > } > Using O() notation, describe how many queries this algorithm makes to > the database for a subtree with n nodes. Why don't you start with an estimate for the actual function of n > that you think represents the number of queries? Then we > can talk about the big-O of that function. - Randy- Hide quoted text - - Show quoted text - Randy If someone knows of a VERY simple big O example that shows how to > determine the number of nodes in a sub tree, I can probably figure out > the solution myself. I just don't know anything about Big O. I'm confused about your question. Determining the number of nodes in a sub tree is not a question about big O notation. There are two different questions: (1) Estimate some property of an algorithm which is a function of n, the problem size, and (2) Write that function in big-O notation. For example, I might have an algorithm which goes through three steps: - initialization, which takes a constant time of A. - some loop which takes a time proportional to n, B*n - some other loop which takes a time C*n^2 for some other constant C. So my runtime is A + B*n + C*n^2. That's the type of estimate I asked you to come up with. Once you have that, you can show that this runtime is O(n^2). What that means is that there is some other constant D such that the parabola D*n^2 is above the function A + B*n + C*n^2 for all values of n large enough. So big-O notation is in a sense about finding simpler functions that act as bounds to more complicated functions. It's a way of comparing how algorithms scale. An algorithm that takes runtime 1000*n^2 seconds and another that takes 0.01 n^2 + 5 seconds are both O(n^2), because they grow as n^2. So are you really asking about how to figure out the big-O bounds on a function that you already know, or are you asking how to figure out the function that represents number of database queries versus tree size? If you're struggling with the second, I suggest working through a couple of small examples first, say with n=2 or 3. - Randy === Subject: Conditions on bijection Let A and X be topological spaces. Let f: A-> X be a continuous map, under what conditions of A and X is f a bijection? === Subject: Re: Conditions on bijection Let A and X be topological spaces. Let f: A-> X be a continuous map, >under what conditions of A and X is f a bijection? The only conditions that would ->guarantee<- that f is a bijection is > if A and X are both spaces with a single point. > Also A = X = nulset, the only space with exactly one open set. Compact, metrizable, connected, discrete, indiscrete,... Enjoy your stay in nothingness. ;-) > Otherwise, you can always take a constant function, which would either > be non-injective (if A has more than one point), or nonsurjective (if > X has more than one point). > It's all set theory. What does topology have to do with it? > Are you sure this is the question you meant to ask? > Riddle of the day. What topological properties distinguish the empty space from a one point space? Though they are not homeomorphic, is different sizes the only reason why? === Subject: Re: Conditions on bijection days. My association with the Department is that of an alumnus. >> Otherwise, you can always take a constant function, which would either >> be non-injective (if A has more than one point), or nonsurjective (if >> X has more than one point). >It's all set theory. What does topology have to do with it? Continuity. >> Are you sure this is the question you meant to ask? >Riddle of the day. What topological properties distinguish the empty >space from a one point space? Though they are not homeomorphic, is >different sizes the only reason why? You cannot make the empty space a pointed topological space. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Conditions on bijection > Otherwise, you can always take a constant function, which would either >> be non-injective (if A has more than one point), or nonsurjective (if >> X has more than one point). >It's all set theory. What does topology have to do with it? Continuity. > Doesn't seem to come into use. >Riddle of the day. What topological properties distinguish the empty >space from a one point space? Though they are not homeomorphic, is >different sizes the only reason why? You cannot make the empty space a pointed topological space. > Oh my gosh, you mean the empty space is pointless? By golly it is! Yet however it's not as pointless as pointless topology, which isn't really pointless with all of its pseudo-point sets and which doesn't work for drunk spaces, ie not sober spaces. === Subject: Re: Conditions on bijection space from a one point space? Though they are not homeomorphic, is > different sizes the only reason why? They have different dimensions: the one-point space has dimension 0, the empty space has dimension -1. Wwll, size is not a technical term, and you could say that dimensionality is a kind of size. Here's another one. The one-point space has the fixed-point property, every continuous selfmap has a fixed point. The empty space does not have this property. === Subject: Re: Conditions on bijection Riddle of the day. What topological properties distinguish the empty > space from a one point space? Though they are not homeomorphic, is > different sizes the only reason why? They have different dimensions: the one-point space has dimension 0, > the empty space has dimension -1. > By who's dimensional dementia? Both are zero dimensional because both have a base of clopen sets. > Here's another one. The one-point space has the fixed-point property, > every continuous selfmap has a fixed point. The empty space does not > have this property. > LOL. That'll be hard to top. === Subject: Re: Conditions on bijection Riddle of the day. What topological properties distinguish the empty > space from a one point space? Though they are not homeomorphic, is > different sizes the only reason why? They have different dimensions: the one-point space has dimension 0, > the empty space has dimension -1. By who's dimensional dementia? The word is whose. I'm following the best authorities on the subject: the classic text Dimension Theory by Witold Hurewicz and Henry Wallman, published by Princeton University Press; my copy is the revised edition of 1948. Quoting from the introduction, p. 4: The definition of dimension we shall adopt in this book (see page 24) is due to Menger and Urysohn. In the formulation of Menger, it reads: a) the empty set has dimension -1, b) the dimension of a space is the least integer n for which every point has arbitrarily small neighborhoods whose boundaries have dimension less than n. It is the opinion of the authors that none of the several other possible definitions of dimension has the immediate intuitive appeal of this one and none leads so elegantly to the existing theory. === Subject: Re: Conditions on bijection Riddle of the day. What topological properties distinguish the empty > space from a one point space? Though they are not homeomorphic, is > different sizes the only reason why? > They have different dimensions: the one-point space has dimension 0, > the empty space has dimension -1. By whose dimensional dementia? > Oh, the inductive dimension. IIRC, all definitions of dimension agree for compact Hausdorff spaces. The empty space seems to be an exception. > The definition of dimension we shall adopt in this book (see page > 24) is due to Menger and Urysohn. In the formulation of Menger, it > reads: > a) the empty set has dimension -1, > b) the dimension of a space is the least integer n for which every > point has arbitrarily small neighborhoods whose boundaries have > dimension less than n. > It is the opinion of the authors that none of the several other > possible definitions of dimension has the immediate intuitive appeal > of this one and none leads so elegantly to the existing theory. > This, nor any of the other definitions of dimension includes all spaces. Does that definition apply to R^N or R^R with Tychonov product topology? A space that amuses me is one constructed from closed unit balls B_n of R^n for all n in N. Take the disjoint union of these balls and for all n, identify a point in B_n with a point in B_(n+1) making B_n 'tangent' to B_(n+1). (The identify point in B_n is not the identify point in B_(n-1) that makes B_(n-1) tangent to B_n.) Every point of this space has a finite local dimension, yet by the above definition, it has no dimension. Is compactness required for there to be a dimension? No, infinite cofinite spaces don't have a dimension. Is compact Hausdorff required? BTW, is Urysohn, the Urysohn of Urysohn's metrization theorem? === Subject: Re: Conditions on bijection Riddle of the day. What topological properties distinguish the empty > space from a one point space? Though they are not homeomorphic, is > different sizes the only reason why? They have different dimensions: the one-point space has dimension 0, > the empty space has dimension -1. Wwll, size is not a technical term, and you could say that > dimensionality is a kind of size. Sorry for the typo, Wwll was meant to be Well. > Here's another one. The one-point space has the fixed-point property, > every continuous selfmap has a fixed point. The empty space does not > have this property. === Subject: Function required Hello all, Few days ago I have posted a discussion, which I don't think that I got (or I am not understand) the answer. browse_thread/thread/e7bd260e071164c1/af20d02326735d4e? lnk=gst&q=maximum+function&rnum=1&hl=en#af20d02326735d4e I want to make the question clear, and to explain what I meant to several replies: My problem is as followed: I got a list of m inequalities with n parameters. The inequalities are of two form: X1*Am,1+X2*Am,2+....+Xn*Am,n>0 and X1*Am,1+X2*Am,2+....+Xn*Am,n<0, Where Ai,j equal to 1 or -1 for all i,j. I need to find the coefficients X1..Xn such that I will have a maximum of correct answers for the inequalities (notice that X1,..,Xn are common for the entire set of inequalities). It means that the coefficients need to satisfy as many inequalities as possible! I need a maximum of correct answer because two inequalities might contradict each other. Notice that the inequalities are strict. if it would ease (and the calculation time will improve), than the inequalities can be >0 and <=0 (NOTICE: not >=0 and <=0!!!, no overlaping) The coefficients can have any real values include negative. I would prefer a certain positive range like [0,..,100] but it should not limit us. you. Is there a known (and efficient) algorithm for problems of my type? (m,n can be big: n several hundreds, m tens of thousands) Is anyone familiar with a library or software which does it? Asi === Subject: Conditions on bijection Let A and X be topological spaces. Let f: A-> X be a continuous map, under what conditions of A and X is f a bijection? === Subject: Re: Conditions on bijection > Let A and X be topological spaces. Let f: A-> X be > a continuous map, under what conditions of A and X > is f a bijection? >> Are you sure this is the question you meant to ask? > Let A and X be topological spaces. Let f: A-> X be > a continuous map, under what conditions of A and X > is f a bijection? Carl, I'm confused. You asked a question, Arturo addressed your question (not copied above), then Arturo asked you a YES/NO question about your question, and in reply (yes, your second post was actually a reply to Arturo, and it was), you asked the same question again, character-by-character. Dave L. Renfro === Subject: Re: Conditions on bijection > Let A and X be topological spaces. Let f: A-> X be > a continuous map, under what conditions of A and X > is f a bijection? >> Are you sure this is the question you meant to ask? > Let A and X be topological spaces. Let f: A-> X be > a continuous map, under what conditions of A and X > is f a bijection? Carl, I'm confused. You asked a question, Arturo addressed > your question (not copied above), then Arturo asked you > a YES/NO question about your question, and in reply (yes, > your second post was actually a reply to Arturo, and it was), > you asked the same question again, character-by-character. Dave L. Renfro I'm sorry but I got an error and thought the message wasn't sent, that's why I sent it again. I forgot to state that f: A-> X is an embedding also, i.e A is homeomorphic to f(A). === Subject: Re: Conditions on bijection days. My association with the Department is that of an alumnus. >I'm sorry but I got an error and thought the message wasn't sent, >that's why I sent it again. >I forgot to state that f: A-> X is an embedding also, i.e A is >homeomorphic >to f(A). This is still not a sensible question. You have a one-to-one continuous map f:A -> X, and you want conditions on A and X that will ensure this is a bijection. The answer is now A and X are finite sets of the same size. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Conditions on bijection days. My association with the Department is that of an alumnus. >I'm sorry but I got an error and thought the message wasn't sent, >>that's why I sent it again. >>I forgot to state that f: A-> X is an embedding also, i.e A is >>homeomorphic >>to f(A). This is still not a sensible question. You have a one-to-one continuous map f:A -> X, and you want conditions >on A and X that will ensure this is a bijection. The answer is now A >and X are finite sets of the same size. And as William hinted, this has little or nothing to do with topology. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Conditions on bijection > Let A and X be topological spaces. Let f: A-> X be > a continuous map, under what conditions of A and X > is f a bijection? >> Are you sure this is the question you meant to ask? > Let A and X be topological spaces. Let f: A-> X be > a continuous map, under what conditions of A and X > is f a bijection? Carl, I'm confused. You asked a question, Arturo addressed > your question (not copied above), then Arturo asked you > a YES/NO question about your question, and in reply (yes, > your second post was actually a reply to Arturo, and it was), > you asked the same question again, character-by-character. I'm sorry but I got an error and thought the message wasn't sent, > that's why I sent it again. > I forgot to state that f: A-> X is an embedding also, i.e A is > homeomorphic > to f(A). So, f is a bijection from A to f(A). If f(A) = X, then f is a bijection from A to X. Still doesn't seem like a very sensible question. Care to explain what you are trying to do? -- === Subject: positive definite matrix and inner product Concerns this question from an old term test: Let V be a real inner product space having a basis B = {v_1, v_2, ... , v_n}. Define a real n x n matrix A = [a_ij] where a_ij = < v_i, v_ j > , i.e., (i,j)th element is inner product of v_i and v_ j. a) Show that A is symmetric. b) Let u, w be vectors in V with u = x_1*v_1+ ... + x_n*v_n and w = y_1*v_1 ... y_n*v_n (where x_i's and y_i's are scalars/reals). Show that < u , w > = x^t * A * y where x = (x_1, ..., x_n)^t and y = (y_1, ...,y_n)^t. c) Prove that A is positive definite. I can do a) and b) with confidence in my solutions, but I'm not quite sure about my solution to c). I think I'm missing something because my solution looks too simply relative to a), b). my solution to part c): by part b), x^t * A * x = = || x || Since for all x<>0, 0 same is true for x^t * A * x, i.e., x^t * A * x > 0 for all x<>0 which by definition implies A is positive semidefinite Solution has to have gaps, since it seems trivial (my solution, that is). Just verifying whether it's a correct solution or not would be Will appreciate any help. === Subject: Re: positive definite matrix and inner product v_2, ... , v_n}. Define a real n x n matrix A = [a_ij] where a_ij = > < > v_i, v_ j > , i.e., (i,j)th element is inner product of v_i and v_ j. a) Show that A is symmetric. b) Let u, w be vectors in V with u = x_1*v_1+ ... + x_n*v_n and w = > y_1*v_1 ... y_n*v_n (where x_i's and y_i's are scalars/reals). > Show that < u , w > = x^t * A * y where x = (x_1, ..., x_n)^t and y > = > (y_1, ...,y_n)^t. c) Prove that A is positive definite. I can do a) and b) with confidence in my solutions, but I'm not quite > sure about my solution to c). I think I'm missing something because > my solution looks too simply relative to a), b). my solution to part c): by part b), x^t * A * x = = || x || Since for all x<>0, 0 same is true for x^t * A * x, > i.e., x^t * A * x > 0 for all x<>0 > which by definition implies A is positive semidefinite Solution has to have gaps, since it seems trivial (my solution, that > is). Just verifying whether it's a correct solution or not would be > Will appreciate any help. The solution is fine. If it looks too simply relative to > a), b), I'd say you need to take another look at your > answers to a and b. In particular, a follows from a single > line of algebra. > Yes, but c) required 0 thinking at all, whereas a) required a few milliseconds. And c) seems to follow directly from b) , which is why I doubted my solution. === Subject: Re: Probability Theory > Say we have 6 blue socks, 4 yellow, and 2 green. We pick 4 socks at random, what is the probability that we get a pair of the same color. We would compose combinations such as (6C2*4C2*2C0)/12C4 + (6C2*4C0*2C2)/12C4 + (6C2*4C1*2C1)/12C4 + (6C0*4C2*2C2)/12C4 + (6C2*4C2*2C0)/12C4 + .... My question is do we repeat values here, such as in this example (6C2*4C2*2C0)/12C4 would repeat twice. So do we just include it once or have it factored in twice? > I hope that you meant this question as a joke! It seems that you may have fooled Randy P.? Bil J === Subject: Re: Probability Theory > Say we have 6 blue socks, 4 yellow, and 2 green. We pick 4 socks at random, what is the probability that we get a pair of the same color. We would compose combinations such as (6C2*4C2*2C0)/12C4 + (6C2*4C0*2C2)/12C4 + (6C2*4C1*2C1)/12C4 + (6C0*4C2*2C2)/12C4 + (6C2*4C2*2C0)/12C4 + .... My question is do we repeat values here, such as in this example (6C2*4C2*2C0)/12C4 would repeat twice. > So do we just include it once or have it factored in twice? There's an easier way to solve this problem (already suggested by someone else). But I'll address your question. You sum probabilities if all terms represent mutually-exclusive events. If two terms represent events that can overlap, then you will be double-counting some possibilities and end up with a value that is too high. However, sometimes that approach turns out to be convenient: add up P(A) + P(B) + P(C) even though P(A) includes cases where B occurs and where it doesn't. Then subtract out P(AB), etc to account for the double counting. Then add P(ABC), etc for the same reason... Look up inclusion-exclusion principle if you're interested. In this case, I'm not sure what your two occurrences of (6C2*4C2*2C0) represent, so I can't comment on whether they are overlapping events. - Randy === Subject: Re: Probability Theory > Say we have 6 blue socks, 4 yellow, and 2 green. We pick 4 socks at random, what is the probability that we get a pair of the same color. We would compose combinations such as We *could*, but consider... how would you go about picking - eyes open, not randomly - four socks from this collection so as *not* to get a pair? You would pick a red, then a yellow, then ... -- Larry Lard The address is real, but unread - please reply to the group === Subject: Re: Probability Theory > Say we have 6 blue socks, 4 yellow, and 2 green. We pick 4 socks at random, what is the probability that we get a pair of the same color. We would compose combinations such as (6C2*4C2*2C0)/12C4 + (6C2*4C0*2C2)/12C4 + (6C2*4C1*2C1)/12C4 + (6C0*4C2*2C2)/12C4 + (6C2*4C2*2C0)/12C4 + .... My question is do we repeat values here, such as in this example (6C2*4C2*2C0)/12C4 would repeat twice. So do we just include it once or have it factored in twice? > Think before calculating! With three colours available, what is the probability that we do *not* get a pair of the same colour? === Subject: Re: Probability Theory Say we have 6 blue socks, 4 yellow, and 2 green. We pick 4 socks at random, what is the probability that we get a pair of the same color. We would compose combinations such as (6C2*4C2*2C0)/12C4 + (6C2*4C0*2C2)/12C4 + (6C2*4C1*2C1)/12C4 + (6C0*4C2*2C2)/12C4 + (6C2*4C2*2C0)/12C4 + .... My question is do we repeat values here, such as in this example (6C2*4C2*2C0)/12C4 would repeat twice. So do we just include it once or have it factored in twice? > Think before calculating! > With three colours available, what is the probability that we do *not* > get a pair of the same colour? Seems to me it's not quite trivial, but you can condition on the first sock. P(two different colors) = P(2nd sock is Y or G | 1st sock is B) P(1st sock is B) + etc. - Randy === Subject: Re: Probability Theory > Say we have 6 blue socks, 4 yellow, and 2 green. We pick 4 socks at random, what is the probability that we get a pair of the same color. > We would compose combinations such as > (6C2*4C2*2C0)/12C4 + (6C2*4C0*2C2)/12C4 + (6C2*4C1*2C1)/12C4 + (6C0*4C2*2C2)/12C4 + (6C2*4C2*2C0)/12C4 + .... > My question is do we repeat values here, such as in this example (6C2*4C2*2C0)/12C4 would repeat twice. So do we just include it once or have it factored in twice? > Think before calculating! > With three colours available, what is the probability that we do *not* > get a pair of the same colour? Seems to me it's not quite trivial, but you can condition on the first > sock. P(two different colors) = P(2nd sock is Y or G | 1st sock is B) P(1st > sock is B) + etc. - Randy I hope that you are just going along with the joke? Bill J === Subject: division by 7 Last month I appeared for an interview with EA sports and they asked me this question. How would you divide a number by 7 without using division operator ? I did by doing a subtraction and keeping a counter that kept a tab on how many times I subtracted. Later, the EA sport guy told me that of course there are can be better technique by using bit operator. Since 7 has a binary representation 111, my guess is that a left shift operation of 3 bits should give the answer, but I couldn't get it to work. Any comments ? === Subject: Re: division by 7 <010220071540018266%anniel@nym.alias.net.invalid > Last month I appeared for an interview with EA sports and they asked > me this question. How would you divide a number by 7 without using division operator ? I did by doing a subtraction and keeping a counter that kept a tab on > how many times I subtracted. Later, the EA sport guy told me that of course there are can be better > technique by using bit operator. Since 7 has a binary representation 111, my guess is that a left shift > operation of 3 bits should give the answer, but I couldn't get it to > work. Any comments ? shift by three multiplies by 8 = 1000 binary, not 7. How about this... 1/7 = (-1)/(1-8) = (-1)(1 + 8 + 8^2 + 8^3 + ...) everything mod 2^N, where N is the word size. So to divide by 7: shift left 3, then again, then again, etc. > add them all together, take the 2s complement. ... which can be combined to essentially one multiplication - standard trick. But this works only for integers in binary representation, whereas the question was about numbers. What was the expected answer for dividing pi by 7? === Subject: Re: Simple Polygon (Triangle) Decomposition On 31 Jan., 07:25, Gerry Myerson I am wondering if it is possible to divide a triangle into three equal > parts. The three parts should be a (any) simple polygon. > They also should be equal (allowed is rotation, translation and > mirroring). > (like someone trying to cut a triangle on paper with a scissor) If the triangle is equilateral it is not hard to see the solution. > Cut from each corner to the center and you have three equal triangles. But at the momemt I am stuck at an isosceles traingle. > I conjecture, that it is not possible to cut it, but mabe I am wrong. Any suggestions? You may find what you want in this paper, or in the book mentioned > in this review: MR1333711 (96f:52032) > Laczkovich, M.(H-EOTVO-AN) > Tilings of triangles. (English summary) > Discrete Math. 140 (1995), no. 1-3, 79--94. > 52C20 (05B45) In Part 1 of his book ref[ How does one cut a triangle?, Center > Excell. Math. Ed., Colorado Springs, CO, 1990; MR1050322 (91f:51002)], > the reviewer was interested in finding all integers $n$ such that every > triangle $T$ can be dissected into $n$ triangles similar to each other > (or congruent to each other). The author here continues this direction > with a special interest in the kinds and quantity of similar (or > congruent) to each other triangles $Delta$ into which the triangle $T$ > can be dissected. Given a non-equilateral triangle $T$, if $T$ can be > dissected into triangles that are all similar to one triangle $Delta$, > which is not similar to $T$, then we include $Delta$ in the set > $S(T)$. The author proves that $S(T)$ contains at most 6 triangles that > are not similar to each other. Given a triangle $T$, if $T$ can be > dissected into triangles that are all congruent to one triangle > $Delta$, which is not similar to $T$, then we include $Delta$ in the > set $C(T)$. The author proves that there are infinitely many isosceles > triangles $T$ such that $C(T)$ includes at least 6 triangles that are > not similar to each other. Finally, the author studies dissections of > equilateral triangles. In the process, he discovers a good number of > ingenious dissections. Reviewed by Alexander Soifer -- > Gerry Myerson (g...@maths.mq.edi.ai) (i -> u for email) I have stumpled upon Laczkovich several times. At http://mathworld.wolfram.com/Dissection.html for example one can see some interesting dissections. But in my case the number of parts have to be exactly three. And this differs a bit from the case above, where the numbers of dissection can very. An equilateral tringle has an infinitive number of solutions. You can (if you wish) decompose it into very complex polygons. To do so, just start 3 paths at every corner simultaneously and head for the center. While heading for the center move as complex as you like as long as every path does the same (relativ) and the paths do not cross and you do not leave the border. After reaching the center, you are finsihed. Just cut along the lines the 3 paths defined. Benjamin === Subject: Re: Simple Polygon (Triangle) Decomposition > On Jan 30, 4:32 am, Benjamin A. Bartsch > It is good to hear that my conjecture may be right. > I know that it can be divided in 4 pieces, > but I am only interested in dividing it into 3 pieces. Benjamin Is it just a conjecture that you came up with or did you hear this as > a problem somewhere? It's pretty interesting, but I don't know how one > could prove it. I'm sure you can't do it in general with three > triangles, but with three polygons it's much harder to show. Greg It comes from an extension I have to do for my diploma thesis. I have to extend the case of splitting a polygon into two congruent pieces from Kimmo Eriksson to the case of three pieces. I have created an algorithm to solve this problem. Until yet the algorithm did not fail on any (simple) polygon as input. Which means, that if (and only if) the polygon is splitable into 3 congruent pieces (simple polygons), then the algorithm does so. Later I have to prove that the algorithm always works. This also proves whether a given triangle can be split or not (just run the algorithm on it). So currently I am searching for (every) different type of tringles which CAN be decomposed into three parts, just to verify the functioning of the algorithm. For example the 30-60-90 tringle G. Myerson suggested is working. Also any equilateral triangle is working. At the moment I am tuning the algorithm for the traingle case. A quadrangle starts making problem (not unsolvable ones). Benjamin === Subject: Re: Quadratic Forms, maximum > of R^n, and A is symmetric What is the maximum of Q(x) subject to someconstrainton ||x||. Say > ||x|| = C (some positive constant)? I know there is a theorem about this (was in the linear algebra book I > resold a year ago), I just don't know how to search for it on web. I > believe the answer is C * largest eigenvalue of A, or something close > to that. Now I understand why some people keep all their textbooks. it is C* max |lambda_i| where the lambda_i are the eigenvalues of the > matrix A. To see this, remember that one can make A diagonal throught > you don't know that theorem, this is one approach to prove it :) ) Can someone confirm the C*largest_eig answer? Even better, can somone > point me to a web reference (I've tried searching with quadraticform > maximumconstraint and didn't get proper results). Maybe there is a > name for this topic/theorem? Also, the definition of semidefinite includes definite matrices? > Correct? yes Semidefinite means (according to something I read) the associated > matrix has nonnegative eigenvalues, which would include the case where > all the eigenvalues are positives (positive definite matrix). Anyone > agree with this? correct Fedor?). Just a quick verification (reasoning behind your answer): So if we have a quadratic form on R^3, say Q(x) = x^t A x And we orthogonally diagonalize A = P D P^t Then we can create a new quadratic form R(y) =y^t D y where y = P^t x So any constraint ||x|| = C corresponds to ||y|| = C So now to get maximum of Q with constraint ||x|| = C, we just find the maximum of R with constraint ||y|| = C? I guess this follows because x = Py and P is orthogonal, thus the transformation does nothing to norm of vector. And then we just pick off the diagonal element of D with greatest magnitude, and multiply this by C? C* max | d_i | is the answer. Correct? === Subject: Re: Quadratic Forms, maximum > of R^n, and A is symmetric > What is the maximum of Q(x) subject to someconstrainton ||x||. Say > ||x|| = C (some positive constant)? > I know there is a theorem about this (was in the linear algebra book I > resold a year ago), I just don't know how to search for it on web. I > believe the answer is C * largest eigenvalue of A, or something close > to that. Now I understand why some people keep all their textbooks. it is C* max |lambda_i| where the lambda_i are the eigenvalues of the > matrix A. To see this, remember that one can make A diagonal throught > you don't know that theorem, this is one approach to prove it :) ) > Can someone confirm the C*largest_eig answer? Even better, can somone > point me to a web reference (I've tried searching with quadraticform > maximumconstraint and didn't get proper results). Maybe there is a > name for this topic/theorem? > Also, the definition of semidefinite includes definite matrices? > Correct? yes > Semidefinite means (according to something I read) the associated > matrix has nonnegative eigenvalues, which would include the case where > all the eigenvalues are positives (positive definite matrix). Anyone > agree with this? correct Fedor?). Just a quick verification (reasoning behind your answer): > So if we have a quadratic form on R^3, say Q(x) = x^t A x And we orthogonally diagonalize A = P D P^t Then we can create a new quadratic form R(y) =y^t D y where y = P^t > x So any constraint ||x|| = C corresponds to ||y|| = C So now to get maximum of Q with constraint ||x|| = C, we just find the > maximum of R with constraint ||y|| = C? > I guess this follows because x = Py and P is orthogonal, thus the > transformation does nothing to norm of vector. And then we just pick off the diagonal element of D with greatest > magnitude, and multiply this by C? C* max | d_i | is the answer. Correct? > yes, that's it :-) === Subject: Re: Secret Technology of THERMATE and 911 Crime the NYT has a story about the new found steel beams at the WTC, 18 feet by 4 feet, underneath the haul road surface, and taken to the Hangar 17 at the Kennedy International Airport, lets hope they stay there and they dont take them away to Asia for melting like they did with the key core beams! NYT,feb,1,07 Search for remains will go beneath an asphalt lot by David Dunlap. === Subject: Re: Semisimple Lie Algebras- <45C230A6.2010804@xs4all.nl >Why is there no Semisimple lie algebra of dimension 4, 5 or 7? > >Is it related to the root system? > > A somewhat unsatisfactory answer is implied by the classification > theorem (Killing, .83lie Cartan). See the copy of a web page of mine below. > Unsatisfactory in the sense that the proof of this theorem has a > brute-force character. It is indeed related to root systems; actually > the proof is an exhaustive enumeration of all possible > finite-dimensional root systems. > 4, 5, 7 just do not appear among the possible numbers of dimensions. Consult any textbook on Lie algebras for more details. brute-force solutions are never unsatisfactory in my eyes they are often the quickest means to a solution and give more structural knowledge in the business i work in (software engineering) the engineers that look for simple or elegant solutions often end up taking days to solve problems that can be solved in hours with some dedication to brute force for this problem in particular we could go about defining a 4-dimensional lie algebra and try to impose semisimplicity but eventually you'll start proving either distinct elements (like basis elements) are the same or some other contradiction that's brute force on one of the cases if you do it for 5 and 7 you'll follow similar steps but different structure will be found along the way eventually you might start seeing that the structure being manipulated is more general and constructive possibly discovering something like the cartan representation or creating some analogue of dynkin diagrams to represent that structure that direction leads directly to the classification theorem and i suspect is similar to what originally happened by attacking the general problem as brute force you generalise and deepen the understanding of these objects ( ps. i like your page ) -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Math Tutors Needed desperately needs tutors proficient in Mathematics (up to calculus), English (up to high school level), some sciences (physics, chemistry, etc), and SAT prep. The pay rate is competitive with other tutoring centers and is flexible depending on the skill set you can offer. The children are well behaved, respectful and are ready and willing to learn, their minds are just waiting to be molded by your able mind. For more information please visit the following website: http://www.monicalearningcenter.com/monicalearning_new/english/main/employme nt.html Or simple visit: http://www.monicalearningcenter.com === Subject: Proving an inequality for the sequence of weighted means I'm not sure if my proof is OK, would like some help please. Let a_n be a sequence of real numbers, w_n a sequence of positive weights and s_n = (Sum(i=1,n) (w(i) a(i))/(Sum(i=1, n) w(i)). If Sum(i=1,oo) w(i) diverges, then lim inf a_n <= lim inf s_n <= lim sup s_n <= lim sup _a_n (lim inf and lim sup are quite confusing to me) The middle inequality holds automatically. My proof for the left one is as follows (the right one is proved similarly): If lin inf a_n = -oo , then there's nothing to prove, right? So, suppose lim inf a_n > -oo. For every q < lim inf a(n), there exists (property of the lim inf) a natural k such that a(n) > q for every n > k. Let S(n) be the sequence of partial sums of w(n) and let w = infimum{w(1)...w(k)}. Then, for n >k we have s_n = (Sum(i=1, k) w(i) a(i) + Sum(i= k+1, n) w(i) a(i))/S(n) > (w S(k) + q (S(n) - S(k))/ (S(n) = ((w - q)S(k) + q S(n)) = ((w - q)S(k))/S(n) + q . Putting y(n) = ((w - q)S(k))/S(n) + q , we get s(n) > y(n) n for n > k. Therefore, lim inf s(n) >= lim inf y(n) Since S(n) diverges and it's terms are positive, S(n) => oo and lim inf y(n) = lim y(n) = 0 + q = q, because k is fixed and doesn't depend on n. Therefore, lim inf s(n) >= q for every q < lim inf a(n). In order for this to be true, we must have lim inf s(n) >= lim inf a(n). These inequalities imply that, if a(n) converges to a, then s(n) converges to a. This is true even if a = oo or -oo, right? But if Sum( w(i)) converges, then I think a(n) may converge to an a and s(n) converges to a s <>a Sharon === Subject: An elementary integral Hello everyone, I have looked very hard, without success so far, for a way to compute int_{-infty}^{infty} e^{-x^2}/cosh(x) dx or more generally int_{-infty}^{infty} e^{alpha x -x^2}/cosh(x) dx Any idea ? === Subject: Re: An elementary integral > Hello everyone, I have looked very hard, without success so far, for a way to compute int_{-infty}^{infty} e^{-x^2}/cosh(x) dx or more generally int_{-infty}^{infty} e^{alpha x -x^2}/cosh(x) dx Any idea ? If alpha = i*pi, then the integral evaluates to pi*exp(-pi^2/4). Evaluating the integral for other values of alpha is harder. For any alpha, there is a recurrence relation: K(alpha) + exp(2*pi^2+i*pi*alpha) K(alpha-2*pi*i) = 2*pi*exp(pi^2/4), where K(alpha) is exp(-i*pi*alpha/2) times the second integral in the original post. All of this is obtained by looking at the contour consisting of the real axis and the real axis shifted by (+-)i*pi. Hope this helps. Igor === Subject: Re: An elementary integral mathman a .8ecrit : >> Hello everyone, >> I have looked very hard, without success so far, for >> a way to compute >> int_{-infty}^{infty} e^{-x^2}/cosh(x) dx >> or more generally >> int_{-infty}^{infty} e^{alpha x -x^2}/cosh(x) dx >> Any idea ? Replace cosh(x) by (exp(x)+exp(-x))/2. You then have two integrals of the form exp(-x^2+kx), with different k's. You should be able to do the rest (Hint: complete the squares of the exponents). Hint: read the question first. === Subject: Re: An elementary integral <5741446.1170369851914.JavaMail.jakarta@nitrogen.mathforum.org Hello everyone, I have looked very hard, without success so far, for > a way to compute int_{-infty}^{infty} e^{-x^2}/cosh(x) dx or more generally int_{-infty}^{infty} e^{alpha x -x^2}/cosh(x) dx Any idea ? Replace cosh(x) by (exp(x)+exp(-x))/2. This doesn't work: it just gives int e^{-x^2}*2/[e^{x} + e^{-x}]. I suppose you could write 1/cosh(x) as 2*e{-x} * sum((-1)^n * e^(-n*x),n=0..infinity), to get a series expansion for the 0-->infinity part of the integral, and you could do something similar for the -infinity-->0 part. R.G. Vickson > You then have two integrals of the form exp(-x^2+kx), with different k's. You should be able to do the rest (Hint: complete the squares of the exponents). === Subject: Multiple, variable number of Unknowns - Need a function/pattern unknowns in a Sum term. The key difficulty is that this is being converted to a program, and the user will determine the number of unknowns by the parameters they choose. I misunderstood a crucial element however which means that unlike the first example I posted, this one can be solved: The equation will look different than I'd previously posted because I made some simplifications and got a bit further along. I'll give sample equations for a 2 unknowns and 3 unknowns version of the equation: (2 unknowns - i, k): Ai = Mi + Mi * y * (Ak * Bk) Ak = Mk + Mk * y * (Ai * Bi) (3 unknowns - i, k, n): Ai = Mi + Mi * y * (Ak * Bk + An * Bn) Ak = Mk + Mk * y * (Ai * Bi + An * Bn) An = Mn + Mn * y * (Ai * Bi + Ak * Bk) Hopefully this gives an idea how this equation is expanded. To add another unknown, just add another A with a subscript, and make sure that the parenthetical part has a sum of (A * B) terms for each subscript except the one in the A. I've solved for Ai for 1, 2 and 3 terms and came up with: Ai = Mi Ai = (Mi * ( 1 + Mk * Bk * y) ) / (1 - Mi * Mk * Bi * Bk * y^2) Ai = Mi * N/D (N = numerator, D = denominator) N = (1 + (Mk * Bk * y) + (Mn * Bn * y) + (Mk * Mn * Bk * Bn * y^2) ) D = (1 - (Mi * Mk * Bi * Bk * y^2) - (Mi * Mn * Bi * Bn * y^2) - (Mk * Mn * Bk * Bn * y^2) - (2 * Mi * Mk * Mn * Bi * Bk * Bn * y^3) ) There looks like there's a pattern, but I used QuickMath to try to solve for the four unknown version and got an absolute mess. I'm not sure if that can be simplified further, but I'm hoping someone can look at this and tell me if there's a function, algorithm, etc, that this is not for a school project, it's for work and I, my boss, and pretty much everyone involved is a bit over their head on this one I think). I'll be perfectly happy to credit anyone who provides help that goes to solving the problem in the final program (an in house piece of work). === Subject: Re: Multiple, variable number of Unknowns - Need a function/pattern unknowns are the 'A' terms. :) > unknowns in a Sum term. The key difficulty is that this is being > converted to a program, and the user will determine the number of > unknowns by the parameters they choose. I misunderstood a crucial > element however which means that unlike the first example I posted, > this one can be solved: The equation will look different than I'd previously posted because I > made some simplifications and got a bit further along. I'll give > sample equations for a 2 unknowns and 3 unknowns version of the > equation: (2 unknowns - i, k): Ai = Mi + Mi * y * (Ak * Bk) > Ak = Mk + Mk * y * (Ai * Bi) (3 unknowns - i, k, n): Ai = Mi + Mi * y * (Ak * Bk + An * Bn) > Ak = Mk + Mk * y * (Ai * Bi + An * Bn) > An = Mn + Mn * y * (Ai * Bi + Ak * Bk) Hopefully this gives an idea how this equation is expanded. To add > another unknown, just add another A with a subscript, and make sure > that the parenthetical part has a sum of (A * B) terms for each > subscript except the one in the A. I've solved for Ai for 1, 2 and 3 terms and came up with: Ai = Mi Ai = (Mi * ( 1 + Mk * Bk * y) ) / (1 - Mi * Mk * Bi * Bk * y^2) Ai = Mi * N/D (N = numerator, D = denominator) N = (1 + (Mk * Bk * y) + (Mn * Bn * y) + (Mk * Mn * Bk * Bn * y^2) ) > D = (1 - (Mi * Mk * Bi * Bk * y^2) - (Mi * Mn * Bi * Bn * y^2) - (Mk * > Mn * Bk * Bn * y^2) - (2 * Mi * Mk * Mn * Bi * Bk * Bn * y^3) ) There looks like there's a pattern, but I used QuickMath to try to > solve for the four unknown version and got an absolute mess. I'm not > sure if that can be simplified further, but I'm hoping someone can > look at this and tell me if there's a function, algorithm, etc, that > this is not for a school project, it's for work and I, my boss, and > pretty much everyone involved is a bit over their head on this one I > think). I'll be perfectly happy to credit anyone who provides help > that goes to solving the problem in the final program (an in house > piece of work). === Subject: Re: My prime counting, understanding forms educational background? Physics, currently working in Patent Office. Badum bum, really: B.Sc Biology, B. Sc Comp Sci - first class with distinction But that's not your question. I'm a whole brain thinker, that's what you're asking. > 50 bucks says you can't even say what my idea is. > The important question is can you? Where would you like to start? How about this: take a bit pattern 100... ... means repeat ad infinitum everything to the left 100... means 100100100100100 and so on Do you agree that this is NOT random? I hope so. Define Pat(1) = 100... Define Pat(n) = lhspat MERGE Pat(n-1) define lhspat = 1 followed by (location of first zero in Pat(n-1) * 2) 0's define MERGE: list lhspat length(Pat(n-1)) times with the first 1 aligned with the first zero of Pat(n-1) - ie 010000100001000 lhspat list Pat(n-1) length(lhspat) times - ie 100100100100100 AND Pat(1) -------------------------------------- AND the results 110100100101100= Pat(2) Ok, lemma 1 Pat(n) is more complicated than Pat(n-1). I'm deliberately not defining complicated here because it will inevitably sidetrack us infinitely. My website offers several definitions though. But if you choose to argue that 100... is equally or less complicated than 110100100101100... I'm going to accuse you of arguing for the sake of arguing. Ok, so who cares about all this stuff overall? Well the first 0 in any Pat(n) is always prime. If you want a rigorous proof, it's an easy one, let me know. If the first zero in Pat(n) is immediately followed by another zero, you have a prime twin. To review, we so far have a recursive self complicating algorithm which is laying down primes. Complication is a lemma at the moment given by complexity(Pat(n)) > complexity(Pat(n-1)) I think that's enough for one post. I leave it for you now, what would you like to know next? === Subject: Re: My prime counting, understanding forms > I don't know what that issue is. But seeing how judgemental and quick > to dismiss the lot of you are with ideas, you'll forgive me if I > dismiss your judgement that JSH has nothing to say. Because your ideas, which you believe to be true, are quickly dismissed, all quick dismissals are untrustworthy? Wow. -- Larry Lard The address is real, but unread - please reply to the group === Subject: Re: Question: Closed form for an integral the integration yields a very long solution. Even worse than that, it requires solving a polynomial equation of the same degree as f(x) is. I have paste the solution that matlab found for a toy problem of f(x)=(a*x^3+b*x^2+c*x+d. Is there any other idea on defining the expression under integration to achieve a straight foward answer, I mean something that doe snot require solving a polynomial equation. >> int (1/(a*x^3+b*x^2+c*x+d)^2,x) ans = ((c*b^2+3*d*a*b-4*a*c^2)/(27*a^2*d^2+4*d*b^3+4*c^3*a-18*d*c*a*b- b^2*c^2)+(-7*c*a*b+2*b^3+9*a^2*d)/ (27*a^2*d^2+4*d*b^3+4*c^3*a-18*d*c*a*b-b^2*c^2)*x-2*a*(-b^2+3*a*c)/ (27*a^2*d^2+4*d*b^3+4*c^3*a-18*d*c*a*b-b^2*c^2)*x^2)/(a*x^3+b*x^2+c*x +d)+2*sum(((9*a^2*d+b^3-4*c*a*b)/ (27*a^2*d^2+4*d*b^3+4*c^3*a-18*d*c*a*b-b^2*c^2)-a*(-b^2+3*a*c)/ (27*a^2*d^2+4*d*b^3+4*c^3*a-18*d*c*a*b-b^2*c^2)*_R)/(3*a*_R^2+2*b*_R +c)*log(x-_R),_R = RootOf(a*_Z^3+b*_Z^2+c*_Z+d)) On Feb 1, 3:55 pm, Gerry Myerson question. Given a non-negative polynomial f(x) where the leading > exponent, i.e. the largest exponent, is even, e.g. > a*x^1000+b*x^999.... Is there any way to manipulate f(x) using algebraic operations and/or > applying elementary functions to create an expression in f(x) that has > the following properties: 1. The expression is non-negative for any x. > 2. The expression has a finite value for any x, except those x's that > make f(x)=0. At such x's, the expression must be infinity (the > infinity must be positive too due to condition 1). > 3. The definite integral of the expression can be expressed in closed > form based on f(x) in a symetric inteval, e.g. [-L,L]. How about 1 over f-squared? Certainly satisfies 1 and 2 above. Partial fractions enables you, > at least in principle, to find the antiderivative. -- > Gerry Myerson (g...@maths.mq.edi.ai) (i -> u for email)- Hide quoted text - - Show quoted text - === Subject: numerator/denominator vs dividend/divisor I am writing some code for a formula = (a + b)/(c+d) I was just wondering what is the more accurate terminology: to refer to (c + d) as the denominator or divisor? Or both are okay? http://dictionary.reference.com/search?q=numerator http://dictionary.reference.com/search?q=denominator http://dictionary.reference.com/search?q=dividend http://dictionary.reference.com/search?q=divisor I also found a related discussion, but I didn't quite get it. === Subject: Re: numerator/denominator vs dividend/divisor > I am writing some code for a formula = (a + b)/(c+d) > I was just wondering what is the more accurate terminology: to refer > to (c + d) as the denominator or divisor? Or both are okay? http://dictionary.reference.com/search?q=numeratorhttp://dictionary.referenc e .com/search?q=denominatorhttp://dictionary.reference.com/search?q=dividendht t p://dictionary.reference.com/search?q=divisor > That depends on whether you are talking about fractions or operations. === Subject: Re: numerator/denominator vs dividend/divisor Am 02.02.2007 00:32 schrieb Bucky: > I am writing some code for a formula = (a + b)/(c+d) > I was just wondering what is the more accurate terminology: to refer > to (c + d) as the denominator or divisor? Or both are okay? http://dictionary.reference.com/search?q=numerator > http://dictionary.reference.com/search?q=denominator > http://dictionary.reference.com/search?q=dividend > http://dictionary.reference.com/search?q=divisor I also found a related discussion, but I didn't quite get it. > I think, that dividend/divisor is used more, if the focus is on the process of division itself, say you talk about the formula. You change to numerator/denominator if your focus is more on the *result* of the formula, as on the rational constant, the rational representation of the number which is the result of your operation. A tiny difference, though. Gottfried Helms === Subject: Re: numerator/denominator vs dividend/divisor on the *result* of the formula, as on the rational constant, > the rational representation of the number which is the > result of your operation. > A tiny difference, though. then, since everyone can get those straight, but dividend/divisor can be confusing which one is which. === Subject: Re: numerator/denominator vs dividend/divisor Am 02.02.2007 07:56 schrieb Bucky: >> You change to numerator/denominator if your focus is more >> on the *result* of the formula, as on the rational constant, >> the rational representation of the number which is the >> result of your operation. >> A tiny difference, though. then, since everyone can get those straight, but dividend/divisor can > be confusing which one is which. > Yes, that may be a good decision. Note, that with latin words it is: ~or is the active part, ~end the passive. say usurpa-tor: that one who usurpates, usurp-end would be then the one who is/is to be usurpated (I never read that word ;-) ) or ac-tor is the one who does some thing and then analoguously ac-tend would then be the one whith whom a thing is made. (there surely are far better examples, sorry for not having some currently) Gottfried Helms === Subject: Re: numerator/denominator vs dividend/divisor > I am writing some code for a formula = (a + b)/(c+d) > I was just wondering what is the more accurate terminology: to refer > to (c + d) as the denominator or divisor? Or both are okay? http://dictionary.reference.com/search?q=numeratorhttp://dictionary.referenc e .com/search?q=denominatorhttp://dictionary.reference.com/search?q=dividendht t p://dictionary.reference.com/search?q=divisor > In the GMP math library, the components of a rational are numer() for the numerator and denom() for the denominator: > import gmpy > r = gmpy.mpq(1,3) > r mpq(1,3) > r.numer() mpz(1) > r.denom() mpz(3) > r = r * 3 > r mpq(1) To use any other terminology would be confusing (in this case). === Subject: oscillation of points Let G(n) be a decreasing sequence of real numbers with limit 0. Define f as follows: f(0) = 1 f(x) = 0 if x irrational f(m/n) = G(n) if x rational What is the oscillation at each point in [0,1]? The oscillation is defined by: Let f be defined and bounded on an interval S. Let T be a subset of S. O(T) = sup{f(x) - f(y) : x,y in T} [oscillation of a set] w(x) = lim_{h-->0} O( B(x, h) / S) [oscillation at point x, where B(x,h) is open ball about x] Now it seems at x = 0, the oscillation must be 1. Because no matter how small we make the ball, we will have 0 and some irrational number, so we will get 1 - 0 = 1. If x irrational, then every open ball about x will contain infinitely many rationals which correspond to G(n). So I'm thinking as we make the ball smaller and smaller the rational numbers nearby are approximating the irrational x, and the denominator of such rationals must be getting large; and since G(n)-->0 as n goes large, the irrational points have 0 oscillation? Does that argument work? Is it similar for rationals? === Subject: Re: oscillation of points > Let G(n) be a decreasing sequence of real numbers with limit 0. > Define f as follows: f(0) = 1 > f(x) = 0 if x irrational > f(m/n) = G(n) if x rational What is the oscillation at each point in [0,1]? For this to be well defined, you have to say a bit more. For example, -2/3 = 2/(-3) = -8/12, etc. The usual convention is m/n is in lowest terms with n > 0. For this to be well defined, you have to say just a bit more. For example, -2/3 = 2/(-3) = -8/12, etc. The usual convention is m/n is in lowest terms with n > 0. The oscillation will be 0 if x is irrational, G(n) if x = m/n (above convention) is rational, and 1 if x = 0. I posted a fair amount of information about this type of function not long ago. See the following post I made on Dec. 13 and the follow-up post in the same thread I made on Dec. 14. Dave L. Renfro === Subject: Re: oscillation of points Let G(n) be a decreasing sequence of real numbers with limit 0. > Define f as follows: f(0) = 1 > f(x) = 0 if x irrational > f(m/n) = G(n) if x rational What is the oscillation at each point in [0,1]? For this to be well defined, you have to say a bit > more. For example, -2/3 = 2/(-3) = -8/12, etc. The > usual convention is m/n is in lowest terms with n > 0. For this to be well defined, you have to say just > a bit more. For example, -2/3 = 2/(-3) = -8/12, etc. > The usual convention is m/n is in lowest terms with n > 0. > The oscillation will be 0 if x is irrational, > G(n) if x = m/n (above convention) is rational, > and 1 if x = 0. I posted a fair amount of information > about this type of function not long ago. See the > following post I made on Dec. 13 and the follow-up > post in the same thread I made on Dec. 14. > preview option when posting? One would think that if the technology was available to preview posts in the last few years, that we'd still have the technology to preview posts this year, but maybe I'm missing something. (Like the fact that previewing posts is only possible in those years whose digits add to less than 9, or something equally inane.) Dave L. Renfro === Subject: JSH: Do we make math in order to earn money? JSH DOES NOT MEAN ANYTHING If we consider it futile, and if we don't waste our time over a word that doesn't mean anything... The first thought that comes to these minds is of a bacteriological order: at least to discover its etymological, historical or psychological meaning. We read in the papers that the negroes of the Kroo race call the tail of a sacred cow: JSH. A cube, and a mother, in a certain region of Italy, are called: JSH. The word for a hobby horse, a children's nurse, a double affirmative in Russian and Romanian, is also: JSH. Some learned journalists see it as an art for babies, other Jesus calling the little children unto him saints see it as a return to an unemotional and noisy primitivism - noise and monotonous. A sensitivity cannot be built on the basis of a word; every sort of construction converges into a boring sort of perfection, a stagnant idea of a golden swamp, a relative human product. A work of Math shouldn't be beauty per se, because it is dead; neither gay nor sad, neither light nor dark; it is to rejoice or maltreat individualities to serve them up the cakes of sainted haloes or the sweat of a meandering chase through the atmosphere. A work of math is never beautiful, by decree, objectively, for everyone. Criticism is, therefore, useless; it only exists subjectively, for every individual, and without the slightest general characteristic. Do people imagine they have found the psychic basis common to all humanity? The attempt of Jesus, and the Bible, conceal, under their ample, benevolent wings: , animals and days. How can anyone hope to order the chaos that constitutes that infinite, formless variation: JSH? The principle: Love thy neighbour is hypocrisy. Know thyself is utopian, but more acceptable because it includes malice. No pity. After the carnage we are left with the hope of a purified humanity. I always speak about myself because I don't want to convince, and I have no right to drag others in my wake, I'm not compelling anyone to follow me, because everyone makes his art in his own way, if he knows anything about the joy that rises like an arrow up to the astral strata, or that which descends into the mines stewn with the flowers of corpses and fertile spasms. Stalactites: look everywhere for them, in creches magnified by pain, eyes as white as angels' hares. Thus JSH was born out of a need for independence, out of mistrust for the community. People who join us keep their freedom. We don't accept any theories. We've had enough of the cubist and futurist academies: laboratories of formal ideas. Do we make math in order to earn money and keep the dear bourgeoisie happy? Rhymes have the smack of money, and inflexion slides along the line of the stomach in profile. Every group of mathemitions has ended up at this bank, straddling various comets. Leaving the door open to the possibility of wallowing in comfort and food. Here we are dropping our anchor in fertile ground. === Subject: integral identity help If f, f^2, g, g^2 are riemann-stieltjes integrable on [a,b] with respect to z, show that: [all integrals are from a to b] 1/2 * int( int( |A|^2 dz(y) ) dz(x) ) = ( int( f(x)^2 dz(x) )*( int( g(x)^2 dz(x) ) - ( int( f(x)g(x) dz(x) )^2 where A is the matrix: [f(x) g(x) ; f(y) g(y)] where semi-colon denotes new row, and |A| means determinant. So it seems somehow they integrated with respect to dz(y) since only dz(x)'s remain. I don't see how this is done. I expanded the determinant and squared it, and tried to do iterative integration, but I wasn't able to get far. === Subject: Re: integral identity help >If f, f^2, g, g^2 are riemann-stieltjes integrable on [a,b] with >respect to z, show that: [all integrals are from a to b] 1/2 * int( int( |A|^2 dz(y) ) dz(x) ) = ( int( f(x)^2 dz(x) )*( int( g(x)^2 dz(x) ) - ( int( f(x)g(x) >dz(x) )^2 where A is the matrix: [f(x) g(x) ; f(y) g(y)] where semi-colon >denotes new row, and |A| means determinant. >So it seems somehow they integrated with respect to dz(y) since only >dz(x)'s remain. I don't see how this is done. I expanded the >determinant and squared it, and tried to do iterative integration, but >I wasn't able to get far. > What they are doing is replacing the integration variable y with x in each term. We can do this since the integration bounds are the same for both x and y (i.e. the domain is a square). We have: int int (f(x)g(y)-g(x)f(y))^2 dx dy = int int [ f(x)^2 g(y)^2 + f(y)^2 g(x)^2 - 2 f(x)g(x)f(y)g(y) ] dx dy = int int f(x)^2 g(y)^2 dx dy + int int f(y)^2 g(x)^2 dx dy - 2 int int f(x) g(x) f(y) g(y) dx dy = (int f(x)^2 dx) (int g(y)^2 dy) + (int f(y)^2 dy) (int g(x)^2 dx) - 2 (int f(x) g(x) dx) (int f(y) g(y) dy) (here is where the substitution takes place) = 2 (int f(x)^2 dx)(int g(x)^2 dx) - 2 (int f(x) g(x) dx)^2 -- Daniel Mayost === Subject: Re: A cubic is a torus >>non-singular irreducible cubic in P^2 over the complex field is >>homeomorphic to a torus in R^3. > It depends a lot on what techniques you know and have available. > For instance, your previous posts make it clear you know why the > points of the cubic can be given the structure of a group in a > natural way, using geometry. Do you know enough about Lie groups > to (1) see why that construction actually makes the cubic into a > Lie group, and (2) see why the only connected 2-dimensional manifold > that can be made into a compact Lie group is the torus? (I see that > I'm also assuming, from earlier posts, that you know why the cubic > is a connected 2-dimensional manifold, and compact.) I would have thought it would be even easier to show that a 2-dimensional _abelian_ compact Lie group was a torus? But I wonder what the connection is between this approach, and my - probably ignorant - suggestion that one consider the elliptic curve C/L arising from a lattice? I thought it was reasonably difficult to show that every elliptic curve over C arises in this way from some lattice L, basically by showing that the j-invariant takes all values. Does the Lie group argument give a simple proof of this? Or does the C/L argument give something more? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: A cubic is a torus >non-singular irreducible cubic in P^2 over the complex field is >homeomorphic to a torus in R^3. > It depends a lot on what techniques you know and have available. >> For instance, your previous posts make it clear you know why the >> points of the cubic can be given the structure of a group in a >> natural way, using geometry. Do you know enough about Lie groups >> to (1) see why that construction actually makes the cubic into a >> Lie group, and (2) see why the only connected 2-dimensional manifold >> that can be made into a compact Lie group is the torus? (I see that >> I'm also assuming, from earlier posts, that you know why the cubic >> is a connected 2-dimensional manifold, and compact.) I would have thought it would be even easier to show that >a 2-dimensional _abelian_ compact Lie group was a torus? Well, I suppose it depends on what theorems one knows (of). It's easy to show (and I think that, if pressed, I could do it) that the fundamental group of a topological group is abelian, so the only surfaces (compact, connected, without boundary) that could be topological groups are the sphere, the real projective plane, the torus, and the Klein bottle. It's true (but not so easy that I think I could do it without a lot of help) that the second homotopy group of a Lie group is trivial. That rules out the sphere and the real projective plane. And a complex plane curve is oriented, so the Klein bottle is out, too. >But I wonder what the connection is between this approach, >and my - probably ignorant - suggestion that one consider >the elliptic curve C/L arising from a lattice? I thought it was reasonably difficult to show that >every elliptic curve over C arises in this way from some lattice L, >basically by showing that the j-invariant takes all values. Does the Lie group argument give a simple proof of this? >Or does the C/L argument give something more? Much more, I think. I make no claim about the lattice inside the universal cover. My proof is just topology; you're doing arithmetic (or algebra, or complex analysis). Lee Rudolph === Subject: An inequality Hi: Could someone help me with proving this inequality: 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 where a and c>=0. I can think of two inequalities which might be useful in proving the above inequality: a^2 + c^2 >= 2*ac and a^3 + 2*c^3 >= 3*a*c^2 but I cannot still prove the first inequality using these last two ones. === Subject: Re: An inequality > Hi: Could someone help me with proving this inequality: 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 where a and c>=0. I can think of two inequalities which might be useful in proving the > above inequality: > a^2 + c^2 >= 2*ac > and > a^3 + 2*c^3 >= 3*a*c^2 but I cannot still prove the first inequality using these last two > ones. > Consider 50*a^3 + 54*c^3 - 135*a*c^2 + 72*c*a^2 >= 0 Let f(x) = 50*x^3 + 72*x^2 - 135 * x + 54. Then using calculus, f(x) has a local maximum of f(x) = (81162 + 4239*sqrt(314))/625 ~= 250.1436 at x = -(24 + 3* sqrt(314))/50 ~= -1.5432, and a local minimum of f(x) = (81162 - 4239*sqrt(314))/625 ~= 9.6748 at x = (-24 + 3* sqrt(314))/50 ~= +0.5832 So for all a/c = x > 0, 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 In fact for all x > -2.6252, f(x) will be positive. === Subject: Re: An inequality > Could someone help me with proving this inequality: 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 where a and c>=0. Let x = a / c, then you have 50 x^3 + 54 + 72 x^2 ge 135 x, which is to say you're trying to decide whether 50 x^3 + 72 x^2 - 135 x + 54 = 0 has any solutions with x > 0. Shouldn't be too hard.... -- === Subject: Inequality of degree 3 Hi: Could someone help me with proving this inequality: 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 where a and c>=0. I can think of two inequalities which might be useful in proving the above inequality: a^2 + c^2 >= 2*ac and a^3 + 2*c^3 >= 3*a*c^2 but I cannot still prove the first inequality using these last two ones. === Subject: Re: Inequality of degree 3 > Hi: Could someone help me with proving this inequality: 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 where a and c>=0. I can think of two inequalities which might be useful in proving the > above inequality: > a^2 + c^2 >= 2*ac > and > a^3 + 2*c^3 >= 3*a*c^2 but I cannot still prove the first inequality using these last two > ones. Rewrite the LHS as 50*a^3 + 18*c^3 + 18*c^3 + 18*c^3 + 72*c*a^2 . Apply the A.M.-G.M. inequality. (We can replace 135 with 145. ) === Subject: Re: Inequality of degree 3 = 135*a*c^2 where a and c>=0. I can think of two inequalities which might be useful in proving the > above inequality: > a^2 + c^2 >= 2*ac > and > a^3 + 2*c^3 >= 3*a*c^2 but I cannot still prove the first inequality using these last two > ones. > Rewrite the LHS as > 50*a^3 + 18*c^3 + 18*c^3 + 18*c^3 + 72*c*a^2 . > Apply the A.M.-G.M. inequality. (We can replace 135 with 145. ) The inequality that I originally stated is derived from the following inequality: 27a^3 + 14b^3 + 54c^3 + 54abc + 9a*b^2 + 18c*b^2 >= 81a*c^2 + 54b*c^2 by putting b=a. Now I don't know how to extend the proof for the two variable case to the three variable case. I am trying to use the A.M.- G.M. inequality and Schur's inequality which is as follows a^3 + b^3 + c^3 + 3abc >= a*b^2 + a*c^2 + b*a^2 + b*c^2 + c*a^2 + c*b^2 === Subject: Re: Inequality of degree 3 >> Hi: >> Could someone help me with proving this inequality: >> 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 >> where a and c>=0. >> I can think of two inequalities which might be useful in proving the >> above inequality: >> a^2 + c^2 >= 2*ac >> and >> a^3 + 2*c^3 >= 3*a*c^2 >> but I cannot still prove the first inequality using these last two >> ones. >> Rewrite the LHS as >> 50*a^3 + 18*c^3 + 18*c^3 + 18*c^3 + 72*c*a^2 . >> Apply the A.M.-G.M. inequality. >> (We can replace 135 with 145. ) The inequality that I originally stated is derived from the following > inequality: 27a^3 + 14b^3 + 54c^3 + 54abc + 9a*b^2 + 18c*b^2 >= 81a*c^2 + 54b*c^2 by putting b=a. Now I don't know how to extend the proof for the two > variable case to the three variable case. I am trying to use the A.M.- > G.M. inequality and Schur's inequality which is as follows a^3 + b^3 + c^3 + 3abc >= a*b^2 + a*c^2 + b*a^2 + b*c^2 + c*a^2 + > c*b^2 > Looks like we can prove 27a^3 + 54c^3 + 54abc + 18c*b^2 >= 81a*c^2 + 54b*c^2. LHS - RHS = 27(a^3 + 2c^3 - 3a*c^2) - 18bc*(3c - 3a - b) = 27(c-a)^2 * (2c+a) - 18bc *(3c - 3a - b) (**) If 3c <= 3a + b, then (**) is >= 0 and we're done. So assume 3c > 3a + b. (**) = 3 * [(3c - 3a - b) + b]^2 * (2c+a) - 18bc* (3c - 3a - b) >= 3 * [4b * (3c - 3a - b)] * (2c+a) - 18bc * (3c - 3a - b) = 3b * (3c - 3a - b) * [4(2c+a) - 6c] = 3b * (3c - 3a - b) * (2c + 4a) >= 0. === Subject: Re: Inequality of degree 3 <%48xh.766$Tx5.0@newsfe11.lga> Hi: > Could someone help me with proving this inequality: > 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 > where a and c>=0. > I can think of two inequalities which might be useful in proving the >> above inequality: >> a^2 + c^2 >= 2*ac >> and >> a^3 + 2*c^3 >= 3*a*c^2 > but I cannot still prove the first inequality using these last two >> ones. >> Rewrite the LHS as >> 50*a^3 + 18*c^3 + 18*c^3 + 18*c^3 + 72*c*a^2 . >> Apply the A.M.-G.M. inequality. > (We can replace 135 with 145. ) The inequality that I originally stated is derived from the following > inequality: 27a^3 + 14b^3 + 54c^3 + 54abc + 9a*b^2 + 18c*b^2 >= 81a*c^2 + 54b*c^2 by putting b=a. Now I don't know how to extend the proof for the two > variable case to the three variable case. I am trying to use the A.M.- > G.M. inequality and Schur's inequality which is as follows a^3 + b^3 + c^3 + 3abc >= a*b^2 + a*c^2 + b*a^2 + b*c^2 + c*a^2 + > c*b^2 Looks like we can prove > 27a^3 + 54c^3 + 54abc + 18c*b^2 >= 81a*c^2 + 54b*c^2. LHS - RHS = 27(a^3 + 2c^3 - 3a*c^2) - 18bc*(3c - 3a - b) > = 27(c-a)^2 * (2c+a) - 18bc *(3c - 3a - b) (**) If 3c <= 3a + b, then (**) is >= 0 and we're done. > So assume 3c > 3a + b. (**) = 3 * [(3c - 3a - b) + b]^2 * (2c+a) - 18bc* (3c - 3a - b) > >= 3 * [4b * (3c - 3a - b)] * (2c+a) - 18bc * (3c - 3a - b) > = 3b * (3c - 3a - b) * [4(2c+a) - 6c] > = 3b * (3c - 3a - b) * (2c + 4a) > >= 0. === Subject: Re: Inequality of degree 3 Hi: > Could someone help me with proving this inequality: > 50*a^3 + 54*c^3 + 72*c*a^2 >= 135*a*c^2 > where a and c>=0. > I can think of two inequalities which might be useful in proving the > above inequality: > a^2 + c^2 >= 2*ac > and > a^3 + 2*c^3 >= 3*a*c^2 > but I cannot still prove the first inequality using these last two > ones. > Rewrite the LHS as > 50*a^3 + 18*c^3 + 18*c^3 + 18*c^3 + 72*c*a^2 . > Apply the A.M.-G.M. inequality. (We can replace 135 with 145. ) The inequality that I originally stated is derived from the following > inequality: 27a^3 + 14b^3 + 54c^3 + 54abc + 9a*b^2 + 18c*b^2 >= 81a*c^2 + 54b*c^2 by putting b=a. Now I don't know how to extend the proof for the two > variable case to the three variable case. I am trying to use the A.M.- > G.M. inequality and Schur's inequality which is as follows a^3 + b^3 + c^3 + 3abc >= a*b^2 + a*c^2 + b*a^2 + b*c^2 + c*a^2 + > c*b^2 Actually the most general case is as follows, prove (9i^2-4i^3)a^3 + (9j^2-4j^3)b^3 + (9k^2-4k^3)b^3 + 3ijk*abc >= (12*j*i^2-9ij)b*a^2 + (12*k*i^2-9ik)c*a^2 + (12*i*j^2-9ij)a*b^2 + (12*k*j^2-9jk)c*b^2 + (12*i*k^2-9ik)a*c^2 + (12*j*k^2-9jk)b*c^2 where a,b and c,i,j,k are non-negative real numbers and i=m1*m2, j=m1*m3, k=m2*m3 and m1+m2+m3=3 for some m1,m2,m3 which are again non- negative. === Subject: Re: I'd like suggestions on how to improve my math 01/18/2007 >I'm 15, like Math and try to study on my own more than required by >school. I'll try to give you some advise based on my experience, but some of the advice will involve questions for you to answer. Also, be aware that everybody is different; books that I found hard you may find easy while books that I found easy you may find hard. Also, you might wind up reading more recent editions than what I read. First, as other posters have mentioned, happiness is a good library. I grew up in Detroit, where the library system was excellent, but I now live in Northern Virginia, where the library system is essentially useless for anybody trying to learn Mathematics on their own. So the first question is: How good is your local library system? Related to that, will your local university allow you to use their library? I was able to audit classes at a local university while I was in high school. Have you explored that option? If you have the necessary Mathematical maturity, have a good[1] school that you can get to and their policy allows it, I would encourage you to take advantage of it. Expect to do all of the homework and to take all of the exams that are required of the other students. Even if auditing courses is not an option, talking to the faculty might help to clarify your options and to come up with other resources. If there's a good school at all close, I'd strongly encourage you to make and maintain contact with the Mathematics department. Have you explored resources on the Internet? Have you talked to your parents? They may know of resources that you don't, or have contacts that could help. I'm not sure what you mean by an elementary level. It is possible that you already know enough Algebra to handle the other areas that use it, or you might need to strengthen your grasp of Group Theory, Ring Theory and Field Theory. You will definitely need Linear Algebra. I'd recommend getting some exposure to Set Theory Topology Real Analysis Complex Analysis Look for text books that are oriented towards Mathematics students rather than Engineering or Physics students. Some books that I found to be easy reading and quite helpful were Apostol, Mathematical Analysis Halmos, Finite Dimensional Vector Spaces Kelley, General Topology I'm sure that others in the news group will suggest alternatives and will suggest texts for areas that I haven't covered. When you're at the point where you can handle graduate-level topics, see whether your local Mathematics department will allow you to attend their colloquia. >So, I'd like suggestions on how I can carry on with math. In general, I'd say that you should do what you enjoy, except where a particular topic is uninteresting but will be needed in order to learn something that you are interested in. Given your interest in Analysis, you might eventually look into Functional Analysis and Differential Geometry. Whatever you study, work enough of the exercises to get a good grasp of the material. Work hard, but not so hard that it stops being fun. Others have told you not to worry about rigor. Well, in my case I was totally unable to learn calculus from a book that ignored rigor; it seemed[2] to be self contradictory and to make no sense. When I read and Epsilon-Delta explanation everything suddenly became simple. So you might find a rigorous book to be easier. Pay no attention to any moron that tells you that women can't do Mathematics; they can and have done world class work or better. Usenet is an unregulated and chaotic set of fora. You will encounter people who are polite but unreliable, people who are reliable but rude, people who are polite and reliable and people who are neither. Try to keep track of who is helpful and who isn't, and, if your software allows it, put the really bad posters in your twit list. There are also some dangerous predators, so don't ever meet with present to ensure your safety. [1] One that has a good graduate school, not just a good undergraduate division. [2] And, in fact, it *was* self contradictory and didn't make sense. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Physics comments from math/logic types? Already discussed on the primary targets (sci.physics,sci.physics.relativity), wondering if those interested in math and logic would have an interest in commenting: http://www.mukesh.ws/physics.html Requires some basic background in physics, particularly waves and light. === Subject: Re: Physics comments from math/logic types? > Already discussed on the primary targets > (sci.physics,sci.physics.relativity), wondering > if those interested in math and logic would have an interest in > commenting: http://www.mukesh.ws/physics.html Requires some basic background in physics, particularly waves and > light. > I find this area very interesting, but I disagree with your approach. In my opinion, you are working in an outdated paradigm of space and energy. First off, waves can propagate in empty space, such as lightwaves. Dimension itself is a suitable medium for wave propagation. All you need is the appropriate mathematics. Harris' Theorem 1.1 actually accomplishes this. Everything else is also solved by this, duality, nonlocality, order-disorder, etc. I cant really blame you because you are doing the same exact thing that everybody else does. You assume that space is like R3 or R4, and this energy stuff is just floating around in space like Turds in a Toilet. I cannot criticize you for believing what everybody else is saying. It is what it is. However - I will say this. Despite the fact that you have been led down a blind alley, detoured toward a dead end, that there is no resolution to your questions without Harris 1.1, I will say that you have done the best that could have been done with what's been given to you. === Subject: Re: Physics comments from math/logic types? if those interested in math and logic would have an interest in > commenting: > ? ?http://www.mukesh.ws/physics.html That answer is no good. Apart from the fact that the original poster > couldn't be bothered to make his point directly, your idea that by Maybe this will be more direct: During the late 1800's, physicists were wondering about the nature of light. All experiments were indicating it was a wave. However, physicists couldn't get their heads around the idea of a wave waving in nothing. Directly because of this inability, various ideas arose in physics: 1) Luminoferous Ether 4) Dual nature of light 5) Special Relativity 6) Minskowskian Geometry 7) General Relativity While not directly related, the Quantum Interpretations should also be viewed as arising from the tradition of mysticism that had taken plant in physics of the inability to understand light propagation in vacuum. The initial comments at http://www.mukesh.ws/physics.html show how light propagation in vacuum is extremely trivial to understand. (At least no more complicated than the Columbus egg.) Any good high school physics student should be able to understand that. Following that, are some comments that show SR, GR, QM interpretations etc to be essentially irrelevant, fundamentally incorrect, and nothing but a bad trip (possibly cocaine related, at the time these ideas came about, cocaine use in high society in the West was not only legal, but very prevalent) in the development of Physics. When Albert Einstein said in 1927 A good joke should not be repeated..., he was not joking at all! === Subject: Re: Physics comments from math/logic types? , [...] > While not directly related, the Quantum Interpretations > should also be viewed as arising from the > tradition of mysticism that had taken plant in > physics of the inability to understand light > propagation in vacuum. The initial comments at http://www.mukesh.ws/physics.html > show how light propagation in vacuum is extremely > trivial to understand. (At least no more complicated > than the Columbus egg.) Any good high school physics > student should be able to understand that. I looked it up. You assume your result with > 1) Consider an empty cubic foot of space. Suppose you stretch > two opposite poles of a magnet at both ends of it. Now this cubic > foot of space has a magnetic field within it. Empty space? Magnetic field? -- Michael Press === Subject: Re: Physics comments from math/logic types? [...] While not directly related, the Quantum Interpretations > should also be viewed as arising from the > tradition of mysticism that had taken plant in > physics of the inability to understand light > propagation in vacuum. The initial comments athttp://www.mukesh.ws/physics.html > show how light propagation in vacuum is extremely > trivial to understand. (At least no more complicated > than the Columbus egg.) Any good high school physics > student should be able to understand that. I looked it up. You assume your result with 1) Consider an empty cubic foot of space. Suppose you stretch > two opposite poles of a magnet at both ends of it. Now this cubic > foot of space has a magnetic field within it. Empty space? Magnetic field? -- > Michael Press Well, the concept of wave propagation in zero - once grasped - is very simple, but the explanation may play havoc with cherished personal word definitions... Maybe this one has a better explanation: http://www.mukesh.ws/transmit2.html This presents the same idea using different words and images. (Scroll down past the quoted text in there...) === Subject: Re: Physics comments from math/logic types? [...] > While not directly related, the Quantum Interpretations > should also be viewed as arising from the > tradition of mysticism that had taken plant in > physics of the inability to understand light > propagation in vacuum. > The initial comments athttp://www.mukesh.ws/physics.html > show how light propagation in vacuum is extremely > trivial to understand. (At least no more complicated > than the Columbus egg.) Any good high school physics > student should be able to understand that. I looked it up. You assume your result with > 1) Consider an empty cubic foot of space. Suppose you stretch > two opposite poles of a magnet at both ends of it. Now this cubic > foot of space has a magnetic field within it. Empty space? Magnetic field? -- > Michael Press Well, the concept of wave propagation in zero - once > grasped - is very simple, but the explanation may play > havoc with cherished personal word definitions... Maybe this one has a better explanation: http://www.mukesh.ws/transmit2.html This presents the same idea using different > words and images. (Scroll down past > the quoted text in there...) > I have to be honest, the links you posted did not convince me. Lets suppose that there's a magnetic field everywhere as described. You would expect empty space to act like it's polarized. Where does this field come from, and can you measure it ? Can it be observed ? How strong is it ? The dimensions of Length and Time constitute the aether. That's the medium of wave propagation. Only makes sense with Harris 1.1. === Subject: Re: Physics comments from math/logic types? couldn't be bothered to make his point directly, your idea that by Huh? > moving symbols according to standard mathematical sign rules we can > resolve metaphysical problems is a non-starter. The metaphysics that > mathematics employs is mapped to its symbols. So mathematics CANNOT > create a new explanatory metaphysics. Harris 1.1 is local to sci.math, and apparently the previous poster was in a rather jocular mood when posting. (Continuing in the same vein, I might respond that invoking the mighty Harris 1.1 would not be appropriate for the trivial and rudimentary material under consideration here.) === Subject: Re: Physics comments from math/logic types? > That answer is no good. Apart from the fact that the original poster > couldn't be bothered to make his point directly, your idea that by Huh? moving symbols according to standard mathematical sign rules we can > resolve metaphysical problems is a non-starter. The metaphysics that > mathematics employs is mapped to its symbols. So mathematics CANNOT > create a new explanatory metaphysics. Harris 1.1 is local to sci.math, and apparently the previous poster > was in a rather jocular mood when posting. (Continuing in the same vein, I might respond that invoking the mighty > Harris 1.1 would not be appropriate for the trivial and rudimentary > material under consideration here.) > Public apology, I forgot to post the disclaimer that my views are not mainstream, and I'm probably FOS if you know what I mean. I should probably clam up - dont go out and get an F on your physics homework on account of what I posted. Modern physics does NOT currently teach that waves can propagate space where dimension is itself the medium of wave propagation. Physics does not teach this, even though this is precisely what I believe and feel can be proven. If you are a highschool student please note that I am full of crap. Even though I'm techically right, I'm still full of crap, and you should just stick to what the book says. === Subject: Re: graph theory question Your explanation agrees perfectly with what I thought! Let's think about the state of the graph after the common vertex (call > it v_0) is > 'erased'; but before any of the edges are connected. It is possible > that the > graph consists of four disconnected subgraphs! It is impossible to > connect four subgraphs with only two edges; the Jordan Curve > Theorem > not withstanding! Bill J I am assuming that each vertex in the entire graph has four edges coming into it(i.e. is of degree 4, I'm sorry that I didn't use standard terminology in the first post), so the scenario you describe can't happen. Your last post led me to a proof though, although it seems to have nothing to do with the Jordan curve theorem. Let us suppose we erase the vertex in question, but keep the edges sort of dangling off into space. Let C1, C2, C3, and C4 denote the components of what's left containing the edges 1, 2, 3, 4 respectively. The fact that every remaining vertex is of degree 4 implies that each Ci is equal to some Cj for some j not equal to i. That is, C1,..., C4 are each equal to some other in the list. The only possibilities in which what's left is not connected to begin with are C1 = C2 != C3 = C4, C1 = C3 != C2 = C4, and C1 = C4 != C2 = C3. It is easy to check all possibilities that we connect the whole mess by either adding two and replies. Greg === Subject: Re: graph theory question > Your explanation agrees perfectly with what I thought! Let's think about the state of the graph after the common vertex (call > it v_0) is > 'erased'; but before any of the edges are connected. It is possible > that the > graph consists of four disconnected subgraphs! It is impossible to > connect four subgraphs with only two edges; the Jordan Curve > Theorem > not withstanding! Bill J I am assuming that each vertex in the entire graph has four edges > coming into it(i.e. is of degree 4, I'm sorry that I didn't use > standard terminology in the first post), so the scenario you describe > can't happen. Your last post led me to a proof though, although it > seems to have nothing to do with the Jordan curve theorem. Let us > suppose we erase the vertex in question, but keep the edges sort of > dangling off into space. Let C1, C2, C3, and C4 denote the components > of what's left containing the edges 1, 2, 3, 4 respectively. The fact > that every remaining vertex is of degree 4 implies that each Ci is > equal to some Cj for some j not equal to i. That is, C1,..., C4 are > each equal to some other in the list. The only possibilities in which > what's left is not connected to begin with are C1 = C2 != C3 = C4, C1 > = C3 != C2 = C4, and C1 = C4 != C2 = C3. It is easy to check all > possibilities that we connect the whole mess by either adding two > and replies. Greg What do you mean when you say that Ci = Cj? Bill === Subject: Re: Fractal dimension of the double integration of a random process? Are you sure that this is really a fractal ? > What is the fractal dimension of the double integration of variants > with a Gaussian/normal distribution? > John -- John Conover, conover@email.rahul.net, http://www.johncon.com/ > === Subject: Re: hi..can yu help me find the solution to the following recurrence? 'cause T(n)=T(n-1)+ log(n) therefore T(n)-T(n-1)=log(n). We have T(n)=[T(n)-T(n-1)]+[T(n-1)-T(n-2)]+[T(n-2)-T(n-3)]+...+[T(3)-T(2)]+T(2) =log(n-1)+log(n-2)+...+log3+T(2) =log[(n-1)!]+T(2)-log2. Obviously you should give the initial condition T(2)=? or some other such as T(3) or T(4) etc. === Subject: Re: hi..can yu help me find the solution to the following recurrence? <13318375.1170384884847.JavaMail.jakarta@nitrogen.mathforum.org 'cause T(n)=T(n-1)+ log(n) therefore > T(n)-T(n-1)=log(n). > We have > T(n)=[T(n)-T(n-1)]+[T(n-1)-T(n-2)]+[T(n-2)-T(n-3)]+...+[T(3)-T(2)]+T(2) > =log(n-1)+log(n-2)+...+log3+T(2) > =log[(n-1)!]+T(2)-log2. > Obviously you should give the initial condition T(2)=? or some other such as T(3) or T(4) etc. S(n) = T(n) - T(n-1) = log(n) T(n) = [T(n)-T(n-1)]+[T(n-1)-T(n-2)]+[T(n-2)-T(n-3)]+...+[T(3)-T(2)] + [T(2) -T(1) ] + [T(1) - T(0) ] + T(0) = log(n) + log(n-1) + ........ + log 1 + theta(1) ( assuming T(0) = theta(1) ) = log (n!) + theta(1) = theta( n logn ) So your solution matches with the ones provided earlier === Subject: probability hello, consider my problem , suppose i have a bankroll .I divide it into a hundred parts and use a system which has the following efficiency . For every 100 games the system wins 75 percent , for every win i make 2.5 % of the stake for every loss i loose 1.25 % of the stake . ------------------------------------------------------ Now my stakes ; i put the stakes in a single line matrix as 1111111.........1111111111111. For every game , i may have two games or more running simultaneously it gives the same average for the system so irrelevant i use the followingmethod to bankroll the games. these are the rules You hve to write down on a paper the matrix and update it after each trade . stake an amount equal to the extreme lhs digit and a extreme rhs digit . if trade successful cancel the two digits if trade failure then add the two numbers on the rhs . and proceed until you see the pot full and then start again . e.g. 11111111111........1111111111 this is your single row matix on a piece of paper . you postion size (stake) is 1 + 1 =2 you win trade so cancell the 1 on right and left . the matrix is 1111...1111 = 98 next trade postion size 1 + 1 =2 suppose loose . then u add the number 2 to rhs viz now u have 1111111......... 11112 = 100 so next trade u postion size 1 + 2 =3 and suppose u loose then 1111111.......1111112 3 =103 so next trade u postion size 1 + 3 and u win so your matrix is 1111.....11112 = 99 .etc etc now suppose u loose your first trades in a row u begin with 111111...........1111111 =100 1111.........1111112 =102 ~2~ 105 ~3~109 ~ 4~114~ 5~120~6~127 so u can observe how u are faring .Also note u are making more than looseing on the trades . ------------------------------------------------------------------------ Now suppose i have a streak of bad luck .what is teh probability that i would be bankrupted before i start to win and finally win .e.g .I would be suppse starting with a loosing trade also in a row or maybe i have a bad patch and start looseing inbetween so what is teh probability of such an event happening. Probability question. === Subject: probability hello, consider my problem , suppose i have a bankroll .I divide it into a hundred parts and use a system which has the following efficiency . For every 100 games the system wins 75 percent , for every win i make 2.5 % of the stake for every loss i loose 1.25 % of the stake . ------------------------------------------------------ Now my stakes ; i put the stakes in a single line matrix as 1111111.........1111111111111. For every game , i may have two games or more running simultaneously it gives the same average for the system so irrelevant i use the followingmethod to bankroll the games. these are the rules You hve to write down on a paper the matrix and update it after each trade . stake an amount equal to the extreme lhs digit and a extreme rhs digit . if trade successful cancel the two digits if trade failure then add the two numbers on the rhs . and proceed until you see the pot full and then start again . e.g. 11111111111........1111111111 this is your single row matix on a piece of paper . you postion size (stake) is 1 + 1 =2 you win trade so cancell the 1 on right and left . the matrix is 1111...1111 = 98 next trade postion size 1 + 1 =2 suppose loose . then u add the number 2 to rhs viz now u have 1111111......... 11112 = 100 so next trade u postion size 1 + 2 =3 and suppose u loose then 1111111.......1111112 3 =103 so next trade u postion size 1 + 3 and u win so your matrix is 1111.....11112 = 99 .etc etc now suppose u loose your first trades in a row u begin with 111111...........1111111 =100 1111.........1111112 =102 ~2~ 105 ~3~109 ~ 4~114~ 5~120~6~127 so u can observe how u are faring .Also note u are making more than looseing on the trades . ------------------------------------------------------------------------ Now suppose i have a streak of bad luck .what is teh probability that i would be bankrupted before i start to win and finally win .e.g .I would be suppse starting with a loosing trade also in a row or maybe i have a bad patch and start looseing inbetween so what is teh probability of such an event happening. Probability question. === Subject: JSH? JSH stands for what? === Subject: Re: JSH? > JSH stands for what? go to James S. Harris dissociated http://209.150.62.151:8080/~jesse/geek/dissocJSH.shtml === Subject: Re: JSH? > JSH stands for what? http://www.amazon.com/Mathematical-Cranks-Spectrum-Underwood-Dudley/dp/08838 55070/sr=8-1/qid=1170439632/ref=pd_bbs_sr_1/104-0421779-7044723?ie=UTF8&s=bo o ks -- David T. Ashley (dta@e3ft.com) http://www.e3ft.com (Consulting Home Page) http://www.dtashley.com (Personal Home Page) http://gpl.e3ft.com (GPL Publications and Projects) === Subject: Re: JSH? > JSH stands for what? JSH is well know for the short proof of FLT that only has a few mistakes left. JSH's personality is best described as a virgin microbe that penetrates with the insistence of air into all the spaces that reason has not been able to fill with words or conventions. Here is where JSH's believers keep his true sayings; http://jstevh.blogspot.com/ I think one of his greatist inspirational speaches on Math is JSH: I WILL GET MY MONEY http://jstevh.blogspot.com/2006/06/jsh-i-will-get-my-money.html One of JSH's privet high holy beliefs; We have done violence to the snivelling tendencies in our natures. Every infiltration of this sort is macerated diarrhoea. To encourage this sort of math is to digest it. What we need are strong straightforward, precise math works which will be forever misunderstood. Logic is a complication. Logic is always false. It draws the superficial threads of concepts and words towards illusory conclusions and centres. Its chains kill, an enormous myriapod that asphyxiates independence. If it were married to logic, math would be living in incest, engulfing, swallowing its own tail, which still belongs to its body, fornicating in itself, and temperament would become a nightmare tarred and feathered with protestantism, a monument, a mass of heavy, greyish intestines. === Subject: Re: JSH? > JSH stands for what? === Subject: Re: JSH? JSH is the last great hope of the free world as we know it. Without JSH, there would be no understanding of continuity of spacetime, randomness would be a mystery, QM wierdness would still be misunderstood, calculus would not work properly in spacetime, http://sciphysicsopenmanuscript.blogspot.com/ JSH is restoring faith in government, curing disease, ending war, healing sick, walking on water, and perhaps most importantly - proving FLT. While some of the things done by James or attributed to him are completely silly, or perhaps maximally irritating, there is one thing which has been attributed to him which actually makes sense, but might very well be completely useless. Harris' Theorem 1.1, which states that The Existence of a Trivial is Indeterminate. === Subject: Re: JSH? Indeterminate. Harris Theorem 1.0: The word trivial is an adjective, not a noun. === Subject: Re: JSH? > Harris' Theorem 1.1, which states that The Existence of a Trivial is > Indeterminate. Harris Theorem 1.0: The word trivial is an adjective, not a noun. > You're right - I have been using it as a noun simply because there are so many thing which are trivial that I simply categorize them all as the trivials. In a sense, this is formally valid, even though my usage extends across broad classes of various kinds of objects. You have many kinds of trivial things: Trivial Numbers Trivial Geometric Objects Also (I have'nt really explored this) Trivial Relationships like functions and morphisms, etc Trivial Properties (I believe randomness existentially indeterminate, i.e. trivial) Anyhow - there are so many potentially trivial things that I simply refer to them collectively as the trivials. The reason that this is formally valid is because these objects are existentially indeterminate. I think that such objects could be anti-unique. Zero bananas is identical to zero oranges, even though apples are not oranges. So, trivial planes and trivial numbers or trivial functions, properties, whatever, might also be indistinguishable by virtue of their nonexistence. But this cannot be absolutely true because they also have an existent aspect which would would render them as distinguishable. To reiterate, a trivial is an identical clone of an object. Violate uniqueness intentionally and you have created a trivial. === Subject: Re: JSH? <24110937.1170386736342.JavaMail.jakarta@nitrogen.mathforum.org>, > JSH stands for what? Just Stop Huffing. It was part of Nancy Reagan's campaign to get kids to stop sniffing glue. === Subject: Re: JSH? > JSH stands for what? JSH is the destruction of the human race as we know it. JSH is the seer of all things big and round. JSH is the past, present, and future. JSH is in your food, drink, and even the air you breath. Without JSH, there is nothingness... absolute, cold, nothingness. There's also some peon math-wannabee whose incompetence at math, logic, and computer programming cannot be fathomed by mere mortals. === Subject: Re: JSH? > JSH stands for what? James n Harris, notorious sci.math nutjob. He has been asked to prefix his post titles with JSH so that those who aren't interested in hearing his rants are spared the indignity of accidentally opening one of his posts. See http://mathforum.org/kb/profile.jspa?userID=376101, for example, or http://mathforum.org/kb/profile.jspa?userID=23927. === Subject: Re: Angle Bisector Theorem <45c0aa5d@news1.ethz.ch> Let ABC be a triangle, A the vertex and BC the base. Let theangle >> bisector > side AB at D. If DC = BE then ABC is isocoles. Steiner-Lehmus theorem. > I found an indirect proof, but I cannot seem to come up with a direct >> proof. Does anyone one know a direct classical Euclidean type proof for >> this theorem? And why is such a simple sounding theorem so damned hard >> to prove by direct means? You will find all the information you ever wanted > to know about this topic in http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/steiner-lehmus > Coxeter's Introduction to Geometry discusses this theorem, > refers to it as the Stiener-Lehmus theorem, and says there are > over 60 proofs. He gives a simple, elegant proof of the more general > theorem that the larger of twoanglebisectorsintersects the > smaller side. The proof is independant of the parallel > postulate, and thus is also valid in hyperbolic geometry. I've often responded to questions about why this theorem is so hard to > prove by saying that it isn't true if you allow the various lengths > involved to be in more general fields, like the complex numbers. After > someone asked for details about this the other day, I took the trouble > to work everything out, and found to my surprise that it isn't even > true when you allow them to be negative reals, which makes the > situation much easier to understand. Here's what's going on. The lengths of the A and C anglebisectorsturn out to be root{bc(a+b+c)(b+c-a)} root{ab(a+b+c)(a+b-c)} > ---------------------- and ---------------------- , > b+c a+b so the equality we're given is bc(b+c-a).(a+b)^2 = ab(a+b-c).(b+c)^2 > or > (c-a){ ac^2 + (a^2+3ab+b^2)c^2 + b^2(a+b) } = 0. Now algebraically, this does not imply that c=a, because > it might be the second factor that vanishes. However, if a,b,c are positive, then so is this second factor, > and so with this assumption we CAN deduce c = a, [...] Hmm. Interesting. Trivia: It just so happens that I'm the one who told Conway about that proof. http://mathforum.org/kb/message.jspa?messageID=1083303&tstart=2565 http://mathforum.org/kb/message.jspa?messageID=1083304&tstart=2565 === Subject: Re: representation of monotone boolean functions | as we all know, the number of monotone Boolean functions on n variables | is less than or equal to 3^binom(n,[n/2]). Please correct me if I'm | wrong. So for encoding of a monotone function binom(n,[n/2])*(log_2 3) | bits should suffice, which is asymptotically less than 2^n. Is there an | encoding that allows evaluating a monotone function in polynomial time | in n but needs asymptotically less than 2^n bits? The answer to this question depends on your model of computation. However, here is a possible encoding: Fix a monotone Boolean function f on n variables. Consider each string of 0s and 1s of length n. Treat 0s as left parentheses and 1s as right parentheses, and decompose the string into maximal correctly parenthesized blocks, separated by the 0s and 1s which cannot be incorporated into these blocks. Color the bits in the blocks red, and color the other bits green. By maximality, no green 0 can be to the left of a green 1. Partition the strings of bits of length n by placing two strings in the same group if they have the same positions colored red and the substrings of red bits in the two strings have the same value. Since no green 0 can be to the left of a green 1, the size of each group will equal 1 more than the number of green bits in each of its strings, and if we order the strings in a group by weight, the substrings of each string formed by the green bits will be 00...0, 10...0, ..., 1...10, 1...11. The values of f on the group can therefore be specified by giving the number of strings in the group it sends to 1. It is fairly easy to show that for each p, there are C(n,p)-C(n,p-1) groups with 2p red positions. We can use ceil(log_2(n-2p+1)) bits to code the value on each such group, giving an encoding of f as a bit string with a total of sum_{0le ple n/2} (C(n,p)-C(n,p-1)) ceil(log_2(n-2p+1)) = O(C(n, floor(n/2)) log n) bits. To evaluate the monotone function on a set of inputs, we view the inputs as a bit string of length n, color it, find which group it is in, and look the value of f up in the appropriate place in the encoding of f. -- David Moews dmoews@fastmail.fm === Subject: Re: Proving lim n * a_n = 0 > Hello I started studying sequences and series abou 2 months ago and I'm not > very secure about some points yet, so I'd like someone to check my > proof. I still have some difficulty to deal with the concepts of lim > inf and lim sup of a sequence. I used to deal more with integers and > prime numbers, so I have some difficulty with limit processes. Show that, if (a_n) is a motonic sequence of real numbers and Sum(n=1, > oo) converges, then lim n a_n = 0. First, let's suppose (a_n) is decreasing. Then, lim a_n =0 and a_n >=0 > for every n, since infimum (a_n) = 0. For each natural k, let S_k = > Sum(n=k, oo) a_n. Since Sum(n=1, oo) converges, each S_k is finite > and lim S_k =0. > Let A_n = a_1 + a_2 +...+ a_n. series is convergent ==> A_(2m) - A_m --> 0 as m -->00 But A_(2m) - A_m = a_(m+1) + a_(m+2) +...+ a_(2m) >= a_(2m) + a_(2m) +...+ a_(2m) = m * a_(2m) , therefore m * a_(2m) --> 0 as n --> 00 and so (2m) * a_(2m) --> 0 as m --> 00 (*) Now, (2m+1) * a_(2m+1) <= (2m+1) * a_(2m) = [(2m) * a_(2m)] * [1 + 1/(2m)], so (2m+1) * a_(2m+1) -->0 as m --> 00. (**) (*) and (**) together give the desired result. > Keeping k fixed and taking into account (a_n) is decreasing, for every > n >=k we have 0 <= a_n ....+ a_n = (n -k +1) a_n <= a_k ....+ a_n <= > S_k. Therefore, 0 <= n a_n <= (k-1) a_n + S_k. Since this holds for > every n >=k (that is, with possibly exception of a finite number of > values of n, it holds for all n) it follows that 0 <= lim sup n a_n <= > lim sup ((k-1) a_n + S_k.). Since a_n -> 0, ,(k-1) a_n + S_k.) - S_k. For this sequence, lim inf, lim and lim sup are therefore the > same number. This leads to 0 <= lim sup n a_n <= S_k. Since these > inequalities hold for every k in {1,2,3....}, we must have 0 <= lim > sup n a_n <= lim S_k = 0 => lim sup n a_n = 0. Since n a_n >= 0 for every n, we also have 0 <= lim inf n a_n < = > limsup n a_n = 0 . Therefore lim inf and lim sup are both 0 , and we > simply have lim n a_n =0. If (a_n) is increasing, we can use a similar reasoning. But since we > already have the previous conclusion, it's enough to apply it to the > decreasing sequence (- a_n) to conclude that lim (n * (-a_n)) = lim (- > n a_n) = -lim (n a_n) = 0, so that lim n a_n =0. Based on this conclusion, I found an easy way to prove the harmonic > series diverges. If a_n =1/n, then, for every n, n a_n =1 and lim n > a_n =1, so that the condition lim n a_n =0, necessary for convergence, > is not satisfied. Therefore, Sum (1/n) diverges. I hadn't seen this > proof before. I agree it's not very informative, other proofs show > much better why this series goes to infinity. Sharon > === Subject: Re: Proving lim n * a_n = 0 > I started studying sequences and series abou 2 months ago and I'm not > very secure about some points yet, so I'd like someone to check my > proof. I still have some difficulty to deal with the concepts of lim > inf and lim sup of a sequence. I used to deal more with integers and > prime numbers, so I have some difficulty with limit processes. Show that, if (a_n) is a motonic sequence of real numbers and Sum(n=1, > oo) converges, then lim n a_n = 0. First, let's suppose (a_n) is decreasing. Then, lim a_n =0 and a_n >=0 > for every n, since infimum (a_n) = 0. For each natural k, let S_k = > Sum(n=k, oo) a_n. Since Sum(n=1, oo) converges, each S_k is finite > and lim S_k =0. Keeping k fixed and taking into account (a_n) is decreasing, for every > n >=k we have 0 <= a_n ....+ a_n = (n -k +1) a_n <= a_k ....+ a_n <= > S_k. Therefore, 0 <= n a_n <= (k-1) a_n + S_k. Since this holds for > every n >=k (that is, with possibly exception of a finite number of > values of n, it holds for all n) it follows that 0 <= lim sup n a_n <= > lim sup ((k-1) a_n + S_k.). Since a_n -> 0, ,(k-1) a_n + S_k.) - S_k. For this sequence, lim inf, lim and lim sup are therefore the > same number. This leads to 0 <= lim sup n a_n <= S_k. Since these > inequalities hold for every k in {1,2,3....}, we must have 0 <= lim > sup n a_n <= lim S_k = 0 => lim sup n a_n = 0. Since n a_n >= 0 for every n, we also have 0 <= lim inf n a_n < = > limsup n a_n = 0 . Therefore lim inf and lim sup are both 0 , and we > simply have lim n a_n =0. If (a_n) is increasing, we can use a similar reasoning. But since we > already have the previous conclusion, it's enough to apply it to the > decreasing sequence (- a_n) to conclude that lim (n * (-a_n)) = lim (- > n a_n) = -lim (n a_n) = 0, so that lim n a_n =0. It looks good to me. Jose Carlos Santos === Subject: Re: Proving lim n * a_n = 0 <52efaaF1ocm5oU1@mid.individual.net > I started studying sequences and series abou 2 months ago and I'm not > very secure about some points yet, so I'd like someone to check my > proof. I still have some difficulty to deal with the concepts of lim > inf and lim sup of a sequence. I used to deal more with integers and > prime numbers, so I have some difficulty with limit processes. Show that, if (a n) is a motonic sequence of real numbers and Sum(n=1, > oo) converges, then lim n a n = 0. First, let's suppose (a n) is decreasing. Then, lim a n =0 and a n >=0 > for every n, since infimum (a n) = 0. For each natural k, let S k = > Sum(n=k, oo) a n. Since Sum(n=1, oo) converges, each S k is finite > and lim S k =0. Keeping k fixed and taking into account (a n) is decreasing, for every > n >=k we have 0 <= a n ....+ a n = (n -k +1) a n <= a k ....+ a n <= > S k. Therefore, 0 <= n a n <= (k-1) a n + S k. Since this holds for > every n >=k (that is, with possibly exception of a finite number of > values of n, it holds for all n) it follows that 0 <= lim sup n a n <= > lim sup ((k-1) a n + S k.). Since a n -> 0, ,(k-1) a n + S k.) - S k. For this sequence, lim inf, lim and lim sup are therefore the > same number. This leads to 0 <= lim sup n a n <= S k. Since these > inequalities hold for every k in {1,2,3....}, we must have 0 <= lim > sup n a n <= lim S k = 0 => lim sup n a n = 0. Since n a n >= 0 for every n, we also have 0 <= lim inf n a n < = > limsup n a n = 0 . Therefore lim inf and lim sup are both 0 , and we > simply have lim n a n =0. If (a n) is increasing, we can use a similar reasoning. But since we > already have the previous conclusion, it's enough to apply it to the > decreasing sequence (- a n) to conclude that lim (n * (-a n)) = lim (- > n a n) = -lim (n a n) = 0, so that lim n a n =0. It looks good to me. > Jose Carlos Santos- Hide quoted text - - Show quoted text - Sharon === Subject: JSH: Adding Backwards mymath/blog/addingbackwards add backwards - to be honest, I never did subtract, except small numbers like 1, 2 or 3. Basically, it's really logical. Since smaller human brains are much more adept at going forward than backward (notice that counting forward is easier than counting backwards, and adding is quicker than subtracting), it follows that any calculation that can use forward patterns would be easier. Adding is also much more accurate. Subtracting forwards is a error introduced in the 1700's by mathematicians that wanted to keep math a secrete. Conceder counting backwards if you are subtracting 1, 2 or 3, but after that use your addition facts. Number families help mathematicians grasp this concept: 8 + 5 = 13 13 - 8 = 5 13 - 5 = 7 Then you can *see* the subtraction problem as a fill-in-the-blank addition problem. This is pretty advanced stuff, and occasionally confuses mathematics graduate students if they aren't ready. Watch for reactions - if your student catches on, run with it. If not, slow down and don't push it, they are not as smart as I am. Example: 16 - 8 = ? Just look at it as 8 + what equals 16... look hard........ get it yet ? The number mathematicians above introduce this concept, although it is flawed for over a hundred years. I like this because, although it takes a little longer, it shows how mathematicians have made their work seem complicated and it is not. Try to learn to look at the relationships between numbers *and* between the different operations, which is going to help them with the rest of math. My research always use lots of manipulative and toys. I teach addition off the doubles, so that helps too: 18 - 9 = ? Using the doubles, some will know that 9 + 9 = 18 and can fill in the blank. The logic here is as follows: We check subtraction by adding the answer to the subtrahend to get the minuend (first number), but we don't check adding by subtracting - reason is because adding is *going forward* and is much more natural, easier and accurate. So, why not just think through the check first (since it is more accurate) and skip the backward stuff? This can be translated to larger problems, where the numbers are over each other: 386 - 143 Conceder how to add up I draw little boxes under each number where the answer goes, one for each digit. Then ask, what + 3 equals 6, fill it in and add up. Next column, what + 4 equals 8, fill in and add up, and finally what + 1 equals 3? This can also be done when you would normally borrow. Remember that anything you can do in math can be undone, simply do it exactly in reverse. 832 - 158 In this case, since you can't take 8 and add anything to get 2, you have to add to get to 12: 4+8 = 12. Just like addition right side up, you put down the 2 (at the top) and carry. Of course, since you are adding up, it follows that you carry DOWN and over to the next number. So the 1 carries over to the 5 . Now, 5 + 1 (the 1 you carried) is 6; so what + 6 equals 13? You fill in the 7, then from the bottom up, 7 + 5 + 1 (carried) = 13, put down the 3 at the top, carry the 1 down and over to the next column. 6 + 1 + 1 carried equals 9, so just put down the 6. At first this seems really confusing, but if you try to think through a couple of borrowing problems like this, AND assuming the person is really proficient in addition, this is much faster and accurate, and when you think about it, isn't this the check you learned to do on a subtraction problem anyway? - === Subject: A picture of xRy Suppose there is a relation R and x,y in A. Let x_A be all x in domain of A. Let y_A be all y in codomain of A. Let x_R be some x in xRy. Let y_R be some y in xRy. Then, there is a path x_R y_A y_R x_A x_R. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy > Suppose there is a relation R and x,y in A. Let x_A be all x in domain > of A. Let y_A be all y in codomain of A. Let x_R be some x in xRy. Let > y_R be some y in xRy. > Then, there is a path x_R y_A y_R x_A x_R. > Domain and codomain are terms used to describe functions. How are you defining them for relations? The domain D and the codomain C are the smallest sets for which R subset DxC? You're stating that x_A = domain of A, y_R = codomain of A. Not recommended notation. In fact disrecommended. The expression x in xRy does not make sense. xRy means (x,y) in R. You stated that x_R is domain of R. How is it that x_R is some x in something? R is a relation within a set S when R subset SxS. xRy means (x,y) in R. How much did you study and consider the definitions before you open your mouth? === Subject: Re: A picture of xRy Suppose there is a relation R and x,y in A. Let x_A be all x in domain > of A. Let y_A be all y in codomain of A. Let x_R be some x in xRy. Let > y_R be some y in xRy. Then, there is a path x_R y_A y_R x_A x_R. Domain and codomain are terms used to describe functions. > How are you defining them for relations? > The domain D and the codomain C are the smallest sets for which > R subset DxC? You're stating that x_A = domain of A, y_R = codomain of A. > Not recommended notation. In fact disrecommended. The expression x in xRy does not make sense. > xRy means (x,y) in R. You stated that x_R is domain of R. > How is it that x_R is some x in something? R is a relation within a set S when R subset SxS. > xRy means (x,y) in R. How much did you study and > consider the definitions before you open your mouth? I've studies the definitons for two days. I already knew that xRy means (x,y) in R. You have misunderstood my writing because I didn't follow conventional notations (since I don't know them). I may be wrong since I have developed *my* version of how xRy could be interpreted. And, at least try to understand my logical statements instead of just saying it is not well-formed. I'll try to explain what I mean. First, the definitions from Zakon, Relation R is defined, R:={(x,y)|(x,y) in R) Domain D_R of R is defined, D_R:={x|(exists y)(x,y) in R} Codomain D'_R of R is defined, D'_R:={y|(exists x)(x,y) in R} We define xRy as, xRy:=(x,y) in R Now my stuff, For any set A and relation R, (forall x in A)(exists b,y | b=y)(b in (a,b) of R implies x in D_R) and (forall y in A)(exists a,x | a=x)(a in (a,b) of R implies y in D_R) Casually, this means that all x,y in A has some (a,b) in R. Interestingly, x in A seems to be associating with b in (a,b) of R, and y in A seems to be associating with a in (a,b) of R. I attempted to provide an explanation for this. I figured that, D_R subset A; D'_R subset A The flow of information in xRy is, (x in A) to (y in (x,y) of R) to (y in A) to (x in (x,y) of R) to (x in A) So, you can see something about my notation that, x_A:=x in A y_R:=y in (x,y) of R y_A:=y in A x_R:=x in (x,y) of R x_A y_R y_A x_R x_A is the shortest return path from x_A to x_A for xRy. I have more innovation but that is as far as relevant to the parent thread. Don't blindly criticize me for using my own notation or for using my own interpretation of xRy, because the sole objective is to learn. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy <87abzwddfz.fsf@laptop.at.pyenos > Suppose there is a relation R and x,y in A. Let x_A be all x in domain > of A. Let y_A be all y in codomain of A. Let x_R be some x in xRy. Let > y_R be some y in xRy. > Then, there is a path x_R y_A y_R x_A x_R. > Domain and codomain are terms used to describe functions. > How are you defining them for relations? > The domain D and the codomain C are the smallest sets for which > R subset DxC? You're stating that x_A = domain of A, y_R = codomain of A. > Not recommended notation. In fact disrecommended. The expression x in xRy does not make sense. > xRy means (x,y) in R. You stated that x_R is domain of R. > How is it that x_R is some x in something? R is a relation within a set S when R subset SxS. > xRy means (x,y) in R. How much did you study and > I already knew that xRy means (x,y) in R. You have misunderstood my > writing because I didn't follow conventional notations (since I don't > know them). > You used xRy in a way you did not explain. If you're using novel notation, then it's your obligation to define what your notation means. > I'll try to explain what I mean. First, the definitions from Zakon, Relation R is defined, > R:={(x,y)|(x,y) in R) > Domain D_R of R is defined, > D_R:={x|(exists y)(x,y) in R} > Codomain D'_R of R is defined, > D'_R:={y|(exists x)(x,y) in R} > We define xRy as, > xRy:=(x,y) in R > They're ease to read and don't cause eye strain. For example, a = b + xy < c is usually read, a=b+xy Now my stuff, For any set A and relation R, > (forall x in A)(exists b,y | b=y)(b in (a,b) of R implies x in D_R) > and > (forall y in A)(exists a,x | a=x)(a in (a,b) of R implies y in D_R) > Don't understand what this means even taking into account D_R is defined. > Casually, this means that all x,y in A has some (a,b) in R. Casually that's a casualty. > Interestingly, x in A seems to be associating with b in (a,b) of R, and > y in A seems to be associating with a in (a,b) of R. I attempted to > provide an explanation for this. I figured that, > D_R subset A; D'_R subset A The flow of information in xRy is, > (x in A) to (y in (x,y) of R) to (y in A) to (x in (x,y) of R) to (x in A) So, you can see something about my notation that, > x_A:=x in A > y_R:=y in (x,y) of R > y_A:=y in A > x_R:=x in (x,y) of R x_A y_R y_A x_R x_A is the shortest return path from x_A to x_A for xRy. Don't blindly criticize me for using my own notation or for using my own > interpretation of xRy, because the sole objective is to learn. > Ok, I'll leave that for others to do. === Subject: Re: A picture of xRy > Suppose there is a relation R and x,y in A. Let x_A be all x in domain > of A. Let y_A be all y in codomain of A. Let x_R be some x in xRy. Let > y_R be some y in xRy. > Then, there is a path x_R y_A y_R x_A x_R. > Domain and codomain are terms used to describe functions. > How are you defining them for relations? > The domain D and the codomain C are the smallest sets for which > R subset DxC? > You're stating that x_A = domain of A, y_R = codomain of A. > Not recommended notation. In fact disrecommended. > The expression x in xRy does not make sense. > xRy means (x,y) in R. > You stated that x_R is domain of R. > How is it that x_R is some x in something? > R is a relation within a set S when R subset SxS. > xRy means (x,y) in R. How much did you study and I already knew that xRy means (x,y) in R. You have misunderstood my > writing because I didn't follow conventional notations (since I don't > know them). You used xRy in a way you did not explain. If you're using novel > notation, then it's your obligation to define what your notation means. I'll try to explain what I mean. First, the definitions from Zakon, Relation R is defined, > R:={(x,y)|(x,y) in R) > Domain D_R of R is defined, > D_R:={x|(exists y)(x,y) in R} > Codomain D'_R of R is defined, > D'_R:={y|(exists x)(x,y) in R} > We define xRy as, > xRy:=(x,y) in R They're ease to read and don't cause eye strain. For example, > a = b + xy < c is usually read, a=b+xy (forall x in A)(exists b,y | b=y)(b in (a,b) of R implies x in D_R) > and > (forall y in A)(exists a,x | a=x)(a in (a,b) of R implies y in D_R) Don't understand what this means even taking into account D_R is defined. Casually, this means that all x,y in A has some (a,b) in R. Casually that's a casualty. Interestingly, x in A seems to be associating with b in (a,b) of R, and > y in A seems to be associating with a in (a,b) of R. I attempted to > provide an explanation for this. I figured that, > D_R subset A; D'_R subset A The flow of information in xRy is, > (x in A) to (y in (x,y) of R) to (y in A) to (x in (x,y) of R) to (x in A) So, you can see something about my notation that, > x_A:=x in A > y_R:=y in (x,y) of R > y_A:=y in A > x_R:=x in (x,y) of R x_A y_R y_A x_R x_A is the shortest return path from x_A to x_A for xRy. This is wrong. But it has been helpful to me. Don't blindly criticize me for using my own notation or for using my own > interpretation of xRy, because the sole objective is to learn. Ok, I'll leave that for others to do. OK, I'll demonstrate my understanding resulting from my own interpretation, by explaning the meaning of R subset AxA. AxA generates pairs of x,y x_1, x_2, ... x_i domain y_1, y_2, ... y_i codomain Now, (x_1, y_1) or (x_1, y_2) ... (x_1, y_i), so i number of ordered pair is possible to generate from x_1 pairing with one of y. And so on until x_i. This describes AxA. R subset AxA means R is a subset of (x,y) pairs generated from AxA operation. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy <87abzwddfz.fsf@laptop.at.pyenos> <87veikbwh0.fsf@laptop.at.pyenos OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. AxA generates pairs of x,y > Gibberish. AxA = { (x,y) | x,y in A } AxA is the set of all ordered pairs (x,y), with both x and y in A. > x_1, x_2, ... x_i domain > y_1, y_2, ... y_i codomain > More nonsense. > Now, (x_1, y_1) or (x_1, y_2) ... (x_1, y_i), so i number of ordered > pair is possible to generate from x_1 pairing with one of y. And so on > until x_i. This describes AxA. R subset AxA means R is a subset of (x,y) pairs generated from AxA > operation. > Coherent nonsense. It means for all r in R, r is in AxA. === Subject: Re: A picture of xRy OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. AxA generates pairs of x,y Gibberish. AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. ~~~~ AxA = { (x,y) | x in A and y in A } also note that AxB = { (x,y) | x in A and y in B } note that AxB != { (x,y) | x,y in (A or B) } x_1, x_2, ... x_i domain > y_1, y_2, ... y_i codomain More nonsense. Now, (x_1, y_1) or (x_1, y_2) ... (x_1, y_i), so i number of ordered > pair is possible to generate from x_1 pairing with one of y. And so on > until x_i. This describes AxA. R subset AxA means R is a subset of (x,y) pairs generated from AxA > operation. Coherent nonsense. It means for all r in R, r is in AxA. If God gave you a dollar every time you call something a nonsense, you would be a rich man ;) -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy <87abzwddfz.fsf@laptop.at.pyenos> <87veikbwh0.fsf@laptop.at.pyenos> <87tzy48wk1.fsf@laptop.at.pyenos AxA = { (x,y) | x in A and y in A } AxB = { (x,y) | x in A and y in B } > Good. Prove AxB / AxC = Ax(B / C) === Subject: Re: A picture of xRy OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. AxA generates pairs of x,y Gibberish. Fine. I didn't use formal definition. AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. x_1, x_2, ... x_i domain > y_1, y_2, ... y_i codomain More nonsense. It is _not_ nonsense. This was in an attempt to describe cartesian product of x in A and x in A. AxA={(x_1, y_1), (x_1, y_2), ... , (x_1, y_i), (x_2, y_1), (x_2, y_2), ... , (x_2, y_i), ... (x_i, y_1), (x_i, y_2), ... (x_i, y_i)} therefore, AxA has i*i number of elements. Now, (x_1, y_1) or (x_1, y_2) ... (x_1, y_i), so i number of ordered > pair is possible to generate from x_1 pairing with one of y. And so on > until x_i. This describes AxA. R subset AxA means R is a subset of (x,y) pairs generated from AxA > operation. Coherent nonsense. It means for all r in R, r is in AxA. I knew that already. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy <87abzwddfz.fsf@laptop.at.pyenos> <87veikbwh0.fsf@laptop.at.pyenos> <873b5obucp.fsf@laptop.at.pyenos > OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. > AxA generates pairs of x,y > Gibberish. > Fine. I didn't use formal definition. > It lacks much quality as informal statement. That is, requires reader to be mind reader. > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. > x_1, x_2, ... x_i domain > y_1, y_2, ... y_i codomain > More nonsense. > It is _not_ nonsense. This was in an attempt to describe cartesian > product of x in A and x in A. > It's failing. > AxA={(x_1, y_1), (x_1, y_2), ... , (x_1, y_i), (x_2, y_1), (x_2, y_2), > ... , (x_2, y_i), ... (x_i, y_1), (x_i, y_2), ... (x_i, y_i)} > Hm, you seem to be thinking there's some connection between the x's, the y's and the A's. You have listed { x1,.. x_i } x { y1,.. y_i } > therefore, AxA has i*i number of elements. > Oh? A has i elements? What if A is infinite? === Subject: Re: A picture of xRy > OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. > AxA generates pairs of x,y > Gibberish. > Fine. I didn't use formal definition. It lacks much quality as informal statement. > That is, requires reader to be mind reader. > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. > x_1, x_2, ... x_i domain > y_1, y_2, ... y_i codomain > More nonsense. > It is _not_ nonsense. This was in an attempt to describe cartesian > product of x in A and x in A. It's failing. AxA={(x_1, y_1), (x_1, y_2), ... , (x_1, y_i), (x_2, y_1), (x_2, y_2), > ... , (x_2, y_i), ... (x_i, y_1), (x_i, y_2), ... (x_i, y_i)} Hm, you seem to be thinking there's some connection between the > x's, the y's and the A's. You have listed > { x1,.. x_i } x { y1,.. y_i } therefore, AxA has i*i number of elements. Oh? A has i elements? What if A is infinite? I meant cartesian product of x in A and y in A. > -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy <87abzwddfz.fsf@laptop.at.pyenos> <87veikbwh0.fsf@laptop.at.pyenos> <873b5obucp.fsf@laptop.at.pyenos> <87sldoaefb.fsf@laptop.at.pyenos > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. > AxA={(x_1, y_1), (x_1, y_2), ... , (x_1, y_i), (x_2, y_1), (x_2, y_2), > ... , (x_2, y_i), ... (x_i, y_1), (x_i, y_2), ... (x_i, y_i)} > Hm, you seem to be thinking there's some connection between the > x's, the y's and the A's. You have listed > { x1,.. x_i } x { y1,.. y_i } > therefore, AxA has i*i number of elements. > Oh? A has i elements? What if A is infinite? I meant cartesian product of x in A and y in A. > You didn't define A and your 'I meant ...' is mumbling, thinking out loud, requires a mind reader to clarify. See above how I defined AxA and described it. Indeed, get a book or find a professor's web page with these basics. Do you have a library card? === Subject: Re: A picture of xRy > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. > AxA={(x_1, y_1), (x_1, y_2), ... , (x_1, y_i), (x_2, y_1), (x_2, y_2), > ... , (x_2, y_i), ... (x_i, y_1), (x_i, y_2), ... (x_i, y_i)} > Hm, you seem to be thinking there's some connection between the > x's, the y's and the A's. Unclear ambiguous statement. 'You seem... ' is mumbling. > You have listed > { x1,.. x_i } x { y1,.. y_i } > therefore, AxA has i*i number of elements. > Oh? A has i elements? What if A is infinite? I meant cartesian product of x in A and y in A. You didn't define A and your 'I meant ...' is mumbling, > thinking out loud, requires a mind reader to clarify. See above how I defined AxA and described it. > Indeed, get a book or find a professor's web page with these basics. > Do you have a library card? -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. > AxA={(x_1, y_1), (x_1, y_2), ... , (x_1, y_i), (x_2, y_1), (x_2, y_2), > ... , (x_2, y_i), ... (x_i, y_1), (x_i, y_2), ... (x_i, y_i)} > Hm, you seem to be thinking there's some connection between the > x's, the y's and the A's. > You have listed > { x1,.. x_i } x { y1,.. y_i } > therefore, AxA has i*i number of elements. > Oh? A has i elements? What if A is infinite? I meant cartesian product of x in A and y in A. You didn't define A and your 'I meant ...' is mumbling, > thinking out loud, requires a mind reader to clarify. See above how I defined AxA and described it. > Indeed, get a book or find a professor's web page with these basics. > Do you have a library card? { x_1, ... x_i } x { y1, ... y_i } is equivalent to (x in A) x (y in A). -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy <87abzwddfz.fsf@laptop.at.pyenos> <87veikbwh0.fsf@laptop.at.pyenos> <873b5obucp.fsf@laptop.at.pyenos> <87sldoaefb.fsf@laptop.at.pyenos> <87irekadd5.fsf@laptop.at.pyenos { x_1, ... x_i } x { y1, ... y_i } is equivalent to (x in A) x (y in A). > Baloney. You haven't defined A. === Subject: Re: A picture of xRy > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. > AxA={(x_1, y_1), (x_1, y_2), ... , (x_1, y_i), (x_2, y_1), (x_2, y_2), > ... , (x_2, y_i), ... (x_i, y_1), (x_i, y_2), ... (x_i, y_i)} > Hm, you seem to be thinking there's some connection between the > x's, the y's and the A's. > You have listed > { x1,.. x_i } x { y1,.. y_i } > therefore, AxA has i*i number of elements. > Oh? A has i elements? What if A is infinite? > I meant cartesian product of x in A and y in A. > You didn't define A and your 'I meant ...' is mumbling, > thinking out loud, requires a mind reader to clarify. See above how I defined AxA and described it. > Indeed, get a book or find a professor's web page with these basics. > Do you have a library card? > { x_1, ... x_i } x { y1, ... y_i } is equivalent to (x in A) x (y in A). { x_1, ... ,x_i } x { y1, ... , y_i } is equivalent to (x in A) x (y in A) for finite i. -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy > OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. > AxA generates pairs of x,y > Gibberish. > Fine. I didn't use formal definition. > It lacks much quality as informal statement. > That is, requires reader to be mind reader. > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. > x_1, x_2, ... x_i domain > y_1, y_2, ... y_i codomain > More nonsense. > It is _not_ nonsense. This was in an attempt to describe cartesian > product of x in A and x in A. > It's failing. > AxA={(x_1, y_1), (x_1, y_2), ... , (x_1, y_i), (x_2, y_1), (x_2, y_2), > ... , (x_2, y_i), ... (x_i, y_1), (x_i, y_2), ... (x_i, y_i)} > Hm, you seem to be thinking there's some connection between the > x's, the y's and the A's. You have listed > { x1,.. x_i } x { y1,.. y_i } > therefore, AxA has i*i number of elements. > Oh? A has i elements? What if A is infinite? I meant cartesian product of x in A and y in A. -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy <87abzwddfz.fsf@laptop.at.pyenos > Suppose there is a relation R and x,y in A. Let x_A be all x in domain > of A. Let y_A be all y in codomain of A. Let x_R be some x in xRy. Let > y_R be some y in xRy. > Then, there is a path x_R y_A y_R x_A x_R. Domain and codomain are terms used to describe functions. > How are you defining them for relations? > The domain D and the codomain C are the smallest sets for which > R subset DxC? You're stating that x_A = domain of A, y_R = codomain of A. > Not recommended notation. In fact disrecommended. The expression x in xRy does not make sense. > xRy means (x,y) in R. You stated that x_R is domain of R. > How is it that x_R is some x in something? R is a relation within a set S when R subset SxS. > xRy means (x,y) in R. How much did you study and > consider the definitions before you open your mouth? I've studies the definitons for two days. I already knew that xRy means (x,y) in R. You have misunderstood my writing because I didn't follow conventional notations (since I don't know them). I may be wrong since I have developed *my* version of how xRy could be interpreted. And, at least try to understand my logical statements instead of just saying it is not well-formed. I'll try to explain what I mean. First, the definitions from Zakon, Relation R is defined, > R:={(x,y)|(x,y) in R) Too circular for my taste. R is simply a subset of AxA > Domain D_R of R is defined, > D_R:={x|(exists y)(x,y) in R} Be careful with this kind of comprehension! This looks like a class when you want a set... > Codomain D'_R of R is defined, > D'_R:={y|(exists x)(x,y) in R} > We define xRy as, > xRy:=(x,y) in R OK Now my stuff, For any set A and relation R, > (forall x in A)(exists b,y | b=y)(b in (a,b) of R implies x in D_R) What is exists b,y | b=y supposed to mean? Where does the a come from and why is x not used in the precedent? > and > (forall y in A)(exists a,x | a=x)(a in (a,b) of R implies y in D_R) Same here Casually, this means that all x,y in A has some (a,b) in R. Interestingly, x in A seems to be associating with b in (a,b) of R, and y in A seems to be associating with a in (a,b) of R. I attempted to provide an explanation for this. I figured that, > D_R subset A; D'_R subset A by definition The flow of information in xRy is, > (x in A) to (y in (x,y) of R) to (y in A) to (x in (x,y) of R) to (x in A) ?? So, you can see something about my notation that, > x_A:=x in A > y_R:=y in (x,y) of R > y_A:=y in A > x_R:=x in (x,y) of R x_A y_R y_A x_R x_A is the shortest return path from x_A to x_A for xRy. I have more innovation but that is as far as relevant to the parent thread. Don't blindly criticize me for using my own notation or for using my own interpretation of xRy, because the sole objective is to learn. === Subject: Re: A picture of xRy > Suppose there is a relation R and x,y in A. Let x_A be all x in domain > of A. Let y_A be all y in codomain of A. Let x_R be some x in xRy. Let > y_R be some y in xRy. > Then, there is a path x_R y_A y_R x_A x_R. > Domain and codomain are terms used to describe functions. > How are you defining them for relations? > The domain D and the codomain C are the smallest sets for which > R subset DxC? > You're stating that x_A = domain of A, y_R = codomain of A. > Not recommended notation. In fact disrecommended. > The expression x in xRy does not make sense. > xRy means (x,y) in R. > You stated that x_R is domain of R. > How is it that x_R is some x in something? > R is a relation within a set S when R subset SxS. > xRy means (x,y) in R. How much did you study and > consider the definitions before you open your mouth? I've studies the definitons for two days. I already knew that xRy means (x,y) in R. You have misunderstood my writing because I didn't follow conventional notations (since I don't know them). I may be wrong since I have developed *my* version of how xRy could be interpreted. And, at least try to understand my logical statements instead of just saying it is not well-formed. I'll try to explain what I mean. First, the definitions from Zakon, Relation R is defined, > R:={(x,y)|(x,y) in R) Too circular for my taste. R is simply a subset of AxA Domain D_R of R is defined, > D_R:={x|(exists y)(x,y) in R} Be careful with this kind of comprehension! > This looks like a class when you want a set... Codomain D'_R of R is defined, > D'_R:={y|(exists x)(x,y) in R} > We define xRy as, > xRy:=(x,y) in R OK > Now my stuff, For any set A and relation R, > (forall x in A)(exists b,y | b=y)(b in (a,b) of R implies x in D_R) What is exists b,y | b=y supposed to mean? > Where does the a come from and why is x not used in the precedent? For me, b=y means the flow of information between b and y is bidirectional. and > (forall y in A)(exists a,x | a=x)(a in (a,b) of R implies y in D_R) Same here > Casually, this means that all x,y in A has some (a,b) in R. Interestingly, x in A seems to be associating with b in (a,b) of R, and y in A seems to be associating with a in (a,b) of R. I attempted to provide an explanation for this. I figured that, > D_R subset A; D'_R subset A by definition > The flow of information in xRy is, > (x in A) to (y in (x,y) of R) to (y in A) to (x in (x,y) of R) to (x in A) ?? This basically means AxA using yours and Elliot's notation (which I assume is the standard). > So, you can see something about my notation that, > x_A:=x in A > y_R:=y in (x,y) of R > y_A:=y in A > x_R:=x in (x,y) of R x_A y_R y_A x_R x_A is the shortest return path from x_A to x_A for xRy. I have more innovation but that is as far as relevant to the parent thread. Don't blindly criticize me for using my own notation or for using my own interpretation of xRy, because the sole objective is to learn. I was essentially describing the flow of information in AxA. a a' x b b' -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A picture of xRy <8764akdccc.fsf@laptop.at.pyenos > This basically means AxA using yours and Elliot's notation (which I > assume is the standard). > It's impossible to even consider that's what it meant. === Subject: Re: A picture of xRy > This basically means AxA using yours and Elliot's notation (which I > assume is the standard). It's impossible to even consider that's what it meant. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Second Derivative of Thin Plate Splines -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Hi. Earlier, I asked about spline techniques for 2d input, and Jentje Goslinga (from here) pointed me towards thin plate splines... At first glance, it worked excellently for me... Visually, everything was smooth, etc, etc. But, I ran into a problem when I needed to take the second derivative of wi*((X - xi)^2 + (Y - yi)^2)*Log[Sqrt[(X - xi)^2 + (Y - yi)^2]] (wrt X), I get: wi + (2*wi*(X - xi)^2)/((X - xi)^2 + (Y - yi)^2) + 2*wi*Log[Sqrt[(X - xi)^2 + (Y - yi)^2]] Which, as Limit X->xi, Y->yi, becomes wi (-Infinity) Are my calculations right? Part of my app, unfortunately, uses the second derivative, so if the second derivative goes to infinity like that, I don't think I can use this method... Is there a way to fix that? (Perhaps, there's something inherent about the way the weights are formed, that those second derivatives would (effectively) cancel themselves out?) Is there a different way of taking the second derivative (at the very least) that would avoid the infinities? Binesh Bannerjee - -- And, you can tell her royal high and mighty Queen Mab that magic or no magic, if she harms you in any way, I'll have her guts for my boot laces. -- Auntie A (Ambrosia), Merlin PGP Key: http://www.hex21.com/~binesh/binesh-public.asc PGP Key fingerprint = 421D B4C2 2E96 B8EE 7190 A0CF B42F E71C 7FC3 AD96 SSH2 Key: http://www.hex21.com/~binesh/binesh-ssh2.pub OpenSSH Key: http://www.hex21.com/~binesh/binesh-openssh.pub BubbleBabble = xibeb-voges-havez-pabaf-debop-cylil-lelyc-viruv-bygeg-zotoh-dixex Fingerprint = 9d:7c:84:5d:80:e3:65:8d:ee:9e:a3:b9:56:0a:e9:ad SSH1 Key: http://www.hex21.com/~binesh/binesh-ssh1.pub CipherKnight Seals: http://www.hex21.com/~binesh/binesh-seal.tar.bz2.cs256 http://www.hex21.com/~binesh/binesh-seal.zip.cs256 http://www.hex21.com/~binesh/binesh-certificate.gif.cs256 Decrypt with CipherSaber2 N=256, Password=WelcomeJedi! (No quotes) -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.5 (GNU/Linux) iD8DBQFFwuortC/nHH/DrZYRAtkIAJ9pYm4Aw2Q5Q9tx6ly75fL2AAQ54QCgrzSY +grGB6BK7z6xuLbW09B2WRg= =BIBH -----END PGP SIGNATURE----- === Subject: Re: Second Derivative of Thin Plate Splines You are using r^2 log(r) which does not have second derivative. If you want second derivative use r^4 log(r). I guess you are using them as radial basis functions. You have to appropriate order polynomial to get positive definiteness. praveen -- http://pc.freeshell.org === Subject: Re: Just saw something quite hurrendous and outright racist on public TV , And what has this to do with mathematics? Nothing other than if this continues, he'll get black listed from > sci.math. OP, post your stuff in newsgroup where it's not Off Topic. How does black-listing work? Does it have to do with the cabal? -- Michael Press === Subject: definition of statistical test for randomness Does anybody know the definition of a statistical test for non randomness over a {0,1}* string ? If the test fails on longer and longer prefixes infinitley you can say for sure the string is not random. Basically what I'm asking is if someone comes to you and says - here is a test that tells that an infinite string is _not_ random, how do you verify that this test is valid and will _not_ exclude random strings. === Subject: Re: definition of statistical test for randomness > Does anybody know the definition of a statistical test for non > randomness over a {0,1}* string ? > If the test fails on longer and longer prefixes infinitley you can say > for sure the string is not random. > Basically what I'm asking is if someone comes to you and says - here > is a test that tells that an infinite string is _not_ random, how do > you verify that this test is valid and will _not_ exclude random > strings. There are several misconceptions in the formulation of that question. A *process* can be random, and if a string is the result of a process then one can ask what evidence it gives in support of the hypothesis that the process is given by some specific stochastic model. specified, it is not random in the new context. (It becomes totally redundant.) The amount of surprise embedded in the string has to be considered in the context of predictability in the context before the string is specified, which is another way of saying it is the source model that the string tests. About all you can meaningfully ask about randomness for a specific string per se, without considering the source, is consider how compressible the string is (Chaitin's model for nonrandomness). There is a theorem that says that you can't find an algorithm that will always answer that question with a reasonable amount of computation. If you try to come up with a definition of randomness that requires uniform distribution for moments of all orders, you end up finding that there are no random strings. (Knuth has something about this in TAOCP, in connection with random number generation.) There are no infinite strings. === Subject: Re: definition of statistical test for randomness >> Does anybody know the definition of a statistical test for non >> randomness over a {0,1}* string ? >> If the test fails on longer and longer prefixes infinitley you can say >> for sure the string is not random. >> Basically what I'm asking is if someone comes to you and says - here >> is a test that tells that an infinite string is _not_ random, how do >> you verify that this test is valid and will _not_ exclude random >> strings. There are several misconceptions in the formulation of that question. > A *process* can be random, and if a string is the result of a process > then one can ask what evidence it gives in support of the hypothesis > that the process is given by some specific stochastic model. specified, it is not random in the new context. (It becomes totally > redundant.) The amount of surprise embedded in the string has to > be considered in the context of predictability in the context before > the string is specified, which is another way of saying it is the source > model that the string tests. ++++++++++++++++++++++++++++++++++++++++ AHA. The source model is the egg, and the egg can't produce the chicken (the string) without a seed, and the seed is just another chicken. Therefore the tests on the string can't specifically test the model, since the string is dependent on a prior string (the seed). The better RNG's use vectors of random numbers as the seed. David Heiser === Subject: Re: definition of statistical test for randomness > Does anybody know the definition of a statistical test for non > randomness over a {0,1}* string ? The usual form of statistical testing for (non) randomness is a classic hypothesis test. The null hypothesis is typically that all strings are equally likely. The test results in a p-value, which is the change of gettng a result at least as extreme given that the null hypothesis is true. We reject if the p-value is too small. The key definition, in my opinion, is that of p-value: http://en.wikipedia.org/wiki/P-value Wikepidia also offers explanations of statistical hypothesis testing, null hypothesis, and statistical significance that you might find helpful. > If the test fails on longer and longer prefixes infinitley you can say > for sure the string is not random. Basically what I'm asking is if someone comes to you and says - here > is a test that tells that an infinite string is _not_ random, how do > you verify that this test is valid and will _not_ exclude random > strings. Any statistical hypothesis test will have some chance of rejecting even if the generator is perfect. If we reject when p <= s, then a perfect generator gets rejected with probability s. Choosing a small s (significance level) reduces the change of rejecting a random source, but allows more bad generators to pass. -- --Bryan === Subject: Re: definition of statistical test for randomness >> Does anybody know the definition of a statistical test for non >> randomness over a {0,1}* string ? >The usual form of statistical testing for (non) randomness >is a classic hypothesis test. The null hypothesis is typically >that all strings are equally likely. The test results in a >p-value, which is the change of gettng a result at least as >extreme given that the null hypothesis is true. We reject if >the p-value is too small. >The key definition, in my opinion, is that of p-value: > http://en.wikipedia.org/wiki/P-value >Wikepidia also offers explanations of statistical hypothesis >testing, null hypothesis, and statistical significance >that you might find helpful. >> If the test fails on longer and longer prefixes infinitley you can say >> for sure the string is not random. >> >> Basically what I'm asking is if someone comes to you and says - here >> is a test that tells that an infinite string is _not_ random, how do >> you verify that this test is valid and will _not_ exclude random >> strings. >Any statistical hypothesis test will have some chance of >rejecting even if the generator is perfect. If we reject >when p <= s, then a perfect generator gets rejected with >probability s. Choosing a small s (significance level) >reduces the change of rejecting a random source, but >allows more bad generators to pass. He was asking how to tell if a string, not a source, is random.Of course I guess you could call a string random if you regarded it as a byte genrator by reading out byte by byte. === Subject: Re: definition of statistical test for randomness [...] > He was asking how to tell if a string, not a source, is random.Of course I > guess you could call a string random if you regarded it as a byte genrator > by reading out byte by byte. I had thought he meant the same thing, the infinite string simply being what the source generates, known past or potential future. His subsequent posts lead me to doubt my interpretation there. -- --Bryan === Subject: Re: definition of statistical test for randomness > Does anybody know the definition of a statistical test for non > randomness over a {0,1}* string ? If the test fails on longer and longer prefixes infinitley you can say > for sure the string is not random. Basically what I'm asking is if someone comes to you and says - here > is a test that tells that an infinite string is _not_ random, how do > you verify that this test is valid and will _not_ exclude random > strings. > Randomness is a matter of definition. The first thing you have to do is to assume a model, that is a set of rules and assumptions about the properties of strings. Only then can you make any statements about the randomness of a given string. There is no such thing as general randomness. === Subject: Re: definition of statistical test for randomness >> Does anybody know the definition of a statistical test for non >> randomness over a {0,1}* string ? >> If the test fails on longer and longer prefixes infinitley you can say >> for sure the string is not random. >> Basically what I'm asking is if someone comes to you and says - here >> is a test that tells that an infinite string is _not_ random, how do >> you verify that this test is valid and will _not_ exclude random >> strings. > > Randomness is a matter of definition. The first thing you have to do > is to assume a model, that is a set of rules and assumptions about the > properties of strings. Only then can you make any statements about the > randomness of a given string. There is no such thing as general > randomness. > +++++++++++++++++++++++++++++++++++++++++++++ Guenther is right on this. I have done some testing of different RNG's and researched the literature. Any requirements are dependent on just what the RNG sequence will be used for. In very general terms the RNG is basically a random sequence of bits, in a designated RNG form. (i.e. floating point, integer, long integer, etc.). Tests on the integer forms are generally are on the bit sequences of sequentially combined RNGs, in a file that may consist of 1 billion sequential bits. For the available tests, floating point numbers have to be converted to 16, 32 or 64 bit unsigned integers to create the file to be tested. Knuth, volume 2, first few chapters is a good beginning view. David Heiser === Subject: Re: definition of statistical test for randomness >> Does anybody know the definition of a statistical test for non >> randomness over a {0,1}* string ? >> If the test fails on longer and longer prefixes infinitley you can say >> for sure the string is not random. >> Basically what I'm asking is if someone comes to you and says - here >> is a test that tells that an infinite string is _not_ random, how do >> you verify that this test is valid and will _not_ exclude random >> strings. > > Randomness is a matter of definition. The first thing you have to do > is to assume a model, that is a set of rules and assumptions about the > properties of strings. Only then can you make any statements about the > randomness of a given string. There is no such thing as general > randomness. +++++++++++++++++++++++++++++++++++++++++++++ > Guenther is right on this. I have done some testing of different RNG's and > researched the literature. Any requirements are dependent on just what the > RNG sequence will be used for. In very general terms the RNG is basically a > random sequence of bits, in a designated RNG form. (i.e. floating point, > integer, long integer, etc.). Tests on the integer forms are generally are > on the bit sequences of sequentially combined RNGs, in a file that may > consist of 1 billion sequential bits. For the available tests, floating > point numbers have to be converted to 16, 32 or 64 bit unsigned integers to > create the file to be tested. Knuth, volume 2, first few chapters is a good beginning view. Particularly section 3.5, _What_is_a_Random_Sequence_? Highly recommended. This essay is part survey of results, part investigation into how to pose and answer the question. One finishes with the conviction that the question cannot be precisely posed, much less answered. Nevertheless we can have a lot of fun looking into it and learning how to generate sequences that are aæpostiori unpredictable for a particular application. -- Michael Press === Subject: Re: definition of statistical test for randomness > Does anybody know the definition of a statistical test for non > randomness over a {0,1}* string ? If the test fails on longer and longer prefixes infinitley you can say > for sure the string is not random. Basically what I'm asking is if someone comes to you and says - here > is a test that tells that an infinite string is _not_ random, how do > you verify that this test is valid and will _not_ exclude random > strings. It will exclude random strings. Since any string , including all ones, as as probable as any othr string if generated by a random process. > Randomness is a matter of definition. The first thing you have to do No, randomness is a matter of process. >> is to assume a model, that is a set of rules and assumptions about the >> properties of strings. Only then can you make any statements about the No about the properties of the generator of the string. >> randomness of a given string. There is no such thing as general >> randomness. Of course Kolmogorov and Chaitin would say that a string is random with respect to a given computer language if the shortest program which can generate that string is equal or greater in length than the string itself. That is a property of the string itself. >+++++++++++++++++++++++++++++++++++++++++++++ >Guenther is right on this. I have done some testing of different RNG's and >researched the literature. Any requirements are dependent on just what the >RNG sequence will be used for. In very general terms the RNG is basically a >random sequence of bits, in a designated RNG form. (i.e. floating point, >integer, long integer, etc.). Tests on the integer forms are generally are >on the bit sequences of sequentially combined RNGs, in a file that may >consist of 1 billion sequential bits. For the available tests, floating >point numbers have to be converted to 16, 32 or 64 bit unsigned integers to >create the file to be tested. >Knuth, volume 2, first few chapters is a good beginning view. >David Heiser === Subject: Re: definition of statistical test for randomness Does anybody know the definition of a statistical test for non > randomness over a {0,1}* string ? If the test fails on longer and longer prefixes infinitley you can say > for sure the string is not random. Basically what I'm asking is if someone comes to you and says - here > is a test that tells that an infinite string is _not_ random, how do > you verify that this test is valid and will _not_ exclude random > strings. It will exclude random strings. Since any string , including all ones, as > as probable as any othr string if generated by a random process. >> Randomness is a matter of definition. The first thing you have to do No, randomness is a matter of process. Yes? Kindly elaborate. > is to assume a model, that is a set of rules and assumptions about the >> properties of strings. Only then can you make any statements about the No about the properties of the generator of the string. That depends on how you pose the question. >> randomness of a given string. There is no such thing as general >> randomness. Of course Kolmogorov and Chaitin would say that a string is random with > respect to a given computer language if the shortest program which can > generate that string is equal or greater in length than the string itself. > That is a property of the string itself. See, its a matter of definition. >+++++++++++++++++++++++++++++++++++++++++++++ >Guenther is right on this. I have done some testing of different RNG's and >researched the literature. Any requirements are dependent on just what the >RNG sequence will be used for. In very general terms the RNG is basically a >random sequence of bits, in a designated RNG form. (i.e. floating point, >integer, long integer, etc.). Tests on the integer forms are generally are >on the bit sequences of sequentially combined RNGs, in a file that may >consist of 1 billion sequential bits. For the available tests, floating >point numbers have to be converted to 16, 32 or 64 bit unsigned integers to >create the file to be tested. >Knuth, volume 2, first few chapters is a good beginning view. >David Heiser === Subject: Re: definition of statistical test for randomness >> Does anybody know the definition of a statistical test for non >> randomness over a {0,1}* string ? >> If the test fails on longer and longer prefixes infinitley you can say >> for sure the string is not random. >> Basically what I'm asking is if someone comes to you and says - here >> is a test that tells that an infinite string is _not_ random, how do >> you verify that this test is valid and will _not_ exclude random >> strings. It will exclude random strings. Since any string , including all ones, as > as probable as any othr string if generated by a random process. > Randomness is a matter of definition. The first thing you have to do No, randomness is a matter of process. +++++++++++++++++++++++++++++++++++++++++++++ No. It is chicken and egg problem, which came first. The properties of the generator are always unknown, except for the linear congruential generators defined in Knuth. The properties of these (the egg) can by found by mathematical analysis of the modulus, multiplier, increment and starting value. All of the more recent generators such as the Mersene Twister are beyond Knuth. Just estimating the period does not presume that it is truely a random process. One has to always test the sequence. One also has to reconcile all the different views of what is a true random process. A process that produces an unpredictable outcome (cryptology) does not guarantee that other requirements (i.e. L'Ecuyer's tests) are met. In this case, randomness is being able to pass all of L'Ecuyer's tests. It also depends on one's belief in the importance of multi-dimensional properties. DAH === Subject: Re: definition of statistical test for randomness >> Does anybody know the definition of a statistical test for non > randomness over a {0,1}* string ? >> If the test fails on longer and longer prefixes infinitley you can say > for sure the string is not random. >> Basically what I'm asking is if someone comes to you and says - here > is a test that tells that an infinite string is _not_ random, how do > you verify that this test is valid and will _not_ exclude random > strings. >> It will exclude random strings. Since any string , including all ones, as >> as probable as any othr string if generated by a random process. > Randomness is a matter of definition. The first thing you have to do >> No, randomness is a matter of process. > +++++++++++++++++++++++++++++++++++++++++++++ > No. It is chicken and egg problem, which came first. The properties of the > generator are always unknown, except for the linear congruential > generators defined in Knuth. The properties of these (the egg) can by > found by mathematical analysis of the modulus, multiplier, increment and > starting value. All of the more recent generators such as the Mersene > Twister are beyond Knuth. Just estimating the period does not presume that > it is truely a random process. One has to always test the sequence. One > also has to reconcile all the different views of what is a true random > process. A process that produces an unpredictable outcome (cryptology) > does not guarantee that other requirements (i.e. L'Ecuyer's tests) are > met. In this case, randomness is being able to pass all of L'Ecuyer's > tests. It also depends on one's belief in the importance of > multi-dimensional properties. > DAH +++++++++++++++++++++++++++++++++++++++++++++++++ I forgot to say, Marsaglia's tests are based on supplying the chicken. L'Ecuyer's tests are based on supplying the egg. Its a matter of opinion, which is adequate / best. DAH > === Subject: Re: definition of statistical test for randomness > Does anybody know the definition of a statistical test for non >> randomness over a {0,1}* string ? >> If the test fails on longer and longer prefixes infinitley you can >> say >> for sure the string is not random. >> Basically what I'm asking is if someone comes to you and says - here >> is a test that tells that an infinite string is _not_ random, how do >> you verify that this test is valid and will _not_ exclude random >> strings. It will exclude random strings. Since any string , including all ones, > as > as probable as any othr string if generated by a random process. > Randomness is a matter of definition. The first thing you have to do No, randomness is a matter of process. >> +++++++++++++++++++++++++++++++++++++++++++++ >> No. It is chicken and egg problem, which came first. The properties of >> the generator are always unknown, except for the linear congruential >> generators defined in Knuth. The properties of these (the egg) can by >> found by mathematical analysis of the modulus, multiplier, increment and >> starting value. All of the more recent generators such as the Mersene >> Twister are beyond Knuth. Just estimating the period does not presume >> that it is truely a random process. One has to always test the sequence. >> One also has to reconcile all the different views of what is a true >> random process. A process that produces an unpredictable outcome >> (cryptology) does not guarantee that other requirements (i.e. L'Ecuyer's >> tests) are met. In this case, randomness is being able to pass all of >> L'Ecuyer's tests. It also depends on one's belief in the importance of >> multi-dimensional properties. >> DAH +++++++++++++++++++++++++++++++++++++++++++++++++ > I forgot to say, Marsaglia's tests are based on supplying the chicken. > L'Ecuyer's tests are based on supplying the egg. Its a matter of > opinion, which is adequate / best. > DAH Au contraire. If a chicken is a procedure that produces a sequence of 32-bit integers (eggs) , then the Diehard battery of tests of randomness, for example, from http://www.csis.hku.hk/~diehard/ may ask for your chicken, which then provides the eggs, or it may just ask for your basket of eggs. George Marsaglia === Subject: Re: Behaviour of this IVP > Oh, I must add y ' (0)=0 to the initial condition. > Does the IVP: y '' = sqrt{y}, y(0)=1 >> have a closed form solution? I doubt it; but maybe there is a tricky way >> to do it. > What does = mean? === Subject: Re: Behaviour of this IVP > Oh, I must add y ' (0)=0 to the initial condition. > Does the IVP: y '' = sqrt{y}, y(0)=1 have a closed form solution? I doubt it; but maybe there is a tricky > way to do it. What does = mean? > It is the LaTex's notation for square root. === Subject: A proof needed (open base) The question 1.1 in Billingsley is to prove that the finitely dimensional open sets form a basis for the space of continuous functions on [0;1], just as they do for R^oo. What i don't see is how to interpret the question. The set of such functions is defined on [0;1] (that's 1 dimension only). Can somebody get me started on this one? -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: A proof needed (open base) On Fri, 2 Feb 2007 11:43:19 +0100, Konrad Viltersten >The question 1.1 in Billingsley is to prove that the finitely >dimensional open sets form a basis for the space of >continuous functions on [0;1], just as they do for R^oo. I doubt very much that this is what the exercise asks. Because any (nonempty) open subset of the space of continuous functions on [0,1] has infinite dimension. the exercise asks, but it's not. _Exactly_ what how does the exercise read, word for word? >What i don't see is how to interpret the question. The set >of such functions is defined on [0;1] (that's 1 dimension >only). [0,1] is not a subset of the space of continuous functions on [0,1]. > Can somebody get me started on this one? Perhaps. First you need to state the question correctly. ************************ David C. Ullrich === Subject: Re: A proof needed (open base) > On Fri, 2 Feb 2007 11:43:19 +0100, Konrad Viltersten > Can somebody get me started on this one? Perhaps. First you need to state the question correctly. The exact formulation is as follows. Show that... The open, finite dimensional sets in R^inf for a basis for the topology. Is this true in C. Of what i might provide as extra information is that C is (in the text, a bit above) mentioned to be the space of continuous functions on [0;1]. I tried to avoid a pure recitation of the question because it's of no use to me and no fun for you if i only post half of the book. Perhaps it was a bad idea in this case. I'm sorry. -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: A proof needed (open base) On Sat, 3 Feb 2007 19:58:26 +0100, Konrad Viltersten >> On Fri, 2 Feb 2007 11:43:19 +0100, Konrad Viltersten >> Perhaps. First you need to state the question correctly. The exact formulation is as follows. Show that... The open, finite dimensional sets in R^inf for a basis >for the topology. Is this true in C. That's exactly what it says in the book? It says Is this true in C., and not Is this true in C? ? Hard to believe. >Of what i might provide as extra information is that C >is (in the text, a bit above) mentioned to be the space >of continuous functions on [0;1]. I tried to avoid a pure recitation of the question >because it's of no use to me and no fun for you if i >only post half of the book. Perhaps it was a bad idea >in this case. I'm sorry. You didn't just restate the question! You changed Is this true? to Show that this is true.. ************************ David C. Ullrich === Subject: Re: A proof needed (open base) > On Sat, 3 Feb 2007 19:58:26 +0100, Konrad Viltersten > On Fri, 2 Feb 2007 11:43:19 +0100, Konrad Viltersten > Can somebody get me started on this one? Perhaps. First you need to state the question correctly. >> The exact formulation is as follows. Show that... >> The open, finite dimensional sets in R^inf for a basis >> for the topology. Is this true in C. That's exactly what it says in the book? It says Is this > true in C., and not Is this true in C? ? Hard to believe. Yes, of course. Typo, sorry. I've also found two more as i read it once again. Must have been tired last time... >> Of what i might provide as extra information is that C >> is (in the text, a bit above) mentioned to be the space >> of continuous functions on [0;1]. >> I tried to avoid a pure recitation of the question >> because it's of no use to me and no fun for you if i >> only post half of the book. Perhaps it was a bad idea >> in this case. I'm sorry. You didn't just restate the question! You changed > Is this true? to Show that this is true.. So, to re-state the question, it will be as follows. The open, finite-dimensional sets in R^inf form a basis for the topology. Is this true in C? C is to be regarded (i guess) as before and the task itself is supposed to seen as show that. I assumed mr. B not to ask to show things if they weren't true but i might have err there. Sorry. So, now, that the misspellings are gone, is there any competent soul wishing to get me started? -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: A proof needed (open base) >On Fri, 2 Feb 2007 11:43:19 +0100, Konrad Viltersten >The question 1.1 in Billingsley is to prove that the finitely >>dimensional open sets form a basis for the space of >>continuous functions on [0;1], just as they do for R^oo. >> > >I doubt very much that this is what the exercise asks. >Because any (nonempty) open subset of the space of >continuous functions on [0,1] has infinite dimension. > > Since this iis a probability text, Billingsley is probably referring to continuous-path stochastic processes. Th topology is most likely the subspace topology of the point-open (also called product) topology on R^[0,1]. BTW, this question is nowhere to be found in chapter 1 of the second edition of Probability and Measure. I wonder to what Billingsley the OP refers. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: A proof needed (open base) > >On Fri, 2 Feb 2007 11:43:19 +0100, Konrad Viltersten >The question 1.1 in Billingsley is to prove that the finitely >>dimensional open sets form a basis for the space of >>continuous functions on [0;1], just as they do for R^oo. >> > >I doubt very much that this is what the exercise asks. >Because any (nonempty) open subset of the space of >continuous functions on [0,1] has infinite dimension. > > Since this iis a probability text, Billingsley is probably referring to > continuous-path stochastic processes. Th topology is most likely the > subspace topology of the point-open (also called product) topology on > R^[0,1]. BTW, this question is nowhere to be found in chapter 1 of the second > edition of Probability and Measure. I wonder to what Billingsley the OP > refers. Based on the subject matter and question number, probably his Convergence of Probability Measures. I'd look but don't have my copy at hand. -- A. === Subject: Re: A proof needed (open base) > The question 1.1 in Billingsley is to prove that the finitely > dimensional open sets form a basis for the space of > continuous functions on [0;1], just as they do for R^oo. What i don't see is how to interpret the question. The set > of such functions is defined on [0;1] (that's 1 dimension > only). Can somebody get me started on this one? > [0,1] is one dimensional. { f:[0,1] -> R } = R^[0,1] is an uncountable product of R's, or of unit intervals if your functions have codomain [0,1]. It's an uncountable dimension space. Not knowing what topology you're putting on the function space and what you mean by finite dimensional open set, I'll draw from topology. A base for S = uncountable product of R's given the Tychonov product topology (not always used for function spaces) are sets with the form of S with finite many of the factors limited to open sets while the rest of the factors are R. === Subject: Re: A proof needed (open base) >> The question 1.1 in Billingsley is to prove that the finitely >> dimensional open sets form a basis for the space of >> continuous functions on [0;1], just as they do for R^oo. >> What i don't see is how to interpret the question. The set >> of such functions is defined on [0;1] (that's 1 dimension >> only). Can somebody get me started on this one? > [0,1] is one dimensional. > { f:[0,1] -> R } = R^[0,1] is an uncountable product of R's, > or of unit intervals if your functions have codomain [0,1]. It's an uncountable dimension space. Not knowing what > topology you're putting on the function space and what > you mean by finite dimensional open set, I'll draw from > topology. Billingsley doesn't explicitly state what metric he uses but judgin from previous examples, one can guess he means the uniform metric. Does this clarify anything? -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: A proof needed (open base) <52gv7iF1nvmblU1@mid.individual.net> === Subject: Re: A proof needed (open base) >> The question 1.1 in Billingsley is to prove that the finitely >> dimensional open sets form a basis for the space of >> continuous functions on [0;1], just as they do for R^oo. >> What i don't see is how to interpret the question. The set >> of such functions is defined on [0;1] (that's 1 dimension >> only). Can somebody get me started on this one? > [0,1] is one dimensional. > { f:[0,1] -> R } = R^[0,1] is an uncountable product of R's, Whoops, that's the subspace S = { f:[0,1] -> R : f continuous } > or of unit intervals if your functions have codomain [0,1]. It's an uncountable dimension space. Not knowing what > topology you're putting on the function space and what > you mean by finite dimensional open set, I'll draw from > topology. | Billingsley doesn't explicitly state what metric he uses | but judgin from previous examples, one can guess he | means the uniform metric. Does this clarify anything? No. I don't like uniform spaces. The uniform metric for R^[0,1] is d(f,g) = sup{ |f(x) - g(x)| : x in [0,1] } generated by the sup norm ||f|| = sup{ |f(x)| : x in [0,1] } ? subspace of the space I consider. That however, was based upon guessing a definition for finite dimensional open set. On the other hand are you guessing what metric he's using, or does he make it clear from the start that all function spaces have the sup norm? ---- === Subject: Re: A proof needed (open base) Konrad- Are you reading Billingsley on your own? Based on your recent posts, I suspect that that is not a good idea for someone with your background. For example, have you studied basic topology and analysis (as summarized in Billingsley's appendix)? Heck, I don't think learning from Billingsley is good for anyone. It is encyclopaedic, but not very clear from a pedagogical viewpoint. It might be OK if used in conjuction with an instructor. (I say might even though that was how I first used it, as a student.) -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: How to prove an inequality involving sequence of weighted means? I'm not sure if my proof is OK, would like some help please. Let a_n be a sequence of real numbers, w_n a sequence of positive weights and s_n = (Sum(i=1,n) (w(i) a(i))/(Sum(i=1, n) w(i)). If Sum(i=1,oo) w(i) diverges, then lim inf a_n <= lim inf s_n <= lim sup s_n <= lim sup _a_n (lim inf and lim sup are quite confusing to me) The middle inequality holds automatically. My proof for the left one is as follows (the right one is proved similarly): If lin inf a_n = -oo , then there's nothing to prove, right? So, suppose lim inf a_n > -oo. For every q < lim inf a(n), there exists (property of the lim inf) a natural k such that a(n) > q for every n > k. Let S(n) be the sequence of partial sums of w(n) and let w = infimum{w(1)...w(k)}. Then, for n >k we have s_n = (Sum(i=1, k) w(i) a(i) + Sum(i= k+1, n) w(i) a(i))/S(n) > (w S(k) + q (S(n) - S(k))/ (S(n) = ((w - q)S(k) + q S(n)) = ((w - q)S(k))/S(n) + q . Putting y(n) = ((w - q)S(k))/S(n) + q , we get s(n) > y(n) n for n > k. Therefore, lim inf s(n) >= lim inf y(n) Since S(n) diverges and it's terms are positive, S(n) => oo and lim inf y(n) = lim y(n) = 0 + q = q, because k is fixed and doesn't depend on n. Therefore, lim inf s(n) >= q for every q < lim inf a(n). In order for this to be true, we must have lim inf s(n) >= lim inf a(n). These inequalities imply that, if a(n) converges to a, then s(n) converges to a. This is true even if a = oo or -oo, right? But if Sum( w(i)) converges, then I think a(n) may converge to an a and s(n) converges to a s <>a Sharon === Subject: Re: How to prove an inequality involving sequence of weighted means? > I'm not sure if my proof is OK, would like some help please. Let a_n be a sequence of real numbers, w_n a sequence of positive > weights and s_n = (Sum(i=1,n) (w(i) a(i))/(Sum(i=1, n) w(i)). If > Sum(i=1,oo) w(i) diverges, then lim inf a_n <= lim inf s_n <= lim sup s_n <= lim sup _a_n (lim inf > and lim sup are quite confusing to me) The middle inequality holds automatically. My proof for the left one > is as follows (the right one is proved similarly): If lin inf a_n = -oo , then there's nothing to prove, right? Right. > So, > suppose lim inf a_n > -oo. For every q < lim inf a(n), there exists > (property of the lim inf) a natural k such that a(n) > q for every n > k. Let S(n) be the sequence of partial sums of w(n) and let w = > infimum{w(1)...w(k)}. Then, for n >k we have s_n = (Sum(i=1, k) w(i) > a(i) + Sum(i= k+1, n) w(i) a(i))/S(n) > (w S(k) + q (S(n) - S(k))/ > (S(n) = ((w - q)S(k) + q S(n)) = ((w - q)S(k))/S(n) + q . Putting > y(n) = ((w - q)S(k))/S(n) + q , we get > s(n) > y(n) n for n > k. Therefore, lim inf s(n) >= lim inf y(n) > Since > S(n) diverges and it's terms are positive, S(n) => oo and lim inf > y(n) = lim y(n) = 0 + q = q, because k is fixed and doesn't depend on > n. Almost. There is a problem when you write that (Sum(i=1, k) w(i) a(i) + Sum(i= k+1, n) w(i) a(i))/S(n) is greater than (w S(k) + q (S(n) - S(k))/S(n). In this expression you should have Sum(i=1,k)a_i instead of S(k). But this doesn't affect your argument. > Therefore, lim inf s(n) >= q for every q < lim inf a(n). In order for > this to be true, we must have lim inf s(n) >= lim inf a(n). These inequalities imply that, if a(n) converges to a, then s(n) > converges to a. This is true even if a = oo or -oo, right? Almost. The language is not correct, but use the limit of a(n) is a and it will be correct. (If the limit of a sequence is +oo or -oo, it does not converge.) Jose Carlos Santos === Subject: Re: How to prove an inequality involving sequence of weighted means? <52gmdgF1otodbU1@mid.individual.net > I'm not sure if my proof is OK, would like some help please. Let a n be a sequence of real numbers, w n a sequence of positive > weights and s n = (Sum(i=1,n) (w(i) a(i))/(Sum(i=1, n) w(i)). If > Sum(i=1,oo) w(i) diverges, then lim inf a n <= lim inf s n <= lim sup s n <= lim sup a n (lim inf > and lim sup are quite confusing to me) The middle inequality holds automatically. My proof for the left one > is as follows (the right one is proved similarly): If lin inf a n = -oo , then there's nothing to prove, right? Right. So, > suppose lim inf a n > -oo. For every q < lim inf a(n), there exists > (property of the lim inf) a natural k such that a(n) > q for every n > k. Let S(n) be the sequence of partial sums of w(n) and let w = > infimum{w(1)...w(k)}. Then, for n >k we have s n = (Sum(i=1, k) w(i) > a(i) + Sum(i= k+1, n) w(i) a(i))/S(n) > (w S(k) + q (S(n) - S(k))/ > (S(n) = ((w - q)S(k) + q S(n)) = ((w - q)S(k))/S(n) + q . Putting > y(n) = ((w - q)S(k))/S(n) + q , we get > s(n) > y(n) n for n > k. Therefore, lim inf s(n) >= lim inf y(n) > Since > S(n) diverges and it's terms are positive, S(n) => oo and lim inf > y(n) = lim y(n) = 0 + q = q, because k is fixed and doesn't depend on > n. Almost. There is a problem when you write that (Sum(i=1, k) w(i) a(i) + Sum(i= k+1, n) w(i) a(i))/S(n) is greater than (w S(k) + q (S(n) - S(k))/S(n). In this expression you should have Sum(i=1,k)a i instead of S(k). But > this doesn't affect your argument. Therefore, lim inf s(n) >= q for every q < lim inf a(n). In order for > this to be true, we must have lim inf s(n) >= lim inf a(n). These inequalities imply that, if a(n) converges to a, then s(n) > converges to a. This is true even if a = oo or -oo, right? Almost. The language is not correct, but use the limit of a(n) is a > and it will be correct. (If the limit of a sequence is +oo or -oo, it > does not converge.) > Jose Carlos Santos Oh, yes, I made an algebraic mistake. Sharon === Subject: calculation of probability hello, I have a system , the system is 75 percent efficient . So out of one hundred events 75 are true and 25 are false . So what is the probability that there would be 13 false events in the first 50 events . and how is the calculation derived . Also what is the probability that there would be a sequence of 4 continous false events in a row and how many such sequences are probable ? and how calculation is derived . thansk . probabailty function . === Subject: Re: calculation of probability On Feb 2, 6:45 am, Probability function > hello, > I have a system , the system is 75 percent efficient . So out > of one hundred events 75 are true and 25 are false . Efficient is not normally used in that way, but OK. > So what is the probability that there would be 13 false > events in the first 50 events . Do you mean the number of false events is >= 13? Or is exactly equal to 13? The probability that there are exactly k true events (100-k false events) is [ 100!/ k! * (100-k)! ] * p^k * (1-p)^(n-k) where p = probability of one true event (0.75 for you). If you want the probability that k <= 87, then add up this value for k=1, 2, ..., 87. Or add up the values k = 88, 89,... ..., 100, which will give you the probability that there are more than 87 true events = probability that the number of false events is < 13. Then subtract it from 1. > and how is the calculation derived . This is the theory of Bernoulli trials. The distribution is called the binomial distribution. It applies any time you want the probability of k events in n trials, where each trial is independent and an event occurs in each trial with the same probability p. Also what is the probability that there would be a sequence of 4 > continous false events > in a row and how many such sequences are probable ? Counting outcomes with sequences is a little more complicated, but it's not too hard to estimate. For instance, you could divide up the 100 trials into 25 independent sets of 4. The probability of any one of those being a run of 4 is 0.25^4 = 1/256. So the probability that there are no runs among those 25 trials is (1/256)^0 * (255/256)^25, and the probability of being at least 1 success is 1 - that amount. But that's only an (under)estimate. It doesn't count all the possible places a sequence of 4 can occur. > and how calculation is derived . Bernoulli trials again. - Randy === Subject: Re: Question about completion of the preHilbert space C^{infty}(R) > ??? You say later that you're talking about the usual inner >> product, presumably meaning >> = int fg >> (at least for real f,g). >> But that integral need not exist, in any sense whatever, >> for functions in C^infinity(R) or C(R). Oops, of course not. Sorry. > Let's say instead you're talking about functions with compact >> support: OK. Would integrable functions work too? Square-integrable functions would be ok. >I >>ask because I think I read somewhere that both are isomorphic to >>L^2(R) with the ordinary Lebesgue measure, >> They are. That's simply the fact that both these spaces are >> dense in L^2(R). >>and I can't see any reason, >>short of actually proving it (which I probably couldn't do if I tried) >>why it would not be isomorphic to L^2(R) with a measure defined on >>some smaller measurable space, for example with the sigma-algebra of >>measurable sets consisting of all Borel sets rather than all Lebesgue- >>measurable sets. >> But that's exactly the same Hilbert space! If f is Lebesgue measurable >> then there exists a Borel measurable g with f = g almost everywhere. reply. -Rotwang ************************ David C. Ullrich === Subject: Re: Question about completion of the preHilbert space C^{infty}(R) <0li3s2hc3nq5k9sfvljnin2asnb1pnkdn9@4ax.com> <09a6s2d94d1cgivn2i8n7qde1vt244hflu@4ax.comOK. Would integrable functions work too? Square-integrable functions would be ok. That's what I meant. Again. I should really get into the habit of engaging my brain before I post. -Rotwang === Subject: Re: Question about completion of the preHilbert space C^{infty}(R) > Hi all. Can anyone tell me off the top of their heads what is the > completion to a Hilbert space of C^{infty}(R), the space of all smooth > functions R to R? With respect to the usual inner product, right?! > How about C(R), the space of continuous functions? No way, since it is not complete. Therefore, it cannot be a completion! > I > ask because I think I read somewhere that both are isomorphic to > L^2(R) Indeed! > with the ordinary Lebesgue measure, and I can't see any reason, > short of actually proving it (which I probably couldn't do if I tried) > why it would not be isomorphic to L^2(R) with a measure defined on > some smaller measurable space, for example with the sigma-algebra of > measurable sets consisting of all Borel sets rather than all Lebesgue- > measurable sets. L^2(R) contains C(R). Since C(R) contains the function of the type cos(n*t) (n >= 0) and sin(n*t) (n >= 1), the closure of C(R) in L^2(R) must also contain those functions. But the only complete subspace of L^2(R) which contains those functions is L^2(R) itself! Jose Carlos Santos === Subject: Zero Dimensional Rings Is it true that in zero dimensional rings (that are say semiprime) if I add two radical ideals I get a radical ideal again? I know that if I add to two prime ideals then I get a radical ideal, in fact I get the whole ring.. but what about if I add the intersection of prime ideals, do I get something of the same form? Jose Capco === Subject: Re: Zero Dimensional Rings On Feb 2, 1:11 pm, Jose Capco Is it true that in zero dimensional rings (that are say semiprime) if > I add two radical ideals I get a radical ideal again? I know that if I > add to two prime ideals then I get a radical ideal, in fact I get the > whole ring.. but what about if I add the intersection of prime ideals, > do I get something of the same form? Jose Capco I have the answer to my question... In fact: Every ideal in a reduced ring with 0 krull dimension is a radical ideals.. .. I'll just give hints of the facts, if people are interested I can give a full proof: 1. A zero krull dimensional reduced rings is a von Neumann regular ring 2. It suffices to prove that any principal ideal is a radical ideal. 3. It suffices to prove that if b^2 is in aA (for our ring A, a and b in A), then b is in A (the rest can be concluded by induction). Jose Capco === Subject: Re: Zero Dimensional Rings Is it true that in zero dimensional rings (that are say semiprime) if > I add two radical ideals I get a radical ideal again? I know that if I > add to two prime ideals then I get a radical ideal, in fact I get the > whole ring.. but what about if I add the intersection of prime ideals, > do I get something of the same form? If you assume the ring R to be Noetherian: Yes. Let be aa and bb two *different* radical ideals in R. I claim that aa + bb = R. W.l.o.g., there is a maximal ideal mm of R containing aa, but not containing bb (R is 0-dimensional). Then there are b in bb and x in mm with b+m=1 (*). Since R/aa is Artinian (use R Noetherian and 0-dimensional), the ideal mm/aa is nilpotent, there is some n>0 with such that mm^n is contained in aa. Take the n-th power of (*) to get an equation b'+m^n=1 with b' in bb. This implies that bb + mm^n = R, thus bb + aa = R. HTH. J. === Subject: Cyclic Decomposition Theorem The Cyclic Decomposition Theorem claims that if V is a finite dimensional vector space, T a linear operator on it and W0 a proper T-admissible subspace of V, then there exist non-zero vectors a_1, ..., a_r in V with T-annihilators p_1, ..., p_r such that (i) V = W0 (+) Z(a_1;T) (+) ... (+) Z(a_r;T) (ii) p_k | p_{k-1}, k = 2, ..., r. The theorem also states that the integer r and the annihilators p_1, ..., p_r are uniquely determined by (i) and (ii). I know that the T-annihilators can be found by looking at the matrix A-xI in (Smith) normal form, where A is the matrix associated with T, but how can I found a_1, ..., a_r? There exist an efficient algorithm? Kiuhnm === > >In message <87k5z4zi0o.fsf@phiwumbda.org>, Jesse F. Hughes > does (forall x in A)(exists x in B) mean > forall x(x in A implies (exists x (x in A & x in B))) > ? >>No, it doesn't mean anything at all. >>You have two bounded quantifiers in a row, without any formula >>following them. (forall x in A)(exists x in B) is not well-formed. >I'm not qualified to comment on the well-formedness or otherwise of the >semantics but I would say that the proposition: (for all x in A)(there exists x in B) would necessarily imply that A is a subset of B; there is an injection >from A to B. (for every element of A there is a unique corresponding >element of B, but not vice-versa). You can say that if you want. And your interpretation > is probably what the OP meant. But in fact > (for all x in A)(there exists x in B) does not > say what you say it says - in fact it says nothing > at all, exactly as Jesse said. You're confusing there exists x in B with x is in B. > When I write (for all x in A)(x is in B) the second > x means the same thing as the x in the for all x. > But if I write (for all x in A)(there exists x in B) > the two x's are completely independent - they have > nothing to do with each other at all. >So if an x is in A, there necessarily exists that x in B. Except that in the given notation the second x is _not_ > that x. > >So the statement: for all x(x in A implies (exists x (x in A & x in B))) is conditionally true on the statement (for all x in A)(there exists x >in B), as is also the statement: for all x(x in A implies (exists x (x in B))) The first mess really doesn't say anything at all, > so it's hard to explain why it doesn't say what the OP > thinks it says. This mess _does_ say something. > Let's figure out what. First, (exists x (x in B)) says exactly B is nonempty. > So the statement in question is for all x(x in A implies (B is nonempty)) This is true for any B which _is_ in fact > nonempty. For example, if A = {1} and > B = {2} then B is nonempty, hence > P implies B is nonempty is true for any > statement P, hence for all x(x in A implies (B is nonempty)) > is true, and hence for all x(x in A implies (exists x (x in B))) > is true. Even though A is not a subset of B. >But I am probably wrong. >-- >Andy Smith > David C. Ullrich (forall x in A)(exists x inB) is not well-formed because it is expecting a condition. (forall x)(x in A implies (exists x (x in A & x in B))) means for all A, B is non-empty. Therefore it is true even when A not subset B. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === <45bf936c.1820389307@news-south.giganews.com> (snip) >I'm not qualified to comment on the well-formedness or otherwise of the >>semantics but I would say that the proposition: >>(for all x in A)(there exists x in B) >>would necessarily imply that A is a subset of B; there is an injection >>from A to B. (for every element of A there is a unique corresponding >>element of B, but not vice-versa). You can say that if you want. And your interpretation >is probably what the OP meant. But in fact >(for all x in A)(there exists x in B) does not >say what you say it says - in fact it says nothing >at all, exactly as Jesse said. You're confusing there exists x in B with x is in B. Ah! Just so. MoeBlee and Mr Frege are quite correct, and I should have left well alone. I only said anything because thought jesse's comment was totally hostile and unhelpful, and your post here should help the OP a lot (as well as illuminating me) -- Andy Smith === >(snip) >I'm not qualified to comment on the well-formedness or otherwise of the >semantics but I would say that the proposition: (for all x in A)(there exists x in B) would necessarily imply that A is a subset of B; there is an injection >from A to B. (for every element of A there is a unique corresponding >element of B, but not vice-versa). >>You can say that if you want. And your interpretation >>is probably what the OP meant. But in fact >>(for all x in A)(there exists x in B) does not >>say what you say it says - in fact it says nothing >>at all, exactly as Jesse said. >>You're confusing there exists x in B with x is in B. Ah! Just so. MoeBlee and Mr Frege are quite correct, and I should have >left well alone. I only said anything because thought jesse's comment >was totally hostile and unhelpful, Huh? Two points: (i) Jesse said No, it doesn't mean anything at all. You have two bounded quantifiers in a row, without any formula following them. (forall x in A)(exists x in B) is not well-formed. which, as he claims, looks to me like a simple statement of fact, followed by an explanation. I gave a more detailed explanation. (ii) I don't see why anyone would find hostility towards this particular OP inappropriate. He's been nothing but obnoxious. >and your post here should help the OP >a lot (as well as illuminating me) > ************************ David C. Ullrich === > (ii) I don't see why anyone would find hostility towards this > particular OP inappropriate. He's been nothing but obnoxious. I've figured that OP is consistently being used to refer to me, especially by you. What is OP? ************************ David C. Ullrich -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === <45bf936c.1820389307@news-south.giganews.com> <87mz3wbw76.fsf@laptop.at.pyenos I've figured that OP is consistently being used to refer to me, > especially by you. What is OP? Original Poster. The guy who first started the thread. There's some other jargon you should learn like LOL IIRC ;-) :-) -) :-( -( === I've figured that OP is consistently being used to refer to me, > especially by you. What is OP? Original Poster. The guy who first started the thread. There's some other jargon you should learn like > LOL > IIRC > ;-) :-) -) Don't you mean :-) > :-( -( Don't you mean :-( > -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === I've figured that OP is consistently being used to refer to me, > especially by you. What is OP? Original Poster. The guy who first started the thread. There's some other jargon you should learn like > LOL > IIRC > ;-) :-) -) > :-( -( > I can program in more than three different languages. Can you? -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === > I can program in more than three different languages. Can you? Wow. If that's not genius, I don't know what is. -- Jesse F. Hughes Mathematicians who read proofs of my results seem to basically lose some part of themselves, like it rips at their souls, and they are no longer quite right in the head. -- James S. Harris, Geek Cthulhu === I can program in more than three different languages. Can you? Wow. If that's not genius, I don't know what is. I usually don't say what I can do. I usually say what I can't do. This was an exception. -- > Jesse F. Hughes > Mathematicians who read proofs of my results seem to basically lose > some part of themselves, like it rips at their souls, and they are no > longer quite right in the head. -- James S. Harris, Geek Cthulhu -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === > >> I can program in more than three different languages. Can you? >> >> Wow. If that's not genius, I don't know what is. > I usually don't say what I can do. I usually say what I can't > do. This was an exception. Well, you shut him up! Three languages! Goodness! I bet he's feeling inadequate. -- Jesse F. Hughes I, like, cry, when I listen to it, it's so good. -- Paris Hilton on her new album === >> I can program in more than three different languages. Can you? >> >> Wow. If that's not genius, I don't know what is. I usually don't say what I can do. I usually say what I can't > do. This was an exception. Well, you shut him up! Three languages! Goodness! I bet he's feeling inadequate. Actually, I feel inadequate and depressed. -- > Jesse F. Hughes I, like, cry, when I listen to it, it's so good. > -- Paris Hilton on her new album -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === > I can program in more than three different languages. Can you? > Wow. If that's not genius, I don't know what is. > I usually don't say what I can do. I usually say what I can't > do. This was an exception. Well, you shut him up! Three languages! Goodness! I bet he's feeling inadequate. > Actually, I feel inadequate and depressed. But I'm happy that I've learned the meaning of R subset AxA. I'm also happy to have realized that each of definitions in math has exact meaning, which means I should find them easy to learn. > -- > Jesse F. Hughes I, like, cry, when I listen to it, it's so good. > -- Paris Hilton on her new album -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === (ii) I don't see why anyone would find hostility towards this > particular OP inappropriate. He's been nothing but obnoxious. > I've figured that OP is consistently being used to refer to me, especially by you. What is OP? Is it 'obnoxious prick'? > ************************ David C. Ullrich -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === > (ii) I don't see why anyone would find hostility towards this > particular OP inappropriate. He's been nothing but obnoxious. > I've figured that OP is consistently being used to refer to me, especially by you. What is OP? > Is it 'obnoxious prick'? Is it 'out of print'? That doesn't make sense. > ************************ > David C. Ullrich -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: expectation of a waiting time between maxima Let X1, X2, .. independent random variables with a continuous distribution function Let N_n be the time of the first record after time n. It is relatively easy to show that P{N_n = n + k} = n * (n + k - l)^-1 * (n + k)^-1 Now I am having trouble showing that the expected waiting time to the next record is infinite: I have E{k} =sum_{k=0}^{k=infty} k*n*(n + k - l)^-1 * (n + k)^-1 How do I show that this sum diverges? Your help is appreciated. === Subject: Re: expectation of a waiting time between maxima > Let X1, X2, .. independent random variables with a continuous distribution function Let N_n be the time of the first record after time n. > It is relatively easy to show that > P{N_n = n + k} = n * (n + k - l)^-1 * (n + k)^-1 Now I am having trouble showing that the expected waiting time to the next record is infinite: > I have > E{k} =sum_{k=0}^{k=infty} k*n*(n + k - l)^-1 * (n + k)^-1 How do I show that this sum diverges? Your help is appreciated. Compare it with the harmonic series. === Subject: Re: Galileo's Paradox and the Project of the Reals On Mon, 22 Jan 2007 20:06:26 -0800, rem642b@yahoo.com (robert maas, see >> The bijection between N and S is an artefact of infinity. Not necessarily so. For example, consider the following bijection >between the positive integers and the positive even integers: > n=0 >lp: n=n+1 > output pair(n, 2*n) > go lp >The pairs output by that program define a bijection. >Every positive integer is a first-element of exactly one pair that > is output, and vice versa. >Every positive even integer is a second-element of exactly one pair > that is output, and vice versa. Whch proves what exactly? >> OK (to make the last 12 or so paragraphs redundant), start with the Peano. That gives you more than just a set. It gives you a structure >consisting of two components, a set, and a function operating on >that set with range in the set. There are a lot of properties >specific to that particular structure, which aren't >true of another structure also containing that same >set but a different function from the set into the same set. You can define an order relation from that structure, then you can >discard the original function, thereby getting an ordered set, >which consists of two components, the set and the order relation. >There are a lot of properties specific to that particular > structure, which aren't true of another > structure also containing that same set but a >different orderRelation on it. The point is that there are things you can say about a >set-with-operation, or a set-with-order (an ordered set), which are >meaningless if you're talking just about the set with nothing else >defined in regard to it. Cantor's set theory deals only with >properties inherent in the set itself (and in relation to other >sets which may have some elements in common or not). Cardinality is >a way to compare sets when you haven't defined any other structure >related to that set. It provides an equivalence relation (two sets >with same cardinality defined as two sets bijectable), and a total >ordering on equivalence classes with respect to that equivalance >relation. The subset relation provides a totally different ordering >on sets, which is only a partial ordering on anything nontrivial. >There's no such thing as density when dealing with a set without >any additional structure. > Of course I'm not using Cantor's definition of 'more'. Are you using **any** definition in terms of sets with no >additional structure? > What I want you to try and do is to take seriously the idea that >> the infinite never ends or stops. What you just said has no meaning whatsoever. On the contrary, it's very simple. You just aren't prepared to consider the ramifications. >> infinite is just the negation of the finite, If you're talking about cardinality of sets with no additional >structure, that is correct. If you're talking about sets with >additional structure, you need to state up front what additional >structure you are considering. > Counting is unfinishable. That statement is not true in general. >It depends very much on what you are counting. >Some counting tasks are finishable and some are not. You know what I mean. >> Cantor's Diagonal Argument does not work Yes it does. Oh no it doesn't. >You're a ing liar. Your standard of argument is really slipping now. Still, it didn't have far to go. >> The diagonal argument ought to show that there are uncountably >> many big whole numbers. You're stupid if you think that. > infinite integers has been a topic in this group It's just a meaningless idea brought up by another troll. >You and the other troll should keep away from each other. >Each of you has enough stupid ideas without importing any from each other. It's quite logical. You go to the right of the decimal point, you get reals; you go to the left of the decimal point, you get infinite integers. Of course, going right we get closer and closer to some finite value, or rather range of values in most cases -- that is the point. I'm not sure why people stopped worrying about sq. rt of 2 being irrational. I feel it has something to tell us about number. Or take pi. Is there some deep incommensurability between the curvilinear measure of the circle and its rectilinear diameter, hiding behind the irrationality of the ration? But of course any particular real that can be instantiated geometrically or algebraicly or however is not the problem. It's the vast bottomless sea of irrationals that are interminable and indeterminate in exactly the way infinite integers are. >> Let's have another go at listing the reals. Begin (binary -- >> binary point at left): >> 00_> 10-> 010-> 110-> 0010-> 0110-> etc. >> This only produces rationals, I hear you say. That's correct. In fact it produces only rationals of the form >n/2**k where n<2**k. It does't produce any rationals with odd >denominator, nor any negative rationals, nor any rationals with >nonzero whole part. > But you haven't gone into warp drive. Now we can carry this out >> to infinity, and it produces every possible number between 0 and 1. No it doesn't. Oh yes it does. Behind you! >You're a ing liar. There was a young chap called Maas, Who could not infinity parse, All of his words, Were like so many turds, That came from his flatulent Arse. === Subject: Re: Galileo's Paradox and the Project of the Reals >> The bijection between N and S is an artefact of infinity. >Not necessarily so. For example, consider the following bijection >between the positive integers and the positive even integers: > n=0 >lp: n=n+1 > output pair(n, 2*n) > go lp >The pairs output by that program define a bijection. >Every positive integer is a first-element of exactly one pair that > is output, and vice versa. >Every positive even integer is a second-element of exactly one pair > that is output, and vice versa. > Whch proves what exactly? That in a finite number of characters, using a well-defined language, I've specified a bijection between the positive integers N and the positive even integers S. Nothing artefact about that. It's right there for you to see in four lines of code. >> What I want you to try and do is to take seriously the idea that >> the infinite never ends or stops. >What you just said has no meaning whatsoever. > On the contrary, it's very simple. You just aren't prepared to > consider the ramifications. It's more like *you* aren't prepared to say anything meaningful, preferring instead to mumble words together randomly. Perhaps you have some idea in your mind, but you just aren't capable of making a sentence to express it. Or perhaps you don't have an idea and are just trying to snow us. It ain't working. I ain't snowed by your gibberish. >> Counting is unfinishable. >That statement is not true in general. >It depends very much on what you are counting. >Some counting tasks are finishable and some are not. > You know what I mean. Nope. You need to learn how to communicate precisely, say what you mean instead of saying something else and expecting us to gusss what you really mean. When you say Counting is unfinishable. it seems to mean Every act of counting is unfinishable., which is false, as I pointed out. Now if you mean something else, such as Some acts of counting are unfinishable, others are't. then you need to say that clearly. If you mean something else entirely, then you haven't even begun to communicate. >> Cantor's Diagonal Argument does not work >Yes it does. > Oh no it doesn't. Cantor's diagonal argument shows that if you assume a bijection between the positive integers and the reals in the interval [0,1], you get a contradiction. Do you disagree that such a contradiction is achieved under that assumption, or do you somehow claim that's not working despite the contradiction having been achived? Achiving that contradiction sure sounds like working to me. It's just like Euclid's proof that no rational number can satisfy the equation x^2 = 2, by assuming such a rational number solution and getting a contradiction. You got a problem with showing something false by deriving a contradiction from it? > You go to the right of the decimal point, you get reals; you go > to the left of the decimal point, you get infinite integers. The way you get reals by a neverending process of generating digits to the right of a decimal point is by explicitly constructing a neverending chain of nested intervals. For example: print .; while true do { print 1; print 4; print 2; print 8; print 5; print 7; } constructs a neverending sequence of digits to the right of the decimal point whose leftmost 73 characters (1 decimal point and 72 digits) look like: .142857142857142857142857142857142857142857142857142857142857142857142857 and which thereby constructs a sequence of nested intervals whose first eleven elements are: [0,1], [.1,.2], [.14,.15], [.142,.143], [.1428],.1429], [.14285],.14286], [.142857],.142858], [.1428571],.1428572], [.14285714],.14285715], [.142857142],.142857143], [.1428571428],.1428571429] The upper and lower bounds of those intervals define a Cauchy sequence which thereby define a real number, and the sets of rationals eventually excluded below and above the intervals respectively define the two halves of a Dedekind cut which thereby define a real number, and those are in fact equal numbers in the sense of the usual mapping between Cauchy sequence equivalance classes and Dedekind cuts. Thus the sequence of nested intervals define a real number in either sense of real number (Cauchy or Dedekind construction). In fact the particular real number constructed there is equal to the rational number 1/7 via the usual mapping between rational numbers and rational reals, a subset of the reals. In what sense do your neverending sequence of digits to left of decimal point generate anything that can meaningfully be called an infinite integer? If you use a metric whereby such sequences generate a Cauchy sequence, you don't get anything that is either infinite in any meaningful sense, nor anything which is an integer in any meaningful sense. For example, if you use the eleven-adic metric, with digits selected from the set {0,1,2,3,4,5,6,7,8,9,A} (with 'A' having the usual hexadecimal-digit meaning as ten just as '9' has the meaning nine), then you can get such sequences as: print .; printToLeft 4; while true do { printToLeft 7; printToLeft 3; } That particular one converges to the rational number 1/3 in the eleven-adic metric. Nothing infinite nor integer about that!! So obviously either you're mistaken when you claim such a value is infinite, or you're mistaken when you claim such a value is in any sense integer, or you're not talking about a p-adic metric to give meaning to your neverending sequence of digits to left of decimal point. I'll give you the benefit of doubt and presume you do *not* mean any p-adic metric, but have some other idea how to give meaning to such a neverending-leftward sequence of digits in the context of considering them numbers. So please tell us what your proposed rule is for adding and multiplying such sequences. Note that arithmetic (add/mul) on nevereding-rightward sequences of digits, using the real (absolute-value) metric/norm, have a well-defined meaning: - As nested intervals: Use interval arithmetic on the sequence of sums or products of corresponding terms. - As Cauchy sequences: Use all four possible combiations of endpoints of corresponding intervals, sum or product thereof. - As Dedekind cuts: Use the nested-interval definition above, and convert to upper/lower Dedekind cut via upper/lower exclusion from interval. All three definitions give the same result. Likewise arithmetic on neverending-leftward sequences of digits, using p-adic metric/norm, have a well-defined meaning: - As Cauchy sequences: Use all four possible combiations of endpoints of corresponding intervals, sum or product thereof. - As digit sequences: Simply compute the successive digits by the standard grade-school algorithm, extended to the left forever, for example with eleven-adic norm, addition is easy: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 :carries 3737373737373737373737373737373737373737373737373737374. :first addend (1/3) 1919191919191919191919191919191919191919191919191919192. :second addend (1/6) --------------------------------------------------------- 5555555555555555555555555555555555555555555555555555556. :sum (1/2) Eleven-adic multiplication isn't much harder: 3737373737373737373737373737373737373737373737373737374. :multiplicand (1/3) 5555555555555555555555555555555555555555555555555555556. :multiplier (1/2) --------------------------------------------------------- 0000000000000000000000000000000000000000000000000000002. :multiplicand * 6. 373737373737373737373737373737373737373737373737373739 :multiplicand * 50. 73737373737373737373737373737373737373737373737373739 :multiplicand * 500. 3737373737373737373737373737373737373737373737373739 :multiplicand * 5000. 737373737373737373737373737373737373737373737373739 :multiplicand * 50000. ----- 19192. :partial sum so-far Continuing to generate :multiplicand * 500...00. and adding, the partial sum eventually looks like: 1919191919191919191919191919191919191919191919191919192. :partial sum so-far Anyway, it's clear you get the 11-adic representation of 1/6. So if you use similar-looking neverending-leftward sequences of digits to represent some new sort of number you invented, how exactly do *you* define addition and multiplication on them?? > Of course, going right we get closer and closer to some finite > value, or rather range of values in most cases -- that is the > point. Actually at any point along the sequence of nested intervals, you get an interval of values. But there's only a single real number that is in every one of those intervals. You *never* get an interval or as you call it range of values in the intersection of all the intervals in the sequence. Every real number except that one is eventually eliminated from some interval in the sequece. There's not a single case where there's a range of values in the intersection of the nested intervals. So your most cases is a complete lie, a ing lie, and you should stop posting such lies. > I'm not sure why people stopped worrying about sq. rt of 2 being > irrational. I feel it has something to tell us about number. Of course it has something to tell you about that number: It's not rational. So you have to **extend** the rationals in some way to include it. It isn't *already* in the rationals you started with. If you have an equation in rationals x^2=2, it has no solution there. It's only in an extended ring beyond the rationals that you can artificially construct a solution to it. It's just like 1/2 which isn't in the integers. If you have a equation 2*x=1 in the integers, it has no solution. It's only by extending the integers that you can artificially construct a number 1/2 which is a solution to that equation. People stopped worrying because worrying causes acid indigestion and heartburn and ulcers and high blood pressure and heart attacks and strokes. It's stupid to worry about the fact that 2*x=1 has no solution in integers, or that x*x=2 has no solution in rationals or in 2-adics either, or that x*x=-1 has no solution in reals. > Or take pi. Is there some deep incommensurability between the > curvilinear measure of the circle and its rectilinear diameter, > hiding behind the irrationality of the ration? It's not hiding. It's right there for everyone to see. There's no polynomial equation with rational coefficients, except the trivial kind such as 5=5, which are satisfied by pi. If 1/2 doesn't satisfy any equation of the form x + n = m (m,k integer), and if sqrt(2) doesn't satisfy any equation of the form k*x + n = m (k,n,m rational, k nonzero), and if cuberoot(2) doesn't satisfy any equation of the form m*x**2 + n*x = k (m,n,k rational, m nonzero), then what's hiding about the fact that there's a real number that doesn't satisfy *any* nontrivial polynomial equation (with rational coefficients) whatsoever? Is there any reason you would expect that for *every* number r which is the limit of a sequence there'd be some nontrivial polynomial that it satisfies?? There's certainly no way to construct such a polynomial or even come close. For example, what polynomial do you think would be satisfied by the limit of this simple-looking series: 1 - 1/2 + 1/3 - 1/4 + 1/5 etc. For example, pick a prime p, and construct the p-adic representation of each partial sum. See if you can find a pattern in the digit sequences that leads you to a polynomial that the limit satisfies. Ain't gonna happen. > any particular real that can be instantiated geometrically or > algebraicly or however is not the problem. It's the vast bottomless > sea of irrationals that are interminable and indeterminate in > exactly the way infinite integers are. The depth of that sea depends very much on whether you accept the Axiom of Choice, or a Turing Oracle, or a Goedel Oracle, to posit the existance of uncomputable real numbers. Any computable real number is neither interminable nor indeterminate, so it's the other real numbers, if you posit any, which you must be talking about?? > There was a young chap called Maas, > Who could not infinity parse, > All of his words, > Were like so many turds, > That came from his flatulent Arse. That doesn't even rhyme. And I can parse infinity just fine if the grammar is appropriate. There once was a guy named Six Letters, Who let numbers get him the betters. He couldn't grok Canter, Nor stop his dumb banter. He knows as much math as some Setters. === Subject: Re: Galileo's Paradox and the Project of the Reals On Mon, 22 Jan 2007 19:29:03 -0800, rem642b@yahoo.com (robert maas, see >>No, don't be like an ostrich! Can't you see that there are some >>sets that can be put in 1-1 correspondence with the reals (such as >>that reals-plus row-index set I defined above), some infinite sets >>which are strictly smaller than the reals (such as the integers, or >>the rationals, or the computable numbers), and some sets which are >>strictly larger than the reals (such as the set of subsets of >>reals)? >> There will be sets, or rather putative sets, for which a >> bijection between them and the naturals has not been demonstrated. >> That's all. You're lying. There's a lot more. For example: >- Sets where the assumption of a bijection results in > contradiction, showing that not only haven't we yet demonstrated > a bijection, but we never will, because it's impossible for > there to be one. >- Sets which are a subset of one for which a bijection has been > demonstrated. >- Sets which contain a subset for which a bijection has been > demonstrated. > It is impossible for the same man to cross the river twice. Are you making that claim on the basis that a river is constantly >flowing hence what you cross today isn't exactly the same water >that you might cross at some other time? If you use that logic, >you aren't the same person who posted what I'm responding to. You talking to me? What river? Two Letters === Subject: Re: Galileo's Paradox and the Project of the Reals >>All I am saying is, if cardinality detects no difference in set size, >>that doesn't mean a difference detected through other means is invalid. >>It means cardinality isn't the whole picture. >> I think cardinality is nonsense. When you say it's nonsense, do you mean that it is logically >inconsistent? Or do you mean that you find it senseless to pursue >because it tells us nothing mathematically useful? Or what, exactly? > I think bijection between two >> putatively unequal sets is an artefact of infinity. This seems to indicate that you /don't/ believe that cardinality is >nonsense in a logical sense: it is an artifact (less pejoratively, >valid logical implication) of the definition of sets which include >infinite sets, and specifically, the existence of power sets of >infinite sets. Instead, you seem to reject the sense of such sets (and therefore, >find that their comparisons via injection and surjection also lack >sense). Again, I would ask: do you think that set of all subsets of the >naturals is logically inconsistent (and therefore /logical/ nonsense), >or is it that you do not see any usefulness (sense) to the definition >of these concepts? > Cardinality is vacuous, would be a better way of saying what I mean. It only means anything if you already believe in advance that the countably infinite is some super-finite size exceeded by the uncountably infinite. Six Letters === Subject: Re: Galileo's Paradox and the Project of the Reals <4596b5fd@news2.lightlink.com> <459e7662@news2.lightlink.com> naturals is logically inconsistent (and therefore /logical/ nonsense), >or is it that you do not see any usefulness (sense) to the definition >of these concepts? > Cardinality is vacuous, would be a better way of saying what I > mean. It only means anything if you already believe in advance that the > countably infinite is some super-finite size exceeded by the uncountably > infinite. > Do you think then that the statement for any set S, there exists an injection from S into the power set P(S); and there does not exist an injection from P(S) into S is a vacuous statement; i.e., that the statement has no meaning as it stands? Assuming you think it /does/ have meaning, do you think it becomes meaningless when we carefully define set S has smaller cardinality than set T (written as |S| < |T|) iff there exists an injection from S into T and no injection from T into S? Note that I have not said anything about whether sets S or T are finite or infinite (or super-finite or whatever else). I have simply defined /exactly/ what I mean by S has smaller cardinality than P(S), i.e. |S| < |P(S)|. Hopefully, it should be obvious my use of smaller than and < here have entirely /different/ meanings than when I use those /same/ terms to compare real numbers such as pi is smaller than 4 or 2 < sqrt(5). It's only when the two contexts get confused that indeed the statement x is smaller than y becomes ambiguous and therefore meaningless. For example, |N| > sqrt(2) is gibberish. === Subject: Re: Galileo's Paradox and the Project of the Reals <4596b5fd@news2.lightlink.com> <459e7662@news2.lightlink.com> mean. It only means anything if you already believe in advance that the > countably infinite is some super-finite size exceeded by the uncountably > infinite And it means just what its definition provides regardless of any beliefs in advance anyone may or may not have about super-finite size. MoeBlee === Subject: Re: Germany issues warrants for 13 American CIA agents > the Germans. Didn't we kick their ass a couple times > already? OL How about we offer them a trade - one CIA knuckle-dragger for one Dachau doctor? :-) I understand that Mengele's old boss is still alive and well in Deutschland, as well as some of his colleagues. -- Vance P. Frickey remove safety from listed Email address to send mail It's easy to fight when everything's right And you're mad with the thrill and the glory; It's easy to cheer when victory's near, And wallow in fields that are gory. It's a different song when everything's wrong, When you're feeling infernally mortal; When it's ten against one, and hope there is none, Buck up, little soldier, and chortle! - Robert William Service, Carry On === Subject: Re: Statistics The population mean (7) represents the national average of third grade students on an exam. The sample median (6.5) represent one class in the US and their scores on the exam. I cannot see how to compare the median to the population mean. === Subject: Re: Statistics > The population mean (7) represents the national average of third grade > students on an exam. The sample median (6.5) represent one class in the US > and their scores on the exam. I cannot see how to compare the median to the population mean. I would imagine that for the national average the median and mean would be roughly the same ie you would expect for the vast number of 3rd grade students that the data would give a smooth curve across all points. Whereas in you class to take the median excludes the influence of either one or two very poor results or very good results. Therefore the mean of the national data is likely to be close to the 50 percentile point - and therefore you can compare your median point to this. I have to point out that if you have got a number of zeros and the rest are doing somewhat better you might wish to use the median, but this will make the picture look somewhat rosier than it actually is. You should actually be not only using the median, but say the quartiles etc. This shows the danger of using a single measure. Nick === Subject: random infinite strings Does anybody know the definition of a statistical test for non randomness over a {0,1}* string ? If the test fails on longer and longer prefixes infinitley you can say for sure the string is not random. Basically what I'm asking is if someone comes to you and says - here is a test that tells that an infinite string is _not_ random, how do you verify that this test is valid and will _not_ exclude random infinite strings. I'm asking the above to try and understand Martin-Lof randomness. === Subject: Re: random infinite strings > Does anybody know the definition of a statistical test for non > randomness over a {0,1}* string ? If the test fails on longer and longer prefixes infinitely you can say > for sure the string is not random. Basically what I'm asking is if someone comes to you and says - here > is a test that tells that an infinite string is _not_ random, how do > you verify that this test is valid and will _not_ exclude random > infinite strings. I'm asking the above to try and understand Martin-Lof randomness. well you can never test an infinite string unless you can travel infinite worldlines in finite proper time but still for infinite strings martin-lof randomness coincides with chatin randomness if the string is not computible it is random so even if a nicely compressible finite prefix is added to a random infinite string it is still a random infinite string and any test on the finite prefix says nothing of the tail however if you are willing to decrease your claim to only random at size N then the series of diehard tests may be of use for you http://www.stat.fsu.edu/pub/diehard/ they are a battery of tests for specific functional patterns that may occur in a finite sequence of course if you were to require a more formal foundation of basis of computations of random complexity at level N the number of tests to run would change at each level however these types of batteries of tests do tend to capture the most common types of patterns -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: random infinite strings > Does anybody know the definition of a statistical test for non > randomness over a {0,1}* string ? If the test fails on longer and longer prefixes infinitely you can say > for sure the string is not random. I think that it is provable that no such test could exist which decides for sure that any string is random or not. > Basically what I'm asking is if someone comes to you and says - here > is a test that tells that an infinite string is _not_ random, how do > you verify that this test is valid and will _not_ exclude random > infinite strings. I think that you can prove that no such perfect test could exist. > I'm asking the above to try and understand Martin-Lof randomness. I've been looking and cant find a definition of Martin-Lof Randomness - please post if so inclined. > well > you can never test an infinite string > unless you can travel infinite worldlines in finite proper time but still > for infinite strings > martin-lof randomness coincides with chatin randomness if the string is not computible it is random so even if a nicely compressible finite prefix > is added to a random infinite string > it is still a random infinite string > and any test on the finite prefix says nothing of the tail however if you are willing to decrease your claim > to only random at size N > then the series of diehard tests may be of use for you http://www.stat.fsu.edu/pub/diehard/ These tests are really beautiful to me. I used to consider stuff like that just statistical junk. I had very little respect for such things, and really a great deal of disdain for some reason. But, the validity of these tests and various other considerations actually strengthens the position that randomness does indeed exist. Yet, you can argue just as effectively that randomness does not exist. of randomness must itself be indeterminate. It's such a maddening thing - randomness - now you see it, now you dont.It's here, it's gone, it's there, it's not - must be existentially indeterminate. > they are a battery of tests for specific functional patterns > that may occur in a finite sequence of course > if you were to require a more formal foundation > of basis of computations of random complexity at level N > the number of tests to run would change at each level however these types of batteries of tests > do tend to capture the most common types of patterns -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar > === Subject: Re: random infinite strings > Does anybody know the definition of a statistical test for non >> randomness over a {0,1}* string ? >> If the test fails on longer and longer prefixes infinitely you can say >> for sure the string is not random. > I think that it is provable that no such test could exist which decides > for > sure that any string is random or not. >> Basically what I'm asking is if someone comes to you and says - here >> is a test that tells that an infinite string is _not_ random, how do >> you verify that this test is valid and will _not_ exclude random >> infinite strings. > I think that you can prove that no such perfect test could exist. >> I'm asking the above to try and understand Martin-Lof randomness. > I've been looking and cant find a definition of Martin-Lof Randomness - > please post if so inclined. >> well >> you can never test an infinite string >> unless you can travel infinite worldlines in finite proper time >> but still >> for infinite strings >> martin-lof randomness coincides with chatin randomness >> if the string is not computible it is random >> so even if a nicely compressible finite prefix >> is added to a random infinite string >> it is still a random infinite string >> and any test on the finite prefix says nothing of the tail >> however if you are willing to decrease your claim >> to only random at size N >> then the series of diehard tests may be of use for you >> http://www.stat.fsu.edu/pub/diehard/ These tests are really beautiful to me. I used to consider stuff like that > just statistical junk. I had very little respect for such things, and > really > a great deal of disdain for some reason. But, the validity of these tests and various other considerations actually > strengthens the position that randomness does indeed exist. Yet, you can argue just as effectively that randomness does not exist. existence > of randomness must itself be indeterminate. It's such a maddening thing - > randomness - now you see it, now you dont.It's here, it's gone, it's > there, it's not - must be existentially indeterminate. +++++++++++++++++++++++++++++++++++++++++++ I have always taken the definition of randomness as the British World War II experience with encripted messages. Randomness is defined as an outcome that appears random, because we have not yet mound the underlying mathematics. So I can always say that randomness is just a matter of ignorance. If you know the seed and the RNG, you can predict the outcoming process. So why is it random? David Heiser > they are a battery of tests for specific functional patterns >> that may occur in a finite sequence >> of course >> if you were to require a more formal foundation >> of basis of computations of random complexity at level N >> the number of tests to run would change at each level >> however these types of batteries of tests >> do tend to capture the most common types of patterns >> -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- >> galathaea: prankster, fablist, magician, liar > > === Subject: Re: random infinite strings <8vmdnZ5H79PvbV7YnZ2dnUVZ_qKknZ2d@comcast.com> > Does anybody know the definition of a statistical test for non >> randomness over a {0,1}* string ? > If the test fails on longer and longer prefixes infinitely you can say >> for sure the string is not random. I think that it is provable that no such test could exist which decides > for > sure that any string is random or not. > Basically what I'm asking is if someone comes to you and says - here >> is a test that tells that an infinite string is not random, how do >> you verify that this test is valid and will not exclude random >> infinite strings. I think that you can prove that no such perfect test could exist. > I'm asking the above to try and understand Martin-Lof randomness. I've been looking and cant find a definition of Martin-Lof Randomness - > please post if so inclined. > well >> æ you can never test an infinite string >> æ unless you can travel infinite worldlines in finite proper time > but still >> æ for infinite strings >> æ martin-lof randomness coincides with chatin randomness > if the string is not computible it is random > so even if a nicely compressible finite prefix >> æ is added to a random infinite string >> it is still a random infinite string >> and any test on the finite prefix says nothing of the tail > however if you are willing to decrease your claim >> æ to only random at size N >> then the series of diehard tests may be of use for you >http://www.stat.fsu.edu/pub/diehard/ These tests are really beautiful to me. I used to consider stuff like that > just statistical junk. I had very little respect for such things, and > really > a great deal of disdain for some reason. But, the validity of these tests and various other considerations actually > strengthens the position that randomness does indeed exist. Yet, you can argue just as effectively that randomness does not exist. existence > of randomness must itself be indeterminate. It's such a maddening thing - > randomness - now you see it, now you dont.It's here, it's gone, it's > there, it's not - must be existentially indeterminate. +++++++++++++++++++++++++++++++++++++++++++ > I have always taken the definition of randomness as the British World War II > experience with encripted messages. Randomness is defined as an outcome that > appears random, because we have not yet mound the underlying mathematics. So I can always say that randomness is just a matter of ignorance. If you know the seed and the RNG, you can predict the outcoming process. So > why is it random? It's not. It's pseudo-random. === Subject: Re: random infinite strings <8vmdnZ5H79PvbV7YnZ2dnUVZ_qKknZ2d@comcast.com> On Feb 2, 6:12 pm, Dr. V I Plankenstein I think that you can prove that no such perfect test could exist. > I'm asking the above to try and understand Martin-Lof randomness. I've been looking and cant find a definition of Martin-Lof Randomness - > please post if so inclined. Basically there are 3 common definitions for an infinite string to be random: 1) Kolmogorov Random 2) Constructive Null Cover Random 3) Martingale Random All definitions have been proven to be equivalent. You can see below: http://en.wikipedia.org/wiki/Algorithmically_random_sequence 3) is actually easiest to understand and this states that an infinite string is considered random if and only if it is impossible to make an infinite amount of money betting on the next bit in the string. 2) was actually the first test given by Martin-Lof in 1966. This is the one I'm trying to understand. I do understand it now but don't know how Martin-Lof came up with such a definition. But to understand I am going to try and look at the proof of 2 and 3 being equivalent. === Subject: continuous function= describe via formulas only A function (real valued of real variable) that's continous on R can always be representable with (can be defined with) formula[e]. Correct? I was thinking about this myself and wondering why I never saw real continuous functions that could 'only' be explained in nonmathematical ways (like english description, or drawing a mapping between sets). Since a function that's continuous requires it to be defined for an INTERVAL of the real numbers, which is uncountably infinite, this means there must be a way to define it with formulae. This is just an intuitive answer I gave myself. I don't think there can be a continuous function that CAN NOT be defined by formulas (a single formula for the whole Real line, or piecewise for different intervals and points). True? === Subject: Re: continuous function= describe via formulas only >A function (real valued of real variable) that's continous on R can >always be representable with (can be defined with) formula[e]. >Correct? No. Well, whether this is correct depends on exactly what you mean by formula, but it's false with typical definitions of that term. >I was thinking about this myself and wondering why I never saw real >continuous functions that could 'only' be explained in nonmathematical >ways (like english description, or drawing a mapping between sets). >Since a function that's continuous requires it to be defined for an >INTERVAL of the real numbers, which is uncountably infinite, this >means there must be a way to define it with formulae. This is just >an intuitive answer I gave myself. I don't think there can be a continuous function that CAN NOT be >defined by formulas (a single formula for the whole Real line, or >piecewise for different intervals and points). True? False. Of course I can't tell you exactly what continuous function gives a counterexample, since there is no formula for the counterexample... ************************ David C. Ullrich === Subject: Re: continuous function= describe via formulas only > A function (real valued of real variable) that's continous on R can > always be representable with (can be defined with) formula[e]. > Correct? I was thinking about this myself and wondering why I never saw real > continuous functions that could 'only' be explained in nonmathematical > ways (like english description, or drawing a mapping between sets). > Since a function that's continuous requires it to be defined for an > INTERVAL of the real numbers, which is uncountably infinite, this > means there must be a way to define it with formulae. This is just > an intuitive answer I gave myself. I don't think there can be a continuous function that CAN NOT be > defined by formulas (a single formula for the whole Real line, or > piecewise for different intervals and points). True? You are on to something, but you have to allow some freedom about what is a formula. For example, is the integral definition, or the infinite product definition, of Gamma function allowed to be a definition by a formula? If you allow only what can be typed using a finite number of characters on the keyboard of your computer, the answer to your question is no, as mentioned by others. If you allow expressions involving (repeated) limits of sequences, and involving the countably infinite alphabet consisting of all rational numbers, you will get all continuous functions (using a modification of Weierstrass's Approximation Theorem), but also an amazing jungle of other, weird functions. Look up counterexamples in Real Analysis; however, they do require substantial theoretical readiness of the reader. === Subject: Re: continuous function= describe via formulas only > A function (real valued of real variable) that's > continous on R can > always be representable with (can be defined with) > formula[e]. > Correct? I was thinking about this myself and wondering why I > never saw real > continuous functions that could 'only' be explained > in nonmathematical > ways (like english description, or drawing a mapping > between sets). > Since a function that's continuous requires it to be > defined for an > INTERVAL of the real numbers, which is uncountably > infinite, this > means there must be a way to define it with formulae. > This is just > an intuitive answer I gave myself. I don't think there can be a continuous function that > CAN NOT be > defined by formulas (a single formula for the whole > Real line, or > piecewise for different intervals and points). > True? > What do YOU mean by formulae? If I write J_0(x) for the Bessel function of order 0 is that a formula describing it? If not WHAT would you accept as a formula for this function. Mmmm, I notice that hagman has pointed out that the cardinality of the set of all continuous functions is greater than that of the set of all strings in some language so no matter what you consider formulae, there must be some continuous function that cannot be described with a formula. Very nice! === Subject: Re: continuous function= describe via formulas only > A function (real valued of real variable) that's continous on R can > always be representable with (can be defined with) formula[e]. > Correct? No. In fact, it's not even possible to represent every constant function by a formula, because there are only countably many formulae, but there are uncountably many real numbers (constants). > I was thinking about this myself and wondering why I never saw real > continuous functions that could 'only' be explained in nonmathematical > ways (like english description, or drawing a mapping between sets). > Since a function that's continuous requires it to be defined for an > INTERVAL of the real numbers, which is uncountably infinite, this > means there must be a way to define it with formulae. This is just > an intuitive answer I gave myself. > I don't think there can be a continuous function that CAN NOT be > defined by formulas (a single formula for the whole Real line, or > piecewise for different intervals and points). True? Even if we limit our attention to functions that take rational values on the rationals (i.e., extensions to R of continuous functions Q -> Q), then there are still uncountably many and therefore there are not enough formulas or even piecewise descriptions to go around. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: continuous function= describe via formulas only > A function (real valued of real variable) that's continous on R can > always be representable with (can be defined with) formula[e]. > Correct? I was thinking about this myself and wondering why I never saw real > continuous functions that could 'only' be explained in nonmathematical > ways (like english description, or drawing a mapping between sets). > Since a function that's continuous requires it to be defined for an > INTERVAL of the real numbers, which is uncountably infinite, this > means there must be a way to define it with formulae. This is just > an intuitive answer I gave myself. I don't think there can be a continuous function that CAN NOT be > defined by formulas (a single formula for the whole Real line, or > piecewise for different intervals and points). True? There are less potential formulae than continuous functions. In fact, this is true even if you include natural language descriptions. There are even more real numbers than formulae/descriptions for real numbers. === Subject: Normal family and f^n -> oo Let f(z) be an entire function in C. Let U be some connected component on which the iterates f^n of our function f(z) form a normal family. Prove that if U is multiply connected, then f^n -> oo on U (Or f^{n_k} -> oo for some subsequence of naturals {n_k} ). I would very appreciate any fo you suggestions. I tried to somehow use maximal modulus principle, but i don't seem to have an idea to deal this problem out. === Subject: Re: Normal family and f^n -> oo > Let f(z) be an entire function in C. Let U be some connected component > on which the iterates f^n of our function f(z) form a normal family. > Prove that if U is multiply connected, then f^n -> oo on U (Or f^{n_k} > -> oo for some subsequence of naturals {n_k} ). U is a connected component of what? My guess is that you mean a connected component of the Fatou set, i.e. the set of points x such that f^n restricted to some neighbourhood of x form a normal family. But based on what I read about Herman rings at , I doubt that a proof is going to be easy. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Normal family and f^n -> oo Let f(z) be an entire function in C. Let U be some > connected component > on which the iterates f^n of our function f(z) form > a normal family. > Prove that if U is multiply connected, then f^n - oo on U (Or f^{n_k} > -> oo for some subsequence of naturals {n_k} ). U is a connected component of what? My guess is that > you mean a > connected component of the Fatou set, i.e. the set of > points x such > that f^n restricted to some neighbourhood of x form a > normal family. > But based on what I read about Herman rings at > , I doubt > that a proof is going > to be easy. It might really be far from easy in the case when f is rational functions. Because it has been unkown examples of Herman rings(and known example for today are not easy at all). But in our case we are interested in the case of entire function. For example it has been known for a long time that for entire functions Herman rings don't exist at all. And i asked it here because it was written in the paper that a simple argument shows that ... what i asked. So i think it is assumed to be easy. > -- > Robert Israel > israel@math.MyUniversitysInitials.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada === Subject: Re: nonsense OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. AxA generates pairs of x,y Gibberish. AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. AxA is certainly *not* equal to { (x,y) | x,y in A }. Consider, AxB = { (x,y) | x in A and x in B} AxB is certainly not equal to { (x,y) | x,y in (A or B) }. If you have any intelligence left, you would see that AxA is not qual to { (x,y) | x,y in A }. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: nonsense > >> OK, I'll demonstrate my understanding resulting from my own >> interpretation, by explaning the meaning of R subset AxA. >> AxA generates pairs of x,y >> Gibberish. >> >> AxA = { (x,y) | x,y in A } >> AxA is the set of all ordered pairs (x,y), with both x and y in A. >AxA is certainly *not* equal to { (x,y) | x,y in A }. Hard to decide whether you're actually trying to learn something or just trolling. Because if you're actually trying to learn something you'd spend more time trying to understand the things you don't get at first instead of just saying they're false. In fact the _definition_ of AxA is {(x,y) : x in A and y in A}. >Consider, > AxB = { (x,y) | x in A and x in B} That was just a typo for { (x,y) | x in A and y in B}, right? >AxB is certainly not equal to { (x,y) | x,y in (A or B) }. No it's not. Nobody said it was. So what? The notation x,y in A _means_ x in A and b in A. While the notation x in A and y in B does _not_ mean x, y in (A or B). For that matter there's no such thing as A or B, since A and B are sets. >If you have any intelligence left, you would see that AxA is not qual to { (x,y) | x,y in A }. ************************ David C. Ullrich === Subject: Re: nonsense > AxA is certainly *not* equal to { (x,y) | x,y in A }. Consider, > AxB = { (x,y) | x in A and x in B} > AxB is certainly not equal to { (x,y) | x,y in (A or B) }. No one claimed it was. Obviously, A x A = { (x,y) | x in A and y in A }. A common shorthand for the condition x in A and y in A is to write x,y in A. That is, x,y in A *means* x in A and y in A. If you have any intelligence left, you would see that AxA is not > qual to { (x,y) | x,y in A }. Well, if I were one to give advice, I might suggest that you shouldn't be too quick on insulting others' grasps of logic and mathematics. But I'm not one to give advice. -- Jesse F. Hughes Yes, I'm one of those arrogant people who tries to be quotable. There is actually at least one person who quotes me often. -- === Subject: Re: nonsense > OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. > AxA generates pairs of x,y Gibberish. AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. AxA is certainly *not* equal to { (x,y) | x,y in A }. Only if you strike out the *not* in your claim. Consider, > AxB = { (x,y) | x in A and x in B} No. Instead AxB = { (x,y) | x in A and y in B} > AxB is certainly not equal to { (x,y) | x,y in (A or B) }. That would be ..in (A union B) and then it may in fact differ from AxB whenever A and B are different. If you have any intelligence left, you would see that AxA is not qual to { (x,y) | x,y in A }. I assume that x,y in A is meant as a shortcut for x in A and y in A. Then what is the problem? === Subject: Re: nonsense > OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. > AxA generates pairs of x,y > Gibberish. > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. AxA is certainly *not* equal to { (x,y) | x,y in A }. Only if you strike out the *not* in your claim. > Consider, > AxB = { (x,y) | x in A and x in B} No. Instead > AxB = { (x,y) | x in A and y in B} AxB is certainly not equal to { (x,y) | x,y in (A or B) }. That would be ..in (A union B) and then it may in fact differ from > AxB whenever A and B are different. > If you have any intelligence left, you would see that AxA is not qual to { (x,y) | x,y in A }. I assume that x,y in A is meant as a shortcut for x in A and y in > A. > Then what is the problem? > I was wrong to say that AxA = { (x,y) | x,y inA } = { (x,y) | x in A and y in A } is not true. After consideration, I realized that you are right. However, I find it ambiguous to you x,y in A in this context, because if it was AxB, then you can't write it using x,y in A. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: nonsense > However, I find it ambiguous to you x,y in A in this context, > because if it was AxB, then you can't write it using x,y in A. So what? That does not make x,y in A ambiguous. Look, in exactly the same way, Paul is short and George is short, can be said more succinctly: Paul and George are short. But I cannot shorten Paul is short and George is tall, by the same method. So what? Paul and George are short, is perfectly clear to any reader. There is no ambiguity. -- Jesse F. Hughes Doesn't pay to lie if you aren't good at it. -- Captain Friday, /City of the Dead/ Adventures by Morse radio show === Subject: Re: nonsense > OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. > AxA generates pairs of x,y > Gibberish. > AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. AxA is certainly *not* equal to { (x,y) | x,y in A }. Only if you strike out the *not* in your claim. > Consider, > AxB = { (x,y) | x in A and x in B} No. Instead > AxB = { (x,y) | x in A and y in B} AxB is certainly not equal to { (x,y) | x,y in (A or B) }. That would be ..in (A union B) and then it may in fact differ from > AxB whenever A and B are different. > If you have any intelligence left, you would see that AxA is not qual to { (x,y) | x,y in A }. I assume that x,y in A is meant as a shortcut for x in A and y in > A. > Then what is the problem? > -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re:giberish OK, I'll demonstrate my understanding resulting from my own > interpretation, by explaning the meaning of R subset AxA. AxA generates pairs of x,y Gibberish. AxA = { (x,y) | x,y in A } > AxA is the set of all ordered pairs (x,y), with both x and y in A. Consider, AxB = { (x,y) | x in A and y in B} AxB is certainly not equal to { (x,y) | x,y in (A or B) } whenever A != B. Therefore, to account for the cases when A != B, it is better to write AxA = { (x,y) | x in A and y in A }. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: giberish > >> OK, I'll demonstrate my understanding resulting from my own >> interpretation, by explaning the meaning of R subset AxA. >> AxA generates pairs of x,y >> Gibberish. >> >> AxA = { (x,y) | x,y in A } >> AxA is the set of all ordered pairs (x,y), with both x and y in A. >Consider, > AxB = { (x,y) | x in A and y in B} >AxB is certainly not equal to { (x,y) | x,y in (A or B) } whenever A != B. Therefore, to account for the cases when A != B, it is better to write > AxA = { (x,y) | x in A and y in A }. That's better if the reader has problems reading simple English. A definition of AxA is not required to account for the case A <> B. ************************ David C. Ullrich === Subject: Re: giberish Consider cartesian product of any two sets. it is true that for A=B, AxB = AxA = { (x,y) | x,y in A} (*) For each cases A!=B, this can't be expressed by (*). Since AxB = { (x,y) | x in A and y in B } is true for any A or B, this is the recommended way to denote AxB. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: giberish > Consider cartesian product of any two sets. it is true that for A=B, > AxB = AxA = { (x,y) | x,y in A} (*) For each cases A!=B, this can't be expressed by (*). Since AxB = { (x,y) | x in A and y in B } is true for any A or B, this is the recommended way to denote AxB. -- > ### Author Logan Lee ### > ### AuthorHomePagehttp://beam.to/pyenos### > ### AuthorQuote I just want to learn. ### The commonly accepted manner is to proceed from the general to the particular, not the other way around as you do above. If there is a reason to recommend AxB = { (x,y) | x in A and y in B } then it is because it is correct, period, and it has nothing to do with the particularities of AxA. === Subject: Re: giberish Consider, AxB = { (x,y) | x in A and y in B} AxB is certainly not equal to { (x,y) | x,y in (A or B) } whenever A != B. Let A=B, AxA = { (x,y) | x in A and y in B} (*) Because (*) accounts for the cases where A != B, it is more general formula that describes cartesian product of any two sets, than AxB = { (x,y) | x,y in A } which is only true when A=B. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: giberish > Consider, > AxB = { (x,y) | x in A and y in B} > AxB is certainly not equal to { (x,y) | x,y in (A or B) } whenever A != B. Let A=B, > AxA = { (x,y) | x in A and y in B} (*) Because (*) accounts for the cases where A != B, it is more general formula that describes cartesian product of any two sets, than > AxB = { (x,y) | x,y in A } > which is only true when A=B. > -- > ### Author Logan Lee ### > ### AuthorHomePagehttp://beam.to/pyenos### > ### AuthorQuote I just want to learn. ### Why don't you acknowledge your mistake and get on with it? This is pure bull. * is only valid if it includes the premise Let A=B otherwise it is undefined gibberish; Where does that B suddenly pop out of? But if the premise is included, then there is nothing general about it and it certainly does not cover the case AxB for A!=B. AxA = { (x,y) | x,y in A } is as correct as it is concise and lacking any corrections by you. === Subject: Re: giberish Consider cartesian product of any two sets. It is true that for A=B, AxB = AxA = { (x,y) | x,y in A} (*) For each cases A!=B, this can't be expressed by (*). Since AxB = { (x,y) | x in A and x in B } is true for any A or B, this is the recommended way to denote AxB. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: giberish > Consider cartesian product of any two sets. It is true that for A=B, > AxB = AxA = { (x,y) | x,y in A} (*) For each cases A!=B, this can't be expressed by (*). Since AxB = { (x,y) | x in A and x in B } is true for any A or B, this is the recommended way to denote AxB. Since AxB = { (x,y) | x in A and y in B } is true for any A or B, this is the recommended way to denote AxB. -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: giberish > Since AxB = { (x,y) | x in A and y in B } is true for any A or B, > this is the recommended way to denote AxB. Man, what did mathematicians do before you arrived on sci.math to grant them recommendations? It's a wonder that the field has progressed thus far without you. Anyway, keep 'em coming! -- Jesse F. Hughes Just tell me this. Is this all there is? After all that fuss about us, Is this all there is? -- Webb Wilder === Subject: Re: giberish Since AxB = { (x,y) | x in A and y in B } is true for any A or B, > this is the recommended way to denote AxB. Man, what did mathematicians do before you arrived on sci.math to > grant them recommendations? It's a wonder that the field has > progressed thus far without you. I'm just here to learn math. Anyway, keep 'em coming! -- > Jesse F. Hughes Just tell me this. > Is this all there is? > After all that fuss about us, > Is this all there is? -- Webb Wilder -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: giberish > >> Since AxB = { (x,y) | x in A and y in B } is true for any A or B, >> this is the recommended way to denote AxB. >> >> Man, what did mathematicians do before you arrived on sci.math to >> grant them recommendations? It's a wonder that the field has >> progressed thus far without you. >I'm just here to learn math. Nobody believes that. Because if that were so you'd ask people to explain the things you're confused about instead of continually stating your confusions as facts that all the rest of us have wrong. >> Anyway, keep 'em coming! >> >> -- >> Jesse F. Hughes Just tell me this. >> Is this all there is? >> After all that fuss about us, >> Is this all there is? -- Webb Wilder ************************ David C. Ullrich === Subject: Claim The Grid's Squares: Game Here is an unoriginal game for 2 players played on a grid drawn on paper. (I hope this game is more fun that the last game I have posted!) Start by drawing an n-by-n grid on paper (with an n of about 6 to 10). Draw the grid large enough so that each square can contain the game-piece (such as a coin) of each player. Play starts with each player marking any one of the squares, a different square for each player, with the player's symbol or filling in a square, using colored pencils or pens, with the player's color. Players each then place their game-piece on their starting square. The game consists of turns where one player moves and then the other. Who moves first in each turn alternates. So the moves are done like this: (player 1, player 2) (player 2, player 1) (player 1, player 2) (player 2, player 1), etc. Before either player moves in a turn, the player who moves second in the turn calls out how many spaces both players will move in the turn. The number of spaces called out is an integer from 1 to (n-1). Players, in the appropriate order, then move their game- pieces either up, down, left, or right the same number of squares on the grid. (Both players must move the same number of positions in a turn, but each can move in their own direction.) If a player cannot move because the number called is too big, and the move would take the player off the grid, then that player simply does not move on that turn. When a player is the first to land his/her game-piece on a square, the player then marks that square with his/her symbol, or fills the square in with her/his color. Players can always land on squares that are marked (by either player), but they can only claim empty squares for their own. Players can't move onto squares where their opponent's game- piece is located. Play continues until every square is filled in, or until a predetermined number of turns have passed. The winner of the game has the most number of squares marked with their symbol or with their color. Leroy Quet === Subject: Re: Claim The Grid's Squares: Game > Here is an unoriginal game for 2 players played on a grid drawn > on paper. > (I hope this game is more fun that the last game I have posted!) Start by drawing an n-by-n grid on paper (with an n of about 6 > to 10). > Draw the grid large enough so that each square can contain > the game-piece (such as a coin) of each player. Play starts with each player marking any one of the squares, > a different square for each player, with the player's symbol > or filling in a square, using colored pencils or pens, with > the player's color. > Players each then place their game-piece on their starting > square. The game consists of turns where one player moves and then > the other. Who moves first in each turn alternates. So the > moves are done like this: > (player 1, player 2) (player 2, player 1) (player 1, player 2) (player > 2, player 1), etc. Before either player moves in a turn, the player who moves > second in the turn calls out how many spaces both players > will move in the turn. The number of spaces called out > is an integer from 1 to (n-1). > Players, in the appropriate order, then move their game- > pieces either up, down, left, or right the same number of > squares on the grid. (Both players must move the same number > of positions in a turn, but each can move in their own > direction.) > If a player cannot move because the number called is too big, > and the move would take the player off the grid, then that > player simply does not move on that turn. When a player is the first to land his/her game-piece on a > square, the player then marks that square with his/her symbol, > or fills the square in with her/his color. Players can always land on squares that are marked (by either > player), but they can only claim empty squares for their own. > Players can't move onto squares where their opponent's game- > piece is located. Play continues until every square is filled in, or until a > predetermined number of turns have passed. The winner of the game has the most number of squares marked > with their symbol or with their color. Leroy Quet A clarification if you please! When the number of moves in one direction would pass the edge of the board, but not in another direction, is a player obligated to move? Or is it allowed to choose the direction in which no move can be made in order to stay on the same square? === Subject: Re: Claim The Grid's Squares: Game > Here is an unoriginal game for 2 players played on a grid drawn > on paper. > (I hope this game is more fun that the last game I have posted!) Start by drawing an n-by-n grid on paper (with an n of about 6 > to 10). > Draw the grid large enough so that each square can contain > the game-piece (such as a coin) of each player. Play starts with each player marking any one of the squares, > a different square for each player, with the player's symbol > or filling in a square, using colored pencils or pens, with > the player's color. > Players each then place their game-piece on their starting > square. The game consists of turns where one player moves and then > the other. Who moves first in each turn alternates. So the > moves are done like this: > (player 1, player 2) (player 2, player 1) (player 1, player 2) (player > 2, player 1), etc. Before either player moves in a turn, the player who moves > second in the turn calls out how many spaces both players > will move in the turn. The number of spaces called out > is an integer from 1 to (n-1). > Players, in the appropriate order, then move their game- > pieces either up, down, left, or right the same number of > squares on the grid. (Both players must move the same number > of positions in a turn, but each can move in their own > direction.) > If a player cannot move because the number called is too big, > and the move would take the player off the grid, then that > player simply does not move on that turn. When a player is the first to land his/her game-piece on a > square, the player then marks that square with his/her symbol, > or fills the square in with her/his color. Players can always land on squares that are marked (by either > player), but they can only claim empty squares for their own. > Players can't move onto squares where their opponent's game- > piece is located. Play continues until every square is filled in, or until a > predetermined number of turns have passed. The winner of the game has the most number of squares marked > with their symbol or with their color. Leroy Quet A clarification if you please! When the number of moves in one direction would > pass the edge of the board, but not in another > direction, is a player obligated to move? Or is it > allowed to choose the direction in which no move > can be made in order to stay on the same square? > In practice, people can play this game any way they wish. :) I, however, intended that a player must move if he/she is able to move, whether she/he wants to or not. Leroy Quet .. === Subject: Infinite Double-Sum Involving Coprimality It has been a while since I have posted my math results. (Some sci.math readers may recall days of olde when I posted result after result after result.) I am uncertain if the following is true, since I am prone to making errors. Maybe someone could confirm the result. For y >= 0, r = integer >= 2, --- --- 1 1 1 / / ( --- - --- ) ------- = --- --- k+y k+m (m+y)^r m>=1 k>=1 GCD(k,m)=1 r/2 1 ----------- + ------- (1+y)^(r+1) (1+y)^r r-1 --- --- 1 mu(j) - - / --------- / z(k,y/j) *z(r+1-k,y/j), 2 --- j^(r+1) --- j>=1 k=2 where z(k,x) = sum{j>=1} 1/(j+x)^k = zeta(k,x) - 1/x^k, where zeta(k,x) is the Hurwitz zeta function. (For this result, I prefer using the variation on the Hurwitz zeta function where the sum starts at j=1 instead of at j=0.) mu(j) is the Mobius (Moebius) function. The above result in linear-mode for those who are fixed-width-font-challenged: sum{m>=1} sum{k>=1, GCD(k,m)=1} (1/(k+y) -1/(k+m)) /(m+y)^r = (r/2)/(1+y)^(r+1) + 1/(1+y)^r - (1/2) sum{j>=1}(mu(j)/j^(r+1)) sum{k=2 to r-1} z(k,y/j) *z(r+1-k,y/ j). The result above is correct at y = 0, at least, since: sum{m>=1} sum{1<=k<=m, GCD(k,m)=1} 1/(k m^r) = r/2 + 1 - 1/(2 zeta(r+1))* sum{k=2 to r-1} zeta(k)*zeta(r+1-k), where zeta(k) = sum{j>=1} 1/j^k, of course. The bottom double-sum of the result above can be rewritten so that we have this: For y >= 0, r = integer >= 2, --- --- 1 1 1 / / ( --- - --- ) ------- = --- --- k+y k+m (m+y)^r m>=1 k>=1 GCD(k,m)=1 r/2 1 ----------- + ------- (1+y)^(r+1) (1+y)^r r-1 --- --- --- 1 1 - - / / / --------------------- . 2 --- --- --- (m+y)^k (n+y)^(r+1-k) k=2 m>=1 n>=1 GCD(m,n)=1 (Notice that m and n are coprime to each other.) In linear-mode: sum{m>=1} sum{k>=1, GCD(k,m)=1} (1/(k+y) -1/(k+m)) /(m+y)^r = (r/2)/(1+y)^(r+1) + 1/(1+y)^r - (1/2) sum{k=2 to r-1} sum{m>=1} sum{n>=1, GCD(m,n)=1} 1/((m+y)^k (n+y)^(r+1-k)). Leroy Quet === Subject: Lie derivative let P be an anti-symmetric tensor of type (0,2) on R^6 in coordinates (x1,x2,x3,y1,y2,y3): P:=<<0 | -y3 | -y2 | x2 | x3 | 0 >, , , <-x2| x1 | 0 | 0 | -y1 | y2 >, <-x3| 0 | x1 | y1 | 0 | -y3>, <0 | -x3 | x2 | -y2 | y3 | 0 >>; and let X a linear vector field (I mean a vector field whose components are linear functions in the coordinates). I'm looking for the most general form of X under the hypothesis that L_X (P) = 0. (Lie derivative) === Subject: errata for MacDonald's Symmetric Fcns & Hall Polynomials? Is there a list of errata available anywhere for Symmertric Functions and Hall Polynomials, Second Edition? Barry Brent === Subject: Re: Harmonic Cantilever <7e7d7$45c20a45$82a1e228$6495@news2.tudelft.nl determine the constant (c), experimentally. The result is shown in the > following picture, for 50 bricks. So far so good. But the experimental guess can be replaced by an exact > least squares minimalization procedure: Surprisingly enough, if we replace this least squares (c) by the Euler- Mascheroni constant gamma = 0.5772156649015328606065120900824024310 .. then the result is almost a best fit as well: gamma.e^x . Why oh why? Han de Bruijn === Subject: Re: Harmonic Cantilever <7e7d7$45c20a45$82a1e228$6495@news2.tudelft.nl > determine the constant (c), experimentally. The result is shown in the > following picture, for 50 bricks. So far so good. But the experimental guess can be replaced by an exact > least squares minimalization procedure: Surprisingly enough, if we replace this least squares (c) by the Euler- > Mascheroni constant gamma = 0.5772156649015328606065120900824024310 .. > then the result is almost a best fit as well: gamma.e^x . Why oh why? And an even better result is obtained with c = exp(-gamma) instead of c = gamma. Part of the explanation is that these numbers don't differ too much: exp(-E) = 5.61459483566885E-0001 = E = 5.77215664901533E-0001 It is thus _conjectured_ that the best continuous fit to the Harmonic Cantilever is (left hand side, as defined before): y = - exp(x - gamma) But .. can it be proved? More or less. We know that y = -N . Then taking the logarithm at both sides results in: ln(N) = x - gamma <==> gamma = sum(k=1..N) 1/k - ln(N) However, the limit of the latter expression, for N -> oo, is precisely the definition of the Euler-Mascheroni constant gamma. So, that's why! In practice, this means that the fit is best for large N and worse for small N, quite in agreement with experience: http://hdebruijn.soo.dto.tudelft.nl/jaar2007/stapelen.jpg Nice, huh? Han de Bruijn === Subject: Re: Harmonic Cantilever <7e7d7$45c20a45$82a1e228$6495@news2.tudelft.nl It is thus _conjectured_ that the best continuous fit to the Harmonic > Cantilever is (left hand side, as defined before): y = - exp(x - gamma) Meanwhile, this has become a theorem, not a conjecture. When combined with the least squares method for finding the best fit, we find another result (quite surprising to me, at least). Let the function E(k) be defined by: E(k) = exp[ sum(m=1..k) 1/m ] Then: exp(-gamma) = lim(N->oo) [ sum(k=1..N) k.E(k) ] / [ sum(k=1..N) E(k)^2 ] And yes, this also is a theorem, not a conjecture. Is it well known? Here comes a program snippet that confirms the theorem experimentally. program HdB; const gamma : double = 0.577215664901532860606512090082402431042; function konstante : double; { Least Squared Best Fit } const N : integer = 10000000; var y : integer; x,u,v,p : double; begin x := 0; u := 0; v := 0; for y := 1 to N do begin p := exp(x); x := x + 1/y; u := u + y*p; v := v + p*p; end; konstante := u/v; end; begin Writeln(exp(-gamma),' = ',konstante); end. Output: 5.61459483566885E-0001 = 5.61459525677516E-0001 Han de Bruijn === Subject: Re: Harmonic Cantilever <7e7d7$45c20a45$82a1e228$6495@news2.tudelft.nl Let the function E(k) be defined by: E(k) = exp[ sum(m=1..k) 1/m ] Then: exp(-gamma) = lim(N->oo) [ sum(k=1..N) k.E(k) ] / [ sum(k=1..N) E(k)^2 ] = lim(N->oo) [ sum(k=1..N) E(k)^2 exp{- sum(m=1..k) 1/m + ln(k)}] / [ sum(k=1..N) E(k)^2 ] This is a weighted mean of functions exp{- sum(m=1..k) 1/m + ln(k)}] where the weights E(k)^2 are positive & uniformly increasing with (k) and eventually approaching infinity. Meaning that the weights E(k)^2, for large (k), are predominant. But exp{- sum(m=1..k) 1/m + ln(k)}] for large (k) is approximately exp(- gamma) . Thus it is seen that the above limit is mainly a mean of exp(- gamma) approximations. I'm pretty sure that this sloppy argument can be made more rigorous when needed. Han de Bruijn === Subject: subfields Suppose K and M are fields such that K is a subset of M. Let a,b be elements in M. Then is the field K(a,b) a subset of M always? or is it necessary that a and b are algebraic over K? === Subject: Re: subfields days. My association with the Department is that of an alumnus. >Suppose K and M are fields such that K is a subset of M. Is it just a subset, or is it actually a subfield? That is, are the operations of K the restrictions of the operations of M to K? I assume so, given your question below. >Let a,b be elements in M. Then is the field K(a,b) a subset of M > always? Depends entirely of what you mean by K(a,b). Usually, when K is a field, we talk about K(a) only in the context of a given field that contains both K and a, in which case we ->DEFINE<- K(a) to be the smallest subfield of the given field which contains both K and a. In that sense, since you are taking as the given field the field M, then ->by definition<- you have that K(a,b) is the smallest subfield of M that contains K, a, and b, and so the answer is yes (once you replace your imprecise use of subset with subfield). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: subfields Suppose K and M are fields such that K is a subset of M. Is it just a subset, or is it actually a subfield? That is, are the > operations of K the restrictions of the operations of M to K? I assume > so, given your question below. Let a,b be elements in M. Then is the field K(a,b) a subset of M > always? Depends entirely of what you mean by K(a,b). Usually, when K is a field, we talk about K(a) only in the context of > a given field that contains both K and a, in which case we ->DEFINE<- > K(a) to be the smallest subfield of the given field which contains > both K and a. In that sense, since you are taking as the given > field the field M, then ->by definition<- you have that K(a,b) is > the smallest subfield of M that contains K, a, and b, and so the > answer is yes (once you replace your imprecise use of subset with > subfield). -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin > magidin-at-member-ams-org To add an example: Q(pi,e) is a subfield of R. === Subject: test Just a test === Subject: Re: A card game probability [...] > Now take our ordering of the deck to be AC, AD, AH, AS, > 2C, 2D, 2H, 2S, ..., QS, KC, KD, KH, KS. The > restricted positions are thirteen 4 x 4 boards on the > main diagonal of the 52 x 52 board. Yes, I now see the connection between the no-hit card problem and the problem of non-attacking rooks that must be placed on a 52x52 board outside of the 13 4x4 boards on the diagonal. It's easy to see that for an 8x8 board, there are (4!)*(4!) ways, for a no-hit probability of (24*24)/(8!) = 1/70. [as I had previously ] > For a rectangular board with n rows and m columns, m <= > n, we have R_{n,m}(X) = sum_k binom(m,k) * (n)_k * X^k where > (n)_k = n * (n-1) * (n-2) * ... * (n-k+1). > Proof: Obviously there is one way of puting 0 rooks on > the board. To put k rooks on we can choose k columns in > binom(m,k) ways, and then place the k rooks in n ways > on the first chosen column, n-1 ways on the second > chosen column, ..., n-k+1 ways on the final chosen > column. [...] So at least as many columns as rows, with one polynomial for each pair (n,m) such that n>=m, and the coefficient of X^k being a count of ways of putting k non-attacking rooks (with other conditions??). It seems that this is a case where computations involving polynomial multiplication, powers of a polynomial, etc. are of great help in solving a combinatorial problem. I haven't understood or read in detail everything you as a non-attacking rook problem, with a restriction on the squares which the rooks are allowed to occupy. David Bernier === Subject: Re: A card game probability <%mLwh.39649$xX6.212941@wagner.videotron.net>, > [...] > Now take our ordering of the deck to be AC, AD, AH, AS, > 2C, 2D, 2H, 2S, ..., QS, KC, KD, KH, KS. The > restricted positions are thirteen 4 x 4 boards on the > main diagonal of the 52 x 52 board. Yes, I now see the connection between the no-hit > card problem and the problem of non-attacking rooks > that must be placed on a 52x52 board outside > of the 13 4x4 boards on the diagonal. It's easy > to see that for an 8x8 board, there are (4!)*(4!) > ways, for a no-hit probability of > (24*24)/(8!) = 1/70. [as I had previously ] For a rectangular board with n rows and m columns, m <= > n, we have R_{n,m}(X) = sum_k binom(m,k) * (n)_k * X^k where > (n)_k = n * (n-1) * (n-2) * ... * (n-k+1). > Proof: Obviously there is one way of puting 0 rooks on > the board. To put k rooks on we can choose k columns in > binom(m,k) ways, and then place the k rooks in n ways > on the first chosen column, n-1 ways on the second > chosen column, ..., n-k+1 ways on the final chosen > column. > [...] > So at least as many columns as rows, with one polynomial > for each pair (n,m) such that n>=m, and the coefficient > of X^k being a count of ways of putting k non-attacking rooks > (with other conditions??). With no other conditions. At this point we analyze only the restricted positions, treating them as a `board'. There is a theorem that can be applied iteratively to calculate the rook polynomial for any (finite) set of lattice squares. Once we have an enumerator for the restricted positions, then we can mark them on the full n x n board and apply the principle of inclusion and exclusion. If we have k rooks in restricted position, there are (n-k)! ways of placing the remaining rooks (with no constraint on their placement). > It seems that this is a case where computations involving > polynomial multiplication, powers of a polynomial, etc. > are of great help in solving a combinatorial problem. I haven't understood or read in detail everything you > as a non-attacking rook problem, with a restriction on > the squares which the rooks are allowed to occupy. -- Michael Press === Subject: Re: Excel and correlations; parts versus whole > We are doing two Pearson correlations serially of two independent (A, > B) and one dependent (C) variables (rankings in all three cases) using > Excel. When we ran the females, we got a (low) correlation of 0.21; > and for males, we got a correlation of 0.22 for A and C. Fine, but > then when we ran the total population, we got a correlation of 0.98 > for A and C. We had the same sort of thing for B : C. Is this logical? > Can the cumulative be lower than either of the parts? >> That's exactly what you would expect if the difference between the >> groups is large and in the same direction on all three variables. Whoops, I see I made a typo: The correlation for the total poulation > was 0.098 (i.e., lower than for either one). But I don't understand > your answer in any case. Why would one expect a correlation for the > whole to be outside the values (i.e., correlations) for the parts? > It seems that that means as we add data, even though the data added > support the picture of the older data, the new calculated correlations Draw the picture. Think of the data for each group as being a more-or- less elliptical cloud of points. The within-group correlations tell you how elongated each ellipse is. If the centroids of the two groups coincide then the overall correlation would be a weighted average of the within-group correlations. But if the centers do not coincide then the overall correlation depends on how far apart the two groups are, and in which direction. It is quit possible to have strong positive within-group correlations but a strong negative overall correlation. > can land any-where. I find this to be extremely disheartening and > seems to make using the test virftually meaningless! === Subject: Re: Excel and correlations; parts versus whole I'm sorry for being so obtuse, but I still don't follow. Let's take a simpler case than my data. Suppose I find that tall girls are heavy and also that tall boys are heavy. I just don't see the logic of it being possible that tall kids are not heavy (i.e., more accurately, the taller one is, regardless of sex, there is no increase [or a lower increase] in the likelihood of one being heavier, even though, USING and tall kids are shown to be heavy (if they are girls). This seems to make a mockery of statistics. What am I not understanding? === Subject: Re: Excel and correlations; parts versus whole > I'm sorry for being so obtuse, but I still don't follow. Let's take a > simpler case than my data. Suppose I find that tall girls are heavy > and also that tall boys are heavy. I just don't see the logic of it > being possible that tall kids are not heavy (i.e., more accurately, > the taller one is, regardless of sex, there is no increase [or a lower > increase] in the likelihood of one being heavier, even though, USING > and tall kids are shown to be heavy (if they are girls). This seems > to make a mockery of statistics. What am I not understanding? To say that height and weight are positively correlated for girls means that taller girls are generally heavier than shorter girls. To say that height and weight are positively correlated for boys means that taller boys are generally heavier than shorter boys. But neither statement says anything about whether girls are generally taller or heavier than boys. In principle, the picture might look like this, in which the correlation is strongly positive within each group but strongly negative overall. (View this in a fixed-width font.) W gg | gggg | gggg | gg | | | | | | | -------------------+-------------------H | | | | | | | bb | bbbb | bbbb | bb === Subject: Re: Excel and correlations; parts versus whole > I'm sorry for being so obtuse, but I still don't follow. Let's take a > simpler case than my data. Suppose I find that tall girls are heavy > and also that tall boys are heavy. I just don't see the logic of it > being possible that tall kids are not heavy (i.e., more accurately, > the taller one is, regardless of sex, there is no increase [or a lower > increase] in the likelihood of one being heavier, even though, USING > and tall kids are shown to be heavy (if they are girls). This seems > to make a mockery of statistics. What am I not understanding? To say that height and weight are positively correlated for girls > means that taller girls are generally heavier than shorter girls. To say that height and weight are positively correlated for boys > means that taller boys are generally heavier than shorter boys. But neither statement says anything about whether girls are generally > taller or heavier than boys. In principle, the picture might look like > this, in which the correlation is strongly positive within each group > but strongly negative overall. (View this in a fixed-width font.) > Tall girls are heavy; tall boys are heavy, but tall kids aren't heavy (or, to be accurate, aren't shown to be as likely to be heavy as EITHER tall boys or tall girls are). W | | | | | | | -------------------+-------------------H | + C C C C C CCCC | CCCCCC | | | bb | bbbb | bbbb gg | b | ggg | ggg | bb === Subject: Geometry problems wanted Someone please give me an interesting geometry problem to work on. I'm bored at work. Greg === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. I'm > bored at work. Bisect a line segment with a compass only -- no straghtedge. Dave === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. I'm > bored at work. Greg Try my current Wedge Rotation problem posed in geometry.puzzles? How it is, or is not spherical trigonometry? Narasimham === Subject: Re: Geometry problems wanted Someone please give me an interesting geometry problem to work on. I'm > bored at work. Greg Try my current Wedge Rotation problem posed in geometry.puzzles? How > it is, or is not spherical trigonometry? Or if you wish to dabble a bit in 3-D,try finding what geometric quantity is represented by sin(a) / sin(A) in a spherical triangle. > Narasimham === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. I'm > bored at work. ... bored at work and nevertheless with reward...? Try to construct the centre of a given circle without help of a ruler/lineal. === Subject: Re: Geometry problems wanted Someone please give me an interesting geometry problem to work on. > I'm bored at work. Greg Probably a bit late now, but if you're bored at work again next week try this one. Given a triangle and a point (not necessarily on the triangle), construct a line through the point that bisects the area of the triangle. === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. I'm > bored at work. Greg Please try my problem in Hyacinthos: http://tech.groups.yahoo.com/group/Hyacinthos/message/14785 Bui Quang Tuan === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. I'm > bored at work. Try proving that if an angle bisector of a triangle is a median to the opposite side, then the triangle is an isosceles triangle. Dave === Subject: Re: Geometry problems wanted > Try proving that if an angle bisector of a triangle is a median to the > opposite side, then the triangle is an isosceles triangle. Dave The angle bisector divides the opposite side in the ratio of the adjacent sides. If it's a median, then that ratio is 1, and the Greg === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. Consider the grid consisting of all the points (i,j) with integer numbers i and j. Turn it anticlockwise by 22.5Á arond the origin (0,0), name the new grid G. Turn it clockwise by 22,5Á and call this grid H. Then you have the two grids G and H overlapping and with angle 45Á between their j-axes. Try to construct a one-to-one correspondence between the points of G and H. Can you provide an upper bound for the distance between the corresponding points? Can you devise a Bijection such that there is a bound less than 1? In our German newsgroup de.sci.mathematik there was a lengthy discussion resulting in a PDF paper by Klaus Nagel, providing a bijection with least upper bound cos(22.5Á) = 0.92388 and a proof. Can you improve on that? Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. I'm > bored at work. Greg Geometry in the wide sense? Check out http://home.no.net/zamunda/grid.htm J K Haugland === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. I'm > bored at work. Greg Try http://mathforum.org/library/problems/elementary.html (the veru first hit I got) === Subject: Re: Geometry problems wanted > Someone please give me an interesting geometry problem to work on. I'm > bored at work. Prove the Steiner-Lehmus theorem directly in the manner of classical Euclidean geometry. It says that if the the angle bisectors of the the base angles of a triangle are equal the triangle is isocoles. If you succeed, publish and you will be famouse for a little while. Bob Kolker === Subject: Re: Geometry problems wanted > >> Someone please give me an interesting geometry problem to work on. I'm >> bored at work. Prove the Steiner-Lehmus theorem directly in the manner of classical > Euclidean geometry. It says that if the the angle bisectors of the the base angles of a > triangle are equal the triangle is isocoles. If you succeed, publish and > you will be famouse for a little while. Bob Kolker > Since I Steiner-Lehmus here: Coxter/Greitzer : Geometry Revisited is an excellent resource to defeat boredom. There' also a pdf/nptes called geometry unbound by some indian, that Here are a some ti get you started: 1.) Show that on a regular pentagon the ratio of any diagonal and a side is the golden ratio (1.618....). 2.) construct the focus of an arbitrary parabola only using ruler and compass. 3.) http://kilian.byethost5.com/mytemp/landen.pdf === Subject: Re: Geometry problems wanted <52hgeeF1ogimgU1@mid.individual.net> 1.) Show that on a regular pentagon the ratio of any diagonal and a side > is the golden ratio (1.618....). This was shown to me a long time ago. Two diagonals from a vertex, together with the opposite side, create a 72-72-36 triangle. If you draw the angle bisector from one of the 72 angles, you make two new triangles, one is again 72-72-36, where the other is 36-36-108. Both are isosceles, so we have several segments which are the same length, and the result follows easily from here. I think Archimedes also proved this in his Book of Lemmas, or at least something that quickly implies this. And it was very nice, what he proved. 2.) construct the focus of an arbitrary parabola only using ruler and > compass. A ruler, so I am allowed to measure distances? Or is it supposed to be just a straightedge? 3.)http://kilian.byethost5.com/mytemp/landen.pdf Extend PQ to the big circle, say it meets at point Q'. The tangent at Q' is parallel to RS, and thus arc RQ' is equal to arc Q'S, and thus it is an angle bisector. above, and whose problems I'm still working on. Greg === Subject: Re: Geometry problems wanted >> 1.) Show that on a regular pentagon the ratio of any diagonal and a side >> is the golden ratio (1.618....). This was shown to me a long time ago. Two diagonals from a vertex, > together with the opposite side, create a 72-72-36 triangle. If you > draw the angle bisector from one of the 72 angles, you make two new > triangles, one is again 72-72-36, where the other is 36-36-108. Both > are isosceles, so we have several segments which are the same length, > and the result follows easily from here. I think Archimedes also > proved this in his Book of Lemmas, or at least something that quickly > implies this. And it was very nice, what he proved. > >> 2.) construct the focus of an arbitrary parabola only using ruler and >> compass. A ruler, so I am allowed to measure distances? Or is it supposed to be > just a straightedge? darn english - i always forget about the difference between ruler and straightedge. There's a construction using only compass and straightedge, however i think rulers (measuring distances) are allowed as well. > >> 3.)http://kilian.byethost5.com/mytemp/landen.pdf Extend PQ to the big circle, say it meets at point Q'. The tangent at > Q' is parallel to RS, and thus arc RQ' is equal to arc Q'S, and thus > it is an angle bisector. > hmm... nice approach > above, and whose problems I'm still working on. Greg > === Subject: Re: Geometry problems wanted <52hgeeF1ogimgU1@mid.individual.net Someone please give me an interesting geometry problem to work on. I'm > bored at work. Prove the Steiner-Lehmus theorem directly in the manner of classical > Euclidean geometry. It says that if the the angle bisectors of the the base angles of a > triangle are equal the triangle is isocoles. If you succeed, publish and > you will be famouse for a little while. Bob Kolker it using calculus, but wouldn't the second proof at this link: http://mathforum.org/library/drmath/view/54109.html be what you're looking for? Greg === Subject: Re: TI 84 vs HP calc? Mail-To-News-Contact: abuse@dizum.com : Could somebody explain what good a calculator is for doing math? Sure, they're good for getting approximate answers to real-world problems. Which has what, exactly, to do with *math*? I'm an engineer, I find approximate answers to real-world problems for a living. But that ain't math. I use tools that math provides, but that's not the same thing at all. >Getting an exact answer like (5+ln(100))/3 is of course important (and I >always demand exact answers from my students) but if you want them to >appreciate what that means (as a length of time, a mass, a population and >so on) then an approximation will do that. Those things that you mention sound a lot more like physics or biology than math to me. -- Michael F. Stemper #include Why doesn't anybody care about apathy? === Subject: Re: TI 84 vs HP calc? >> : Could somebody explain what good a calculator is for doing math? >> Sure, they're good for getting approximate answers to real-world problems. Which has what, exactly, to do with *math*? a couple of points regarding this: Math for mathematicians is not the only math taught at universities (in fact it is the smaller part probably). And even pure math classes do entertain real world problems (usually not as a main subject though). And even in a pure math context unrelated to real word problems do approximation (i.e. floating point numbers) have their value, in particular does it make comparisons (<,>) much easier. Also think of problems like computing digits for irrational number etc. There are many areas in (applied) math that might deal with real world problems (and data) directly, such as statistics for instance. Another thing is, that mathematical theorems often arises from studying data (and finding patterns and structure in it) and that data can be produced with calculators. Also don't think of calculator as the classic thing from the late 70s/early 80s. Modern calculators are essentially small desktop computers/handhelds which are specialized on math,they do feature programming languages,spreadssheets, CAS and other math software. I'm an engineer, I find approximate answers to real-world problems for > a living. But that ain't math. I use tools that math provides, but that's > not the same thing at all. > Well applying math on real world problem is a grey area or an interface, that you can approach from 2 sides - from the engineering/science area in which the problem originates and from the (applied) math side which provides formalism and models that can be used for the problem. So nobody says engineering is math, but the math in used in engineering is still math :) and doesn't stop being math just because it used by non mathematician. >> Getting an exact answer like (5+ln(100))/3 is of course important (and I >> always demand exact answers from my students) but if you want them to >> appreciate what that means (as a length of time, a mass, a population and >> so on) then an approximation will do that. Those things that you mention sound a lot more like physics or biology > than math to me. > === Subject: Re: TI 84 vs HP calc? : Which has what, exactly, to do with *math*? What is math? You seem to be implying it's the manipulation of symbols without regard for real-world application (if any) and while I can appreciate that on some level it doesn't do a lot of good when teaching engineers or business majors, for example. When someone mentions a math class, it should be relatively clear that this includes classes like algebra and calculus (and statistics, probability, etc.) in which real-world problems are used to put the mathematics in context. Given that the real-world problems are part of the math class it makes sense to get some approximate answers to make some real-world sense of the problems at hand. Justin === Subject: Re: TI 84 vs HP calc? : I would find out which calculator will be used in classes at my school > : and buy another one... : ... then you'll be able to learn something they don't know ;-) Learning things that other people don't know is all well and good but > coping with a calculator that (essentially) none of your friends will have > and your teacher will very probably not be able to help you with will make > things worse. My students have enough trouble remembering the parenthesis for a > calculation like 5/(1+ln 3). Well let's hope your students _do not_ enter 5/(1+ln 3) in any calculator (particularly not a TI). :-) Though correct (and still somewhat popular) as a math notation, i find find it quite problematic from a didactic perspective and i've seen it causing problems for students over and over. I picture the idea of them trying to figure > this out in RPN and it makes me wince! Justin === Subject: Re: TI 84 vs HP calc? > >> So from a math students perspective I would buy whatever calculator you >> might need for your written exams I've been out of school for a long time, but I have trouble seeing what > use one would have for a calculator in a *math* class. I was a EE, and > used a slide rule [1] for my physics and engineering classes, but never > needed one for a math class, and never saw anybody else using one in > them, either. Could somebody explain what good a calculator is for doing math? > [1] Calculators started coming out mid-way through my enrollment, but > at $400.00 for four functions, they were well beyond the reach of most > of us. Well I see a couple of reasons why calculators could be use for math classes/exams (emphasis is on could though). 1.) problems does contain (real) (numerical) data that cannot easily computed by hand. Meaning the problem sets may not be constructed in such a way anymore, that all required (numerical) computations are easily done by hand 2.) graphing/tesing/educated guessing of solutions before attempting the rigorous proof. 3.) Modern calculators do not only have numerical and graphing capabilities but they can symbolic math as well (i.e. the come with an integrated CAS) In some ways those capabilities may also change the kind of problems you can do in math exams, since many of the classical (symbolic) math problems for exams such as computation of limits, integrals, induction for simple formulas, binomial and trigonometric identities,symbolic integration, system of equaltities,odes can now be solved by your calculator directly. As a result schools need to restructure their curricula and exams or (as some do) simply ban advanced calculators. === Subject: Re: TI 84 vs HP calc? <200702011745.l11HjhBH101764@walkabout.empros.com> : Could somebody explain what good a calculator is for doing math? If the calculator is Maple, plenty of uses, judging from the posts in this group. But I guess this is not quite an answer to the question you were asking. Sure, they're good for getting approximate answers to real-world problems. Getting an exact answer like (5+ln(100))/3 is of course important (and I > always demand exact answers from my students) but if you want them to > appreciate what that means (as a length of time, a mass, a population and > so on) then an approximation will do that. Ok, but not to beat a dead horse, why not get them to learn how to estimate ln(100) in their heads? Easy ones like that are IMHO not the best examples to make your point. Justin === Subject: Re: TI 84 vs HP calc? : Ok, but not to beat a dead horse, why not get them to : learn how to estimate ln(100) in their heads? Easy ones : like that are IMHO not the best examples to make your : point. Okay, so what is ln(100) to three decimal digits then? Justin === Subject: Re: TI 84 vs HP calc? <200702011745.l11HjhBH101764@walkabout.empros.com> : Ok, but not to beat a dead horse, why not get them to > : learn how to estimate ln(100) in their heads? Easy ones > : like that are IMHO not the best examples to make your > : point. Okay, so what is ln(100) to three decimal digits then? Justin Why do you insist on three digits? My point was just that it's useful to be aware that ln of 100 is exactly twice ln of 10, and the latter number is a very useful one to remember, to *some* degree of accuracy. As it happens, I remember it to two digits. Though now that you twist my arm, I think I remember a third (and yup, I remembered it right). I'm really not trying get into a big argument over this and am sorry if I came off as combative. === Subject: Re: TI 84 vs HP calc? : Why do you insist on three digits? My point was just that it's useful : to be aware that ln of 100 is exactly twice ln of 10, and the latter : number is a very useful one to remember, to *some* degree of : accuracy. As it happens, I remember it to two digits. : Though now that you twist my arm, I think I remember a third (and yup, : I remembered it right). I also didn't mean to start a fight, so I mean this all in good humor and with respect! For example, if a student is studying a problem such as: The half-life of element X is 10.6 days exactly. If there are 7 grams to start with, when will there be 5 grams remaining? Give an exact answer and also calculate this to the nearest minute. While I appreciate the ability to approximate in ones head, and I do, there's a point at which the time and complications involved in such an approximation becomes wasteful. Justin === Subject: Re: understanding the curl operator >> For a smooth vector field F: R^3 -> R^3, >> curl(F) gives another vector field. >> Suppose v is a vector in R^3, and we define >> G: R^3 -> R^3 by >> G(x) = v. >> Then, I think curl(F+G) = curl(F) >> because the partials del_x (G_y), and all the other 8 similar partials >> are zero. > Yes, the curl is a type of derivative. (the circulation) > >> I've seen, it seems that any vector n R^3 can be a curl value for >> some vector field. Sure, Just rotate the vector field and its vectors representing the curls > must rotate too(think of a plane and then the normal vector) > >> If we have a vector field F such that >> F(w).(0,0,1) = 0 for all w in R^3 , then F has a zero >> z-coordinate everywhere, and if >> F(w+ t(0,0,1)) = F(w) for all w in R^3 and t in R, then >> F varies only with the x and y coordinates. your second statement follows directly from the first. If F(w)*z = 0 then > that means F(w) is orthogonal to z and so must line in the xy plane(z is the > z unit vector) if it lies in the xy plane then it is dependent on at most x > and y. > >> Then I think curl(F).(1,0,0) = 0 everywhere, and also >> curl(F).(0,1,0) = 0 everywhere, so curl(F) is a multiple of >> (0,0,1) everywhere. If I got the conditions on F in the >> preceding paragraph right, F can be imagined as >> a natural trivial version of some 2D vector field. Yes. But not always. There are F's that are 2D but do not fit your > requirements. Just take F(x,y,z) = This 2D vector field lieing entirely on the xy plane. Now rotate it so that > it no longer as any 0 components(assuming f and g are not 0). You have a 2D > vector field that doesn't look 2D. > >> Informally, the curl of a 2D vector field takes on >> 1-dimensional values, and the curl of a 3D vector field >> takes on 3-dimensional values. That doesn't make much sense. Actually, the curl is always 3D as that is how > its defined. One does not have the cross product for any other dimensions > than 3(1D and 2D are embedded in 3D). There is a generalizd cross product > but I doubt your talking about that. I guess it's hard to see the circulation at a point w in a picture of a vector field F because, in general, the vector field at that point is non-zero. But if we subtract F(w) from F(.), then at w this gives a vector field with the same curl value, and I suppose the field vectors near w tend to rotate about some axis, in one direction or the other, and at some rate; the curl would indicate the axis, the direction and the magnitude of the circulation. It's probably easier to see circulation in a diagram of field vectors when the field has zero divergence at w. I have heard of the 7D cross-product. I'm more interested in understanding the circulation concept. And 3 dimensions is enough for now. David Bernier P.S. I'm wondering how circulation relates to angular momentum per unit of volume for a small box of fluid of unit density (this is physics...) >> So in 4 dimensions, if there is a natural operator to >> describe what the curl of a 3D vector field describes in >> 3D, what are the possibilities? (if there are any...) There is no curl in 4D Curl F = Grad X F. the X = cross product is defined only for 3D. You can embed them in higher spaces but then thats cheating. Check out for the full explination and such: http://uk.arxiv.org/PS_cache/math/pdf/0204/0204357.pdf === Subject: Vector space bases How hard is it to prove the following fact: If V is a vector space, and B1, B2 are bases for V, then there is a bijection from B1 to B2. In the linear algebra text I used in school, they only prove this for finite-dimensional vector spaces. This proof is not hard, but I don't see any way to even begin to adapt it to the infinite-dimensional case. Just to clarify, I am refering to algebraic bases here. Can anyone recommend a linear algebra book in which they deal with infinite-dimensional vector spaces? === Subject: Re: Vector space bases > How hard is it to prove the following fact: If V is a vector space, > and B1, B2 are bases for V, then there is a bijection from B1 to B2. > In the linear algebra text I used in school, they only prove this for > finite-dimensional vector spaces. This proof is not hard, but I don't > see any way to even begin to adapt it to the infinite-dimensional > case. Just to clarify, I am refering to algebraic bases here. > Can anyone recommend a linear algebra book in which they deal with > infinite-dimensional vector spaces? > Not a linear algebra book, but the proof is here: Hewitt & Stromberg, REAL AND ABSTRACT ANALYSIS Basically: since each vector in one base involves only finitely many vectors when expanded in the other base, prove that the cardinal of each base is at most aleph_0 times the cardinal of the other. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: A very vague question about sets. > If we have a general set X, is it reasonable to say Let x > be an object not in X.? Is there a way to get hold of such > an object? What about a non-Axiom-of-Choice way? We can show it with just the axiom schema of separation. Theorem: AxEy ~yex. Proof: ~ExAy yex, as follows: Suppose ExAy yex. So let Ay yex. By the axiom schema of separation, EzAy(yez <-> (yex & ~yey)). So, since Ay yex, we have EzAy(yez <-> ~yey). So let Ay(yez <-> ~yey). So zez <-> ~zez, which is a contradiction, so ~ExAy yex. But ~ExAy yex <-> AxEy ~yex. MoeBlee === Subject: Re: A very vague question about sets. > If we have a general set X, is it reasonable to say Let x > be an object not in X.? Is there a way to get hold of such > an object? What about a non-Axiom-of-Choice way? Let x = {y in X: y is not in y}. === Subject: Re: A very vague question about sets. > If we have a general set X, is it reasonable to say Let x >> be an object not in X.? Is there a way to get hold of such >> an object? What about a non-Axiom-of-Choice way? > Let x = {y in X: y is not in y}. Assuming regularity, this always results in x = {}, independently of > whether {} happens to be a member of X or not. No. Assuming regularity, it always results in x = X. In any case, assuming *no* axioms other than separation (which provides for the existence of x), x is not a member of X. === Subject: Re: A very vague question about sets. <25167425.1170341466148.JavaMail.jakarta@nitrogen.mathforum.org> G.Frege has a good answer. If you need an object a not in X, use a = X. (E.g., the set of complex numbers is not a complex number) === Subject: Re: JSH: Duck comes back from dead, again > http://news.bbc.co.uk/1/hi/world/americas/6309159.stm?ls Duck comes back from dead, again ... If You would like to help cover the Duck's medical costs, You can purchase a t-shirt here : http://cart.garnetandgold.com/garnet3/detail.cfm?auto=5358 === Subject: Axioms for classical propositional logic Tarski's axioms for plane geometry: http://en.wikipedia.org/wiki/Tarski%27s axioms I think what I like best about this system is that the role of each axiom is so clear. Especially the later ones: The Lower Dimension axiom creates a plane. The Upper Dimension axiom constrains the theory to the plane. Euclid's parallel postulate makes the plane flat. The axiom schema of Continuity makes it dense. The Segment Construction axiom makes it infinite. ¬©ukasiewicz's axioms for classical propositional logic are amazing, too, but it isn't as clear to me what each one does. Is there a nice paper on this? For me specifically it might be more enlightening, if less elegant, to have *more* axioms, each of which does less work and thus has a more specific role. Does that make any sense? Do such axiom-sets exist? -j === Subject: Re: Axioms for classical propositional logic On 2 Feb 2007 13:46:53 -0800, Jason Orendorff Lukasiewicz's axioms for classical propositional logic are amazing, > too, but it isn't as clear to me what each one does. [...] For me specifically it might be more enlightening, if less elegant, to > have *more* axioms, each of which does less work and thus has a more > specific role. Does that make any sense? > Sure. Actually, very much! The choice of the initial formulas can be made in quite different ways. One has taken great pains, in particular, to get by with the smallest possible number of axioms, and in this respect the limit of what is possible has indeed been reached. The purpose of logical investigations is better served, however, when we separate, as in the axiomatics for geometry, various groups of axioms from one another, such that each group gives expression to the role of one logical operation. The following list then emerges: I Axioms of implication II a) Axioms for & II b) Axioms for v III Axioms of negation. This system of axioms generates through application of the rules all valid formulas of propositional logic. Paul Bernays, Problems of Theoretical Logic (1927) Do such axiom-sets exist? > Yes, see quote above. A more recent one is the one by Kleene (1952): A -> (B -> A). (A -> B) -> ((A -> (B -> C)) -> (A -> C)). A -> (B -> A & B). A & B -> A. A & B -> B. A -> A v B. B -> A v B. (A -> C) -> ((B -> C) -> (A v B -> C)). (A -> B) -> ((A -> ~B) -> ~A). ~~A -> A. You see, it also follows the pattern mentioned by Bernays. F. -- E-mail: infosimple-linede === Subject: Re: Axioms for classical propositional logic > Tarski's axioms for plane geometry: > http://en.wikipedia.org/wiki/Tarski%27s axioms I think what I like best about this system is that the role of each > axiom is so clear. Especially the later ones: The Lower Dimension > axiom creates a plane. The Upper Dimension axiom constrains the > theory to the plane. Euclid's parallel postulate makes the plane > flat. The axiom schema of Continuity makes it dense. The Segment > Construction axiom makes it infinite. ¬©ukasiewicz's axioms for classical propositional logic are amazing, > too, but it isn't as clear to me what each one does. Is there a > nice paper on this? That's probably because, while geometry axioms define something you can put your hand on, axioms for propositional logic are extremely tautologic since the whole subject is on tautologies. You can build your axioms in a redundant way such that they describe the obvious properties of and,or,not etc. one by one, e.g. (p&q) -> p (p&q) <-> (q&p) p -> pvq pvq <-> qvp ~~p <-> p etc. not to forget corresponding rules of inference. === Subject: Re: Foundation. <200701151820.l0FIKYcw093728@walkabout.empros.comIf x = {{x}}, then we have {x}ex and xe{x}. So Exy(yex & xey). I'm not a set theorist, nor do I play one on TV. But, I always thought > that the way ZFC was structured, you had to prove that a given item > existed and was a set before you could say anything else about it. ZFC sits on top of first-order logic, which allows reasoning like MoeBlee's. The original question was fair; the statement ~ E x . x = {{x}} is provable in ZFC. Or so I can only assume. -j === Subject: Re: Foundation. > ZFC has the axiom of foundation, which says that every set x has an > epsilon-minimal element y (i.e. the intersection of y and x is empty). and what is that epsilon-minimal element y. What do you mean what is it?? For a given x, there exists at least > one y such that yex & y/x=0. There is no requirement as to what such > a y is other than what is entailed by yex & y/x=0. if there are no infinite descending membership chains, and there are > no ur-elements, then y is the empty set. Am I right? It is not entailed that y is the empty set. It is utterly trivial to > give examples of x an y such that yex & x/y=0 & ~y=0. MoeBlee Again you didn't understand what I have said. And you have the right this time. I will try to clarify what I meant. let me define in an informal manner what I call Lineage. A lineage is a membership descent chain of a set. example lets take the set x= { {},{{}}} Here we have two lineages of x. Lineage 1:) x > { } . Lineage 2:) x > {{}} > {}. Lets take the set y= { {},{{}},{ {},{{}} } } Lineage 1:) y > {} Lineage 2:) y > {{}} > {} Lineage 3:) y > { {}, {{}} } > {} Lineage 4:) y > { {}, {{}} } > {{}} > {} Since there is no infinitelly descending membership chaim (i.e lineage). then each lineage should end by a set. And obviouselly since we don't have ur-elements in ZFC. then the ending set at each lineage from any set in ZFC should be { }. So every set in ZFC can be broken down by lineages finally into empty sets. However each set has a different heiarchy. for the example above we can say that all four lineages of y defines the lineage heirachy or simply Lineage Tree of y. So in that way sets differ by their lineage tree(heiarchy of membership). But the end result per each lineage is the same. It is the empty set Am I right or wrong? Zuhair === Subject: Re: Foundation. > let me define in an informal manner what I call > Lineage. A lineage is a membership descent chain of a set. example lets take the set x= { {},{{}}} Here we have two lineages of x. Lineage 1:) x > { } . > Lineage 2:) x > {{}} > {}. membership descent chain is not a definition. Don't just give examples. Define it. > So every set in ZFC can be broken down by lineages finally into empty > sets. Even after you give a mathematical definition of 'broken down by lineages', you'll not be able to prove the above assertion (as long as I understand what it is you're trying to say). ZFC does not have an axiom that ensures that all sets are built from the empty set. Even though we might not be able to show an example of a set that is not built from the empty set, we still have not a proof that such a set is precluded. After you've learned some basic logic and set theory, then you should look up 'the constructible universe', which is related to the kinds of things you're wondering about. ZFC does not itself have an axiom that all sets are in the constructible universe. MoeBlee === Subject: Re: Foundation. > Even after you give a mathematical definition of 'broken down by > lineages', you'll not be able to prove the above assertion (as long as > I understand what it is you're trying to say). ZFC does not have an > axiom that ensures that all sets are built from the empty set. It's pretty obvious what zuhair has in mind: let's say a lineage of a set A is a maximal subset of the transitive closure of A totally ordered by the membership relation. It is then a theorem of ZFC that every lineage starts with the empty set. In this sense the axioms of ZFC do ensure that all sets are built from the empty set. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Foundation. Even after you give a mathematical definition of 'broken down by > lineages', you'll not be able to prove the above assertion (as long as > I understand what it is you're trying to say). ZFC does not have an > axiom that ensures that all sets are built from the empty set. It's pretty obvious what zuhair has in mind: let's say a lineage of a > set A is a maximal subset of the transitive closure of A totally > ordered by the membership relation. It is then a theorem of ZFC that > every lineage starts with the empty set. In this sense the axioms of > ZFC do ensure that all sets are built from the empty set. So I am right! Zuhair -- > Aatu Koskensilta (aatu.koskensi...@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Is continuum completely filled up? Which is the first, actual numbers, or representations of numbers? Seems like you just admitted that a representation of a number is not the > same thing as an actual number. > I used the word actual number in deference to your assertion. The concept is our function used to perceive real world. It is based on our experience. So that they have their correspondent in real world like numbers, but I can't find correspondent of infinite set. As for > computable numbers we have its representations as a result of geometrical or > algebraic operations. They are represented individually in concrete way too. I have never claimed that numbers could not be represented. > Its representation is regarded as itself, however you might claim that they are different and numbers actually exist. I have claim about uncomputable numbers having representations. > Ozaki Tosiaki === Subject: Re: Is continuum completely filled up? >> But not the empty set? How can we have a physical example of a nonempty >> set unless there is a physical empty set? > The empty set is the same thing as an empty sack.. We communicate abstract > concept with concrete symbol. Wrong. There are lots of empty sacks, but there is only one empty set. > Proof is by the axiom of extensionality. > This axiom says (A),(B) : A=B ((C) : CeA <--> CeB) That is, the empty sets you use many times are the same things and empty sacks are the empty set too. We regard a ball is the same, even though times are different. There are two different ways to represent the set. But these are the same. 1. By explicitely lists its elements. 2. By staing property of its elements. > But note that there can only be one empty set for the entire universe. >> Good luck in finding it. > What makes it possible is only your belief. What makes it possible is the axioms of set theory. > The axioms of set theory is your belief. I don't oppose so much ZFC, but its interpletation may be sometimes different from you. > Ozaki Toshiaki === Subject: Re: Is continuum completely filled up? > But not the empty set? How can we have a physical example of a > nonempty > set unless there is a physical empty set? >> The empty set is the same thing as an empty sack.. We communicate > abstract >> concept with concrete symbol. >> Wrong. There are lots of empty sacks, but there is only one empty set. >> Proof is by the axiom of extensionality. > This axiom says > (A),(B) : A=B ((C) : CeA <--> CeB) > That is, the empty sets you use many times are the same things and empty > sacks are the empty set too. We regard a ball is the same, even though times > are different. I have a bunch of empty sacks in the storeroom. Are you saying that they are really all the same sack? You continue to confuse abstract objects with mere representations of those objects. Representations can be physical objects, but sets are not. Since all you ever do is assert without evidence, I am tired of playing this game. Bye. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Is continuum completely filled up? I have a bunch of empty sacks in the storeroom. Are you saying that they > are really all the same sack? > We can deal with them as the same thing. The same thing has various use or classification according to our idea. This idea may be represented in our mind or as symbol in our sentences. But concept itself doesn't exist. What actually exist, is only real world including our self. Concepts and relationships between them are changes of our brains. Common concept is a someting which we share incommon in ourselves. Without symbol {,} ZFC might be able to do nothing or has not existed without symbols. Infinite set has its symbol, but doesn't have its real brepresentation. How can I confuse them? You continue to confuse abstract objects with mere representations of > those objects. Representations can be physical objects, but sets are > not. Since all you ever do is assert without evidence, I am tired of playing > this game. Bye. > Ozaki Toshiaki === Subject: Re: Is continuum completely filled up? <4EVN+x3M+0rFFwxx@phoenixsystems.demon.co.uk> <1TM+rzi1j2sFFwJc@phoenixsystems.demon.co.uk> Do you really believe the existence of N, or is it only a matter of > difinition? > Do you really believe the existence of 3, or is it only a matter of > definition? >> I believe the existence of 3 as three pieces of pebble, line as drawed >> line, >> and infinite space. But as for the latter two cases, even though they >> are >> consisted of infinitely many points, we can not know all of them. If 0 >> line >> up infinitely, then there is a 1 acompanying infinite 0 too. >> I wasn't asking about the existence of three objects. I was asking >> about >> the existence of an abstract concept, the natural number 3. My point is >> that this number does not have physical existence at all, and therefore >> its existence is just as much a matter of belief as is an infinite >> cardinal such as aleph_0. > That's really a poor view. What do you see here in the brackets:( # # > # )? You see objects of the form #? Maybe you see three of them? You > are unable to comprehend that you only can see three if you see three? > Let's make it even simpler. Do you believe that the empty set has > physical existence? I don't. It's an abstract object. The empty set is > not the same thing as an empty sack. Your point of view was, that all things which have no physical > existence are a matter of believe. That's wrong. > If you see a red colour there is red (light with 400 nm wavelength). > If you see three things, there is three (multiples of same types of > the amount three). Three is. Three is the equivalent class of all > three objects as red is the equivalence class of all red objects. > What do you think about this view? > You say, the existence of abstract objects is a matter of believe. Do > you really think like this? Is the existence of the univers a matter > of believe? Is the existence of the colour red a matter of believe? Is > the existence of humanity a matter of believe? > The existence of humanity can be verified in the physical world. The > existence of a natural number cannot. We can only see instances of the > number, not the number itself. If there are no numbers, all is one. You can't distinuish anything > from anything other. The concept of multiples exists. Since we exist > (cogito ergo sum). > Do you think trees exists? But tree is just an abstract concept. > There is no object trees. There is only one (concret) tree here, one > tree there, another one on the left, another on the right, .... But > there is nowhere a trees. We can verify the existence of natural numbers since if they are not, > we are not. Maybe you are not but the rest of us are. If there are no numbers in concrete existence, you are me and we all are one. Since if there are no numbers there is only one or nothing (but consider: cogito ergo sum). There is a qualitative difference between the sort of existence enjoyed > by physical objects and that enjoyed by mental images like numbers. You see different sorts of existences? I do so too. A natural number has very much more existence than e.g. aleph_0.# Albrecht S. Storz === Subject: Re: Is continuum completely filled up? > There is a qualitative difference between the sort of existence enjoyed > by physical objects and that enjoyed by mental images like numbers. You see different sorts of existences? I do so too. A natural number > has very much more existence than e.g. aleph_0.# A natural number has a very different sort of existence from the rock onto which your mother undoubtedly dropped you head first as a child. === Subject: Re: Is continuum completely filled up? > There is a qualitative difference between the sort of existence enjoyed > by physical objects and that enjoyed by mental images like numbers. You see different sorts of existences? I do so too. A natural number > has very much more existence than e.g. aleph_0.# A natural number has a very different sort of existence from the rock > onto which your mother undoubtedly dropped you head first as a child. Good one. -- === Subject: Re: Is continuum completely filled up? Do you think trees exists? But tree is just an abstract concept. > There is no object trees. There is only one (concret) tree here, one > tree there, another one on the left, another on the right, .... But > there is nowhere a trees. We can see and touch actual trees. We cannot see and touch actual > numbers, but only representations of numbers. > Which is the first, actual numbers, or representations of numbers? As for computable numbers we have its representations as a result of geometrical or algebraic operations. They are represented individually in concrete way too. > Ozaki Tosiaki === Subject: Re: Is continuum completely filled up? >> Do you think trees exists? But tree is just an abstract concept. >> There is no object trees. There is only one (concret) tree here, one >> tree there, another one on the left, another on the right, .... But >> there is nowhere a trees. >> We can see and touch actual trees. We cannot see and touch actual >> numbers, but only representations of numbers. > Which is the first, actual numbers, or representations of numbers? Seems like you just admitted that a representation of a number is not the same thing as an actual number. > As for > computable numbers we have its representations as a result of geometrical or > algebraic operations. They are represented individually in concrete way too. I have never claimed that numbers could not be represented. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Is continuum completely filled up? : > And instance of nonempty set can be verified in the physical world. No, you do not find _sets_ in the physical world, only their elements > (if they are non-sets). > Mathematician represents sets by the symbols like {},{a},{a,b}... As in the case of numbers, how do we share our concept without physical objects or symbols? Ozaki Tosiaki === Subject: Re: Is continuum completely filled up? > And instance of nonempty set can be verified in the physical world. But not the empty set? How can we have a physical example of a nonempty > set unless there is a physical empty set? > The empty set is the same thing as an empty sack.. We communicate abstract concept with concrete symbol. But note that there can only be one empty set for the entire universe. > Good luck in finding it. > What makes it possible is only your belief. Good luck for you too. Ozaki Toshiaki === Subject: Re: Is continuum completely filled up? > And instance of nonempty set can be verified in the physical world. But not the empty set? How can we have a physical example of a nonempty > set unless there is a physical empty set? > The empty set is the same thing as an empty sack.. We communicate abstract concept with concrete symbol. But note that there can only be one empty set for the entire universe. > Good luck in finding it. > What makes it possible is only your belief. Good luck for you too. Ozaki Toshiaki === Subject: Re: Is continuum completely filled up? >> And instance of nonempty set can be verified in the physical world. >> But not the empty set? How can we have a physical example of a nonempty >> set unless there is a physical empty set? > The empty set is the same thing as an empty sack.. We communicate abstract > concept with concrete symbol. Wrong. There are lots of empty sacks, but there is only one empty set. Proof is by the axiom of extensionality. >> But note that there can only be one empty set for the entire universe. >> Good luck in finding it. > What makes it possible is only your belief. What makes it possible is the axioms of set theory. > Good luck for you too. > Ozaki Toshiaki -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Is continuum completely filled up? <4EVN+x3M+0rFFwxx@phoenixsystems.demon.co.uk> <1TM+rzi1j2sFFwJc@phoenixsystems.demon.co.uk> > Do you really believe the existence of N, or is it only a matter of >> difinition? >> Do you really believe the existence of 3, or is it only a matter of >> definition? > I believe the existence of 3 as three pieces of pebble, line as drawed line, > and infinite space. But as for the latter two cases, even though they are > consisted of infinitely many points, we can not know all of them. If 0 line > up infinitely, then there is a 1 acompanying infinite 0 too. > I wasn't asking about the existence of three objects. I was asking about > the existence of an abstract concept, the natural number 3. My point is > that this number does not have physical existence at all, and therefore > its existence is just as much a matter of belief as is an infinite > cardinal such as aleph_0. >> That's really a poor view. What do you see here in the brackets:( # # >> # )? You see objects of the form #? Maybe you see three of them? You >> are unable to comprehend that you only can see three if you see three? >> Let's make it even simpler. Do you believe that the empty set has >> physical existence? I don't. It's an abstract object. The empty set is >> not the same thing as an empty sack. > Your point of view was, that all things which have no physical > existence are a matter of believe. That's wrong. No, I don't mean to say that. > If you see a red colour there is red (light with 400 nm wavelength). >> If you see three things, there is three (multiples of same types of >> the amount three). Three is. Three is the equivalent class of all >> three objects as red is the equivalence class of all red objects. >> What do you think about this view? >> You say, the existence of abstract objects is a matter of believe. Do >> you really think like this? Is the existence of the univers a matter >> of believe? Is the existence of the colour red a matter of believe? Is >> the existence of humanity a matter of believe? >> The existence of humanity can be verified in the physical world. The >> existence of a natural number cannot. We can only see instances of the >> number, not the number itself. > If there are no numbers, all is one. You can't distinuish anything > from anything other. The concept of multiples exists. Since we exist > (cogito ergo sum). I didn't say there are no numbers. I said numbers are abstract concepts. > Is it your position that abstract concepts do not exist? Do you think trees exists? But tree is just an abstract concept. > There is no object trees. There is only one (concret) tree here, one > tree there, another one on the left, another on the right, .... But > there is nowhere a trees. We can see and touch actual trees. We cannot see and touch actual > numbers, but only representations of numbers. > Look at this: # # # # # # # # # Is this just a representation of an actual number (number 9) - or is this a number? I say: It is a number and a representation of many numbers (e.g. of 9 or of 512)! Is a tree a tree or only a representation of a tree (and our sloppy way of talking is just ignoring this aspect)? Albrecht S. Storz === Subject: Re: Is continuum completely filled up? Which is the first, actual numbers, or representations of numbers? Seems like you just admitted that a representation of a number is not the > same thing as an actual number. > I used the word actual number in deference to your assertion. The concept is our function used to perceive real world. It is based on our experience. So that they have their correspondent in real world like numbers, but I can't find correspondent of infinite set. As for > computable numbers we have its representations as a result of geometrical or > algebraic operations. They are represented individually in concrete way too. I have never claimed that numbers could not be represented. > Its representation is regarded as itself, however you might claim that they are different and numbers actually exist. I have claim about uncomputable numbers having representations. > Ozaki Tosiaki === Subject: Re: Is continuum completely filled up? physical existence ... ... i.e. we can't _see_ them. (At least not with our eyes. :-) F. All what you can't see (with your eyes) doesn't exist? (e.g. air doesn't exist, space doesn't exist, ...?) Albrecht S. Storz === Subject: Re: Is continuum completely filled up? > As I have already explained, sets are abstract objects. They do not have > physical existence ... ... i.e. we can't _see_ them. (At least not with our eyes. :-) F. All what you can't see (with your eyes) doesn't exist? (e.g. air > doesn't exist, space doesn't exist, ...?) Albrecht S. Storz Then numbers cannot exist, since, while we can see numerals, we cannot see numbers. === Subject: Re: Is continuum completely filled up? <1TM+rzi1j2sFFwJc@phoenixsystems.demon.co.uk> > Do you really believe the existence of N, or is it only a matter of > difinition? > Do you really believe the existence of 3, or is it only a matter of > definition? >> I believe the existence of 3 as three pieces of pebble, line as drawed line, >> and infinite space. But as for the latter two cases, even though they are >> consisted of infinitely many points, we can not know all of them. If 0 line >> up infinitely, then there is a 1 acompanying infinite 0 too. >> I wasn't asking about the existence of three objects. I was asking about >> the existence of an abstract concept, the natural number 3. My point is >> that this number does not have physical existence at all, and therefore >> its existence is just as much a matter of belief as is an infinite >> cardinal such as aleph_0. > That's really a poor view. What do you see here in the brackets:( # # > # )? You see objects of the form #? Maybe you see three of them? You > are unable to comprehend that you only can see three if you see three? > Let's make it even simpler. Do you believe that the empty set has > physical existence? I don't. It's an abstract object. The empty set is > not the same thing as an empty sack. >> Your point of view was, that all things which have no physical >> existence are a matter of believe. That's wrong. > No, I don't mean to say that. >> If you see a red colour there is red (light with 400 nm wavelength). > If you see three things, there is three (multiples of same types of > the amount three). Three is. Three is the equivalent class of all > three objects as red is the equivalence class of all red objects. > What do you think about this view? > You say, the existence of abstract objects is a matter of believe. Do > you really think like this? Is the existence of the univers a matter > of believe? Is the existence of the colour red a matter of believe? Is > the existence of humanity a matter of believe? > The existence of humanity can be verified in the physical world. The > existence of a natural number cannot. We can only see instances of the > number, not the number itself. >> If there are no numbers, all is one. You can't distinuish anything >> from anything other. The concept of multiples exists. Since we exist >> (cogito ergo sum). > I didn't say there are no numbers. I said numbers are abstract concepts. >> Is it your position that abstract concepts do not exist? > Do you think trees exists? But tree is just an abstract concept. >> There is no object trees. There is only one (concret) tree here, one >> tree there, another one on the left, another on the right, .... But >> there is nowhere a trees. > We can see and touch actual trees. We cannot see and touch actual >> numbers, but only representations of numbers. Look at this: > # # # # # # # # # > Is this just a representation of an actual number (number 9) - or is > this a number? I say: It is a number and a representation of many > numbers (e.g. of 9 or of 512)! It is a representation of a number. The actual number 9 is a set, defined > recursively as follows: > [...] This is just one interpretation but not the whole truth. Nine is a set (e.g. in ZFC) or nine is nine (a ninefold multiple of an unity) . Albrecht S. Storz === Subject: Re: Is continuum completely filled up? > Look at this: > # # # # # # # # # > Is this just a representation of an actual number (number 9) - or is > this a number? I say: It is a number and a representation of many > numbers (e.g. of 9 or of 512)! It is a representation of a number. The actual number 9 is a set, defined > recursively as follows: > [...] This is just one interpretation but not the whole truth. More of it than Albrecht has hold of. === Subject: Re: Is continuum completely filled up? <1TM+rzi1j2sFFwJc@phoenixsystems.demon.co.uk> Look at this: > # # # # # # # # # > Is this just a representation of an actual number (number 9) - or is > this a number? I say: It is a number and a representation of many > numbers (e.g. of 9 or of 512)! > It is a representation of a number. The actual number 9 is a set, defined > recursively as follows: [...] This is just one interpretation but not the whole truth. More of it than Albrecht has hold of. I don't think that I am aware of the whole truth. But you too, I'm basically sure. Surely 9 is a set, but not only a set. It is much more. There is nothing more to say. Albrecht S. Storz === Subject: Re: Is continuum completely filled up? > > Look at this: > # # # # # # # # # > Is this just a representation of an actual number (number 9) - or is > this a number? I say: It is a number and a representation of many > numbers (e.g. of 9 or of 512)! > It is a representation of a number. The actual number 9 is a set, > defined > recursively as follows: > > [...] > This is just one interpretation but not the whole truth. More of it than Albrecht has hold of. I don't think that I am aware of the whole truth. But you too, I'm > basically sure. > Surely 9 is a set, but not only a set. It is much more. Depends on whose definition of 9 one is looking at. There is nothing more to say. But Storz will continue to say things anyway. Albrecht S. Storz === Subject: Re: Is continuum completely filled up? Look at this: > # # # # # # # # # > Is this just a representation of an actual number (number 9) - or is > this a number? I say: It is a number and a representation of many > numbers (e.g. of 9 or of 512)! > It is a representation of a number. The actual number 9 is a set, > defined > recursively as follows: > [...] > This is just one interpretation but not the whole truth. > More of it than Albrecht has hold of. I don't think that I am aware of the whole truth. But you too, I'm > basically sure. > Surely 9 is a set, but not only a set. It is much more. Depends on whose definition of 9 one is looking at. Contrary to Dave Seaman I don't think that there is only one meaning of the number 9 (as he declared it as a set). My only point was to clear this up. There is nothing more to say. But Storz will continue to say things anyway. Be sure. Why not? Albrecht S. Storz === Subject: Re: Is continuum completely filled up? <1TM+rzi1j2sFFwJc@phoenixsystems.demon.co.uk> <8$dYFAe$UiwFFwKP@phoenixsystems.demon.co.uk> As I have already explained, sets are abstract objects. They do not have >>physical existence. Is a tree a tree or only a representation of a tree (and our sloppy > way of talking is just ignoring this aspect)? >A tree is a tree. It's a physical object. On a tangent, are you sure about that? You have a perception of a tree > through senses of touch and vision, but actually the only things that > you have to work with are mental abstract objects such as tree cat > etc. and in that sense there is no difference between the abstract > concept of number 9 and a tree. Confusing perception with reality is > to mistake a map for the territory. If I bump into a tree, I have no doubt that something solid is there. I > can't bump into a number. If you bump nine times into a tree you would have no doubt that the nineth time was much harder than the first time. If you bump nine times in something hard you would not know (without any other information about the hard object) if it was a hard number or something else. Albrecht S. Storz === Subject: Re: Is continuum completely filled up? Look at this: # # # # # # # # # [...] is this a number? No. It's a list of symbols. There are two possibilities: either it is a list of symbols as it is a number - since both opinions are just interpretations - or it is more a number than anything else, cause as a number it needs the least numbers of assumptions to interpret it (as number 9). Gedankenexperiment: If you and a extraterrestrian alien would look on the sketch. Which interpretation would be the one on which you both could agree the easiest way? > Of course it's possible to count the > occurrences of '#' in this list: there are 9 occurrences of '#' in the > list. But certainly this list _is not_ the number 9 (nor any other > number). Hint: We _can't_ look at numbers. No one has ever _seen_ one. I say: It is a number [...]! Good for you. F. Albrecht S. Storz === Subject: Re: Is continuum completely filled up? Mail-To-News-Contact: abuse@dizum.com >> Look at this: >> # # # # # # # # # >> [...] is this a number? >> No. It's a list of symbols. There are two possibilities: either it is a list of symbols as it is a >number - since both opinions are just interpretations - Or it is: text decoration to mark a scene change in a book (such as seen in _Alice Through the Looking-Glass_) a row in soduku a hopscotch layout space for nine games of tic-tac-toe a plan for the layout of shrubberies in a formal garden -- Michael F. Stemper #include If it's tourist season, where do I get my license? === Subject: Re: Is continuum completely filled up? <200702021820.l12IKJ7E120554@walkabout.empros.com >> Look at this: >> # # # # # # # # # >> [...] is this a number? >> No. It's a list of symbols. There are two possibilities: either it is a list of symbols as it is a >number - since both opinions are just interpretations - Or it is: > text decoration to mark a scene change in a book (such as seen > in _Alice Through the Looking-Glass_) a row in soduku a hopscotch layout space for nine games of tic-tac-toe a plan for the layout of shrubberies in a formal garden > ... Yes. just interpretations. Albrecht S. Storz === Subject: Re: Is continuum completely filled up? If there are no numbers, all is one. You can't distinuish anything > from anything other. The concept of multiples exists. Since we exist > (cogito ergo sum). > Do you think trees exists? But tree is just an abstract concept. > There is no object trees. There is only one (concret) tree here, one > tree there, another one on the left, another on the right, .... But > there is nowhere a trees. When I was a kid I climbed up an abstract concept and fell off. Fortunately I did not break any bones. I only bruised my ribs. Some abstract concept that. As to numbers, they surely do exist: up in our heads. If every sentient being in the Cosmos died off there would be no numbers. We can verify the existence of natural numbers since if they are not, > we are not. We think about them a lot but they are not Out There. They are In Here. Abstractions are a light show put on by our brains. Bob Kolker === Subject: L'Hopital's thm lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by L'Hopital's thm) = lim{x->inf} {(5/2) e^(-x)} = 0 However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It oscillates) What condition violates the assumptions of L'Hopital's thm? === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) This is not an indeterminant form but just an undefined limit. L'Hoptials rule applies only when we have an indeterminant form. The rule applies only to indeterminant forms of oo - oo, 0/0, oo*0, oo/oo, etc... We can even reduce the problem down to e^(-x)/cos(x). (this problem exhibts the same issues as your original problem) Why is this undefined? If you can't understand it and why L'Hopitals rule doesn't apply then you can't understand the more complex case. Obviously applying rules haphazardly is not a good idea. Its best to think about limits in terms of what they mean and not some artifical or abstract rules. if we say lim e^(-x)/cos(x) = L (x->oo) then we mean that there exists a N and e such that if x > N then |e^(-x)/cos(x) - L| < e In fact there is no L that does this. If you think there is then find it. If you give me an L that you claim that works all I have to do is choose x' a little larger that makes the denominator close to 0 which will cause the whole thing to be larger than e. I can always do this beause there exists a zero of cos(x) always close to x. that is, even if cos(x) != 0 then there exists a y > x such that cos(y) = 0 with y close to x. e.g., suppose you claim that L = and N = 100 and that gives e = 0.001 so you claim that |e^(-x)/cos(x)| < 0.001 for all x > 100 This is clearly not the case because I can choose y = 2*Pi*16 + Pi/2 + e^2 y ~ 102 > N but |e^(-y)/cos(y)| > e I could be more rigorous and find y based on n and e that work for any N. In this way I could always counter any claim that you have found N and L that gets within the limit L +- e for all x > N. I did it for a specific case above but it can be done for all N(well, all N larger than some number) So you were right on your second idea about the tan thing where it oscillates and causes problems. The problem is that you cannot divide by tan(x) when tan(x) is zero. This just means that you have to be more rigorous but ultimate it works out to be the same(in this case, not necessarily in all cases). By dividing by tan(x) you might get superfluous results so you have to be more careful but it does exhibit the right behavior of the original function and so its a start. Its not a proof just as much as I have not proved anything but it gets you into looking at what is going on. When you can come up with a y = f(N,L,e) and |e^(-y)/cos(y) - L| > e then you have solved the problem because now you have a full proof way of showing that the |e^(-x)/cos(x) - L| < e cannot be true for all x > N. (just take x = y). Hope this helps a little, Jon === Subject: Re: L'Hopital's thm >> lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >> = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by >> L'Hopital's thm) >> = lim{x->inf} {(5/2) e^(-x)} = 0 >> However, >> lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >> = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It >> oscillates) >This is not an indeterminant form but just an undefined limit. L'Hoptials >rule applies only when we have an indeterminant form. >The rule applies only to indeterminant forms of oo - oo, 0/0, oo*0, oo/oo, >etc... It is indeed an indeterminate form: it is 0/0. It doesn't matter that we can cancel the e^{-2x} with the e^{-x}. L'Hospital does not require that anything be canceled. >We can even reduce the problem down to e^(-x)/cos(x). (this problem exhibts >the same issues as your original problem) L'Hospital does not apply here because the denominator does not tend to 0. This is not 0/0 as is the original ratio. >Why is this undefined? If you can't understand it and why L'Hopitals rule >doesn't apply then you can't understand the more complex case. Obviously >applying rules haphazardly is not a good idea. The reason L'Hospital does not apply, is because the derivative of the denominator vanishes in all neighborhoods of the limit point (oo). L'Hospital requires that the derivative of the denominator not vanish in some deleted neighborhood of the limit point. Rob Johnson take out the trash before replying === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) >>This is not an indeterminant form but just an undefined limit. L'Hoptials >>rule applies only when we have an indeterminant form. >>The rule applies only to indeterminant forms of oo - oo, 0/0, oo*0, oo/oo, >>etc... It is indeed an indeterminate form: it is 0/0. It doesn't matter that > we can cancel the e^{-2x} with the e^{-x}. L'Hospital does not require > that anything be canceled. > it does matter a great deal. simplifying brings out the true character. In that cause you can just make up all kinds of problems by multiplying the top and bottom by functions h(x) -> 0 as x->oo such that h(x)*f(x)/(h(x)*g(x)) is indeterminant of 0/0. What good does that do? Obviously it then hides the fact that g(x)(f could be the issue too but lets assume a fairly simple problem) is the issue.. x/x*f(x)/g(x) introduces a singularity at 0 and causes extraneous solutions. is it not the case that e^(-2x)*(cosx + 2sin(x))/(e^(-x)*(cosx + sin(x)) = e^(-x)*(cosx + 2sin(x))/(cosx + sin(x))?? >>We can even reduce the problem down to e^(-x)/cos(x). (this problem >>exhibts >>the same issues as your original problem) L'Hospital does not apply here because the denominator does not tend > to 0. This is not 0/0 as is the original ratio. This is obviously false the original denominator is e^(-x)*(cosx +sin(x)). Your telling me this does not approach 0 as x->oo? 0 <= |e^(-x)*(cos(x) + sin(x))| <= 2*|e^(-x)| ????? but surely |e^(-x)| < e = 1/x and surely 1/x -> 0 as x->oo. >Why is this undefined? If you can't understand it and why L'Hopitals rule >>doesn't apply then you can't understand the more complex case. Obviously >>applying rules haphazardly is not a good idea. The reason L'Hospital does not apply, is because the derivative of the > denominator vanishes in all neighborhoods of the limit point (oo). > L'Hospital requires that the derivative of the denominator not vanish > in some deleted neighborhood of the limit point. > Ok, this might be true but its much easier to see when you simplify the original problem. In that case one can ignore L'Hopitals rule and just show directly that the limit doesn't exist. Ofcourse this simplification can result in extraneous solutions but its easier to prove directly which results in a general proof in the original problem. Obviously this is not not why the OP asked for but several have already mentioned this. Its not at all obvious why this is true though and no one has given a satisfactory reason why. J ust saying that it doesn't work because of sign variation about 0 isn't good enough. You have to show why this is true and breaks the proof in most L'Hopitals proof's. Unfortunately as it stands most proofs that I have soon are invalid as they don't mention this. Jon === Subject: Re: L'Hopital's thm >> lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >> = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by >> L'Hopital's thm) >> = lim{x->inf} {(5/2) e^(-x)} = 0 >> However, >> lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >> = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It >> oscillates) >This is not an indeterminant form but just an undefined limit. L'Hoptials >rule applies only when we have an indeterminant form. >The rule applies only to indeterminant forms of oo - oo, 0/0, oo*0, oo/oo, >etc... >> It is indeed an indeterminate form: it is 0/0. It doesn't matter that >> we can cancel the e^{-2x} with the e^{-x}. L'Hospital does not require >> that anything be canceled. > >it does matter a great deal. simplifying brings out the true character. In >that cause you can just make up all kinds of problems by multiplying the top >and bottom by functions h(x) -> 0 as x->oo such that h(x)*f(x)/(h(x)*g(x)) is indeterminant of 0/0. What good does that do? Obviously it then hides the fact that g(x)(f could >be the issue too but lets assume a fairly simple problem) is the issue.. >x/x*f(x)/g(x) introduces a singularity at 0 and causes extraneous solutions. is it not the case that e^(-2x)*(cosx + 2sin(x))/(e^(-x)*(cosx + sin(x)) = e^(-x)*(cosx + 2sin(x))/(cosx + sin(x))?? I do not advocate adding factors to the numerator and denominator, I simply said that L'Hospital does not require the removal of common factors. If it did, it would be much harder to apply in some cases. If we expand our dictionary of functions, there are common factors which can be canceled in most 0/0 indeterminate forms. The beauty of L'Hospital, is you don't need to seek out and cancel common factors, unless L'Hospital fails. If L'Hospital fails, some rearrangement of factors may be needed, or perhaps L'Hospital should be applied again. >We can even reduce the problem down to e^(-x)/cos(x). (this problem >exhibts >the same issues as your original problem) >> L'Hospital does not apply here because the denominator does not tend >> to 0. This is not 0/0 as is the original ratio. This is obviously false the original denominator is e^(-x)*(cosx +sin(x)). >Your telling me this does not approach 0 as x->oo? 0 <= |e^(-x)*(cos(x) + sin(x))| <= 2*|e^(-x)| ????? but surely |e^(-x)| < e = 1/x and surely 1/x -> 0 as x->oo. I said that L'Hospital does not apply to e^{-x}/cos(x) since the denominator, cos(x), does not tend to 0 as x -> oo. I said that the original ratio was of the form 0/0. This seems to imply that the denominator e^{-x} (cos(x) + sin(x)) tends to 0. >Why is this undefined? If you can't understand it and why L'Hopitals rule >doesn't apply then you can't understand the more complex case. Obviously >applying rules haphazardly is not a good idea. >> The reason L'Hospital does not apply, is because the derivative of the >> denominator vanishes in all neighborhoods of the limit point (oo). >> L'Hospital requires that the derivative of the denominator not vanish >> in some deleted neighborhood of the limit point. > >Ok, this might be true but its much easier to see when you simplify the >original problem. In that case one can ignore L'Hopitals rule and just show >directly that the limit doesn't exist. Ofcourse this simplification can >result in extraneous solutions but its easier to prove directly which >results in a general proof in the original problem. Obviously this is not >not why the OP asked for but several have already mentioned this. Its not at >all obvious why this is true though and no one has given a satisfactory >reason why. J >ust saying that it doesn't work because of sign variation about 0 isn't good >enough. You have to show why this is true and breaks the proof in most >L'Hopitals proof's. Unfortunately as it stands most proofs that I have soon >are invalid as they don't mention this. They probably simply divide the equation [f(x) - f(y)] g'(c) = [g(x) - g(y)] f'(c) by g'(c) and [g(x) - g(y)] without worrying about whether g'(c) might be 0. Rob Johnson take out the trash before replying === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, >> lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) >>This is not an indeterminant form but just an undefined limit. >>L'Hoptials >>rule applies only when we have an indeterminant form. >>The rule applies only to indeterminant forms of oo - oo, 0/0, oo*0, >>oo/oo, >>etc... It is indeed an indeterminate form: it is 0/0. It doesn't matter that > we can cancel the e^{-2x} with the e^{-x}. L'Hospital does not require > that anything be canceled. >it does matter a great deal. simplifying brings out the true character. >>In >>that cause you can just make up all kinds of problems by multiplying the >>top >>and bottom by functions h(x) -> 0 as x->oo such that >>h(x)*f(x)/(h(x)*g(x)) is indeterminant of 0/0. >>What good does that do? Obviously it then hides the fact that g(x)(f >>could >>be the issue too but lets assume a fairly simple problem) is the issue.. >>x/x*f(x)/g(x) introduces a singularity at 0 and causes extraneous >>solutions. >>is it not the case that e^(-2x)*(cosx + 2sin(x))/(e^(-x)*(cosx + sin(x)) = >>e^(-x)*(cosx + 2sin(x))/(cosx + sin(x))?? I do not advocate adding factors to the numerator and denominator, I > simply said that L'Hospital does not require the removal of common > factors. If it did, it would be much harder to apply in some cases. > If we expand our dictionary of functions, there are common factors > which can be canceled in most 0/0 indeterminate forms. The beauty of > L'Hospital, is you don't need to seek out and cancel common factors, > unless L'Hospital fails. If L'Hospital fails, some rearrangement of > factors may be needed, or perhaps L'Hospital should be applied again. > I NEVER SAID IT DID! All I said was that if you do remove the common factors then the answer is obvious. I never said you had to remove them. >>We can even reduce the problem down to e^(-x)/cos(x). (this problem >>exhibts >>the same issues as your original problem) L'Hospital does not apply here because the denominator does not tend > to 0. This is not 0/0 as is the original ratio. >>This is obviously false the original denominator is e^(-x)*(cosx +sin(x)). >>Your telling me this does not approach 0 as x->oo? >>0 <= |e^(-x)*(cos(x) + sin(x))| <= 2*|e^(-x)| ????? >>but surely |e^(-x)| < e = 1/x and surely 1/x -> 0 as x->oo. I said that L'Hospital does not apply to e^{-x}/cos(x) since the > denominator, cos(x), does not tend to 0 as x -> oo. > Obviously then. That example was to show the behavor of his original problem. I never said L'Hopitals rule applied to that problem. Actually I never said it applied to any problem. > I said that the original ratio was of the form 0/0. This seems to > imply that the denominator e^{-x} (cos(x) + sin(x)) tends to 0. > Ok, But I never said it didn't. All I said is that the original problem is very similar to e^(-x)/cos(x). I never said that e^(-x)/cos was of indeterminant form. >>Why is this undefined? If you can't understand it and why L'Hopitals >>rule >>doesn't apply then you can't understand the more complex case. >>Obviously >>applying rules haphazardly is not a good idea. The reason L'Hospital does not apply, is because the derivative of the > denominator vanishes in all neighborhoods of the limit point (oo). > L'Hospital requires that the derivative of the denominator not vanish > in some deleted neighborhood of the limit point. >Ok, this might be true but its much easier to see when you simplify the >>original problem. In that case one can ignore L'Hopitals rule and just >>show >>directly that the limit doesn't exist. Ofcourse this simplification can >>result in extraneous solutions but its easier to prove directly which >>results in a general proof in the original problem. Obviously this is not >>not why the OP asked for but several have already mentioned this. Its not >>at >>all obvious why this is true though and no one has given a satisfactory >>reason why. J >>ust saying that it doesn't work because of sign variation about 0 isn't >>good >>enough. You have to show why this is true and breaks the proof in most >>L'Hopitals proof's. Unfortunately as it stands most proofs that I have >>soon >>are invalid as they don't mention this. They probably simply divide the equation [f(x) - f(y)] g'(c) = [g(x) - g(y)] f'(c) by g'(c) and [g(x) - g(y)] without worrying about whether g'(c) might > be 0. > Yeah, maybe. if one takes an example such as e^(-2*x)/(e^(-x)*(cos(x) + e)^2) then L'Hopitals rule gives the wrong answer. In this case the limit is 0 but L'Hopitals rule says its undefined(atleast according to maple) when |e| > 1. When |e| <= 1 then the true limit is undefined and and both give the correct result. In this case it seems to depend not on g'(x) but on g(x). (not one the extrema but the zero's). g'(x) = 0 an infinity number of times near oo for any e but in some cases L'Hopitals rule works and in others it doesn't. What seems to be the defining moment is when e = +-1. Here we get either an infinite number of zero's near infinity or we don't. So its not g' that matters but g itself. I guess its when they divide by g(x) they are assuming its not 0 in a small enough neighborhood around the limit point. If g(x) has infinitly many zero around it then it won't work. === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) What condition violates the assumptions of L'Hopital's thm? L'hospital required g'(x)!=0 in f'(x)/g'(x) _for all_ x in the open interval that is looked at http://en.wikipedia.org/wiki/L%27hospital and g'(x)=[-2 e^(-x)(sin x)] does not not meet that requirement, since you would need g'(x)!=0 for all x in (a,oo) which not true. === Subject: Re: L'Hopital's thm >lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >= lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by >L'Hopital's thm) >= lim{x->inf} {(5/2) e^(-x)} = 0 >However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >= lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It >oscillates) What condition violates the assumptions of L'Hopital's thm? The problem is one that is not often mentioned: the rule applies only if the denominator function does not vanish infinitely often toward the limit. Here, the denominator vanishes infinitely often on the way to infinity. Rob Johnson take out the trash before replying === Subject: Re: L'Hopital's thm >lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >= lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by >L'Hopital's thm) >= lim{x->inf} {(5/2) e^(-x)} = 0 >However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >= lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It >oscillates) What condition violates the assumptions of L'Hopital's thm? The problem is one that is not often mentioned: the rule applies only > if the denominator function does not vanish infinitely often toward > the limit. Here, the denominator vanishes infinitely often on the > way to infinity. That's a good way to see the above limit doesn't exist without using LHR. However, LHR applies in all cases where lim f'(x)/g'(x) exists; there's no need to worry about anything else to make it rigorous. In the above example, g'(x) = 0 on a sequence -> oo, so f'(x)/g'(x) is not even defined in a deleted neighborhood of oo, so it can't possibly have a limit there. LHR therefore does not apply. === Subject: Re: L'Hopital's thm >lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >>= lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by >>L'Hopital's thm) >>= lim{x->inf} {(5/2) e^(-x)} = 0 >>However, >>lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >>= lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It >>oscillates) >>What condition violates the assumptions of L'Hopital's thm? >> >> The problem is one that is not often mentioned: the rule applies only >> if the denominator function does not vanish infinitely often toward >> the limit. Here, the denominator vanishes infinitely often on the >> way to infinity. That's a good way to see the above limit doesn't exist without >using LHR. However, LHR applies in all cases where lim >f'(x)/g'(x) exists; there's no need to worry about anything else >to make it rigorous. In the above example, g'(x) = 0 on a >sequence -> oo, so f'(x)/g'(x) is not even defined in a deleted >neighborhood of oo, so it can't possibly have a limit there. LHR >therefore does not apply. In this problem, the confusing point (in my opinion) is that the sin(x) in f'(x) and g'(x) cancel and leave a ratio that does have a limit as x -> oo. However, in the proof of LHR, we arrive at the result [f(x) - f(y)] g'(c) = [g(x) - g(y)] f'(c) [1] for SOME c between x and y. To get the final result, we have to divide by g'(c) and g(x) - g(y), so we must be assured that neither is zero. If g(x) - g(y) is not identically 0 near y, we can choose x where g(x) - g(y) does not vanish and get a useful result, even if g(x) - g(y) does vanish in every deleted neighborhood of y. However, if g'(c) vanishes in every deleted neighborhood of y, [1] tells us nothing since we cannot choose the c at which [1] is satisfied. That is, the c at which [1] is satisfied, may be a c at which g'(c) = 0. This is why we need to know that g'(c) does not vanish anywhere in some deleted neighborhood of y. Rob Johnson take out the trash before replying === Subject: Re: L'Hopital's thm lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) What condition violates the assumptions of L'Hopital's thm? >> The problem is one that is not often mentioned: the rule applies only >> if the denominator function does not vanish infinitely often toward >> the limit. Here, the denominator vanishes infinitely often on the >> way to infinity. That's a good way to see the above limit doesn't exist without > using LHR. However, LHR applies in all cases where lim > f'(x)/g'(x) exists; there's no need to worry about anything else > to make it rigorous. In the above example, g'(x) = 0 on a > sequence -> oo, so f'(x)/g'(x) is not even defined in a deleted > neighborhood of oo, so it can't possibly have a limit there. LHR > therefore does not apply. LHR = ? === Subject: Re: L'Hopital's thm LHR = ? Lhospital's rule -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: L'Hopital's thm >> LHR = ? Lhospital's rule > haha - is it really just that? :) I thought he was referring to some other theorem. Is that a common abbreviation/acronym? === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) What condition violates the assumptions of L'Hopital's thm? I really don't know... L'Hopital's rule is sometimes stated without adequately specifying conditions, as at . Nonetheless, look at the sentence below the first graph there. Unfortunately, that sentence is not quite accurate, but it still indicates why the rule does not apply in your problem. implicitly understood, humbug! David === Subject: Re: L'Hopital's thm >> lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >> = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by >> L'Hopital's thm) >> = lim{x->inf} {(5/2) e^(-x)} = 0 >> However, >> lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} >> = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It >> oscillates) >> What condition violates the assumptions of L'Hopital's thm? >> I really don't know... L'Hopital's rule is sometimes stated without adequately specifying > conditions, as at . Nonetheless, look at the sentence below the first graph there. > Unfortunately, that sentence is not quite accurate, but it still indicates > why the rule does not apply in your problem. implicitly understood, > humbug! David Actually the mathworld entry seems indeed a bit unfortunate, wikipedia springer, planetmath have an explicit and precise formulation: http://planetmath.org/encyclopedia/ProofOfDeLHopitalsRule.html http://eom.springer.de/L/l058340.htm http://en.wikipedia.org/wiki/L%27hospital I found that on many math topics wikipedia and planetmath actually surpass mathworld (being the oldest of those 4) by now. Springer has a solid/high quality. In practice it is probably a good idea to use all 4. === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) What condition violates the assumptions of L'Hopital's thm? I really don't know... L'Hopital's rule is sometimes stated without adequately specifying > conditions, as at . Nonetheless, look at the sentence below the first graph there. > Unfortunately, that sentence is not quite accurate, but it still indicates > why the rule does not apply in your problem. implicitly understood, > humbug! The condition is that f'(x)/g'(x) has a limit as x -> a (a in R or a infinite). That implies this quotient is well defined in a deleted neighborhood of a. That implies the denominator is nonzero in that neighborhood; that is certainly implicitly understood. And that's all there is; the sign change business described at Wolfram is a not just a red herring, it's false: in that example, g'(x) = 0 infinitely many times, so you can't even get LHR off the ground. === Subject: Re: L'Hopital's thm <20070202190607.716$2i@newsreader.com lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) What condition violates the assumptions of L'Hopital's thm? I really don't know... L'Hopital's rule is sometimes stated without adequately specifying > conditions, as at . Nonetheless, look at the sentence below the first graph there. > Unfortunately, that sentence is not quite accurate, but it still indicates > why the rule does not apply in your problem. implicitly understood, > humbug! David Thnak you, all~ === Subject: Re: L'Hopital's thm <20070202190607.716$2i@newsreader.com > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) > What condition violates the assumptions of L'Hopital's thm? > I really don't know... L'Hopital's rule is sometimes stated without adequately specifying > conditions, as at . Nonetheless, look at the sentence below the first graph there. > Unfortunately, that sentence is not quite accurate, but it still indicates > why the rule does not apply in your problem. implicitly understood, > humbug! David > Thnak you, all~- [Micro] .9eÀå .81¯.bd .bc®¨ .bc .9e± .89± .89 - - [Micro] .9eÀå .81¯.bd .bc®¨ .bcü± .89 - Excuse me,,, - -; Could I read something about why the infinite changes of sign make problem ?? Please link it. And something more. Exactly what makes problem ? The infinite zeros or the infinite changes of sign??(of g'(x)) (The g'(x)>=0 that has infinite zeros can be thought) === Subject: Re: L'Hopital's thm > Excuse me,,, -_-; Could I read something about why the infinite changes of sign make > problem ?? Please link it. And something more. Exactly what makes problem ? The infinite zeros or the infinite > changes of sign??(of g'(x)) > (The g'(x)>=0 that has infinite zeros can be thought) > Well the problem is g'(x)=0 (which is a requirement in the proof of the theorem). But since we are are looking at differentiable and therefore continuous functions an infinite sign change implies infintely many g(x)=0 cases, which rules out the use of l'hospital. For detailed understanding it might be best to review the proof itsself. Planetmath and Wikipedia both have a proof of the theorem as well or alternatively you'll find it correctly stated and proven in any good calculus textbook. http://planetmath.org/encyclopedia/ProofOfDeLHopitalsRule.html http://en.wikipedia.org/wiki/L%27hospital === Subject: Re: L'Hopital's thm On 2 Feb 2007 16:27:59 -0800, Kenshin L'Hopital's rule is sometimes stated without adequately specifying >> conditions, as at . >> Nonetheless, look at the sentence below the first graph there. >> Unfortunately, that sentence is not quite accurate, but it still indicates >> why the rule does not apply in your problem. implicitly understood, >> humbug! >> David >> Thnak you, all~- [Micro].9eË .8c .81øü.bc¨¬ .b9.9e±.89±.89 - >> - [Micro].9eË .8c .81øü.bc¨¬ .bc .9f±.89 - Excuse me,,, - -; Could I read something about why the infinite changes of sign make >problem ?? Please link it. And something more. Exactly what makes problem ? The infinite zeros or the infinite >changes of sign??(of g'(x)) >(The g'(x)>=0 that has infinite zeros can be thought) At the MathWorld link that David provided, one of the references listed is: Boas, R. P. Counterexamples to L'Hopital's Rule. Amer. Math. Monthly 93, 644-645, 1986 questions: === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) I see you got this expression in terms of tan by multiplying top and bottom by cos(x). This is valid only if cos(x) is never zero over the domain in which you are taking the limit. Ooops! Leave it in terms of sin and cos and there's no problem with the expression at all. What condition violates the assumptions of L'Hopital's thm? None, you simply did an invalid manipulation. > === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) I see you got this expression in terms of tan by multiplying > top and bottom by cos(x). This is valid only if cos(x) is > never zero over the domain in which you are taking the > limit. Ooops! Leave it in terms of sin and cos and there's no problem > with the expression at all. What condition violates the assumptions of L'Hopital's thm? None, you simply did an invalid manipulation. > - [Micro] .9eÀå .81¯.bd .bc®¨ .bcü± .89 - So, what is the limit?? 0? or doesn't exist?? I think whether cos x could be zero or not is not important. It oscillates because of the neighborhood of x such that tanx=-1... Sorry for mistakes that I could make. === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) I see you got this expression in terms of tan by multiplying > top and bottom by cos(x). This is valid only if cos(x) is > never zero over the domain in which you are taking the > limit. Ooops! Leave it in terms of sin and cos and there's no problem > with the expression at all. > What condition violates the assumptions of L'Hopital's thm? None, you simply did an invalid manipulation. > - [Micro] .9eÀå .81¯.bd .bc®¨ .bcü± .89 - So, what is the limit?? 0? or doesn't exist?? I think whether cos x could be zero or not is not important. It > oscillates because of the neighborhood of x such that tanx=-1... > Sorry for mistakes that I could make. The limit doesn't exist. === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) I see you got this expression in terms of tan by multiplying > top and bottom by cos(x). This is valid only if cos(x) is > never zero over the domain in which you are taking the > limit. Ooops! Leave it in terms of sin and cos and there's no problem > with the expression at all. Ooops myself. For some reason I looked at sin(x) + cos(x) and thought it couldn't be zero. Never mind!! The correct answer was given by Bart Goddard. > What condition violates the assumptions of L'Hopital's thm? None, you simply did an invalid manipulation. === Subject: Re: L'Hopital's thm > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [-5 e^(-2x) ( sin x)] / [-2 e^(-x)(sin x)] } (by > L'Hopital's thm) > = lim{x->inf} {(5/2) e^(-x)} = 0 > However, > lim{x->inf} { [e^(-2x)(cos x + 2sin x)] / [e^(-x)(cos x + sin x)]} > = lim{x->inf} { [e^(-x) (1+2tan x)]/ [1+tan x] } doesn't exist. (It > oscillates) I see you got this expression in terms of tan by multiplying > top and bottom by cos(x). This is valid only if cos(x) is > never zero over the domain in which you are taking the > limit. Ooops! Leave it in terms of sin and cos and there's no problem > with the expression at all. Ooops myself. For some reason I looked at sin(x) + cos(x) > and thought it couldn't be zero. Never mind!! The correct answer was given by Bart Goddard. > What condition violates the assumptions of L'Hopital's thm? None, you simply did an invalid manipulation. > - [Micro] .9eÀå .81¯.bd .bc®¨ .bcü± .89 -- [Micro] .9eÀå .81¯.bd .bc®¨ .bc .9e± .89± .89 - - [Micro] .9eÀå .81¯.bd .bc®¨ .bcü± .89 - : ) All right, but I can't understand the answer of Bart === Subject: Re: L'Hopital's thm > What condition violates the assumptions of L'Hopital's thm? The fact that tan(x) is not defined in a neighborhood of infinity. You pass through infinitely many asymptotes of tan(x) on your way to taking the limit. -- The man without a .sig === Subject: Re: L'Hopital's thm infinity. You pass through infinitely many asymptotes > of tan(x) on your way to taking the limit. -- > The man without a .sig Why does 'the fact' violates the assumptions of L'Hopital's thm? I think, my book says,(in this case of 0/0) i) f'(x) and g'(x) exist in (a,inf) ii) K := lim{x->inf} [f'(x)/g'(x)] exists iii) lim{x->inf}f(x) = 0, lim{x->inf}g(x) = 0 Then lim{x->0} [f(x)/g(x)] = K I apply the thm just to y=e^(-2x)(cos x + 2sin x) and y=e^(-x)(cos x + sin x). === Subject: Re: L'Hopital's thm On 2 Feb 2007 15:24:02 -0800, Kenshin What condition violates the assumptions of L'Hopital's thm? >> The fact that tan(x) is not defined in a neighborhood of >> infinity. You pass through infinitely many asymptotes >> of tan(x) on your way to taking the limit. >> -- >> The man without a .sig Why does 'the fact' violates the assumptions of L'Hopital's thm? >I think, my book says,(in this case of 0/0) i) f'(x) and g'(x) exist in (a,inf) He just explained why (i) fails here! >ii) K := lim{x->inf} [f'(x)/g'(x)] exists >iii) lim{x->inf}f(x) = 0, lim{x->inf}g(x) = 0 Then lim{x->0} [f(x)/g(x)] = K I apply the thm just to y=e^(-2x)(cos x + 2sin x) and y=e^(-x)(cos x + >sin x). > ************************ David C. Ullrich === Subject: Re: L'Hopital's thm > On 2 Feb 2007 15:24:02 -0800, Kenshin > What condition violates the assumptions of L'Hopital's thm? >> The fact that tan(x) is not defined in a neighborhood of >> infinity. You pass through infinitely many asymptotes >> of tan(x) on your way to taking the limit. >> -- >> The man without a .sig Why does 'the fact' violates the assumptions of L'Hopital's thm? >I think, my book says,(in this case of 0/0) i) f'(x) and g'(x) exist in (a,inf) He just explained why (i) fails here! No he didn't. See the original problem. >ii) K := lim{x->inf} [f'(x)/g'(x)] exists >iii) lim{x->inf}f(x) = 0, lim{x->inf}g(x) = 0 Then lim{x->0} [f(x)/g(x)] = K I apply the thm just to y=e^(-2x)(cos x + 2sin x) and y=e^(-x)(cos x + >sin x). ************************ David C. Ullrich === Subject: Re: L'Hopital's thm What condition violates the assumptions of L'Hopital's thm? The fact that tan(x) is not defined in a neighborhood of > infinity. You pass through infinitely many asymptotes > of tan(x) on your way to taking the limit. -- > The man without a .sig Why does 'the fact' violates the assumptions of L'Hopital's thm? > I think, my book says,(in this case of 0/0) i) f'(x) and g'(x) exist in (a,inf) g'(x) must be also nonzero in some neighborhood of inf. That is the condition that is violated. > ii) K := lim{x->inf} [f'(x)/g'(x)] exists > iii) lim{x->inf}f(x) = 0, lim{x->inf}g(x) = 0 Then lim{x->0} [f(x)/g(x)] = K I apply the thm just to y=e^(-2x)(cos x + 2sin x) and y=e^(-x)(cos x + > sin x). === Subject: OT .. the future Middle East http://img157.imageshack.us/img157/5949/groundzeroocean9cx.jpg === Subject: Re: Size of equivalence class of Cauchy sequences expansion with a 5 in position i if i is in S, and a 6 if i is not in > S. This shows that the cardinality of the infinite subsets of N is at > most that of R. To go the other way, take the real numbers in (0,1), expressed in > binary expansion, with the terminating expansion if there is a choice > of expansion. Now, for each such real number r, we define the infinite subset S_r of > N as follows: n is in S_r if and only if (i) n is odd; or > (ii) n is even, and the (n/2)-th digit of the binary expansion of r > is equal to 1. (In other words, we take the characteristic function of the decimal > expansion of r, then map it into the even integers, and then take > subsets that contain all odd natural numbers and those corresponding > to the characteristic function; this ensures we have an infinite > subset). All such S_r are infinite, and distinct real numbers have > distinct expansions, and thus distinct S_r. So we have embeddings going each way, and we can use Cantor-Bernstein > to establish a bijection; I do not think we need AC for any of the > above constructions. (And, of course, we can get bijections from (0,1) to all of R without > too much trouble either. >The set of sequences of rationals is equnimumerous with the set of >functions from N into N. And I know there are proofs that the set of >functions from N into N is equinumerous with R, but those seem to be >pretty advanced proofs. Is there a simple proof I might understand? There is an obvious injection from 2^N to N^N, induced by the > embedding 2-->N. Now, pick your favorite embedding from N to 2^N. This induces an > embedding from N^N to (2^N)^N. There is a natural bijection from > (2^N)^N to 2^(N x N), namely, given f in (2^N)^N, we define the > function g:N x N --> 2 by g(a,b) = f(b)(a) (note that f(b) is an > element of 2^N). Now pick your favorite bijection from N to N xN. > This induces a bijection from 2^(NxN) to 2^N by precomposition. So > we have: emb. bij. bij. > N^N ------> (2^N)^N -----> 2^(NxN) ----> 2^N which gives an embedding from N^N to 2^N. Now we have embeddings going > both ways, so as above we can obtain a bijection through > Cantor-Bernstein. I see what you did there, but I want to mull over a couple of things on this one. It hadn't occurred to me that this one would be so simple. MoeBlee === Subject: Re: Quatification? I think you mean Quetification. Where *is* Leroy, anyway? Lee Rudolph === Subject: Re: Quatification? > I have a question about quantification both universal and existential. > I noticed that x and {x} are quantified by the same manner. > which is strange from the intuitive point of veiw, since x doesn't > exist > in the same manner as {x} exist. Exist in the same manner is, little doubt, a private notion of yours > unrelated to basic mathematical logic. The difference between the existance > of x and {x} appears to me like the difference between the existance > of > a ur-elements and the set that contains it. Iin a single sorted theory, urelements and sets are both members of > the same universe for a structure for the language of the theory. One > can have a many-sorted theory, but it is not required just to have > urelements. In a similar manner {x} > has > a different kind of existance {{x}} has. so how they are all > quantified > in the same way? Because, in a single-sorted theory, for any given mapping of the set > of variables into a universe of a structure for the language, the term > mapping function based on that mapping of the variables maps 'x' and > '{{x}}' into that same universe. If you don't know what I'm talking about, it's because you don't know > basic of mathematical logic. The rest of your post (not quoted here) > is just more nonsense built on your lack of familiarity with basic > mathemtatical logic. what I am trying to say is that ur-elements should be quantified > by a different universal and existential quantifiers than those used > to quantify over the sets that contains them. Then use a 2-sorted language. In a similar manner, the sets that contains ur-elements should be > quantified using different universal and existential quantifiers than > those > used to quanitfy over the sets that contains them(i.e set of sets of > ur-elements). First, are you under the impression that the variable 'x' can only > have urelements as values? No. If so, you are under a mistaken impression. > Second, you just assert what SHOULD be; your should carries no > reason for anyone to adhere to it. Third, if sets of sets of sets > of...ad infinitum are to have different quantifiers, then the number > of different kinds of quantifers will be pretty big, and it is not > clear what advantage would be had. Fourth, aside from just basic > predicate logic, you should learn about the cumulative hierarchy and > also type theories and also stratification theories and things like > that. why not? Of course I should and I will. so for example if we denote variables ranging over ur-elements by > L0,were L is any english letter like x,y,z,..,etc and variables > ranging over sets that contain ur-elements by L1 . Then we should > quantify using appropriate quantifiers, for example: E1x1 A0y0 ( y0 e x1 <-> P0(y0) ). were P0 is a predicate in first > order language > in one free variable, at ur-predicate level(i.e it only holds for ur- > elements). What do you mean by a 'predicate in first order language'? Do you mean > a formula? yes. a formula in first order language. example let P0(y0)<->y0 is a man. then x1 is the set of all men. of course P1(y1)<-> y1 is a set of men. and we can have E2x2 A1y1 ( y1 e x2 <-> P1(y1) ). and here x2 would be the set of all sets of men. and so forth, endlesselly. Indeed, endless. Just as your confusions are endless. well they might be confusional indeed, because this idea only came into my mind for one minute or so, and I spend about 2 minutes writing it? what I wanted to say is that a sentence of the form Ex P(x). were P is a formula in first order language. is not so precise. Example Ex: x is a man. this is not so precise statement. because it doesn't specify the type of existance we are quantifying x by. There Exist x such that x is a man. This is not a precise statement. One should specify in what kind of existance we are saying x is existing. Also here x has range over a very wide values, that can be ur-elements or not. I don't like this. it is not precise. There is a great difference between Existance of men, cats, etc.. and between the existance of sets that contains them. A set has a pritty different kind of existance than a man. when I say a man exist, it means that I can sense this MAN , or effects of this man in such a peculiar and unique manner. But a set cannot be sensed in the same manner weather directly or indirectly. Therefore I prefer the following. E0 x : x is a man here E0 means exists in the objective world we live in, or more preciselly in the domain of ur-elements. Not only this, it makes not sense to quantify over such a big variable like x,that might range over ur-elements as well as non ur-elements. it is better to limit x to be a variable that range over ur-elements only and we denote it as x0. so x0 is defined as a variable that ranges over ur-elements only. so the most precise sentence would be: E0 x0 : x0 is a man. Now to generalize this. a formula in first order language that is true only for ur-elements would be also symbolized as P0 so we have: E0 x0 : P0(x0) of course not every P can be differentiated(spectrated(i.e form a spectrum). into P0 P1 P2 ..... Lets take for example P that can be differentiated into P0P1P2..... example P<->a man here P0 <-> a man P1<-> a set of men P2<-> a set of sets of men P3<-> a set of sets of sets of men. etc..... you see that Pi is different from Pj if ~i=j. so here man is said to be a spectrated predicate. while for an example of a predicate that cannot have these levels is P(y)<->y=y here y=y is not spectrated, since P0=P1=P2 etc.... to be clearer I mean P0 hold for y when y is a ur-element But the very same P0 also holds for sets of these ur-elements, since evey ur- element is equal to itself, so does any set at any level. unlike the formula y is a man, were P1 is different from P0. i.e formula P0 doesn't hold for x1 values (i.e sets of ur-elements), so a set of men is never a man. I think that the Existance of sets themselfs is different according to the levels I showed. so if x2 is a variable that range over level2 sets:ie. sets that can contain sets as elements. then x0 e x2 or x1 e x2 are acceptable statements, but xi e xj were ~(j>i). is unacceptable statement. I understand that this would lead to change of a lot of concept. For example there would be many kinds of empty sets. defined as below A0x0 E1y1 : ~x0ey1, here y1 is empty set at first level. while: A0x0A1x1 E2 y2: ~((x0 or x1)ey2 here y2 is empty set at second level. However if a finite set theory is constructed were there is a upper level n, beyond which there is not heigher level of existance. then it is clear that A0x0 A1x1 A2x3.......A(n-1)x(n-1) Enyn: ~(x0 or x1 or x2 or.....x(n-1))eyn. here yn would be The absolute empty set of the highest level. that cannot have any x inside it. At such n-highest level of existance theorum, we can indeed have a universe which would be zn as below: A0x0 A1x1 A2x3.......A(n-1)x(n-1) Enzn: (x0 or x1 or x2 or.....x(n-1))ezn. This would be the set of all sets. However if we had infinte existance theory, were for every level of existance i there is i+1 level of existance, then there would be no such a set of all sets. anyhow, all of this might be confusional as you said. But it looks more precise and meaningful to me. I thinks Qanitifiers, sets , predicates all should be stratified ( spectrated ). A non stratified predicate like ~yey, ~y=y, y=y, y is an ordinal, |y|>n , etc.... I think these predicates should be analysed in a specific manner, since these would not doubt cause a lot of problems. But as you said I am not to make such a big statements now. As I told you many times I will come after one year after reading these books you told me about, and by then we will see if my thoughts remain the same, and we'll have a disscussion. Zuhair MoeBlee === Subject: Re: Quatification? > what I wanted to say is that a sentence of the form Ex P(x). were P is > a formula in first order language. is not so precise. Example Ex: x is a man. this is not so precise statement. It's not in set theory. But anyway... > because it doesn't specify the type of existance we are quantifying x > by. You arbirtrarily require a type of existence. It's not necessary for first order set theory. > Also here x has range over a very wide values, that can be ur-elements > or not. I don't like this. it is not precise. In usual formulations of Z set theory, there can be no urelements. In formulations of Z set theoryy in which there are urelements, there is a precise definition of 'urelement'. There's nothing imprecise. > There is a great difference between Existance of men, cats, etc.. and > between the existance of sets that contains them. A set has a pritty > different kind of existance than a man. > when I say a man exist, it means that I can sense this MAN , or > effects of this man in such a peculiar and unique manner. > But a set cannot be sensed in the same manner weather directly or > indirectly. And none of that impairs set theory. > Therefore I prefer the following. E0 x : x is a man here E0 means exists in the objective world we live in, or more > preciselly in the domain of ur-elements. We can easily have a 2-sorted language, with variables for urelements and variables for sets. It's easily done. But it doesn't have very much (if any) substantive difference with a set theory with urelements but in a 1-sorted language. 2-sorted, in this instance, is a notational convenience (or inconvenience, depending on one's attitude). > Not only this, it makes not sense to quantify over such a big variable > like x,that might range over ur-elements as well as non ur-elements. > it is better to limit x to be a variable that range over ur-elements > only and we denote it as x0. 1-soterd or 2-sorted. We can do it either way. There's not much (if any) significant different. > so x0 is defined as a variable that ranges over ur-elements only. so the most precise sentence would be: E0 x0 : x0 is a man. I thought we were talking about ZFC. 'is a man' is not a rendering of a defined predicate symbol of ZFC. > Now to generalize this. a formula in first order language that is true > only for ur-elements would be also symbolized as P0 so we have: E0 x0 : P0(x0) Which is equivalent to: Ex(x is an urelement & Px). And the rest of your sorting (not quoted here) is unneeded complication. You'll see that type theories and stratification theories and things like that are formulated to avoid Russell paradox kinds of contradictions. But ZFC does that by restricting comprehension. With such restricted comprehension, there's nothing gained by heaping onto ZFC extra complications as to types or stratifications. If you want a type theory or stratification theory, then it would make more sense to start from scratch with a whole new theory rather than needlessly heaping complications onto ZFC. MoeBlee === Subject: Re: Quatification? what I wanted to say is that a sentence of the form Ex P(x). were P is > a formula in first order language. is not so precise. Example Ex: x is a man. this is not so precise statement. It's not in set theory. But anyway... because it doesn't specify the type of existance we are quantifying x > by. You arbirtrarily require a type of existence. It's not necessary for > first order set theory. Also here x has range over a very wide values, that can be ur-elements > or not. I don't like this. it is not precise. In usual formulations of Z set theory, there can be no urelements. In > formulations of Z set theoryy in which there are urelements, there is > a precise definition of 'urelement'. There's nothing imprecise. There is a great difference between Existance of men, cats, etc.. and > between the existance of sets that contains them. A set has a pritty > different kind of existance than a man. > when I say a man exist, it means that I can sense this MAN , or > effects of this man in such a peculiar and unique manner. > But a set cannot be sensed in the same manner weather directly or > indirectly. And none of that impairs set theory. Therefore I prefer the following. E0 x : x is a man here E0 means exists in the objective world we live in, or more > preciselly in the domain of ur-elements. We can easily have a 2-sorted language, with variables for urelements > and variables for sets. It's easily done. But it doesn't have very > much (if any) substantive difference with a set theory with urelements > but in a 1-sorted language. 2-sorted, in this instance, is a > notational convenience (or inconvenience, depending on one's > attitude). Not only this, it makes not sense to quantify over such a big variable > like x,that might range over ur-elements as well as non ur-elements. > it is better to limit x to be a variable that range over ur-elements > only and we denote it as x0. 1-soterd or 2-sorted. We can do it either way. There's not much (if > any) significant different. so x0 is defined as a variable that ranges over ur-elements only. so the most precise sentence would be: E0 x0 : x0 is a man. I thought we were talking about ZFC. 'is a man' is not a rendering of > a defined predicate symbol of ZFC. Now to generalize this. a formula in first order language that is true > only for ur-elements would be also symbolized as P0 so we have: E0 x0 : P0(x0) Which is equivalent to: Ex(x is an urelement & Px). And the rest of your sorting (not quoted here) is unneeded > complication. You'll see that type theories and stratification theories and things > like that are formulated to avoid Russell paradox kinds of > contradictions. But ZFC does that by restricting comprehension. With > such restricted comprehension, there's nothing gained by heaping onto > ZFC extra complications as to types or stratifications. If you want a > type theory or stratification theory, then it would make more sense to > start from scratch with a whole new theory rather than needlessly > heaping complications onto ZFC. MoeBlee I am talking about a concept that is prior to ZFC or any set theory. It is even prior to logical languages. There are different types of existance, this is pritty intuitive and it is unquestionable. not only between sets and ur-elements as you understood. it is their even between sets themselfs. A set that contains sets that contains ur-elements is different from A set that contains ur-elements . And by different I mean different existance not essence as you might interpret. Although I don't need to go into that, since it is un-necessary,but let me explain it in this way: what I mean is that a set like { {x} } is made of a substance that is different from a set like {y}, were x and y here stands for ur-elements. and more the symbole {{x}} is erronous symbole, in reality we should give a symbole that is different from {or} to describe something like {{x}} which is intended to mean a set that only contains a set that only contains the ur-element x. ( I am deliberatly fixing x to be a ur-element here, I don't mean it is used like that in general). Abetter notation should be for example: [{x0}]. so I used the symboles [and] instead of { and }. to mean a second level set(i.e. a set that EXIST in a second level manner). even better one can use 0,1,2,3,4,.... as superscripts over { OR BESIDE { to denote its level of existance. example {{x0}0}1. I am talking about the existance of a set. It is not a matter of convenience as you think, it is much deeper than this. Anyhow we'll have a discussion about that after one year. cya then. Zuhair === Subject: Re: Quatification? > I am talking about a concept that is prior to ZFC or any set theory. > It is even prior to logical languages. Fine. And, so, as I said, there's no reason for your requirements to impinge on such theories as Z set theorie. > There are different types of existance, this is pritty intuitive and > it is unquestionable. not only between sets and ur-elements as you > understood. it is their even between sets themselfs. > A set that contains sets that contains ur-elements is different from A > set that contains ur-elements . And by different I mean different > existance not essence as you might interpret. Let me know when you can show that any of that has any bearing on mathematics, set theory, or mathematical logic. > Although I don't need to go into that, since it is un-necessary,but > let me explain it in this way: what I mean is that a set like { {x} } > is made of a substance that is different from a set like {y}, were x > and y here stands for ur-elements. > and more the symbole {{x}} is erronous symbole, in reality we should > give a symbole that is different from {or} to describe something > like {{x}} which is intended to mean > a set that only contains a set that only contains > the ur-element x. ( I am deliberatly fixing x to be a ur-element here, > I don't mean it is used like that in general). Abetter notation should be for example: [{x0}]. > so I used the symboles [and] instead of { > and }. to mean a second level set(i.e. a set that > EXIST in a second level manner). even better one can use 0,1,2,3,4,.... as superscripts over { OR > BESIDE { to denote its level of existance. example {{x0}0}1. I am talking about the existance of a set. > It is not a matter of convenience as you think, it is much deeper than > this. Yes, apparently it is much deeper. MoeBlee === Subject: Re: Quatification? I am talking about a concept that is prior to ZFC or any set theory. > It is even prior to logical languages. Fine. And, so, as I said, there's no reason for your requirements to > impinge on such theories as Z set theorie. If I can prove what I said, then you can kiss ZFC bye bye. And a better theory based on a better language shall replace it. Perhaps There are different types of existance, this is pritty intuitive and > it is unquestionable. not only between sets and ur-elements as you > understood. it is their even between sets themselfs. > A set that contains sets that contains ur-elements is different from A > set that contains ur-elements . And by different I mean different > existance not essence as you might interpret. Let me know when you can show that any of that has any bearing on > mathematics, set theory, or mathematical logic. Although I don't need to go into that, since it is un-necessary,but > let me explain it in this way: what I mean is that a set like { {x} } > is made of a substance that is different from a set like {y}, were x > and y here stands for ur-elements. > and more the symbole {{x}} is erronous symbole, in reality we should > give a symbole that is different from {or} to describe something > like {{x}} which is intended to mean > a set that only contains a set that only contains > the ur-element x. ( I am deliberatly fixing x to be a ur-element here, > I don't mean it is used like that in general). Abetter notation should be for example: [{x0}]. > so I used the symboles [and] instead of { > and }. to mean a second level set(i.e. a set that > EXIST in a second level manner). even better one can use 0,1,2,3,4,.... as superscripts over { OR > BESIDE { to denote its level of existance. example {{x0}0}1. I am talking about the existance of a set. > It is not a matter of convenience as you think, it is much deeper than > this. Yes, apparently it is much deeper. MoeBlee- Hide quoted text - - Show quoted text - === Subject: Re: Quatification? > If I can prove what I said, then you can kiss ZFC bye bye. And a > better theory based on a better language shall replace it. Perhaps I have no doubt that you'll transform the very meaning of our human existence. MoeBlee === Subject: Complex, Real Analysis books (Dover) I'm looking for a book on Real Analysis and one on Complex Analysis at the undergraduate level published by Dover. They do not have to be overly gentle introductions, but they should be rigorous introductions for undergrad level (i.e., not applied analysis books and should construct the real/complex number systems respectively). Can I please have some recommendations? There are too many for me to even start looking into myself, so I need to narrow down the search. I tried [P.S. Is there a webpage list of recommendations Dover math books for all major subject matters? ] === Subject: Re: Complex, Real Analysis books (Dover) >I'm looking for a book on Real Analysis and one on Complex Analysis at >the undergraduate level published by Dover. They do not have to be >overly gentle introductions, but they should be rigorous introductions >for undergrad level (i.e., not applied analysis books and should >construct the real/complex number systems respectively). Can I please have some recommendations? There are too many for me to >even start looking into myself, so I need to narrow down the search. >I tried >[P.S. Is there a webpage list of recommendations Dover math books for >all major subject matters? ] For Complex Analysis the answer is easy: read Knopp first. However, in this field there are lots of good books - Saks and Zygmund is my favorite. === Subject: Re: Complex, Real Analysis books (Dover) Complex Analysis with Applications, by Richard A. Silverman (Dover) ISBN 0-486-64762-5 -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Complex, Real Analysis books (Dover) I'm looking for a book on Real Analysis and one on Complex Analysis at > the undergraduate level published by Dover. They do not have to be > overly gentle introductions, but they should be rigorous introductions > for undergrad level (i.e., not applied analysis books and should > construct the real/complex number systems respectively). Go to Doverbooks web site. Bob Kolker === Subject: Re: Complex, Real Analysis books (Dover) I'm looking for a book on Real Analysis and one on Complex Analysis at > the undergraduate level published by Dover. Introductory Real Analysis by A.N. Kolmogorov is a Dover book. It's very extensive, and gives a fabulous treatment of generalized functions and Lesbegue integration. This is an introductory textbook, but it is very rigorous in its treatment of the subject. It is a Russian book translated into English. Incidentally, Kolmogorov in the original Russian is the standard text used by logicians as a source of theorems of real analysis that can be translated into second-order arithmetic. Why do you only want Dover books? >They do not have to be > overly gentle introductions, but they should be rigorous introductions > for undergrad level (i.e., not applied analysis books and should > construct the real/complex number systems respectively). Can I please have some recommendations? There are too many for me to > even start looking into myself, so I need to narrow down the search. > I tried > [P.S. Is there a webpage list of recommendations Dover math books for > all major subject matters? ] http://store.doverpublications.com/by-subject-science-and-mathematics-mathem atics.html === Subject: Re: Complex, Real Analysis books (Dover) > Why do you only want Dover books? Don't know about the OP, but what about price? Typical Dover math textbook: $15 or less Typical modern math textbook: $60 or more Now is the modern text better? Perhaps, perhaps not. Is it 4 times better?????? Definitely not!!!!! Now when you choose a book for yourself, maybe price is not a great consideration. But when you choose a book to require your students to buy, don't you owe it to them at least to CONSIDER the price? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Complex, Real Analysis books (Dover) I'm looking for a book on Real Analysis and one on Complex Analysis at > the undergraduate level published by Dover. They do not have to be > overly gentle introductions, but they should be rigorous introductions > for undergrad level (i.e., not applied analysis books and should > construct the real/complex number systems respectively). Can I please have some recommendations? There are too many for me to > even start looking into myself, so I need to narrow down the search. > I tried > [P.S. Is there a webpage list of recommendations Dover math books for > all major subject matters? ] Shilov maybe. (I don't recall whether he goes into a construction of the reals. But there's a real nice free PDF file on the Internet that has the equivalence class of Cauchy sequences method. I'll try to find the link if you're interested.) MoeBlee === Subject: Laplace's rule of succession In Bayesian statistics, Laplace's rule of succession attempts to solve the problem of how we can predict that the sun will rise tomorrow, given its past frequency of rising. Definitions: 1. Let p be the long-run frequency, as observed. 2. Let n be the total number of trials. 3. Let s be the number of *successes* among these trials, so that n - s is the number of failures. The rule of succession states that the probability of the next success is given by the *expected value of a normalized likelihood function*. The likelihood function is p^s * (1 - p)^(n - s). Normalized with the integral S_{0 to 1}(p^s * (1 - p)^(n - s)) dp, one obtains as the expected value (s + 1)/(n + 2) for the probability of the next success. Thus, if all we know is that the sun has risen 2000 times, the probability of its rising again is 2001/2002. Now, I have a question. What's so special about this likelihood function? It seems to be formulated completely ad hoc. If the sample space were all possible successions, the probability of the next success would simply be 1/2. So what gives? The figure p^s * (1 - p)^(n - s) is the probability that the successes and failures after n trials will happen again, *with the frequency they have already happened*--that is, the probability of s successes along with n - s failures. But to me this is obviously circular. It *assumes* that the long-run *frequency* of successes *is*, in fact, the *probability* of success. And what about the following case in which success and failure follow a pattern: 1 1 0 1 1 0 1 1 0 1? Isn't the likelihood of the next success greater than 1/2? I would like to understand more of the philosophical theory behind the your help. === Subject: convergence a.e. / in measure Show by example that in an infinite measure space, functions can converge almost everywhere without converging in measure. Can anybody help? === Subject: Re: convergence a.e. / in measure appliedstat a .8ecrit : > Show by example that in an infinite measure space, functions can converge almost everywhere without converging in measure. > Can anybody help? On R with the Lebesgue measure, f_n(x) = 1 if n < x < n+1, 0 elsewhere. They converge to 0 everywhere but not in measure. JMA === Subject: Laplace's rule of succession In Bayesian statistics, Laplace's rule of succession attempts to solve the problem of how we can predict that the sun will rise tomorrow, given its past frequency of rising. Definitions: 1. Let p be the long-run frequency, as observed. 2. Let n be the total number of trials. 3. Let s be the number of *successes* among these trials, so that n - s is the number of failures. The rule of succession states that the probability of the next success is given by the *expected value of a normalized likelihood function*. The likelihood function is p^s * (1 - p)^(n - s). Normalized with the integral S_{0 to 1}(p^s * (1 - p)^(n - s)) dp, one obtains as the expected value (s + 1)/(n + 2) for the probability of the next success. Thus, if all we know is that the sun has risen 2000 times, the probability of its rising again is 2001/2002. Now, I have a question. What's so special about this likelihood function? It seems to be formulated completely ad hoc. If the sample space were all possible successions, the probability of the next success would simply be 1/2. So what gives? The figure p^s * (1 - p)^(n - s) is the probability that the successes and failures after n trials will happen again, *with the frequency they have already happened*--that is, the probability of s successes along with n - s failures. But to me this is obviously circular. It *assumes* that the long-run *frequency* of successes *is*, in fact, the *probability* of success. And what about the following case in which success and failure follow a pattern: 1 1 0 1 1 0 1 1 0 1? Given the pattern, isn't the likelihood of the next success greater than (s + 1)/(n + 2) = 3/4? I would like to understand more of the philosophical theory behind the your help. === Subject: Re: Laplace's rule of succession > In Bayesian statistics, Laplace's rule of succession attempts to solve > the problem of how we can predict that the sun will rise tomorrow, > given its past frequency of rising. Definitions: 1. Let p be the long-run frequency, as observed. > 2. Let n be the total number of trials. > 3. Let s be the number of *successes* among these trials, so that n - > s is the number of failures. The rule of succession states that the probability of the next success > is given by the *expected value of a normalized likelihood function*. > The likelihood function is p^s * (1 - p)^(n - s). Normalized with the integral S_{0 to 1}(p^s * (1 - p)^(n - s)) dp, one > obtains as the expected value (s + 1)/(n + 2) for the probability of the next success. Thus, if all we know is that > the sun has risen 2000 times, the probability of its rising again is > 2001/2002. Now, I have a question. What's so special about this likelihood > function? It seems to be formulated completely ad hoc. If the sample > space were all possible successions, the probability of the next > success would simply be 1/2. So what gives? The figure p^s * (1 - p)^(n - s) is the probability that the successes > and failures after n trials will happen again, *with the frequency > they have already happened*--that is, the probability of s successes > along with n - s failures. But to me this is obviously circular. It > *assumes* that the long-run *frequency* of successes *is*, in fact, > the *probability* of success. And what about the following case in which success and failure follow > a pattern: 1 1 0 1 1 0 1 1 0 1? Given the pattern, isn't the > likelihood of the next success greater than (s + 1)/(n + 2) = 3/4? I would like to understand more of the philosophical theory behind the > your help. Assuming the trials are independent and we have a uniform prior on p, then I believe you have written down the posterior for p. Whether you believe this answer depends on whether you believe the assumptions (i.e., the prior and the independence). Are you familiar with the principles of Bayesian statistics? -- === Subject: Re: Laplace's rule of succession then I believe you have written down the posterior for p. Whether you > believe this answer depends on whether you believe the assumptions > (i.e., the prior and the independence). Are you familiar with the > principles of Bayesian statistics? The posterior for p depends on the likelihood function, which seems to be pulled out of nowhere. Where do you get the likelihood function? My original post had some mistakes. I've just sent my new one. === Subject: Re: Laplace's rule of succession > Assuming the trials are independent and we have a uniform prior on p, > then I believe you have written down the posterior for p. Whether you > believe this answer depends on whether you believe the assumptions > (i.e., the prior and the independence). Are you familiar with the > principles of Bayesian statistics? The posterior for p depends on the likelihood function, which seems to > be pulled out of nowhere. Where do you get the likelihood function? Suppose you have some data with density f. And, suppose that f depends on some parameter p. Let's write f(z;p) for the probability of observing a value of z for the data given that the parameter has value p. Then the likelihood is just f considered as a function of p with the actual observed value of the data inserted for z. The posterior distribution of p (i.e., the distribution of p given that the data equals z) is the likelihood times the prior. The likelihood you gave is for independent Bernoulli trials with probability p. If you don't think the data comes from independent Bernoulli trials, then don't use that likelihood. It is a bit odd for you to say the likelihood is pulled out of nowhere. The more usual complaint is that the prior is pulled out of nowhere. (It isn't actually pulled out of nowhere, but until you do a real problem, you don't see where it comes from.) -- === Subject: Re: Laplace's rule of succession be pulled out of nowhere. Where do you get the likelihood function? Suppose you have some data with density f. And, suppose that f depends > on some parameter p. Let's write f(z;p) for the probability of observing > a value of z for the data given that the parameter has value p. Then the > likelihood is just f considered as a function of p with the actual > observed value of the data inserted for z. The posterior distribution of > p (i.e., the distribution of p given that the data equals z) is the > likelihood times the prior. I assume that in our case, p is the probability of success, z is either 1 or 0 for success or failure, and f(z; p) is the probability of the success of one trial, which, as you say, depends on a parameter. *What* parameter? > The likelihood you gave is for independent Bernoulli trials with > probability p. If you don't think the data comes from independent > Bernoulli trials, then don't use that likelihood. If the Bernoulli trials are independent, how does a sequence of 1, 1, 1 determine a probability of 4/5, instead of 1/2 (again), for the next 1? Doesn't that contradict the independence of the trials? > It is a bit odd for you to say the likelihood is pulled out of nowhere. > The more usual complaint is that the prior is pulled out of nowhere. (It > isn't actually pulled out of nowhere, but until you do a real problem, > you don't see where it comes from.) Where does it come from? === Subject: Re: Laplace's rule of succession be pulled out of nowhere. Where do you get the likelihood function? Suppose you have some data with density f. And, suppose that f depends > on some parameter p. Let's write f(z;p) for the probability of observing > a value of z for the data given that the parameter has value p. Then the > likelihood is just f considered as a function of p with the actual > observed value of the data inserted for z. The posterior distribution of > p (i.e., the distribution of p given that the data equals z) is the > likelihood times the prior. I assume that in our case, p is the probability of success, z is either 1 or 0 for success or failure, and f(z; p) is the probability of the success of one trial, which, as you say, depends on a parameter. *What* parameter? > The likelihood you gave is for independent Bernoulli trials with > probability p. If you don't think the data comes from independent > Bernoulli trials, then don't use that likelihood. If the Bernoulli trials are independent, how does a sequence of 1, 1, 1 determine a probability of 4/5, instead of 1/2 (again), for the next 1? Doesn't that contradict the independence of the trials? > It is a bit odd for you to say the likelihood is pulled out of nowhere. > The more usual complaint is that the prior is pulled out of nowhere. (It > isn't actually pulled out of nowhere, but until you do a real problem, > you don't see where it comes from.) Where does it come from? === Subject: Re: Laplace's rule of succession Suppose you have some data with density f. And, suppose that f depends > on some parameter p. Let's write f(z;p) for the probability of observing > a value of z for the data given that the parameter has value p. Then the > likelihood is just f considered as a function of p with the actual > observed value of the data inserted for z. The posterior distribution of > p (i.e., the distribution of p given that the data equals z) is the > likelihood times the prior. I assume that in our case, p is the probability of success, z is > either 1 or 0 for success or failure, and f(z; p) is the probability > of the success of one trial, which, as you say, depends on a > parameter. *What* parameter? z should be a vector. Its length is the number of trials you have. The parameter is p. That's why I used a semicolon. > The likelihood you gave is for independent Bernoulli trials with > probability p. If you don't think the data comes from independent > Bernoulli trials, then don't use that likelihood. If the Bernoulli trials are independent, how does a sequence of 1, 1, > 1 determine a probability of 4/5, instead of 1/2 (again), for the next > 1? Doesn't that contradict the independence of the trials? Our knowledge of p changes as we collect data. Suppose initially we think p is uniform on [0,1]. Suppose our experiment is to do three trials. Then the probability that the experiment will produce data ofp successes and f failures is p^s (1-p)^f. If we do the experiment and get three successes, then the conditional density of p is p^3 / int_0^1 p^3 = 4 p^3. The mean of this is int_0^1 4 p^4 = 4/5. So, that is our best estimate of p after seeing the three trials. > It is a bit odd for you to say the likelihood is pulled out of nowhere. > The more usual complaint is that the prior is pulled out of nowhere. (It > isn't actually pulled out of nowhere, but until you do a real problem, > you don't see where it comes from.) Where does it come from? The prior comes from our knowledge/belief about the problem before seeing the data. If you are asking me whether the sun will come up each day, I wouldn't use a uniform prior for p, since I'm pretty sure it will. On the other hand, I don't think that the sun coming up each day is well modeled as independent Bernoulli trials. If the sun didn't come up tomorrow, I'd be rather surprised if it did come up two days from now. -- === Subject: Re: Laplace's rule of succession 1 determine a probability of 4/5, instead of 1/2 (again), for the next > 1? Doesn't that contradict the independence of the trials? Our knowledge of p changes as we collect data. Suppose initially we > think p is uniform on [0,1]. Suppose our experiment is to do three > trials. Then the probability that the experiment will produce data ofp > successes and f failures is p^s (1-p)^f. But how do we know there *is* a p? In other words, how do we know that there is a *fixed probability* of success or failure that remains in effect for all trials? And what's wrong with my sample space interpretation, which gives a different answer? If we include all possible outcomes in the sample space, the probability remains 1/2. Do you see where I'm coming from? > It is a bit odd for you to say the likelihood is pulled out of nowhere. > The more usual complaint is that the prior is pulled out of nowhere. (It > isn't actually pulled out of nowhere, but until you do a real problem, > you don't see where it comes from.) Where does it come from? The prior comes from our knowledge/belief about the problem before > seeing the data. Is it knowledge, or mere belief? === Subject: Re: Laplace's rule of succession 1 determine a probability of 4/5, instead of 1/2 (again), for the next > 1? Doesn't that contradict the independence of the trials? Our knowledge of p changes as we collect data. Suppose initially we > think p is uniform on [0,1]. Suppose our experiment is to do three > trials. Then the probability that the experiment will produce data ofp > successes and f failures is p^s (1-p)^f. But how do we know there *is* a p? In other words, how do we know that there is a *fixed probability* of success or failure that remains in effect for all trials? And what's wrong with my sample space interpretation, which gives a different answer? If we include all possible outcomes in the sample space, the probability remains 1/2. Do you see where I'm coming from? > It is a bit odd for you to say the likelihood is pulled out of nowhere. > The more usual complaint is that the prior is pulled out of nowhere. (It > isn't actually pulled out of nowhere, but until you do a real problem, > you don't see where it comes from.) Where does it come from? The prior comes from our knowledge/belief about the problem before > seeing the data. Is it knowledge, or mere belief? === Subject: Re: Laplace's rule of succession > If the Bernoulli trials are independent, how does a sequence of 1, 1, > 1 determine a probability of 4/5, instead of 1/2 (again), for the next > 1? Doesn't that contradict the independence of the trials? Our knowledge of p changes as we collect data. Suppose initially we > think p is uniform on [0,1]. Suppose our experiment is to do three > trials. Then the probability that the experiment will produce data ofp > successes and f failures is p^s (1-p)^f. But how do we know there *is* a p? In other words, how do we know that > there is a *fixed probability* of success or failure that remains in > effect for all trials? If you don't believe that, then don't use that model. Not much sense in estimating a parameter that you don't believe exists. > And what's wrong with my sample space interpretation, which gives a > different answer? If we include all possible outcomes in the sample > space, the probability remains 1/2. Do you see where I'm coming from? Why should all points in the sample space have equal probability? On the other hand, if you know that all points in the sample space are equally likely, then there isn't much point in estimating the probabilities. You can still do a Bayesian analyis, but your prior for p should have Prob(p = 1/2) = 1. Then the posterior for p will be the same, regardless of what data you observe. > The prior comes from our knowledge/belief about the problem before > seeing the data. Is it knowledge, or mere belief? For a scientist, they are the same. I don't think many religions advocate using Bayesian statistics. Is there a reason your are posting your messages twice? -- === Subject: Re: Cantor Confusion > Here is the definition of the set N 1 > 2 > 3 > ... Here is the definition of the sum of all elements of N 1 > 23 > 456 > ... That's pretty impressive. Did you come up with that all by yourself? -- === Subject: Re: Cantor Confusion On Thu, 1 Feb 2007 13:24:10 -0500, >> >> I can't tell whether the union of all lengths is finite or not. >> As the union of all lengths is obviously equivalent to the union of all >> finite ordinals and the union of all finite ordinals is clearly not >> finite, WM can't tell much. >> >> The union of all lengths is not the sum of all lengths, which, >> according to your standpoint might be infinite. But the union of all >> lengths is a length. You may say that it is infinite or not. If it is >> infinite, then it corresponds to an infinite natural number. If it is >> not infinite, then the set N does not exist other than as a >> potentially infinite (= finite but unbounded) set. It is amazing what you can write if you fail to define your terms. If we >read it quickly, it sounds vaguely like a logical argument. Of course, >it is actually nonsense. Did you have to practice for a long time to >develop this technique? David, I can readily imagine you attended a Special Ed. course for math majors at MIT. Tell us were there any actual course requirements beyond the liturgical basis for modern math or was the degree simply a pro forma social promotion. ~v~~ === Subject: Re: Cantor Confusion <$6rsRmghMfuFFwrV@phoenixsystems.demon.co.uk> <$EiMOojF$huFFwoB@phoenixsystems.demon.co.uk> <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> (snip) > I am exercised by the idea of space filling curves, which is the same >> issue, covering something by something of a lower classical dimension, >> and so maybe the recursive fractal definition of reals does cover my >> concept of line. No. This is a different problem. Even Euclid treated lines as consisting >of points and planes as consisting of lines. A space-filling curve is a >is a continuous function from [0,1] to [0,1]^2 that is onto. The >surprising feature is that the function is continuous. > I asked some time before on the other thread, and didn't really get a >> straight answer from DM, If you don't get a straight answer, maybe you didn't ask a straight >question. Your question was about changing the space-filling curve >construction, but it wasn't clear how you wanted to change it. > about a fractal scenario where case 1: we >> start of with the interval 0-1. On each iteration, each interval is >> divided in two, and we associate a real number with the mid-point of >> each section. After an infinite number of iterations, all the reals in >> [0,1] are covered, no problem. Do you mean that at stage n+1, we divide all the intervals from stage n >in half? So, at each stage we have twice as many intervals as before? If >we do this for all natural numbers n, we only get a countable number of >intervals (all together), so the mid-points of the intervals most >certainly do not include all real numbers in [0,1]. The only numbers you >get are numbers with a finite binary expansion. > OK, I understand that argument. You are saying that I can't generate a number with an infinite decimal expansion by a finite limit process, it just converges to some subset of Q. I can't just loosely say an infinite number of iterations, it's not a number. But in that case, I misunderstand how the space filling curve iteration can cover all the points in the plane - if you take a cross-section of the plane along a line at e.g. 45 degrees, on iteration 1 it has 2 points covered, and then doubling each iteration. Doesn't the same argument then apply - you can't generate something infinite from a finite iteration? -- Andy Smith === Subject: Re: Cantor Confusion > (snip) > I am exercised by the idea of space filling curves, which is the same >> issue, covering something by something of a lower classical dimension, >> and so maybe the recursive fractal definition of reals does cover my >> concept of line. No. This is a different problem. Even Euclid treated lines as consisting >of points and planes as consisting of lines. A space-filling curve is a >is a continuous function from [0,1] to [0,1]^2 that is onto. The >surprising feature is that the function is continuous. > I asked some time before on the other thread, and didn't really get a >> straight answer from DM, If you don't get a straight answer, maybe you didn't ask a straight >question. Your question was about changing the space-filling curve >construction, but it wasn't clear how you wanted to change it. > about a fractal scenario where case 1: we >> start of with the interval 0-1. On each iteration, each interval is >> divided in two, and we associate a real number with the mid-point of >> each section. After an infinite number of iterations, all the reals in >> [0,1] are covered, no problem. Do you mean that at stage n+1, we divide all the intervals from stage n >in half? So, at each stage we have twice as many intervals as before? If >we do this for all natural numbers n, we only get a countable number of >intervals (all together), so the mid-points of the intervals most >certainly do not include all real numbers in [0,1]. The only numbers you >get are numbers with a finite binary expansion. OK, I understand that argument. You are saying that I can't generate a > number with an infinite decimal expansion by a finite limit process, I don't think I'm saying that because I don't know what you mean. What is a finite limit process? Can you give an example? > it > just converges to some subset of Q. I can't just loosely say an > infinite number of iterations, it's not a number. Of course you can say an infinite number of iterations. For example, we can have an iteration for each natural number. That's certainly an infinite number of iterations. > But in that case, I misunderstand how the space filling curve iteration > can cover all the points in the plane - if you take a cross-section of > the plane along a line at e.g. 45 degrees, on iteration 1 it has 2 > points covered, and then doubling each iteration. Doesn't the same > argument then apply - you can't generate something infinite from a > finite iteration? I don't know what a finite iteration is. The iterations are not the final curve. Let f_n be the curve at the n-th step. Then f_n does not cover the unit square. However, we can prove that for all x in [0,1], f_n(x) converges to a point in the unit square. So, we can define f(x) = lim_{n->oo} f_n(x). And, we can prove that f is continuous and its image is the unit square. Also, note that f_n is typically composed of line segments. So, its intersection with your 45 degree line may be an entire segment. -- === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl ... > Case 2: a variant on that, we have colours black and white, and start > off with the interval 0-1 painted white. On each iteration, each > interval is subdivided, with the first sub-segment of each sub-interval > painted black if its parent interval was white, or vice-versa. Then, > after an infinite number of iterations, is each real point painted black > or white? And, if we decide to delete all white points after an infinite > number of iterations, would that make any difference? Some numbers will change between black and white indefinitely. So you can not ask whether there is an end colour. This is quite similar to the physical question about a fly going from train to train to an imminent disaster (head on collision). The question not being how many miles the fly did travel, but what direction the fly did travel when the disaster did occur. > Or is that all completely irrelevant? Not irrelevant, but unanswerable. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl ... > 1) Is there a set with cardinality greater than N but less than R ? The Continuum Hypothesis states that there isn't, but it is not provable. And so some people use it as an axiom, others do not. > 2) I am (dimly aware) that mathematicians talk about an infinite range > of cardinalities. Is that just an abstract concept, or can you point to > some set and say e.g. that has a higher cardinality than the reals? That has already been shown by Cantor. In one of his proofs he did show (in effect) that given a set K, the set of subsets of K has a cardinality that is strictly larger than the cardinality of K itself. The most succinct proof of this is due to Hessenberg. The proof is fairly simple. Consider a set K and its powerset (i.e. set of subsets) P(k). Assume that there is an injection f: K -> P(K) (those clearly do exist, consider f(k) = {k}). Now we set out to prove that f (as given) can not be a surjection. Clearly for each 'k' in K, f(k) is a subset of K, so we can consider those 'k' where 'k' is an element of f(k) and those where 'k' is not an element of f(k). That defines two subsets of K, call them Y (the 'k' in that set map to subsets that contain 'k') and N. Now the question is, is there a 'k' that maps to N? (1) Assume 'k' in N, but by the definition of N, the elements do not map to a subset that contains themselve, so 'k' not in N. (2) Assume 'k' in Y, but by the definition of Y, the elements do map to a subset that contains themselve, and N does not contain 'k', so 'k' can not be in Y. As 'k' is in neither Y nor N, there is no 'k' that maps to N and so the map is not surjective. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion 1) Is there a set with cardinality greater than N but less than R ? The Continuum Hypothesis states that there isn't, but it is not provable. > And so some people use it as an axiom, others do not. > 2) I am (dimly aware) that mathematicians talk about an infinite range > of cardinalities. Is that just an abstract concept, or can you point to > some set and say e.g. that has a higher cardinality than the reals? That has already been shown by Cantor. In one of his proofs he did show > (in effect) that given a set K, the set of subsets of K has a cardinality > that is strictly larger than the cardinality of K itself. The most > succinct proof of this is due to Hessenberg. The proof is fairly simple. > Consider a set K and its powerset (i.e. set of subsets) P(k). Assume > that there is an injection f: K -> P(K) (those clearly do exist, consider > f(k) = {k}). Now we set out to prove that f (as given) can not be a > surjection. Clearly for each 'k' in K, f(k) is a subset of K, so we can > consider those 'k' where 'k' is an element of f(k) and those where 'k' is not > an element of f(k). That defines two subsets of K, call them Y (the 'k' > in that set map to subsets that contain 'k') and N. Now the question is, > is there a 'k' that maps to N? > (1) Assume 'k' in N, but by the definition of N, the elements do not map > to a subset that contains themselve, so 'k' not in N. > (2) Assume 'k' in Y, but by the definition of Y, the elements do map to > a subset that contains themselve, and N does not contain 'k', so > 'k' can not be in Y. > As 'k' is in neither Y nor N, there is no 'k' that maps to N and so the map > is not surjective. Alas, this proof has nothing to do with cardinality of sets. === Subject: Re: Cantor Confusion > ... > 1) Is there a set with cardinality greater than N but less than R ? The Continuum Hypothesis states that there isn't, but it is not provable. > And so some people use it as an axiom, others do not. > 2) I am (dimly aware) that mathematicians talk about an infinite range > of cardinalities. Is that just an abstract concept, or can you point to > some set and say e.g. that has a higher cardinality than the reals? That has already been shown by Cantor. In one of his proofs he did show > (in effect) that given a set K, the set of subsets of K has a cardinality > that is strictly larger than the cardinality of K itself. The most > succinct proof of this is due to Hessenberg. The proof is fairly simple. > Consider a set K and its powerset (i.e. set of subsets) P(k). Assume > that there is an injection f: K -> P(K) (those clearly do exist, consider > f(k) = {k}). Now we set out to prove that f (as given) can not be a > surjection. Clearly for each 'k' in K, f(k) is a subset of K, so we can > consider those 'k' where 'k' is an element of f(k) and those where 'k' is > not > an element of f(k). That defines two subsets of K, call them Y (the 'k' > in that set map to subsets that contain 'k') and N. Now the question is, > is there a 'k' that maps to N? > (1) Assume 'k' in N, but by the definition of N, the elements do not map > to a subset that contains themselve, so 'k' not in N. > (2) Assume 'k' in Y, but by the definition of Y, the elements do map to > a subset that contains themselve, and N does not contain 'k', so > 'k' can not be in Y. > As 'k' is in neither Y nor N, there is no 'k' that maps to N and so the map > is not surjective. Alas, this proof has nothing to do with cardinality of sets. Alas for WM, it does have something to do with cardinalities of sets. It proves that given any set, S, and its power set P(S), the set of all subsets of S, then, according to Cantor's definition of order for cardinalities, Card(S) < Card(P(S)). That WM doesn't like it is irrelevant and illogical. === Subject: Re: Cantor Confusion > (snip) > I had previously imagined that the Continuum Hypothesis related to > whether the set of reals (as 0 dimensional points, each with no > neighbouring point) could cover the line, with dimension 1 (Which seemed > like a good question). >>You are asking whether R and the real line are the same set. In order >>for that question to have a nontrivial answer, you must have some meaning >>in mind for the real line other than R. What is your definition? > Well until recently my naive view was that points and lines were > different entities, and you might as well ask how many metres in an acre > as how many points in a line. I would have said (in your terminology) > that there is clearly an injection from the reals into the line, but > that you can never cover it. But I doubt now if that is true, or if the > question is a sensible one to start with. I don't follow. Is it your claim that: (1) A line does not have any points as members at all, or (2) A line has points as members, but perhaps there aren't enough points to fill it? Your metres-in-an-acre analogy seems to suggest (1), but otherwise you seem to be saying (2). If (2) is your claim, then can you show me a point on the line that is not represented by a real number? If you can do this, then evidently the line to you must be something other than simply the set of real numbers, as it is to most people. Can you elaborate on this? -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Cantor Confusion <$6rsRmghMfuFFwrV@phoenixsystems.demon.co.uk> <$EiMOojF$huFFwoB@phoenixsystems.demon.co.uk> <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> (snip) > I had previously imagined that the Continuum Hypothesis related to >> whether the set of reals (as 0 dimensional points, each with no >> neighbouring point) could cover the line, with dimension 1 (Which seemed >> like a good question). You are asking whether R and the real line are the same set. In order >for that question to have a nontrivial answer, you must have some meaning >in mind for the real line other than R. What is your definition? > Well until recently my naive view was that points and lines were >> different entities, and you might as well ask how many metres in an acre >> as how many points in a line. I would have said (in your terminology) >> that there is clearly an injection from the reals into the line, but >> that you can never cover it. But I doubt now if that is true, or if the >> question is a sensible one to start with. I don't follow. Is it your claim that: (1) A line does not have any points as members at all, or > (2) A line has points as members, but perhaps there aren't > enough points to fill it? Your metres-in-an-acre analogy seems to suggest (1), but otherwise you >seem to be saying (2). If (2) is your claim, then can you show me a point on the line that is >not represented by a real number? If you can do this, then evidently >the line to you must be something other than simply the set of real >numbers, as it is to most people. Can you elaborate on this? Well I meant 2) - (and I can't find any location in the acre that is not covered by a metre line either). And I can't answer your question . Yes, of course, the moment that you specify a real number you have defined a location on the line but I am suffering from dimensional anxiety, particularly when you observe that any point on the line has no nearest neighbour. If I was a point sized entity wanting to walk from 0 to 1 without falling down a crack, how would I do it? (I know that is not a particularly sensible question, just illustrating my mis-perspective). > -- Andy Smith === Subject: Re: Cantor Confusion >> (snip) > I had previously imagined that the Continuum Hypothesis related to >> whether the set of reals (as 0 dimensional points, each with no >> neighbouring point) could cover the line, with dimension 1 (Which seemed >> like a good question). You are asking whether R and the real line are the same set. In order >for that question to have a nontrivial answer, you must have some meaning >in mind for the real line other than R. What is your definition? > Well until recently my naive view was that points and lines were >> different entities, and you might as well ask how many metres in an acre >> as how many points in a line. I would have said (in your terminology) >> that there is clearly an injection from the reals into the line, but >> that you can never cover it. But I doubt now if that is true, or if the >> question is a sensible one to start with. I don't follow. Is it your claim that: (1) A line does not have any points as members at all, or > (2) A line has points as members, but perhaps there aren't > enough points to fill it? Your metres-in-an-acre analogy seems to suggest (1), but otherwise you >seem to be saying (2). >If (2) is your claim, then can you show me a point on the line that is >not represented by a real number? If you can do this, then evidently >the line to you must be something other than simply the set of real >numbers, as it is to most people. Can you elaborate on this? > Well I meant 2) - (and I can't find any location in the acre that is not > covered by a metre line either). And I can't answer your question . Yes, > of course, the moment that you specify a real number you have defined a > location on the line but I am suffering from dimensional anxiety, > particularly when you observe that any point on the line has no nearest > neighbour. If I was a point sized entity wanting to walk from 0 to 1 > without falling down a crack, how would I do it? (I know that is not a > particularly sensible question, just illustrating my mis-perspective). Where does one find a crack between points not already filled by a point? The least upper bound property says that if any non-empty set of reals is bounded above, then there is a particular real number as its least upper bound filling any crack between that set and the set of reals greater than all of the members of that set. Similarly for sets of reals bounded below. So where is there room for any crack between points, which is not filled in by a least upper bound or greatest lower bound? === Subject: Re: Cantor Confusion > (snip) > I had previously imagined that the Continuum Hypothesis related to > whether the set of reals (as 0 dimensional points, each with no > neighbouring point) could cover the line, with dimension 1 (Which seemed > like a good question). >>You are asking whether R and the real line are the same set. In order >>for that question to have a nontrivial answer, you must have some meaning >>in mind for the real line other than R. What is your definition? > Well until recently my naive view was that points and lines were > different entities, and you might as well ask how many metres in an acre > as how many points in a line. I would have said (in your terminology) > that there is clearly an injection from the reals into the line, but > that you can never cover it. But I doubt now if that is true, or if the > question is a sensible one to start with. >>I don't follow. Is it your claim that: >> (1) A line does not have any points as members at all, or >> (2) A line has points as members, but perhaps there aren't >> enough points to fill it? >>Your metres-in-an-acre analogy seems to suggest (1), but otherwise you >>seem to be saying (2). >>If (2) is your claim, then can you show me a point on the line that is >>not represented by a real number? If you can do this, then evidently >>the line to you must be something other than simply the set of real >>numbers, as it is to most people. Can you elaborate on this? > Well I meant 2) - (and I can't find any location in the acre that is not > covered by a metre line either). And I can't answer your question . Yes, > of course, the moment that you specify a real number you have defined a > location on the line but I am suffering from dimensional anxiety, > particularly when you observe that any point on the line has no nearest > neighbour. If I was a point sized entity wanting to walk from 0 to 1 > without falling down a crack, how would I do it? (I know that is not a > particularly sensible question, just illustrating my mis-perspective). Zeno was bothered by this when he talked about an arrow in flight being know that the arrow is never actually at rest during flight. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Cantor Confusion <$6rsRmghMfuFFwrV@phoenixsystems.demon.co.uk> <$EiMOojF$huFFwoB@phoenixsystems.demon.co.uk> <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> In message , Dave Seaman >> (snip) > I had previously imagined that the Continuum Hypothesis related to >> whether the set of reals (as 0 dimensional points, each with no >> neighbouring point) could cover the line, with dimension 1 (Which seemed >> like a good question). >>You are asking whether R and the real line are the same set. In order >for that question to have a nontrivial answer, you must have some meaning >in mind for the real line other than R. What is your definition? > Well until recently my naive view was that points and lines were >> different entities, and you might as well ask how many metres in an acre >> as how many points in a line. I would have said (in your terminology) >> that there is clearly an injection from the reals into the line, but >> that you can never cover it. But I doubt now if that is true, or if the >> question is a sensible one to start with. I don't follow. Is it your claim that: (1) A line does not have any points as members at all, or > (2) A line has points as members, but perhaps there aren't > enough points to fill it? Your metres-in-an-acre analogy seems to suggest (1), but otherwise you >seem to be saying (2). >If (2) is your claim, then can you show me a point on the line that is >not represented by a real number? If you can do this, then evidently >the line to you must be something other than simply the set of real >numbers, as it is to most people. Can you elaborate on this? >> Well I meant 2) - (and I can't find any location in the acre that is not >> covered by a metre line either). And I can't answer your question . Yes, >> of course, the moment that you specify a real number you have defined a >> location on the line but I am suffering from dimensional anxiety, >> particularly when you observe that any point on the line has no nearest >> neighbour. If I was a point sized entity wanting to walk from 0 to 1 >> without falling down a crack, how would I do it? (I know that is not a >> particularly sensible question, just illustrating my mis-perspective). Zeno was bothered by this when he talked about an arrow in flight being >know that the arrow is never actually at rest during flight. But the positions are reversed? I would be very happy to consider distance along a line as a continuous function x and rate of change of position dx/dt etc. But by asserting (or, as I understand it, defining) the line is an infinite collection of points one reintroduces the conceptual granularity that bothered Zeno? You just start with points and define the real line as the set of all of them. I have an image of a line as a continuous thing of point width, and it is trying to marry up the perception of continuity with the set of real points that is difficult for me. I mentioned fractals because they are a clear example of e.g. how you can get infinite volume from a finite surface area, etc. so I thought that the essentially recursive structure of the reals (you can map the real line into the space between any two points, indefinitely) was some sort of fractal explanation of how the line could be simultaneously pointlike and continuous. But doubtless, as with everything else, my problem is getting my head to a point where I see things in the correct perspective. -- Andy Smith === Subject: Re: Cantor Confusion > In message , Dave Seaman >> (snip) > I had previously imagined that the Continuum Hypothesis related to >> whether the set of reals (as 0 dimensional points, each with no >> neighbouring point) could cover the line, with dimension 1 (Which seemed >> like a good question). >>You are asking whether R and the real line are the same set. In order >for that question to have a nontrivial answer, you must have some meaning >in mind for the real line other than R. What is your definition? > Well until recently my naive view was that points and lines were >> different entities, and you might as well ask how many metres in an acre >> as how many points in a line. I would have said (in your terminology) >> that there is clearly an injection from the reals into the line, but >> that you can never cover it. But I doubt now if that is true, or if the >> question is a sensible one to start with. I don't follow. Is it your claim that: (1) A line does not have any points as members at all, or > (2) A line has points as members, but perhaps there aren't > enough points to fill it? Your metres-in-an-acre analogy seems to suggest (1), but otherwise you >seem to be saying (2). >If (2) is your claim, then can you show me a point on the line that is >not represented by a real number? If you can do this, then evidently >the line to you must be something other than simply the set of real >numbers, as it is to most people. Can you elaborate on this? >> Well I meant 2) - (and I can't find any location in the acre that is not >> covered by a metre line either). And I can't answer your question . Yes, >> of course, the moment that you specify a real number you have defined a >> location on the line but I am suffering from dimensional anxiety, >> particularly when you observe that any point on the line has no nearest >> neighbour. If I was a point sized entity wanting to walk from 0 to 1 >> without falling down a crack, how would I do it? (I know that is not a >> particularly sensible question, just illustrating my mis-perspective). Zeno was bothered by this when he talked about an arrow in flight being >know that the arrow is never actually at rest during flight. > But the positions are reversed? I would be very happy to consider > distance along a line as a continuous function x and rate of change of > position dx/dt etc. But by asserting (or, as I understand it, defining) > the line is an infinite collection of points one reintroduces the > conceptual granularity that bothered Zeno? > You just start with points and define the real line as the set of all of > them. I have an image of a line as a continuous thing of point width, > and it is trying to marry up the perception of continuity with the set > of real points that is difficult for me. Suppose at time zero you start walking in a straight line at constant speed. At any time, you are at some point. And, you passed through each point at some time. And, the time you were at a point is the same as the distance you had travelled to get to that point. And, we can measure time (and distance) using real numbers. So, if you think lines aren't made up of points, then time isn't made up of instants. -- === Subject: Re: Cantor Confusion On Fri, 2 Feb 2007 14:09:42 -0500, >> In message , Dave Seaman [. . .] >> You just start with points and define the real line as the set of all of >> them. I have an image of a line as a continuous thing of point width, >> and it is trying to marry up the perception of continuity with the set >> of real points that is difficult for me. Suppose at time zero you start walking in a straight line at constant >speed. At any time, you are at some point. So we're doing parables instead of math now? > And, you passed through each >point at some time. Or you didn't. Might we dispense with the metaphors already? > And, the time you were at a point is the same as the >distance you had travelled to get to that point. Oh gloria!! We are indeed saved. > And, we can measure >time (and distance) using real numbers. And what are these magic things you call real numbers pray tell? > So, if you think lines aren't >made up of points, then time isn't made up of instants. So if time isn't made up of instants lines aren't made up of points? Rather a curious concept I must say but correct as far as it goes. ~v~~ === Subject: Re: Cantor Confusion <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> Your post appears to have been deleted off the NG server, or else my NG reader can't cope with this thread. Piecing it together: > You just start with points and define the real line as the set of all of > them. I have an image of a line as a continuous thing of point width, > and it is trying to marry up the perception of continuity with the set > of real points that is difficult for me. >>Suppose at time zero you start walking in a straight line at constant >>speed. At any time, you are at some point. >And, you passed through each >>point at some time. >And, the time you were at a point is the same as the >>distance you had travelled to get to that point. >And, we can measure >>time (and distance) using real numbers. >So, if you think lines aren't >>made up of points, then time isn't made up of instants. > OK, I understand that, and that is all self-consistent. Both distance and time along conceptual axes are equivalent, but (my naive view) would have been that both were 1-dimensional entities, and that time was no more made up of connected but non-adjacent instants than the line is made up of connected but non-adjacent points. -- Andy Smith === Subject: Re: Cantor Confusion > In message , Dave Seaman >>Zeno was bothered by this when he talked about an arrow in flight being >>know that the arrow is never actually at rest during flight. > But the positions are reversed? I would be very happy to consider > distance along a line as a continuous function x and rate of change of > position dx/dt etc. But by asserting (or, as I understand it, defining) > the line is an infinite collection of points one reintroduces the > conceptual granularity that bothered Zeno? What is your proposed alternate definition of a line? I don't know what you mean by granularity. Seems to me there would be granularity only in the discrete case, where each point has an immediate neighbor. Obviously, that doesn't hold on the real line. Hence, no granularity. > You just start with points and define the real line as the set of all of > them. I have an image of a line as a continuous thing of point width, > and it is trying to marry up the perception of continuity with the set > of real points that is difficult for me. I mentioned fractals because > they are a clear example of e.g. how you can get infinite volume from a > finite surface area, etc. so I thought that the essentially recursive > structure of the reals (you can map the real line into the space between > any two points, indefinitely) was some sort of fractal explanation of > how the line could be simultaneously pointlike and continuous. But > doubtless, as with everything else, my problem is getting my head to a > point where I see things in the correct perspective. Continuity is a term that applies to functions, not to lines. A line is connected, but it has the property that the removal of any single point disconnects the line. A plane, for example, does not have that property. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Cantor Confusion <$EiMOojF$huFFwoB@phoenixsystems.demon.co.uk> <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> In message , Dave Seaman Zeno was bothered by this when he talked about an arrow in flight being >know that the arrow is never actually at rest during flight. >> But the positions are reversed? I would be very happy to consider >> distance along a line as a continuous function x and rate of change of >> position dx/dt etc. But by asserting (or, as I understand it, defining) >> the line is an infinite collection of points one reintroduces the >> conceptual granularity that bothered Zeno? What is your proposed alternate definition of a line? > Well I am tempted to say the locus of intersection of two planes, but I can see that just defers the question to what is a plane?. Doubtless you would like me to respond that it is a connected linear sequence of points, and I am sure that you are right. >I don't know what you mean by granularity. Seems to me there would be >granularity only in the discrete case, where each point has an immediate >neighbor. Obviously, that doesn't hold on the real line. Hence, no >granularity. Yes, I understand, I was speaking figuratively - the whole issue is about the infinite set of real points being connected. > You just start with points and define the real line as the set of all of >> them. I have an image of a line as a continuous thing of point width, >> and it is trying to marry up the perception of continuity with the set >> of real points that is difficult for me. I mentioned fractals because >> they are a clear example of e.g. how you can get infinite volume from a >> finite surface area, etc. so I thought that the essentially recursive >> structure of the reals (you can map the real line into the space between >> any two points, indefinitely) was some sort of fractal explanation of >> how the line could be simultaneously pointlike and continuous. But >> doubtless, as with everything else, my problem is getting my head to a >> point where I see things in the correct perspective. Continuity is a term that applies to functions, not to lines. A line is >connected, but it has the property that the removal of any single point >disconnects the line. A plane, for example, does not have that property. > Yes, understood, I should have said connected. -- Andy Smith === Subject: Re: Cantor Confusion <$EiMOojF$huFFwoB@phoenixsystems.demon.co.uk> <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> (snip everything else) At root I think my problem comes down to achieving a suitably Zen-like perspective on the following apparently incompatible statements: 1) The real line is made up of an ordered and infinite set of points, and is connected. 2) No point on the real line has an adjacent point. -- Andy Smith === Subject: Re: Cantor Confusion > (snip everything else) At root I think my problem comes down to achieving a suitably Zen-like > perspective on the following apparently incompatible statements: 1) The real line is made up of an ordered and infinite set of points, > and is connected. 2) No point on the real line has an adjacent point. Do you have the same problem with the following? 1) The real numbers consist of an infinite number of numbers that are ordered and connected. 2) No real number has an adjacent real number. Of course, the real problem is that you are relying on your intuition. The first thing a mathematician learns is to adjust their intuition to match the facts. -- === Subject: Re: Cantor Confusion <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> The first thing a mathematician learns is to adjust their intuition to > match the facts. True of mathematicians, logicians, scientists, and realists, all of whom deal with some kind of demonstrably truths derived from simpler truths, whether they are abstract or physically concrete. On the other hand, those that do not make this adjustment are generally of the more philosophical or religious bent, who deal with beliefs and rules only loosely based on (or not based on at all) provable logical or physical truths. Thus it's pretty accurate to say that mathematical cranks are more religious in their fervent beliefs than true mathematicians. Indeed, mathematicians can be downright agnostic about many things (e.g., the CH). === Subject: Re: Cantor Confusion On Fri, 2 Feb 2007 16:51:39 -0500, >> (snip everything else) >> >> At root I think my problem comes down to achieving a suitably Zen-like >> perspective on the following apparently incompatible statements: >> >> 1) The real line is made up of an ordered and infinite set of points, >> and is connected. >> >> 2) No point on the real line has an adjacent point. Do you have the same problem with the following? 1) The real numbers consist of an infinite number of numbers that are >ordered and connected. An infinite number of numbers? And what exactly does that mean when it's at home? >2) No real number has an adjacent real number. It doesn't? And why perchance not pray tell? The last I heard from Bob Kolker the real number line is densely pointy.I mean unless you really mean to indicate the real number line is sparsely pointy in which case you really need to take it up with Bob since I, like mathematicians, have no preference in the matter because I use lines to define points. >Of course, the real problem is that you are relying on your intuition. Whereas you prefer to rely on someone elses intuition. >The first thing a mathematician learns is to adjust their intuition to >match the facts. Which facts exactly did you have in mind? True facts or just any old facts? ~v~~ === Subject: Re: Cantor Confusion <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> perspective No Zen-like perspective is required. Knowing the axioms and defintions, though, does help. MoeBlee === Subject: Re: Cantor Confusion > At root I think my problem comes down to achieving a suitably Zen-like >> perspective No Zen-like perspective is required. Knowing the axioms and >defintions, though, does help. As does knowing Zen. ~v~~ === Subject: Re: Cantor Confusion <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> At root I think my problem comes down to achieving a suitably Zen-like > perspective >>No Zen-like perspective is required. Knowing the axioms and >>defintions, though, does help. As does knowing Zen. I read Zen and the Art of Motorcycle Maintenance once, which I recall was mildly interesting & entertaining. In Western society I think that it is common parlance to describe something as Zen-like to imply either, that there exists a deep resolution of some apparently irreconcilable statements, or that consideration of some suitably impossible conundrum may allow some enlightenment on a related problem. I meant no disrespect to any religious beliefs that you may hold ... -- Andy Smith === Subject: Re: Cantor Confusion >No Zen-like perspective is required. Knowing the axioms and >>defintions, though, does help. As does knowing Zen. I read Zen and the Art of Motorcycle Maintenance once, which I recall > was mildly interesting & entertaining. In Western society I think that it is common parlance to describe > something as Zen-like to imply either, that there exists a deep > resolution of some apparently irreconcilable statements, or that > consideration of some suitably impossible conundrum may allow some > enlightenment on a related problem. I meant no disrespect to any religious beliefs that you may hold ... An apt way of describing Lester's knowledge of math. -- === Subject: Re: Cantor Confusion On Sat, 3 Feb 2007 20:21:17 -0500, >No Zen-like perspective is required. Knowing the axioms and >defintions, though, does help. >>As does knowing Zen. >> >> I read Zen and the Art of Motorcycle Maintenance once, which I recall >> was mildly interesting & entertaining. >> >> In Western society I think that it is common parlance to describe >> something as Zen-like to imply either, that there exists a deep >> resolution of some apparently irreconcilable statements, or that >> consideration of some suitably impossible conundrum may allow some >> enlightenment on a related problem. >> >> I meant no disrespect to any religious beliefs that you may hold ... An apt way of describing Lester's knowledge of math. Also sprach the Anita Bryant of faith based mathematics. ~v~~ === Subject: Re: Cantor Confusion > (snip everything else) > At root I think my problem comes down to achieving a suitably Zen-like > perspective on the following apparently incompatible statements: > 1) The real line is made up of an ordered and infinite set of points, > and is connected. > 2) No point on the real line has an adjacent point. I don't understand why you think those two statements are incompatible. If any point on the real line actually *had* an adjacent point, then the line would be disconnected precisely at the gap between those two points. Hence, connectedness is incompatible with the existence of adjacent points. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Cantor Confusion > (snip everything else) At root I think my problem comes down to achieving a suitably Zen-like > perspective on the following apparently incompatible statements: 1) The real line is made up of an ordered and infinite set of points, > and is connected. 2) No point on the real line has an adjacent point. I don't understand why you think those two statements are incompatible. > If any point on the real line actually *had* an adjacent point, then the > line would be disconnected precisely at the gap between those two points. > Hence, connectedness is incompatible with the existence of adjacent > points. I suppose he is thinking of points as having size, e.g., like little marbles. Of course, they aren't like that. -- === Subject: Re: Cantor Confusion <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> (snip everything else) > At root I think my problem comes down to achieving a suitably Zen-like > perspective on the following apparently incompatible statements: > 1) The real line is made up of an ordered and infinite set of points, > and is connected. > 2) No point on the real line has an adjacent point. I don't understand why you think those two statements are incompatible. > If any point on the real line actually *had* an adjacent point, then the > line would be disconnected precisely at the gap between those two points. > Hence, connectedness is incompatible with the existence of adjacent > points. I suppose he is thinking of points as having size, e.g., like little > marbles. Of course, they aren't like that. > Just to screw with Andy's head... IIRC, Rudy Rucker in White Light talks about a book with an uncountable number of pages. I think it was an Encyclopedia of Curves, containing an image of every curve in R^2, spread across two adjacent pages, and of course drawn in lovingly accurate detail. The pages were numbered; but to number the pages, they used an infinitely long string. This was written at the bootom of the page, with the first line in 1pt type, the second line in 1/2 pt type, and so on with each succesive line being written in a type size 1/2 the size of the previous line. You can certainly open this book to any particular page; but the pages are so thin that no matter how hard you try, you always end up flipping a bunch of pages, not a single page. (Which is only to say.: Pages cannot be planes, any more than marbles can be points!). === Subject: Re: Cantor Confusion On Fri, 2 Feb 2007 16:49:09 -0500, >> (snip everything else) >> >> At root I think my problem comes down to achieving a suitably Zen-like >> perspective on the following apparently incompatible statements: >> >> 1) The real line is made up of an ordered and infinite set of points, >> and is connected. >> >> 2) No point on the real line has an adjacent point. >> >> I don't understand why you think those two statements are incompatible. >> If any point on the real line actually *had* an adjacent point, then the >> line would be disconnected precisely at the gap between those two points. >> Hence, connectedness is incompatible with the existence of adjacent >> points. I suppose he is thinking of points as having size, e.g., like little >marbles. Of course, they aren't like that. Of course they aren't. Infintesimals have size not points. Your brain is like little marbles. ~v~~ === Subject: Re: Cantor Confusion <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> perspective on the following apparently incompatible statements: 1) The real line is made up of an ordered and infinite set of points, > and is connected. 2) No point on the real line has an adjacent point. I don't understand why you think those two statements are incompatible. > If any point on the real line actually *had* an adjacent point, then the > line would be disconnected precisely at the gap between those two points. > Hence, connectedness is incompatible with the existence of adjacent > points. >>I suppose he is thinking of points as having size, e.g., like little >>marbles. Of course, they aren't like that. > No, I think of points as 0-dimensional, with no size. They must necessarily have zero size, otherwise you couldn't have an infinite number of points in a finite distance. The issue was whether you could actually cover the line with points. But I can see that my Euclidean view of a line and dimensionality of points, lines and planes is not sustainable, and I can't see anything wrong with your definition of the line, other than my mis-intuition; -- Andy Smith === Subject: Re: Cantor Confusion <$EiMOojF$huFFwoB@phoenixsystems.demon.co.uk> <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> infinite volume from a finite surface area should of course have read infinite surface area from a finite volume -- Andy Smith === Subject: Re: Cantor Confusion > infinite volume from a finite surface area > should of course have read > infinite surface area from a finite volume Are you familiar with Gabriel's Horn? That's the curve y = log x for x >= 1, rotated about the x-axis. It has finite volume and infinite surface area, but it is not a fractal. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Cantor Confusion <$EiMOojF$huFFwoB@phoenixsystems.demon.co.uk> <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> <+V84fQs4y0wFFwpK@phoenixsystems.demon.co.uk> > infinite volume from a finite surface area > should of course have read > infinite surface area from a finite volume Are you familiar with Gabriel's Horn? That's the curve y = log x for x >>= 1, rotated about the x-axis. It has finite volume and infinite >surface area, but it is not a fractal. > No I wasn't, but that's quite fun. (I guess you meant to write 0 infinite volume from a finite surface area > should of course have read > infinite surface area from a finite volume >>Are you familiar with Gabriel's Horn? That's the curve y = log x for x >= 1, rotated about the x-axis. It has finite volume and infinite >>surface area, but it is not a fractal. > No I wasn't, but that's quite fun. (I guess you meant to write 0 for the defining curve) No, actually I meant y = 1/x for 1 <= x < oo. Log x is the antiderivative. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Cantor Confusion <$EiMOojF$huFFwoB@phoenixsystems.demon.co.uk> <622tCB$Tw0vFFwT5@phoenixsystems.demon.co.uk> <1iAKSznLBxwFFwaz@phoenixsystems.demon.co.uk> <0kbqjRrnI0wFFwqi@phoenixsystems.demon.co.uk> <+V84fQs4y0wFFwpK@phoenixsystems.demon.co.uk> Are you familiar with Gabriel's Horn? That's the curve y = log x for x >= 1, rotated about the x-axis. It has finite volume and infinite >>surface area, but it is not a fractal. >No I wasn't, but that's quite fun. (I guess you meant to write 0for the defining curve) No, that doesn't work either. I think you meant the surface of revolution of y=1/x for x>=1. Nice trumpet! -- Andy Smith === Subject: Re: Cantor Confusion Continued: > > Why do you accept then that the union of sets of nodes (finite trees) > establishes an infinite set of nodes (an infinite tree) without such > an infinite set (infinite tree) being contained in the union? Because that is something different. The union of sets of paths > establishes an infinite set of paths. But that infinite set of > paths does not contain an infinite path, because none of the > constituent sets contains one. So the union of finite chains of nodes like the following 0 12 6543 does not contain an infinite chain of nodes of the form 0 12 6543 789... ??? Frankly, I do not see much difference to: The union of finite paths contains an infinite path. To translate to sets of nodes: the union of the sets of nodes establishes > an infinite set of nodes that does not contain an infinite node. > More formal, define > a path p_k as {n_0, n_1, n_2, n_k-1}, where the n_i are nodes. The sets > of paths from the finite trees contain p_k for all k. The union of the > trees contains a path {n_0, n_1, n_2, ...}, but the union of the sets of > paths does consist of: > { {n_0, n_1, n_2, ..., n_k-1} | n_0 is a node, k in N} > and so does not contain {n_0, n_1, n_2, ...}. So there is *no* infinite > path in the union of the sets of paths from finite trees. > > How then does this infinite path manage to enter the union of the > finite trees? Because the set of paths of the union of finite trees is *not* contained in > the union of the sets of finite paths. And it is not contained in any of the finite trees. Why is it said to be contained in the union of all finite trees? > Because I need the identity of path-lengths and numbers for my > argument. > > Well, it does not hold. If you measure the lengths as you did define, the > union of all lengths is not a length, because it is not a natural number. > > If all natural numbers are finite, their union as defined for > pathlengths cannot be infinite. Why not? Because there cannot be more than infinitely many natural numbers. All those infinitely many numbers, however, are allegedly finite. > Except you assert that an infinite union requires a number of infinite > size. It is not a natural number. So it cannot be in the set of natural numbers. > If T(oo) is constructed as the union of all finite trees T(k), > then every path in P is a path which is in the union > P(1) U P(2) U ... of the elements of P_C. > > Prove it. The union of P(i) contains finite paths only, *by the > definition*. > > The infinite tree contains finite trees only. Is it finite on that > behalf? > > No. The union of finite sets can be infinite. But the union of sets > with finite elements does not contain an infinite element. > > Finite trees are sets with finite elements, namely the finite number > of nodes. That number is not an element of a tree. The elements of the tree are > *nodes* (by your own definition). The union of the finite trees contains > finite elements only (i.e. nodes at a finite distance from the root). > But there are infinite sequences of such elements. Don't you accept that a finite tree simutaneously represents a finite chain of nodes? And that an infinite tree represents an infinite chain of nodes? > And, as > your P(i) are sets with finite elements, their union does not contain > an infinite element. On the other hand, their union *is* an infinite > set. And so it is an infinite set of finite elements. > > The cardinal number is infinite, each pathlength is finite. Right. But you said that there is an infinite pathlength corresponding to an infinite set of paths. > The same you can require for the tree. A finite tree is nothing but a > sequence of nodes (if you count them as Cantor did count the rational > numbers: > > 0 > 12 > 6543 > 789...) > > What was your argument distinguishing infinite unions of finite trees > and finite paths, respectively? None. I distinguish unions of finite trees and unions of *sets* of > finite paths. Above you are using P(1) U P(2) U ..., where each > P(i) is a *set* of finite paths. What about unions of sets of finite trees? Would they work completely different from unions of sets of paths? In particular, would the union of sets of finite trees not contain or establish the inifnite tree? > Wrong. Infinite paths do not have a path-length that is a natural number. > > Pathlengths ARE natural numbers. For finite paths. In all cases, by definition. === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl > Why do you accept then that the union of sets of nodes (finite trees) > establishes an infinite set of nodes (an infinite tree) without such > an infinite set (infinite tree) being contained in the union? > > Because that is something different. The union of sets of paths > establishes an infinite set of paths. But that infinite set of > paths does not contain an infinite path, because none of the > constituent sets contains one. > > So the union of finite chains of nodes like the following > 0 > 12 > 6543 > does not contain an infinite chain of nodes of the form > 0 > 12 > 6543 > 789... > ??? your own interpretation to it. > Frankly, I do not see much difference to: The union of finite paths > contains an infinite path. But that is true. What I state again, again and again, but what you misread each and every time is: the union of sets of finite paths does not contain an infinite path. > How then does this infinite path manage to enter the union of the > finite trees? > > Because the set of paths of the union of finite trees is *not* contained in > the union of the sets of finite paths. > > And it is not contained in any of the finite trees. Why is it said to > be contained in the union of all finite trees? Because an infinite path is a union of paths. And when you construct unions of *sets* of paths you are not constructing unions of paths. > If all natural numbers are finite, their union as defined for > pathlengths cannot be infinite. > > Why not? > > Because there cannot be more than infinitely many natural numbers. All > those infinitely many numbers, however, are allegedly finite. And there are infinitely many such. And they are, indeed, all finite. The path lengths however do not map to natural numbers, which you assert. That is valid for finite paths, but not for infinite paths. Let's, for once, go to a proper definition (assuming all are ordered sets): given a path {n_1, n_2, ..., n_k} then the path length is |{n_1, n_2, ..., n_k}| = k. Clearly for finite paths the path length is finite, and so a natural number. However, when we construct the path: p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... we get as path: {n_1, n_2, n_3, n_4, ...} the path length in this case is *not* a natural number. It is the cardinality of N. > Except you assert that an infinite union requires a number of infinite > size. > > It is not a natural number. > > So it cannot be in the set of natural numbers. Right. > That number is not an element of a tree. The elements of the tree are > *nodes* (by your own definition). The union of the finite trees contains > finite elements only (i.e. nodes at a finite distance from the root). > But there are infinite sequences of such elements. > > Don't you accept that a finite tree simutaneously represents a finite > chain of nodes? And that an infinite tree represents an infinite chain > of nodes? Oh, it does contain them as subsets, yes. > And, as > your P(i) are sets with finite elements, their union does not contain > an infinite element. On the other hand, their union *is* an infinite > set. And so it is an infinite set of finite elements. > > The cardinal number is infinite, each pathlength is finite. > > Right. > > But you said that there is an infinite pathlength corresponding to an > infinite set of paths. But that set of paths is *not* the union of the P(i). Why do you think that the union of sets with finite elements does contain an infinite element? > What was your argument distinguishing infinite unions of finite trees > and finite paths, respectively? > > None. I distinguish unions of finite trees and unions of *sets* of > finite paths. Above you are using P(1) U P(2) U ..., where each > P(i) is a *set* of finite paths. > > What about unions of sets of finite trees? Would they work completely > different from unions of sets of paths? In particular, would the union > of sets of finite trees not contain or establish the inifnite tree? The union of sets of finite trees would neither contain, nor establish, an infinite tree. It would not contain one, because none of the constituent sets contains an infinite tree. It would not establish one, because the union is a set of trees, not a tree. A tree is a set of nodes, a set of trees is a set of sets of nodes. > Wrong. Infinite paths do not have a path-length that is a > natural number. > > Pathlengths ARE natural numbers. > > For finite paths. > > In all cases, by definition. *What* definition? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion So the union of finite chains of nodes like the following > 0 > 12 > 6543 > does not contain an infinite chain of nodes of the form > 0 > 12 > 6543 > 789... > ??? your own interpretation to it. > Frankly, I do not see much difference to: The union of finite paths > contains an infinite path. But that is true. So you say: The union of finite trees contains an infinite tree. The union of finite chains contains an infinite chain. The union of finite paths contains an infinite path. > What I state again, again and again, but what you > misread each and every time is: the union of sets of finite paths > does not contain an infinite path. So you say the union of finite paths p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... contains the infinite path p(oo) = {0,0,0,...}. But the union of finite sets of finite paths {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... does not contain the union of finite paths p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... I would say: The union of all finite sets of finite path contains all finite paths. The union of these finite paths can be consructed (it is a set of nodes) and is realized in the binary tree by its nodes. It contains all infinite paths of the infinite tree. The tree contains the union of all finite sets of finite path as well as the union of these paths. > How then does this infinite path manage to enter the union of the > finite trees? > > Because the set of paths of the union of finite trees is *not* contained in > the union of the sets of finite paths. > > And it is not contained in any of the finite trees. Why is it said to > be contained in the union of all finite trees? Because an infinite path is a union of paths. And when you construct > unions of *sets* of paths you are not constructing unions of paths. The union of a set of sets of elements is a set of elements, not a set of sets of elements. The union of one ot more sets of paths, as I defined it, can be a path. > If all natural numbers are finite, their union as defined for > pathlengths cannot be infinite. > > Why not? > > Because there cannot be more than infinitely many natural numbers. All > those infinitely many numbers, however, are allegedly finite. And there are infinitely many such. And they are, indeed, all finite. > The path lengths however do not map to natural numbers, which you assert. > That is valid for finite paths, but not for infinite paths. The finite natural numbers map on the finite pathlengths. The finite segments of natural numbers map on the finite pathlengths. If there is an infinite pathlength, then there must be an infinite number as well as an infinite segment of natural numbers. You say, the former does not exist in N while the latter does exist in the form of the whole set N. That is a contradiction. You see that there is no infinite pathlength possible without an infinite pathlength. > Let's, for > once, go to a proper definition (assuming all are ordered sets): > given a path {n_1, n_2, ..., n_k} then the path length is > |{n_1, n_2, ..., n_k}| = k. > Clearly for finite paths the path length is finite, and so a natural > number. However, when we construct the path: > p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... > we get as path: > {n_1, n_2, n_3, n_4, ...} > the path length in this case is *not* a natural number. It is the > cardinality of N. The pathlength is not a natural number but it is a number (by definition), namely omega. You see that there is no infinite set N without an infinte number in it. > But you said that there is an infinite pathlength corresponding to an > infinite set of paths. But that set of paths is *not* the union of the P(i). Above you said that the set of all paths p(n) contains the infinite path p(oo). > What was your argument distinguishing infinite unions of finite trees > and finite paths, respectively? > > None. I distinguish unions of finite trees and unions of *sets* of > finite paths. Above you are using P(1) U P(2) U ..., where each > P(i) is a *set* of finite paths. > > What about unions of sets of finite trees? Would they work completely > different from unions of sets of paths? In particular, would the union > of sets of finite trees not contain or establish the inifnite tree? The union of sets of finite trees would neither contain, nor establish, > an infinite tree. It would not contain one, because none of the > constituent sets contains an infinite tree. It would not establish > one, because the union is a set of trees, not a tree. The union of a set of sets of elements is a set of elements, not a set of sets of elements. The union of a set of trees, as I defined it, is a tree. > A tree is a > set of nodes, a set of trees is a set of sets of nodes. And the union of two or more sets of nodes is a set of nodes. Therefore the union of a set of trees, as I defined it, is a tree. > Wrong. Infinite paths do not have a path-length that is a > natural number. > > Pathlengths ARE natural numbers. > > For finite paths. > > In all cases, by definition. *What* definition? Definition: Map the pathlength x on the number x. === Subject: Re: Cantor Confusion > If there is an infinite pathlength, then there must be an infinite > number It is hard to believe that anyone could actually believe this is correct. Quite remarkable. > Wrong. Infinite paths do not have a path-length that is a > natural number. > > Pathlengths ARE natural numbers. > > For finite paths. > > In all cases, by definition. *What* definition? Definition: Map the pathlength x on the number x. Hard to believe that anyone could consider this to be a definition. Quite remarkable. -- === Subject: Re: Cantor Confusion On Sat, 3 Feb 2007 11:38:54 -0500, [. . .] >> *What* definition? >> >> Definition: Map the pathlength x on the number x. Hard to believe that anyone could consider this to be a definition. >Quite remarkable. No more remarkable than what you consider to be a definition. ~v~~ === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl ... > Frankly, I do not see much difference to: The union of finite paths > contains an infinite path. > > But that is true. > > So you say: > The union of finite trees contains an infinite tree. > The union of finite chains contains an infinite chain. > The union of finite paths contains an infinite path. Indeed. > > What I state again, again and again, but what you > misread each and every time is: the union of sets of finite paths > does not contain an infinite path. > > So you say the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > contains the infinite path p(oo) = {0,0,0,...}. > > But the union of finite sets of finite paths > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > does not contain the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... Again, indeed. > I would say: The union of all finite sets of finite path contains all > finite paths. Right. As I have stated all the time. And so the set of paths in the complete tree is *not* a subset of the union of all finite sets of finite paths, something you have stated repeatedly. > The union of these finite paths can be consructed (it is > a set of nodes) and is realized in the binary tree by its nodes. It > contains all infinite paths of the infinite tree. > > The tree contains the union of all finite sets of finite path as well > as the union of these paths. Right. But you *use* that the union of all finite sets of finite paths *contains* the union of these paths. > And it is not contained in any of the finite trees. Why is it said to > be contained in the union of all finite trees? > > Because an infinite path is a union of paths. And when you construct > unions of *sets* of paths you are not constructing unions of paths. > > The union of a set of sets of elements is a set of elements, not a set > of sets of elements. What definition of union do you use? Ah, I am seeing now. The axiom of sumsets. But I am *not* talking about a single set, I am talking about *multiple* sets. To rephrase your wording so that they are applicable: > The union of sets of sets of elements is not a set of elements, but a set > of sets of elements. To know why the axiom of sumsets is also the axiom of union, you have to consider the following: given two sets A and B by the axiom of pairing there exists a set {A, B} by the axiom of union there exists a set U{A, B} consisting of the elements of A and B. That set is normally called the union of A and B. I allow that the terminology is a bit confusing. But U applied to a single set of sets yields the set of the elements. But U applied to multiple sets does not. And you are applying it to multiple sets, so it does not. > The union of one ot more sets of paths, as I defined it, can be a > path. But is not in the union of the sets of paths. Pray follow the logic. > Because there cannot be more than infinitely many natural numbers. All > those infinitely many numbers, however, are allegedly finite. > > And there are infinitely many such. And they are, indeed, all finite. > The path lengths however do not map to natural numbers, which you assert. > That is valid for finite paths, but not for infinite paths. > > The finite natural numbers map on the finite pathlengths. > The finite segments of natural numbers map on the finite pathlengths. > > If there is an infinite pathlength, then there must be an infinite > number as well as an infinite segment of natural numbers. Why must there be an infinite number? But rest assured, there may be an infinite segment (depending on how you define segment). > You say, the former does not exist in N while the latter does exist in > the form of the whole set N. > > That is a contradiction. You see that there is no infinite pathlength > possible without an infinite pathlength. This is gibberish. I only state that infinite paths have infinite pathlength, and that there is no natural number that maps to. What is the contradiction? > Let's, for > once, go to a proper definition (assuming all are ordered sets): > given a path {n_1, n_2, ..., n_k} then the path length is > |{n_1, n_2, ..., n_k}| = k. > Clearly for finite paths the path length is finite, and so a natural > number. However, when we construct the path: > p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... > we get as path: > {n_1, n_2, n_3, n_4, ...} > the path length in this case is *not* a natural number. It is the > cardinality of N. > > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. No, I do not see it. You state that the pathlength is a number. Let me grant that. So omega is a number. But I do not see why N should contain omega, because by definition it is the set of *natural* numbers, and omega is not a natural number. It all boils down to the same story. Let A be the set of natural numbers plus omege. A is infinite (I think you agree). Remove omega from A. What do you have now? I state that you now have N. Apparently you disagree. Why? I also state that because A is infinite, N is also infinite. But for some reasons you state that N is finite. For what reasons? > But you said that there is an infinite pathlength corresponding to an > infinite set of paths. > > But that set of paths is *not* the union of the P(i). > > Above you said that the set of all paths p(n) contains the infinite > path p(oo). Yes. But not that the union of the P(i) contains an infinite path. > The union of sets of finite trees would neither contain, nor establish, > an infinite tree. It would not contain one, because none of the > constituent sets contains an infinite tree. It would not establish > one, because the union is a set of trees, not a tree. > > The union of a set of sets of elements is a set of elements, not a set > of sets of elements. But I am not talking about the union of a set. I am talking about the union of set*s*. Lets analyse. Given two sets A and B: by pairing {A, B} exists by union U{A, B} exists. The definition of A U B is U{A, B}. > The union of a set of trees, as I defined it, is a tree. I was talking about the union of set*s* of trees. > A tree is a > set of nodes, a set of trees is a set of sets of nodes. > > And the union of two or more sets of nodes is a set of nodes. Right. > Therefore the union of a set of trees, as I defined it, is a tree. And I never did contradict this. > Wrong. Infinite paths do not have a path-length that is a > natural number. > > Pathlengths ARE natural numbers. > > For finite paths. > > In all cases, by definition. > > *What* definition? > > Definition: Map the pathlength x on the number x. Fine, I do the mapping. How do I get that x is a *natural* number? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion ... > Frankly, I do not see much difference to: The union of finite paths > contains an infinite path. > > But that is true. > > So you say: > The union of finite trees contains an infinite tree. > The union of finite chains contains an infinite chain. > The union of finite paths contains an infinite path. Indeed. > How can a union of finite elements contain an infinite element? > > What I state again, again and again, but what you > misread each and every time is: the union of sets of finite paths > does not contain an infinite path. > > So you say the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > contains the infinite path p(oo) = {0,0,0,...}. > > But the union of finite sets of finite paths > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > does not contain the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... Again, indeed. But that is incorrect! The union of finite sets of finite paths {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... is the set of all finite paths {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } which obviously contains the union of finite paths of the special form p(0), p(1), p(2), .... as a subset. If you say, as you do above, that the union of finite paths contains an infinite path, p(0) U p(1) U p(2) U ... contains the infinite path p(oo) then this is also true for this subset. It is impossible that you loose p(oo) because the set of all finite paths {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } contains some more paths than he union of special finite paths p(0), p(1), p(2), .... > I would say: The union of all finite sets of finite path contains all > finite paths. Right. As I have stated all the time. And so the set of paths in the > complete tree is *not* a subset of the union of all finite sets of > finite paths, something you have stated repeatedly. The union of all finite paths of a special kind contains the due infinite path. The union of all finite paths contains all infinite paths. > The union of these finite paths can be consructed (it is > a set of nodes) and is realized in the binary tree by its nodes. It > contains all infinite paths of the infinite tree. > > The tree contains the union of all finite sets of finite path as well > as the union of these paths. Right. But you *use* that the union of all finite sets of finite paths > *contains* the union of these paths. If the union p(0), p(1), p(2), ....contains p(oo) then the union p(0), p(1), q(1), p(2), q(2), r(2),s(2), ... cannot miss it. > And it is not contained in any of the finite trees. Why is it said to > be contained in the union of all finite trees? > > Because an infinite path is a union of paths. And when you construct > unions of *sets* of paths you are not constructing unions of paths. > > The union of a set of sets of elements is a set of elements, not a set > of sets of elements. What definition of union do you use? Ah, I am seeing now. The axiom > of sumsets. But I am *not* talking about a single set, I am talking > about *multiple* sets. To rephrase your wording so that they are > applicable: > The union of sets of sets of elements is not a set of elements, but a set > of sets of elements. > To know why the axiom of sumsets is also the axiom of union, you have to > consider the following: > given two sets A and B > by the axiom of pairing there exists a set {A, B} > by the axiom of union there exists a set U{A, B} consisting of the > elements of A and B. That set is normally called the union of A and B. > I allow that the terminology is a bit confusing. But U applied to a > single set of sets yields the set of the elements. But U applied to > multiple sets does not. And you are applying it to multiple sets, so > it does not. Then apply it twice. First you get the single set of all paths, second you get the set of elements. The tree guarantees the existence of these unions. > If there is an infinite pathlength, then there must be an infinite > number as well as an infinite segment of natural numbers. Why must there be an infinite number? Because a union of finite pathlengths cannot lead to an infinite pathength. The infinite union of pathlengths 1,1,1,... 2,2,2,... 3,3,3 4,4 5 yields a finite pathlength, namely 5. > But rest assured, there may be an > infinite segment (depending on how you define segment). It is necessary. As an infinite natural number would be necessary to obtain an actually infinite set of numbers. > You say, the former does not exist in N while the latter does exist in > the form of the whole set N. > > That is a contradiction. You see that there is no infinite pathlength > possible without an infinite pathlength. This is gibberish. I only state that infinite paths have infinite pathlength, > and that there is no natural number that maps to. What is the contradiction? There cannot be an infinite pathlength. There cannot be an infinite segment {1,2,3,...n} with a last element unless the last element is infinite. It all boils down to the same story. Let A be the set of natural numbers > plus omege. A is infinite (I think you agree). Remove omega from A. > What do you have now? I state that you now have N. Apparently you > disagree. Why? I also state that because A is infinite, N is also > infinite. But for some reasons you state that N is finite. For what > reasons? I state that N is potentially infinite. There is no number countaing all natural numbers. But if there was a number counting all natural numbers, then it could not be infinite unless there waqs an infinite natural number because the natural numbers count themselves. As long as we are always counting another finite number, the counted set of nunmbers is finite. > But you said that there is an infinite pathlength corresponding to an > infinite set of paths. > > But that set of paths is *not* the union of the P(i). > > Above you said that the set of all paths p(n) contains the infinite > path p(oo). Yes. But not that the union of the P(i) contains an infinite path. The union of the P(i) is the set of all paths. It includes the set of all paths of the form p(n). Why should it not contain p(oo) if the set of all paths of the form p(n) alone contains p(oo)? > The union of sets of finite trees would neither contain, nor establish, > an infinite tree. It would not contain one, because none of the > constituent sets contains an infinite tree. It would not establish > one, because the union is a set of trees, not a tree. > > The union of a set of sets of elements is a set of elements, not a set > of sets of elements. But I am not talking about the union of a set. I am talking about the > union of set*s*. Lets analyse. Given two sets A and B: > by pairing {A, B} exists > by union U{A, B} exists. > The definition of A U B is U{A, B}. And that is a single set containing every element which is in A or in B or in both. The union of some sets of paths is one set of paths. The union of all paths is one set of elements. The tree unions all sets of paths. > The union of a set of trees, as I defined it, is a tree. I was talking about the union of set*s* of trees. How many trees must be in the sets? The union of two trees is a tree. Two trees is a set of two trees. The union of tree paths and 12 paths and 17 paths is a set of nodes, perhaps this set includes even one or more paths. The union of {three paths} and {12 paths} and {17 paths} is a set of paths. These paths are unioned in the tree to give a set of nodes, perhaps including even one or more paths. > Wrong. Infinite paths do not have a path-length that is a > natural number. > > Pathlengths ARE natural numbers. > > For finite paths. > > In all cases, by definition. > > *What* definition? > > Definition: Map the pathlength x on the number x. Fine, I do the mapping. How do I get that x is a *natural* number? Because there are no other numbers of elements (nodes). What you call omega is nothing but the possibility to extend the path of n nodes by another node to n+1 nodes. === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl > So you say: > The union of finite trees contains an infinite tree. > The union of finite chains contains an infinite chain. > The union of finite paths contains an infinite path. > > Indeed. > > How can a union of finite elements contain an infinite element? Where in the above is there a union of finite elements? Unions are about *sets*, not about elements. > So you say the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > contains the infinite path p(oo) = {0,0,0,...}. > > But the union of finite sets of finite paths > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > does not contain the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > > Again, indeed. > > But that is incorrect! > > The union of finite sets of finite paths > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > is the set of all finite paths > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } Indeed. > which obviously contains the union of finite paths of the special form > p(0), p(1), p(2), .... > as a subset. Wrong. There is no obviously here. The union of sets of elements does *not* contain the union of elements. This is abundantly clear from the first eight chapters of your book. > If you say, as you do above, that the union of finite paths contains > an infinite path, > p(0) U p(1) U p(2) U ... contains the infinite path p(oo) > then this is also true for this subset. Let's get back to the finite. A = {{a, b}, {a, c}, {a, d}} and B = {{a, d}, {b, d}, {c, d}}. A U B consists of: {{a, b}, {a, c}, {a, d}, {b, d}, {c, d}}. It does *not* contain unions of the sets that are elements of A or B. > It is impossible that you loose p(oo) because the set of all finite > paths > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > contains some more paths than he union of special finite paths > p(0), p(1), p(2), .... In what way does {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } contain: p(0) U p(1) U p(2) U ...? > I would say: The union of all finite sets of finite path contains all > finite paths. > > Right. As I have stated all the time. And so the set of paths in the > complete tree is *not* a subset of the union of all finite sets of > finite paths, something you have stated repeatedly. > > The union of all finite paths of a special kind contains the due > infinite path. > The union of all finite paths contains all infinite paths. *Sets* of paths. Remember? > Right. But you *use* that the union of all finite sets of finite paths > *contains* the union of these paths. > > If the union p(0), p(1), p(2), ....contains p(oo) then the union > p(0), p(1), q(1), p(2), q(2), r(2),s(2), ... > cannot miss it. What *sets* of paths are you using? > To know why the axiom of sumsets is also the axiom of union, you have to > consider the following: > given two sets A and B > by the axiom of pairing there exists a set {A, B} > by the axiom of union there exists a set U{A, B} consisting of the > elements of A and B. That set is normally called the union of A and B. > I allow that the terminology is a bit confusing. But U applied to a > single set of sets yields the set of the elements. But U applied to > multiple sets does not. And you are applying it to multiple sets, so > it does not. > > Then apply it twice. First you get the single set of all paths, second > you get the set of elements. Which is a set of nodes. So you are not taking the union but the union of the union. Let us analyse. A set of paths is a set of sets of nodes. So let us consider the basic two trees (level 1 and level 2). The sets of paths are: L1: {{0, 1}, {0, 2}} L2: {{0, 1, 3}, {0, 1, 4}, {0, 2, 5}, {0, 2, 6}}. Two sets. Unite them once: {{0, 1}, {0, 2}, {0, 1, 3}, {0, 1, 4}, {0, 2, 5}, {0, 2, 6}}. Unite this one, and we get: {0, 1, 2, 3, 4, 5, 6}. I do not see a path there. > If there is an infinite pathlength, then there must be an infinite > number as well as an infinite segment of natural numbers. > > Why must there be an infinite number? > > Because a union of finite pathlengths cannot lead to an infinite > pathength. > > The infinite union of pathlengths > 1,1,1,... > 2,2,2,... > 3,3,3 > 4,4 > 5 > > yields a finite pathlength, namely 5. Yes. Infinite unions can give finite things. But that is not a contradcition. > But rest assured, there may be an > infinite segment (depending on how you define segment). > > It is necessary. As an infinite natural number would be necessary to > obtain an actually infinite set of numbers. You think so, but provide no proof. > This is gibberish. I only state that infinite paths have infinite > pathlength, and that there is no natural number that maps to. What > is the contradiction? > > There cannot be an infinite pathlength. Oh. > There cannot be an infinite segment {1,2,3,...n} with a last element > unless the last element is infinite. Sure. But again that is not in contradiction to anything I state. > It all boils down to the same story. Let A be the set of natural numbers > plus omege. A is infinite (I think you agree). Remove omega from A. > What do you have now? I state that you now have N. Apparently you > disagree. Why? I also state that because A is infinite, N is also > infinite. But for some reasons you state that N is finite. For what > reasons? > > I state that N is potentially infinite. There is no number countaing > all natural numbers. There is no *natural* number containing all natural numbers. How you conclude from this that there is no number containing all natural numbers escapes me. Pray give a proper mathematical definition of number. > But if there was a number counting all natural numbers, then it could > not be infinite unless there waqs an infinite natural number because > the natural numbers count themselves. As long as we are always > counting another finite number, the counted set of nunmbers is finite. Note how you sometimes use natural number and othertimes number as if they were the same? > Above you said that the set of all paths p(n) contains the infinite > path p(oo). > > Yes. But not that the union of the P(i) contains an infinite path. > > The union of the P(i) is the set of all paths. It includes the set of > all paths of the form p(n). Why should it not contain p(oo) if the set > of all paths of the form p(n) alone contains p(oo)? I think we are talking at cross purposes. And are misunderstanding each other about what the meaning is of the term the set of all paths p(n). > The union of a set of sets of elements is a set of elements, not a set > of sets of elements. > > But I am not talking about the union of a set. I am talking about the > union of set*s*. Lets analyse. Given two sets A and B: > by pairing {A, B} exists > by union U{A, B} exists. > The definition of A U B is U{A, B}. > > And that is a single set containing every element which is in A or in > B or in both. > The union of some sets of paths is one set of paths. > The union of all paths is one set of elements. Right. So the union of all paths is a set of nodes. But I do not think that set is a path. That set is a tree (by your definition). > The union of a set of trees, as I defined it, is a tree. > > I was talking about the union of set*s* of trees. > > How many trees must be in the sets? > The union of two trees is a tree. Two trees is a set of two trees. I was talking about the union of set*s* of trees. The union of a set containing (as elements) two trees and a set containing (as element) two other trees, is a set containing (as element) two to four trees. > The union of tree paths and 12 paths and 17 paths is a set of nodes, > perhaps this set includes even one or more paths. > > The union of {three paths} and {12 paths} and {17 paths} is a set of > paths. These paths are unioned in the tree to give a set of nodes, > perhaps including even one or more paths. This does not make sense at all. > Definition: Map the pathlength x on the number x. > > Fine, I do the mapping. How do I get that x is a *natural* number? > > Because there are no other numbers of elements (nodes). > What you call omega is nothing but the possibility to extend the path > of n nodes by another node to n+1 nodes. Makes no sense again. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion > How can a union of finite elements contain an infinite element? WM again conflates the size of the sets being unioned with how many sets are being unioned. A union of infinitely many finite sets can be an infinite set just as easily as a finite union of infinite sets. > > What I state again, again and again, but what you > misread each and every time is: the union of sets of finite paths > does not contain an infinite path. > > So you say the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > contains the infinite path p(oo) = {0,0,0,...}. WM deliberately misreads one more time. That is NOT what was said. nested sequence of paths to form a path in the WM psuedo-union of the containing trees. But the union of two or more sets of paths is merely a set of paths and not itself a path, and more than the union of two sets of dogs is a dog. > Right. As I have stated all the time. And so the set of paths in the > complete tree is *not* a subset of the union of all finite sets of > finite paths, something you have stated repeatedly. The union of all finite paths of a special kind contains the due > infinite path. The union of sets of dogs is not a dog. > I state that N is potentially infinite. If N represents a set, then it is either finite or not finite. If it is not finite, then it is actually not finite, not merely potentially not finite. === Subject: Re: Cantor Confusion > So the union of finite chains of nodes like the following > 0 > 12 > 6543 > does not contain an infinite chain of nodes of the form > 0 > 12 > 6543 > 789... > ??? your own interpretation to it. > Frankly, I do not see much difference to: The union of finite paths > contains an infinite path. But that is true. So you say: > The union of finite trees contains an infinite tree. > The union of finite chains contains an infinite chain. > The union of finite paths contains an infinite path. What I state again, again and again, but what you > misread each and every time is: the union of sets of finite paths > does not contain an infinite path. > So you say the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > contains the infinite path p(oo) = {0,0,0,...}. But the union of finite sets of finite paths > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > does not contain the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... If p(n) = {node_0, node_1, node_2,...node_n} = { node_k : k in N and k <= n} then Union_{n in N} p(n) = {node_0, node_1, node_2, ...} = { node_n : n in N} is an endless sequence of nodes, which is a path, but Union_{n in N} {p(n)} = { p(0), p(1), p(2),...} = { p(n) : n in N } is an endless sequence of paths, which is not a path. Note that { node_n : n in N} != { p(n) : n in N } and { node_n : n in N} is a path but { p(n) : n in N } is a set of paths and not a path. === Subject: Re: Cantor Confusion [...] >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. As omega is not member of N I don't see that. F. N. -- xyz === Subject: Re: Cantor Confusion <45c31c0b$0$97214$892e7fe2@authen.yellow.readfreenews.net> On 2 Feb., 12:10, Franziska Neugebauer [...] >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. As omega is not member of N I don't see that. The total pathlength is the union of all single pathlengths. The total pathlength cannot be omega without omega being a pathlengt (in the union). But omega cannot be in the union of finite pathlengths. Hence there is no actually infinite pathlength. Infinite pathlength means only that every finite pathlength is surpassed by another finite pathlength. === Subject: Re: Cantor Confusion > On 2 Feb., 12:10, Franziska Neugebauer [...] >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. As omega is not member of N I don't see that. The total pathlength is the union of all single pathlengths. If all those separate pathlengths are finite ordinal numbers and there are more than any finite number of them, then the union is the first limit ordinal. > The total > pathlength cannot be omega without omega being a pathlengt (in the > union). False! It is quite possible for a union not to be any one of the sets being unioned. But omega cannot be in the union of finite pathlengths. It can be, and is, everywhere except possibly in WM's weird world. > Hence > there is no actually infinite pathlength. It can be, and is, everywhere except possibly in WM's weird world. > Infinite pathlength means > only that every finite pathlength is surpassed by another finite > pathlength. Which requires a path with no end. === Subject: Re: Cantor Confusion > On 2 Feb., 12:10, Franziska Neugebauer > [...] > However, when we construct the path: > p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... > we get as path: > {n_1, n_2, n_3, n_4, ...} > the path length in this case is *not* a natural number. It is the > cardinality of N. >> The pathlength is not a natural number but it is a number (by >> definition), namely omega. You see that there is no infinite set N >> without an infinte number in it. >> As omega is not member of N I don't see that. The total pathlength is the union of all single pathlengths. The path-length of the infinite path is the supremum of the set of paths-lengths of all finite paths. The latter is the supremum of the set of natural numbers. This supremum exists. > The total pathlength cannot be omega without omega being a pathlengt Omega is the path-length of the infinite path. > (in the union). Non sequitur. This is merely your claim X is not finite -> there must be an x in X which is infinite > But omega cannot be in the union of finite pathlengths. Straw man. Nobody claimed that omega is (a member) in N. > Hence there is no actually infinite pathlength. Non sequitur. > Infinite pathlength means only that every finite pathlength is > surpassed by another finite pathlength. This is the wrong meaning in the framwork of contemporary set theory. Perhaps WH may demonstrate that you are also in error with respect to a potentially infinity framework. F. N. -- xyz === Subject: Re: Cantor Confusion <45c31c0b$0$97214$892e7fe2@authen.yellow.readfreenews.net> <45c5e0e9$0$97220$892e7fe2@authen.yellow.readfreenews.net> On 4 Feb., 14:34, Franziska Neugebauer On 2 Feb., 12:10, Franziska Neugebauer > [...] > However, when we construct the path: > p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... > we get as path: > {n_1, n_2, n_3, n_4, ...} > the path length in this case is *not* a natural number. It is the > cardinality of N. > The pathlength is not a natural number but it is a number (by >> definition), namely omega. You see that there is no infinite set N >> without an infinte number in it. > As omega is not member of N I don't see that. The total pathlength is the union of all single pathlengths. The path-length of the infinite path is the supremum of the set of > paths-lengths of all finite paths. The latter is the supremum of the > set of natural numbers. This supremum exists. No. What does it consist of? A supremum exists in many places but not here. The simplest reason is that omega - n = omega for all n in N. The total pathlength cannot be omega without omega being a pathlengt Omega is the path-length of the infinite path. The pathlengths is measured *and marked* by nodes. This is the advantage of the tree. While you may insist that there is an infinite umber of nodes we know that there is no node number omega. (in the union). Non sequitur. This is merely your claim Non sequitur. This is merely your claim. The pathlengths is measured *and marked* by nodes. Infinite pathlength means only that every finite pathlength is > surpassed by another finite pathlength. This is the wrong meaning in the framwork of contemporary set theory. What do you think is the correct meaning with respect to node numbers? === Subject: Re: Cantor Confusion > On 4 Feb., 14:34, Franziska Neugebauer The path-length of the infinite path is the supremum of the set of > paths-lengths of all finite paths. The latter is the supremum of the > set of natural numbers. This supremum exists. No. What does it consist of? A supremum exists in many places but not > here. The simplest reason is that > omega - n = omega > for all n in N. WM does not seem to have much of a grasp of what a supremum is. Among ordinals, a supremum of a set of ordinals is an ordinal which is at least as large as any member of the set and is at least as small as any other ordinal having this property. For every n in omega, n < omega, and for any ordinal, k, not in omega, omega <= k. Thus omega is, according to the definition of supremum, the supremum of the set of ordinals < omega. > The total pathlength cannot be omega without omega being a pathlengt Omega is the path-length of the infinite path. The pathlengths is measured *and marked* by nodes. Only for finite paths, and not merely by nodes, by edges connected in a chain. > This is the > advantage of the tree. While you may insist that there is an infinite > umber of nodes we know that there is no node number omega. But the number of nodes is more thatn every finite natural or ordinal number. The pathlengths is measured *and marked* by nodes. Only for finite paths, and not merely by nodes, by edges connected in a chain. > Infinite pathlength means only that every finite pathlength is > surpassed by another finite pathlength. This is the wrong meaning in the framwork of contemporary set theory. What do you think is the correct meaning with respect to node numbers? That the length of the ultimate path is greater than every node number. === Subject: Re: Cantor Confusion > On 4 Feb., 14:34, Franziska Neugebauer > On 2 Feb., 12:10, Franziska Neugebauer [...] >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is >> the cardinality of N. > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set > N without an infinte number in it. > As omega is not member of N I don't see that. >> The total pathlength is the union of all single pathlengths. >> The path-length of the infinite path is the supremum of the set of >> paths-lengths of all finite paths. The latter is the supremum of the >> set of natural numbers. This supremum exists. No. What does it consist of? It consists of (i.e. contains as members) every and all natural numbers, each of which is finite. > A supremum exists in many places but not here. Not where? > The simplest reason is that omega - n = omega for all n in N. Where did you get that from? Reference? EB? F. N. -- xyz === Subject: Re: Cantor Confusion [...] >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. As omega is not member of N I don't see that. F. N. Also, every natural is either even or odd but not both. is omega even or odd? === Subject: Re: Cantor Confusion >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. As omega is not member of N I don't see that. Also, every natural is either even or odd but not both. > is omega even or odd? You are missing a WM axiom: Path lengths are natural numbers (even for infinite paths). -- === Subject: Re: Cantor Confusion >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. > As omega is not member of N I don't see that. Also, every natural is either even or odd but not both. > is omega even or odd? You are missing a WM axiom: Path lengths are natural numbers (even for > infinite paths). On the contrary, I don't miss it a bit! === Subject: Re: Cantor Confusion >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. > As omega is not member of N I don't see that. > Also, every natural is either even or odd but not both. > is omega even or odd? You are missing a WM axiom: Path lengths are natural numbers (even for > infinite paths). On the contrary, I don't miss it a bit! But, if you assume it, you can prove so many more things! -- === Subject: Re: Cantor Confusion >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. > As omega is not member of N I don't see that. > Also, every natural is either even or odd but not both. > is omega even or odd? > You are missing a WM axiom: Path lengths are natural numbers (even for > infinite paths). On the contrary, I don't miss it a bit! But, if you assume it, you can prove so many more things! if one asssumes 1 = 2, one can prove all sorts of curious things, but that does not justify assuming 1 = 2. === Subject: Re: Cantor Confusion æ> If you have a set and you add something to this set (in unary > æ> representation), then you have a super set of the former. The former > æ> is a subset of the latter. This is not defining what I asked for. æWhat are the elements of the sets > you are talking about? The type of element is not restricted other than by the requirement that it be part of the real world. æ> æ> Bizarre. æSo 3 is the set of all existing (what does that mean) sets > æ> æ> with 3 elements, > æ> æ æ> æ> The number 3 is the set of all sets of 3 elements (unless they were > æ> æ> existing, they would not have 3 elements). But 3 is also here III. It > æ> æ> is every set of 3 elements. > æ æ> You failed to respond to what followed. æIf you *now* define it as every > æ> set of 3 elements, that is a circular definition. æ3 is any set of 3 > æ> elements looks quite circular to me. > æ æ> It is circular. But please do not forget that I do not *define* a > æ> natural number! I leave that to Peano etc. I look for its *existence*. So it was not a definition at all. æNote also that the Peano axioms to > *not* talk about representation. Therefore we have to investigate whether such a representation exists. Note that I use the definition of a number by Peano in order to look for the existence of that number. æ> æ> I do not ask what possible answers there are, I only ask how you would > æ> æ> like to propose it. æAnd, pray extend to 111111111. > æ> æ æ> æ> This is a set of 9 distinguishable elements, it is a set representing > æ> æ> the number 9. And: It is the nunmber 9. Every set of 9 elements and > æ> æ> avery set of sets of 9 elements is the number 9. > æ æ> So you are now going to equivalence classes of sets with the same number > æ> of elements? æOr what? æAs you write it your definition is still circular. > æ æ> Of course it is. You must start at some point and eventually you will > æ> come back. Circular reasoning cannot be avoided. But that is not a > æ> problem when existence is concerned. The definition of a number is > æ> given on p. 3 and p 130 of my book. A definition that uses circular reasoning is not a definition. æIf a write: > æ æ a foo is a foo The definition of a number is given as non-circular as possible: p. 3: 1) 1 ist eine nat.9frliche Zahl. 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. 3) Es gibt keine Zahl mit dem Nachfolger 1. 4) Aus a' = b' folgt a = b. 5) Jede Menge M von nat.9frlichen Zahlen, welche die Zahl 1 und zu jeder Zahl a ? M auch den Nachfolger a' enth.8alt, enth.8alt alle nat.9frlichen Zahlen. p. 130: 1) 1 in M. 2) If n in M then n + 1 in M. 3) IN ist Durchschnitt aller Mengen M, die (1) und (2) erf.9fllen. To investigate the *existence* of number 3 we can use the definition of 3. === Subject: Re: Cantor Confusion > > If you have a set and you add something to this set (in unary > representation), then you have a super set of the former. The former > is a subset of the latter. > > This is not defining what I asked for. What are the elements of the sets > you are talking about? > > The type of element is not restricted other than by the requirement > that it be part of the real world. We were specifically talking about the sets I, II, III, etc. Or about 4 and 5. I ask you how you define them as sets. But let me elaborate more on this below: > Bizarre. So 3 is the set of all existing (what does that mean) > sets with 3 elements, If 4 is the set of all existing (whatever that does mean) sets with 4 elements and 5 is the set of all existing sets with 5 elements, we find immediately that 4 is *not* a subset of 5. With this definition of numbers as sets subsetting does not give what you wish. You should (at the least) be consistent. > It is circular. But please do not forget that I do not *define* a > natural number! I leave that to Peano etc. I look for its *existence*. > > So it was not a definition at all. Note also that the Peano axioms to > *not* talk about representation. > > Therefore we have to investigate whether such a representation exists. > Note that I use the definition of a number by Peano in order to look > for the existence of that number. What does the latter sentence *mean*? But indeed, for most numbers some representations do not exist, while other representations do exist. The existence of a number is independent of the existence of a representation. For instance, for most rational numbers a decimal representation does not exist. > Of course it is. You must start at some point and eventually you will > come back. Circular reasoning cannot be avoided. But that is not a > problem when existence is concerned. The definition of a number is > given on p. 3 and p 130 of my book. > > A definition that uses circular reasoning is not a definition. If a write: > a foo is a foo > > The definition of a number is given as non-circular as possible: p. 3: > 1) 1 ist eine nat.9frliche Zahl. > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. in N is wrong here; N is not defined. A more proper version is: 2) Jede nat.9frliche Zahl a had einen bestimmten Nachfolger a', und dass ist auch ein nat.9frliche Zahl. > 3) Es gibt keine Zahl mit dem Nachfolger 1. > 4) Aus a' = b' folgt a = b. > 5) Jede Menge M von nat.9frlichen Zahlen, welche die Zahl 1 und zu jeder > Zahl a in M auch den Nachfolger a' enth.8alt, enth.8alt alle nat.9frlichen > Zahlen. Yes, I know Peano pretty well. But this is *not* a circular definition. I wonder why you think it is circular. > p. 130: > > 1) 1 in M. > 2) If n in M then n + 1 in M. I prefer the successor of n, rather than n + 1 here. > 3) IN ist Durchschnitt aller Mengen M, die (1) und (2) erf.9fllen. That is an alternative definition, again not circular. > To investigate the *existence* of number 3 we can use the definition > of 3. And *in this context* the definition of 3 is: succ(succ(1)) no set at all. And the existence of 3 follows from the first set of axioms you gave. Your definition is circular, this one is not. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion > If you have a set and you add something to this set (in unary > representation), then you have a super set of the former. The former > is a subset of the latter. > > This is not defining what I asked for. What are the elements of the sets > you are talking about? > > The type of element is not restricted other than by the requirement > that it be part of the real world. We were specifically talking about the sets I, II, III, etc. Or about 4 and > 5. I ask you how you define them as sets. But let me elaborate more on > this below: > Bizarre. So 3 is the set of all existing (what does that mean) > sets with 3 elements, If 4 is the set of all existing (whatever that does mean) sets with 4 > elements and 5 is the set of all existing sets with 5 elements, we find > immediately that 4 is *not* a subset of 5. With this definition of > numbers as sets subsetting does not give what you wish. 4 is the set of all sets with 4 elements. 4 is also every set with 4 elements. In unary representation or by using peanuts or matchsticks 4 is a subset of 5. > It is circular. But please do not forget that I do not *define* a > natural number! I leave that to Peano etc. I look for its *existence*. > > So it was not a definition at all. Note also that the Peano axioms to > *not* talk about representation. > > Therefore we have to investigate whether such a representation exists. > Note that I use the definition of a number by Peano in order to look > for the existence of that number. What does the latter sentence *mean*? But indeed, for most numbers some > representations do not exist, while other representations do exist. The > existence of a number is independent of the existence of a representation. This point of view has lead to the present mess-math. > For instance, for most rational numbers a decimal representation does not > exist. Correct, for instance for 1/7. > Of course it is. You must start at some point and eventually you will > come back. Circular reasoning cannot be avoided. But that is not a > problem when existence is concerned. The definition of a number is > given on p. 3 and p 130 of my book. > > A definition that uses circular reasoning is not a definition. If a write: > a foo is a foo > > The definition of a number is given as non-circular as possible: p. 3: > 1) 1 ist eine nat.9frliche Zahl. > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. in N is wrong here; N is not defined. N is already in (1), because (1) is identical to 1 is in N. A more proper version is: > 2) Jede nat.9frliche Zahl a had einen bestimmten Nachfolger a', und > dass ist auch ein nat.9frliche Zahl. What is every natural number if N is not defined? > 3) Es gibt keine Zahl mit dem Nachfolger 1. > 4) Aus a' = b' folgt a = b. > 5) Jede Menge M von nat.9frlichen Zahlen, welche die Zahl 1 und zu jeder > Zahl a in M auch den Nachfolger a' enth.8alt, enth.8alt alle nat.9frlichen > Zahlen. Yes, I know Peano pretty well. But this is *not* a circular definition. I > wonder why you think it is circular. I did not say it is circular. I said as non-circular as possible. If it is not circular in your opinion, then be happy. (In fact every definition is circular, because every language is. You cannot explain anything without already using some unexplained wording. If you want to explaine the unexplained by the words already explained, you get circular. That is unavoidable. Cp. N being undefined but appearing in the Peano axioms. But that is not the point here.) > p. 130: > > 1) 1 in M. > 2) If n in M then n + 1 in M. > I prefer the successor of n, rather than n + 1 here. I think that everybody able to read and understand these lines will know what + 1 means while the successor is not immediately clear. (The successor of n coul be n+2 or 2*n or 10*n or ....) > 3) IN ist Durchschnitt aller Mengen M, die (1) und (2) erf.9fllen. That is an alternative definition, again not circular. It is better than Peano, because one does not use natural numbers, but is circular though as little as possible. You only need to know +1 and the intersection of sets. (And what a set is, and what 1 is, and so on.) > To investigate the *existence* of number 3 we can use the definition > of 3. And *in this context* the definition of 3 is: > succ(succ(1)) > no set at all. And the existence of 3 follows from the first set of > axioms you gave. The existence does not follow from axioms. Axioms state something in an arbitrary way. They can even state the famous pink elephant. Whether it exists remains to be investigated. But for this sake we need not define again what a pink elephant is. > Your definition is circular, this one is not. I do not give a definition but I look for the existence of the number already defined. === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl ... > If 4 is the set of all existing (whatever that does mean) sets with 4 > elements and 5 is the set of all existing sets with 5 elements, we find > immediately that 4 is *not* a subset of 5. With this definition of > numbers as sets subsetting does not give what you wish. > > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. That can not be true. 4 can not be all kinds of different things at once, at least not in mathematics. With this kind of definition 4 contains 4. > Therefore we have to investigate whether such a representation exists. > Note that I use the definition of a number by Peano in order to look > for the existence of that number. > > What does the latter sentence *mean*? But indeed, for most numbers some > representations do not exist, while other representations do exist. The > existence of a number is independent of the existence of a representation. > > This point of view has lead to the present mess-math. > > For instance, for most rational numbers a decimal representation does not > exist. > > Correct, for instance for 1/7. And for computable numbers some representation does exist. > The definition of a number is given as non-circular as possible: p. 3: > 1) 1 ist eine nat=FCrliche Zahl. > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. > > in N is wrong here; N is not defined. > > N is already in (1), because (1) is identical to 1 is in N. N is not *defined*. Whether you use it in (1) is irrelevant, what is relevant is that it is not defined. > A more proper version is: > 2) Jede nat.9frliche Zahl a had einen bestimmten Nachfolger a', und > dass ist auch ein nat.9frliche Zahl. > > What is every natural number if N is not defined? This is a recursive definition of natural numbers. By (1) we have one natural number, by (2), from that single natural number we get a lot of other natural numbers. > 3) Es gibt keine Zahl mit dem Nachfolger 1. > 4) Aus a' = b' folgt a = b. > 5) Jede Menge M von nat=FCrlichen Zahlen, welche die Zahl 1 und zu jed= > er > Zahl a in M auch den Nachfolger a' enth=E4lt, enth=E4lt alle nat=FC= > rlichen > Zahlen. > > Yes, I know Peano pretty well. But this is *not* a circular definition. = > I > wonder why you think it is circular. > > I did not say it is circular. I said as non-circular as possible. If > it is not circular in your opinion, then be happy. (In fact every > definition is circular, because every language is. You cannot explain > anything without already using some unexplained wording. If you want > to explaine the unexplained by the words already explained, you get > circular. That is unavoidable. Cp. N being undefined but appearing in > the Peano axioms. But that is not the point here.) This is getting phylosofical. In mathematics a definition is circular if in the definition one of the deciding features is the term you want to define. So a definition as you gave: 3 is the set of all sets of 3 elements is a perfect example of a circular definition. The reason is that it states precisely nothing about what 3 is. It is not better than the definition: 3 is 3. In order to know whether a particular set fits in 3 you have to know what 3 is, and to know that you have to know whether that particular set does fit. > p. 130: > > 1) 1 in M. > 2) If n in M then n + 1 in M. > I prefer the successor of n, rather than n + 1 here. > > I think that everybody able to read and understand these lines will > know what + 1 means while the successor is not immediately clear. > (The successor of n coul be n+2 or 2*n or 10*n or ....) Yes, indeed, and that is the crux. Abstraction. When you use the '+ 1' notation you are already assuming the existence of addition. When you do not, you can properly define addition (and all other operations) using the Peano axioms. You are *defining* the natural numbers, presumably without any knowledge about what natural numbers even are. And I may note that with '2*n' you can get a set that is isomorphic (with respect to all operations) to the natural numbers, only the naming is different. Consider the set of powers of 2, (your 2*n case) define: a '+' b = 2^[ log_2(a) + log_2(b) + 1 ] a '*' b = 2^[ log_2(a) * log_2(b) + log_2(a) + log_2(b) ] Call this set K. You may verify that the rings: R(N, +, *) and R(K, '+', '*') are isomorphic. > And *in this context* the definition of 3 is: > succ(succ(1)) > no set at all. And the existence of 3 follows from the first set of > axioms you gave. > > The existence does not follow from axioms. Axioms state something in > an arbitrary way. They can even state the famous pink elephant. The non-existence of a pink elephant in the realm of the axiom that states that there is a pink elephant is not mathematics but philosophy. Axioms state what things exist (or do not exist) in their realm. In geometry there is the parallel axiom (that you also use as example for the axiom of choice in your book). In two forms of geometry there exists a line through a point parallel to a given line not through that point. In one form of geometry such a line does not exist. (Exist meaning exist mathematically here.) The axiom states existence or non-existence, and that is all. > Whether it exists remains to be investigated. Your existence is not a mathematical existence. > Your definition is circular, this one is not. > > I do not give a definition but I look for the existence of the number > already defined. So when I ask you for a definition you do not give a definition? Yes, I have been thinking that all along. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion ... > If 4 is the set of all existing (whatever that does mean) sets with 4 > elements and 5 is the set of all existing sets with 5 elements, we find > immediately that 4 is *not* a subset of 5. With this definition of > numbers as sets subsetting does not give what you wish. > > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. That can not be true. 4 can not be all kinds of different things at once, > at least not in mathematics. What is 4 in mathematics? > With this kind of definition 4 contains 4. Of course. Cardinal number = ordinal number. 4 contains 4 elements (even in set theory). But 4 contains IIII --- not 0,1,2,3. Henry VIII had 7 predecessors --- only including him they were 8 Henries. > Therefore we have to investigate whether such a representation exists. > Note that I use the definition of a number by Peano in order to look > for the existence of that number. > > What does the latter sentence *mean*? But indeed, for most numbers some > representations do not exist, while other representations do exist. The > existence of a number is independent of the existence of a representation. > > This point of view has lead to the present mess-math. > > For instance, for most rational numbers a decimal representation does not > exist. > > Correct, for instance for 1/7. And for computable numbers some representation does exist. But this representation does not necessarily enable us to determine the trichotomy relation with numbers which are really numbers. > The definition of a number is given as non-circular as possible: p. 3: > 1) 1 ist eine nat=FCrliche Zahl. > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. > > in N is wrong here; N is not defined. > > N is already in (1), because (1) is identical to 1 is in N. N is not *defined*. Whether you use it in (1) is irrelevant, what is > relevant is that it is not defined. > A more proper version is: > 2) Jede nat.9frliche Zahl a had einen bestimmten Nachfolger a', und > dass ist auch ein nat.9frliche Zahl. > > What is every natural number if N is not defined? This is a recursive definition of natural numbers. By (1) we have one > natural number, by (2), from that single natural number we get a lot of > other natural numbers. Why do you say N is wrong in (2) but not in (1)? > This is getting phylosofical. In mathematics a definition is circular if > in the definition one of the deciding features is the term you want to > define. So a definition as you gave: > 3 is the set of all sets of 3 elements > is a perfect example of a circular definition. !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IT IS NOT A DEFINITION !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > I think that everybody able to read and understand these lines will > know what + 1 means while the successor is not immediately clear. > (The successor of n coul be n+2 or 2*n or 10*n or ....) Yes, indeed, and that is the crux. Abstraction. When you use the > '+ 1' notation you are already assuming the existence of addition. > When you do not, you can properly define addition (and all other > operations) using the Peano axioms. You are *defining* the natural > numbers, presumably without any knowledge about what natural numbers > even are. And I may note that with '2*n' you can get a set that is > isomorphic (with respect to all operations) to the natural numbers, > only the naming is different. Consider the set of powers of 2, > (your 2*n case) define: > a '+' b = 2^[ log 2(a) + log 2(b) + 1 ] > a '*' b = 2^[ log 2(a) * log 2(b) + log 2(a) + log 2(b) ] > Call this set K. You may verify that the rings: > R(N, +, *) and R(K, '+', '*') > are isomorphic. I do know that. But I don't want some isomorphic sets. I want to define *the natural numbers* which are I II III ... > And *in this context* the definition of 3 is: > succ(succ(1)) > no set at all. And the existence of 3 follows from the first set of > axioms you gave. > > The existence does not follow from axioms. Axioms state something in > an arbitrary way. They can even state the famous pink elephant. The non-existence of a pink elephant in the realm of the axiom that > states that there is a pink elephant is not mathematics but philosophy. > Axioms state what things exist (or do not exist) in their realm. No. This belief is the core of mathmess > Whether it exists remains to be investigated. Your existence is not a mathematical existence. This form of existence is the only possible existence. > Your definition is circular, this one is not. > > I do not give a definition but I look for the existence of the number > already defined. So when I ask you for a definition you do not give a definition? Yes, > I have been thinking that all along. I gave two definitions: Peano and that with +1 which is very close to Dedekind's attitude. (I don't know whether he actually created it, but I know that he would have liked it with +1 as a primitive). But the axioms do not establish any existence, in particular not when you apply Dedekind's definition of what a number is. === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl ... > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. > > That can not be true. 4 can not be all kinds of different things at once, > at least not in mathematics. > > What is 4 in mathematics? succ(succ(succ(succ(0))), according to the Peano axioms. > With this kind of definition 4 contains 4. > > Of course. Cardinal number = ordinal number. 4 contains 4 elements > (even in set theory). Depends on the model. > But 4 contains IIII --- not 0,1,2,3. What *is* IIII. You never have defined it. You really do not like definitions, as they pin down the real meaning. > Henry VIII had 7 predecessors --- only including him they were 8 > Henries. You are shifting position again? When I asked you about what basic way, III c IV c V, you answered that I had to continue with IIII, IIIII, etc. > Correct, for instance for 1/7. > > And for computable numbers some representation does exist. > > But this representation does not necessarily enable us to determine > the trichotomy relation with numbers which are really numbers. Perhaps. How do you establish trichotomy between 1/13 and 1/64? Are you really going to base-26 to establish that? That would be pathetic. > 1) 1 ist eine nat.9frliche Zahl. > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. ... > This is a recursive definition of natural numbers. By (1) we have one > natural number, by (2), from that single natural number we get a lot of > other natural numbers. > > Why do you say N is wrong in (2) but not in (1)? Where in (1) is N? I do not see N at all. > This is getting phylosofical. In mathematics a definition is circular if > in the definition one of the deciding features is the term you want to > define. So a definition as you gave: > 3 is the set of all sets of 3 elements > is a perfect example of a circular definition. > > !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IT IS NOT A > DEFINITION !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! You stated that when I asked you for a definition. So what is happening here? > I think that everybody able to read and understand these lines will > know what + 1 means while the successor is not immediately clear. > (The successor of n coul be n+2 or 2*n or 10*n or ....) > > Yes, indeed, and that is the crux. Abstraction. When you use the > '+ 1' notation you are already assuming the existence of addition. > When you do not, you can properly define addition (and all other > operations) using the Peano axioms. You are *defining* the natural > numbers, presumably without any knowledge about what natural numbers > even are. And I may note that with '2*n' you can get a set that is > isomorphic (with respect to all operations) to the natural numbers, > only the naming is different. Consider the set of powers of 2, > (your 2*n case) define: > a '+' b = 2^[ log_2(a) + log_2(b) + 1 ] > a '*' b = 2^[ log_2(a) * log_2(b) + log_2(a) + log_2(b) ] > Call this set K. You may verify that the rings: > R(N, +, *) and R(K, '+', '*') > are isomorphic. > > I do know that. But I don't want some isomorphic sets. I want to > define *the natural numbers* which are > I > II > III > .=2E. And elsewhere you are using 1, 2, 3, 4, VIII, etc. Pray remain consistent. > The non-existence of a pink elephant in the realm of the axiom that > states that there is a pink elephant is not mathematics but philosophy. > Axioms state what things exist (or do not exist) in their realm. > > No. This belief is the core of mathmess Ah, so the parallel axiom, and all Euclidean axioms are the core of mathmess. > Whether it exists remains to be investigated. > > Your existence is not a mathematical existence. > > This form of existence is the only possible existence. As you do not define your form of existence, it is impossible to talk about it, at least mathematically. > Your definition is circular, this one is not. > > I do not give a definition but I look for the existence of the number > already defined. > > So when I ask you for a definition you do not give a definition? Yes, > I have been thinking that all along. > > I gave two definitions: Peano and that with +1 which is very close > to Dedekind's attitude. (I don't know whether he actually created it, > but I know that he would have liked it with +1 as a primitive). But > the axioms do not establish any existence, in particular not when you > apply Dedekind's definition of what a number is. Existence is a mathematical thing when you can establish it by axioms or through theorems based on axioms. Anything else is merely phylosophy. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion >> >> What is 4 in mathematics? > succ(succ(succ(succ(0)))), according to the Peano axioms. > Right. @WM: In axiomatic set theory we (usually) have the definitions: 0 := {} , 1 := {0} , 2 := {0,1} , 3 := {0,1,2} , 4 := {0,1,2,3} , and succ(x) := x u {x} . Hence 4 = succ(succ(succ(succ(0)))) , as required. F. === Subject: Re: Cantor Confusion What is 4 in mathematics? > In axiomatic set theory it's (usually) the set {0, 1, 2, 3}. Moreover it's usually considered a natural number, hence: 4 e N, 4 e Z, 4 e Q, 4 e R, 4 e C, since N c Z c Q c R c C. But 4 contains IIII --- not 0,1,2,3. > No. As you can see, 4 has exactly the elements 0,1,2,3. IIII is only an element of 4 if IIII is either 0,1,2 or 3. Henry VIII had 7 predecessors --- only including him they were 8 > Henries. > Was Henry VIII a natural number? >> ... for computable numbers some representation does exist. > But this representation does not necessarily enable us to determine > the trichotomy relation with numbers which are really numbers. > All numbers are real, but some numbers are more real than others. (Animal Farm) I want to define *the natural numbers* [the following way] I > II > III > ... > Rules for the construction of these sequences of symbols: Rule 1: We may construct I. Rule 2: If n is constructed, we may construct nI. >> The non-existence of a pink elephant in the realm of the axiom that >> states that there is a pink elephant is not mathematics but philosophy. >> Axioms state what things exist (or do not exist) in their realm. No. > Yes. Whether it exists remains to be investigated. > Your existence is not a mathematical existence. > This form of existence is the only possible existence. > No. F. === Subject: Re: Cantor Confusion What is 4 in mathematics? In axiomatic set theory it's (usually) the set {0, 1, 2, 3}. That is circular. What is 3, what is 2, wat is 1, what is 0, what is the empty set? > But 4 contains IIII --- not 0,1,2,3. No. As you can see, 4 has exactly the elements 0,1,2,3. IIII is only > an element of 4 if IIII is either 0,1,2 or 3. That is set theory, not mathematics. > Henry VIII had 7 predecessors --- only including him they were 8 > Henries. Was Henry VIII a natural number? No, but VIII is. > ... for computable numbers some representation does exist. But this representation does not necessarily enable us to determine > the trichotomy relation with numbers which are really numbers. All numbers are real, but some numbers are more real than > others. (Animal Farm) As far as I remember, there is not reality in question but equality. That is in fact an important point for numbers. I want to define *the natural numbers* [the following way] No, you misinterpret me. The definition is done with more or less success by Peano or Dedekind. The success is measured by comparing how closely the numbers, intuitively known as natural numbers, sketched here: > I > II > III > ... are reprodced. The result is tha Peano is not so good. > Rules for the construction of these sequences of symbols: Rule 1: We may construct I. > Rule 2: If n is constructed, we may construct nI. Yes. This definition is better than that by Peano. >> The non-existence of a pink elephant in the realm of the axiom that >> states that there is a pink elephant is not mathematics but philosophy. >> Axioms state what things exist (or do not exist) in their realm. No. Yes. No. Whether it exists remains to be investigated. > Your existence is not a mathematical existence. This form of existence is the only possible existence. No. No to no. === Subject: Re: Cantor Confusion > What is 4 in mathematics? In axiomatic set theory it's (usually) the set {0, 1, 2, 3}. That is circular. What is 3, what is 2, wat is 1, what is 0, what is > the empty set? > But 4 contains IIII --- not 0,1,2,3. No. As you can see, 4 has exactly the elements 0,1,2,3. IIII is only > an element of 4 if IIII is either 0,1,2 or 3. That is set theory, not mathematics. Only to those like WM for whom set theory and mathematics are separate and non-overlapping, To mathematicians, set theory is a part of mathematics. No, you misinterpret me. Since WM deliberately misinterprets virtually everyone, he is hardly in a position to object to being misinterpreted himself. === Subject: Re: Cantor Confusion What is 4 in mathematics? > In axiomatic set theory it's (usually) the set >> {0, 1, 2, 3}. > What is 3, what is 2, what is 1, what is 0, what is > the empty set? > The empty set is the only set which does not have any elements. Its existence (in ZFC) is guaranteed by the axiom of subsets (Aussonder- ung). You know, modern math is based on _axioms_, not some sort of hand waving (you seem to prefer). Usually the empty set is denoted with {}. Then we may define (in an entirely non-circular way): 0 := {} 1 := {0} 2 := {0,1} 3 := {0,1,2} 4 := {0,1,2,3} That is set theory, not mathematics. > Well, the last time I've checked it set theory was mathematics. F. === Subject: Re: Cantor Confusion That is set theory, not mathematics. Well, the last time I've checked it set theory was mathematics. I guess you haven't read WM's book. -- === Subject: Re: Cantor Confusion On Sun, 4 Feb 2007 15:35:53 -0500, I guess you haven't read WM's book. > Right. And I certainly never will. F. === Subject: Re: Cantor Confusion > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. elements, since these are the elements of 4. It also follows that there is only one set with four elements, namely four. So 4=1. You should write a book about this. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page. === Subject: Re: Cantor Confusion > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. elements, since these are the elements of 4. It also follows that there > is only one set with four elements, namely four. So 4=1. You should > write a book about this. He did! Just ask him! === Subject: Re: Cantor Confusion 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. elements, since these are the elements of 4. It also follows that there > is only one set with four elements, namely four. So 4=1. You should > write a book about this. > Dik is even reviewing it, I think. === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl ... > elements, since these are the elements of 4. It also follows that there > is only one set with four elements, namely four. So 4=1. You should > write a book about this. > > Dik is even reviewing it, I think. I am. I am now halfway chapter 8 and only found one serious error (in chapter 7). And one place where I have serious doubts (in chapter 8, but I have to look thoroughly at that). But until this point it is an excellent review about the history about the thinking about the infinite. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl > ... > elements, since these are the elements of 4. It also follows that there > is only one set with four elements, namely four. So 4=1. You should > write a book about this. > > Dik is even reviewing it, I think. > > I am. I am now halfway chapter 8 and only found one serious error (in > chapter 7). And one place where I have serious doubts (in chapter 8, but > I have to look thoroughly at that). But until this point it is an > excellent review about the history about the thinking about the infinite. Having read chapter 8 now completely, there are two errors there. But it remains an excellent review about the history of thinking. I just started chapter 9, and, well, I will not say anymore now. I need a few days for the remaining chapters. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion [...] I am now halfway chapter 8 and only found one serious error (in > chapter 7). And one place where I have serious doubts (in chapter 8, but > I have to look thoroughly at that). But until this point it is an > excellent review about the history about the thinking about the infinite. > Man soll den Tag nicht vor dem Abend loben. F. -- E-mail: infosimple-linede === Subject: Re: Cantor Confusion , > Man soll den Tag nicht vor dem Abend loben I looked this up and found the meaning. How does mid-sentence capitalization work? Is this verse? -- Michael Press === Subject: Re: Cantor Confusion >> Man soll den Tag nicht vor dem Abend loben. > I looked this up and found the meaning. > How does mid-sentence capitalization work? > The substantives in German are usually written this way. Is this verse? > A saying. F. -- E-mail: infosimple-linede === Subject: Re: Cantor Confusion >> >> Man soll den Tag nicht vor dem Abend loben. >> > I looked this up and found the meaning. > How does mid-sentence capitalization work? > > The substantives in German are usually written this way. > > Is this verse? > > A saying. And indeed, it is also a saying in Dutch, when translated. And this was quite appropriate as a response to my earlier posting (which it was). But we do not have such capitals mid-sentence. In Dutch: Men moet de dag niet prijzen voor het avond is. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion ... > elements, since these are the elements of 4. Is this intended to be a serious statement? > It also follows that there > is only one set with four elements, namely four. With respect to the property set having elements one can in fact take this position. There is only one set with four elements. Therefore a set of four elements is the same as every set of all sets of four elements. > You should > write a book about this. That book is available, for instance from Amazon or from the German National Library or directly from the publisher http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&IS BN=3-8322-5587-7 > > Dik is even reviewing it, I think. I am. I am now halfway chapter 8 and only found one serious error (in > chapter 7). And one place where I have serious doubts (in chapter 8, but > I have to look thoroughly at that). But until this point it is an > excellent review about the history about the thinking about the infinite. I am afraid, in chapter 7 (which gives an uncritical introduction to set theory) there is a multitude of serious errors. === Subject: Re: Cantor Confusion >> ... >> elements, since these are the elements of 4. Is this intended to be a serious statement? Of course I am serious in pointing out that this is a consequence of what you have written, and had you not snipped your statement from which this is derived, evryone could see this. >> It also follows that there >> is only one set with four elements, namely four. With respect to the property set having elements one can in fact > take this position. There is only one set with four elements. So for example {1,2,3,4} = {7,42,666,1984}? > Therefore a set of four elements is the same as every set of all > sets of four elements. Good to know. > You should >> write a book about this. That book is available, Who would have guessed this! And in it you prove that 1=4? > for instance from Amazon or from the German > National Library or directly from the publisher > http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&IS B N=3-8322-5587-7 I also continuously get paper spam by this publisher. Now I know what kind of customers they have. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page. === Subject: Re: Cantor Confusion ... >> elements, since these are the elements of 4. Is this intended to be a serious statement? Of course I am serious in pointing out that this is a consequence of > what you have written, and had you not snipped your statement from which > this is derived, evryone could see this. I did not snip anything here. (I did not answer you.) But what ever you may have thought, it must be mad. > It also follows that there >> is only one set with four elements, namely four. With respect to the property set having elements one can in fact > take this position. There is only one set with four elements. So for example {1,2,3,4} = {7,42,666,1984}? With respect to the property set having n elements (= cardinal number), yes, of course. Cardinal number is 4 in both cases. Therefore a set of four elements is the same as every set of all > sets of four elements. Good to know. > This is new for you? > You should >> write a book about this. That book is available, Who would have guessed this! And in it you prove that 1=4? Usually I do no longer read your writings. But I have looked at your conclusion: quote: > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. elements, since these are the elements of 4. quote end. I do not see a valid conclusion. Regard, WM === Subject: Re: Cantor Confusion Who would have guessed this! And in it you prove that 1=4? Usually I do no longer read your writings. I wonder how we can get WM to do this for everyone. -- === Subject: Re: Cantor Confusion On Sun, 4 Feb 2007 15:39:43 -0500, And in [your book] you prove that 1=4? [CS] > >> Usually I do no longer read your writings. [WM] > I wonder how we can get WM to do this for everyone. > You have to be interested in mathematics as such if replying to the nonsense WM produces, and n o t in _arguing_ with WM (like Virgil and/or F. Neugebauer). (The reason is simple: He likes to argue, but he hates _real_ math; or at least he's not interested in it.) F. === Subject: Re: Cantor Confusion > On Sun, 4 Feb 2007 15:39:43 -0500, And in [your book] you prove that 1=4? [CS] > >> Usually I do no longer read your writings. [WM] I wonder how we can get WM to do this for everyone. You have to be interested in mathematics as such if replying to the > nonsense WM produces, and n o t in _arguing_ with WM (like Virgil > and/or F. Neugebauer). (The reason is simple: He likes to argue, but > he hates _real_ math; or at least he's not interested in it.) That makes sense. In my later replies to WM, I would translate what he math, so he won't reply to me anymore. If you just argue with him using words, he can interpret the words any way he wishes (instead of the way you intend). -- === Subject: Re: Cantor Confusion > Who would have guessed this! And in it you prove that 1=4? Usually I do no longer read your writings. I wonder how we can get WM to do this for everyone. It is not his reading that I object to but his writing. === Subject: Re: Cantor Confusion > Who would have guessed this! And in it you prove that 1=4? > Usually I do no longer read your writings. I wonder how we can get WM to do this for everyone. It is not his reading that I object to but his writing. If people stop replying to him, I doubt he will continue to post. Of course, it means letting him have the last word. Or, you can reply to him, but you have to be more mathematical (symbols and such). He won't reply to that, apparently. -- === Subject: Re: Cantor Confusion On Sun, 4 Feb 2007 17:50:03 -0500, If people stop replying to him, I doubt he will continue to post. > But David, Virgil LOVES to argue with WM (it's not just the math). F. === Subject: Re: Cantor Confusion > On Sun, 4 Feb 2007 17:50:03 -0500, If people stop replying to him, I doubt he will continue to post. But David, Virgil LOVES to argue with WM (it's not just the math). Well, then that's fine! -- === Subject: Re: Cantor Confusion >> ... >> elements, since these are the elements of 4. > Is this intended to be a serious statement? Of course I am serious in pointing out that this is a consequence of > what you have written, and had you not snipped your statement from which > this is derived, evryone could see this. I did not snip anything here. (I did not answer you.) But what ever > you may have thought, it must be mad. >> It also follows that there >> is only one set with four elements, namely four. > With respect to the property set having elements one can in fact > take this position. There is only one set with four elements. So for example {1,2,3,4} = {7,42,666,1984}? WM declares absolutely that There is only one set with four elements. But that would require {1,2,3,4} = {7,42,666,1984}, which is false. === Subject: Re: Cantor Confusion > If 4 is the set of all existing (whatever that does mean) sets with 4 > elements and 5 is the set of all existing sets with 5 elements, we find > immediately that 4 is *not* a subset of 5. With this definition of > numbers as sets subsetting does not give what you wish. 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. In unary representation or by using peanuts or matchsticks 4 > is a subset of 5. The set of all sets with exactly 4 elements is not a subset of the set of all sets with exactly 5 elements since the set of all sets with exactly 5 elements does not contain any sets with less that five elemeNts. WM must have completely lost touch with reality to think otherwise. But indeed, for most numbers some > representations do not exist, while other representations do exist. The > existence of a number is independent of the existence of a representation. This point of view has lead to the present mess-math. The mess is only in the minds of those who imagine it into existence. For those of us who perceive no mess there is no mess. For instance, for most rational numbers a decimal representation does not > exist. Correct, for instance for 1/7. Wrong! That is one for which a decimal representation does exist. > Yes, I know Peano pretty well. But this is *not* a circular definition. I > wonder why you think it is circular. I did not say it is circular. I said as non-circular as possible. If > it is not circular in your opinion, then be happy. (In fact every > definition is circular, because every language is. When one accepts some words as primitives and undefined, definitions based only on them are not circular at all. > I think that everybody able to read and understand these lines will > know what + 1 means while the successor is not immediately clear. > (The successor of n coul be n+2 or 2*n or 10*n or ....) Successor is a primitive, but the Peano statements require it to have certain properties. > 3) IN ist Durchschnitt aller Mengen M, die (1) und (2) erf.9fllen. That is an alternative definition, again not circular. It is better than Peano, because one does not use natural numbers, but > is circular though as little as possible. You only need to know +1 > and the intersection of sets. (And what a set is, and what 1 is, and > so on.) In Peano, you do not have to know what any of the things are, all you have to know is that they work according to the rules stated. The existence does not follow from axioms. Axioms state something in > an arbitrary way. They can even state the famous pink elephant. > Whether it exists remains to be investigated. But for this sake we > need not define again what a pink elephant is. Your definition is circular, this one is not. I do not give a definition but I look for the existence of the number > already defined. But without a definition there can be no testing of existence. How can you tell if a floop exists if you can't even tell whether something is or is not a floop? If you don't know at all what you're looking for, you can't ever tell whether you've found it or not. === Subject: Re: Cantor Confusion > Continued: > > You are correct. > The union of all natural numbers is not a natural number. Therefore > you think it (the cardinality) can be infinite. > > Well, it is certainly not finite, so it must be infinite (translation: > not finite). > > Not finite is correct. But you translate it as actually existing set, > i.e., finished infinity. That is incorrect. Not finite is identical to infinite. But the axiom of infinity states that > it exists. Not finite is identical to infinite. The axiom of infinity, however, finishes infinity. > The union of all paths is a path in many cases. In such a case it is > simultaneousy a union of natural numbers which is a natural number. > > The length of finite paths are natural numbers. The length of a > finite union is a natural number. The length of an infinite union is > not a natural number. Why do you think it is? > > Because there is nothing else in the union but natural numbers. > You cannot get an apple if you collect nuts. The union of all finite natural numbers is not a natural number. Why > do you think it is? The union of paths is a path. The projection of a finite number on a pathlength remains a finite number. The projection of all finite numbers on a pathlength remains a finite number. > Nevertheless, if you assume that the length of the union of all paths > (not the cardinal number !) is infinite, then you advocate infinite > number sizes are required for an infinite union of numbers. That is quite confusing as stated. It is only unfamiliar to think about that. > But indeed, the infinite union of > all natural numbers (considered as sets) is not a natural number. Maybe. But the natural numbers consideed as paths remain natural numbers in any case. > There is no infinite union of finite paths unless there is an infinite > path. > Yes. And so what? There is an infinite number, more than one actually. > But it is *not* a member of the set of finite numbers, as the infinite > path is *not* a member of the set of finite paths. And natural numbers > are (through the definitions) the numbers in the set of finite numbers, > so there is no natural number that is infinite. And on the other hand, > the infinite paths do not have a natural number as path-length. > > On the other hand many people assert that there is an infinite union > of finite pathlengths without containing or establishing an infinite > pathlength. Eh? Who? Everybody who says that there is an infinite set of finite pathlengths. === Subject: Re: Cantor Confusion > The union of trees implies the union of outmost > left-hand side paths for example. This union is a path. > > Indeed, but it is *not* in the union of the sets of paths of finite trees. > In the above example, that set does contain *all* the paths {0}, {0, 1}, > {0, 1, 2}, {0, 1, 2, 3}, etc., but *not* the path {0, 1, 2, 3, ...}, > because the last path is in *none* of the constituent sets. > > This is the same with the infinite tree. Why do you accept the > existence of the infinite tree? Why is it the same? The infinite tree *does* contain the path > {0, 1, 2, 3, ...}. But that path is *not* in the union of the sets > of finite paths. And that path is ot contained in the union of sets of finite trees. > The union of all finite paths of type p(r) establishes the infinite > path r. > The union of all finite paths of all finite trees establishes all > infinite paths. Yes. I do not contradict *that*. The set of the union of paths is > *not* the union of the sets of paths. No, it is only a subset. === Subject: Re: Cantor Confusion <45bb7541$0$97231$892e7fe2@authen.yellow.readfreenews.net> <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > You can try to utter again and agan this nonsense, but after a while > aI will cease to reply. Every such thing including only natural > numbers is covered by induction. All natural numbers are subject to > induction. True, you can use induction to prove something about > any natural number. However, the question is: Can > you use induction to prove something about a set > of natural numbers? This question has a trivial answer: > If the set of all natural numbers is nothing than all natural numbers, > then yes. > If the set of all natural numbers is more than all natural numbers, > then no. No there are two questions. i: Is X true for every element of the set A? ii: Is X true for the set A. The two questions are different. The answer to question ii can be No, even when the answer to question i is yes. Your trivial answer only applies to question i. There are two types of sets of natural numbers I: sets of natural numbers that are not (potentially) infinite II: sets of natural numbers that are (potentially) infinite Induction can only prove things about sets of type I. So you believe that type II sets have some esoteric supplement? No. The answer to question ii can be No, even if there is no esoteric supplement (i.e. the answer to question i is Yes). We can prove that every number is a number while a set of several > numbers is not a number. As you point out there is a difference between the elements of a set and the set. > There is no reason to distinguish between > finite and infinite sets. By running induction longer and longer you can get more and more > sets of type I. By running induction for an infinite time you can > get all > sets of type I. No. By two steps of induction, namely > P(1) > P(n) ==> P(n+1) > you can prove P for all natural numbers and for all sets of natural > number (except quantitative statements as I mentioned above). No you can prove things for all elements of all sets of natural numbers. There is a difference between proving that something is true of every element in a set A and proving that something is true of a set A. Your strong belief in the inaccesible infinite uttered above is your > (and some other people's) personal opinion but has nothing to do with > mathematics. But you will never get a set of type II. (You may get > a collection of sets whose union contains the same elements > as a set of type II, but you will never get a single set of type > II). Small wonder. There is no set of type II. No. We have already agreed that potentially infinite sets exist. Whether these sets actually exist, that is whether every element in the set can be said to exist at once, is of no interest here. The set of all natural numbers, the union of all natural numbers, N, > is a set of type II. You cannot use induction to prove anything about > the union of all natural numbers. Small wonder. This set does not exist, No, the potentially infinite set of natural numbers exists. We are not interested in the question of whether this set actually exists. (Whether or not the set actually exists, the question Is the maximum element of N fixed (i.e. can it change if you change the subset of N that actually exists)? has the answer No.) - William Hughes === Subject: Re: Cantor Confusion <45bb7541$0$97231$892e7fe2@authen.yellow.readfreenews.net> <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > You can try to utter again and agan this nonsense, but after a while > aI will cease to reply. Every such thing including only natural > numbers is covered by induction. All natural numbers are subject to > induction. > True, you can use induction to prove something about > any natural number. However, the question is: Can > you use induction to prove something about a set > of natural numbers? This question has a trivial answer: > If the set of all natural numbers is nothing than all natural numbers, > then yes. > If the set of all natural numbers is more than all natural numbers, > then no. No there are two questions. i: Is X true for every element of the set A? > ii: Is X true for the set A. The two questions are different. > The answer to question ii can be No, even when > the answer to question i is yes. That difference concerning quantity is correct. But this concerns finite sets as well as infinite sets. > Your trivial answer only applies to question i. And only this question is of interest. For instance: Is every segment of the set of finite natural numbers a finite set? > So you believe that type II sets have some esoteric supplement? No. The answer to question ii can be No, even if > there is no esoteric supplement (i.e. the answer to > question i is Yes). Yes. That is not in question, but the difference of answers for finite and infinite sets. > We can prove that every number is a number while a set of several > numbers is not a number. As you point out there is a difference between > the elements of a set and the set. > Yes. That is not in question. What I dispute is that the difference depends on the cardinal number of the set. > There is no reason to distinguish between > finite and infinite sets. > By running induction longer and longer you can get more and more > sets of type I. By running induction for an infinite time you can > get all > sets of type I. No. By two steps of induction, namely > P(1) > P(n) ==> P(n+1) > you can prove P for all natural numbers and for all sets of natural > number (except quantitative statements as I mentioned above). No you can prove things for all elements of all sets > of natural numbers. There is a difference between > proving that something is true of every element in a set A > and proving that something is true of a set A. Yes. That is not in question. What I dispute is that the difference (induction is possible or is not) depends on the cardinal number of the set. === Subject: Re: Cantor Confusion <45bb7541$0$97231$892e7fe2@authen.yellow.readfreenews.net> <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > You can try to utter again and agan this nonsense, but after a while > aI will cease to reply. Every such thing including only natural > numbers is covered by induction. All natural numbers are subject to > induction. > True, you can use induction to prove something about > any natural number. However, the question is: Can > you use induction to prove something about a set > of natural numbers? > This question has a trivial answer: > If the set of all natural numbers is nothing than all natural numbers, > then yes. > If the set of all natural numbers is more than all natural numbers, > then no. No there are two questions. i: Is X true for every element of the set A? > ii: Is X true for the set A. The two questions are different. > The answer to question ii can be No, even when > the answer to question i is yes. That difference concerning quantity is correct. But this concerns > finite sets as well as infinite sets. Your trivial answer only applies to question i. And only this question is of interest. For instance: Is every segment > of the set of finite natural numbers a finite set? > So you believe that type II sets have some esoteric supplement? No. The answer to question ii can be No, even if > there is no esoteric supplement (i.e. the answer to > question i is Yes). Yes. That is not in question, but the difference of answers for finite > and infinite sets. > We can prove that every number is a number while a set of several > numbers is not a number. As you point out there is a difference between > the elements of a set and the set. Yes. That is not in question. What I dispute is that the difference > depends on the cardinal number of the set. It may or may not. For instance the question Is the cardinal number of the set a (finite) natural number (if the set does not have a cardinal number then the answer is no)? certainly depends on the cardinal number of the set. Sets of type I have a cardinal which is a (finite) natural number. Sets of type II do not. > There is no reason to distinguish between > finite and infinite sets. > By running induction longer and longer you can get more and more > sets of type I. By running induction for an infinite time you can > get all > sets of type I. > No. By two steps of induction, namely > P(1) > P(n) ==> P(n+1) > you can prove P for all natural numbers and for all sets of natural > number (except quantitative statements as I mentioned above). No you can prove things for all elements of all sets > of natural numbers. There is a difference between > proving that something is true of every element in a set A > and proving that something is true of a set A. Yes. That is not in question. What I dispute is that the difference > (induction is possible or is not) depends on the cardinal number of > the set. Induction on the elements of A is always possible given any set A of natural numbers. However, induction may not be able to prove that something is true of A For example, every one of the elements of A is finite. This can be shown by induction. However, this does not show that the set A is not (potentially) infinite. - William Hughes === Subject: Re: Cantor Confusion <45bb7541$0$97231$892e7fe2@authen.yellow.readfreenews.net> <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > Yes. That is not in question. What I dispute is that the difference > depends on the cardinal number of the set. It may or may not. For instance the question Is the cardinal number of the set a (finite) natural number > (if the set does not have a cardinal number then the > answer is no)? certainly depends on the cardinal number of the set. Sets of type > I have a cardinal which is a (finite) natural number. Sets of type > II do not. My questions are: What numbers are subject to complete induction? What is the set the elements of which are subject to complete induction? > Yes. That is not in question. What I dispute is that the difference > (induction is possible or is not) depends on the cardinal number of > the set. Induction on the elements of A > is always possible given any set A of > natural numbers. However, induction may not > be able to prove that something is true of A Why should it? But it can be able. For instance it can prove that every set of natural numbers is finite while the size of sets of natural numbers is unbounded from above. For example, every one of the elements > of A is finite. This can be shown by induction. > However, this does not show that the set > A is not (potentially) infinite. It does. Of course every set of natural numbers is finite, as proved by induction over all initial segments. (Potential infinity.) This simple truth has only been veiled by the arbitrary assumption that there is an actual infinity. As this assertion cannot be proved, some have argue that induction was not valid for the whole set. === Subject: Re: Cantor Confusion > Induction on the elements of A > is always possible given any set A of > natural numbers. However, induction may not > be able to prove that something is true of A Why should it? But it can be able. For instance it can prove that > every set of natural numbers is finite while the size of sets of > natural numbers is unbounded from above. Not so. Induction can only prove every set of naturals which is bounded above by a natural is finite, but that does not, according to the axiom of infinity, exhaust the possible sets of naturals. For example, every one of the elements > of A is finite. This can be shown by induction. > However, this does not show that the set > A is not (potentially) infinite. It does. Then lets see this alleged proof. What Induction says is precisely: Given a set, S, of natural numbers such that (a) S contains the first (smallest) natural, and (b) whenever a particular natural is a member of S, the successor of that natural is also a member of S, then every natural is a member of S. And induction says nothing else. > Of course every set of natural numbers is finite, as proved > by induction over all initial segments. That only proves that such initial segments as have a last element are finite. It says nothing at all about any initial segment which does not have a last element, and there is one. === Subject: Re: Cantor Confusion <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net> every set of natural numbers is finite while the size of sets of > natural numbers is unbounded from above. Not so. Induction can only prove every set of naturals which is bounded > above by a natural is finite, but that does not, according to the axiom > of infinity, exhaust the possible sets of naturals. There is no natural number in this set which is not covered by induction. So, which number is missing to exhaust N? > What Induction says is precisely: > Given a set, S, of natural numbers such that > (a) S contains the first (smallest) natural, and > (b) whenever a particular natural is a member of S, > the successor of that natural is also a member of S, > then every natural is a member of S. And induction says nothing else. Induction says: If P holds for n then it holds for n+1. This is the main property of an infinite set as the set described by the axiom of infinity originally was. Only the dubious interpretation of esoteric infinity removes this set from the realm of induction and from the realm of mathematics too. > Of course every set of natural numbers is finite, as proved > by induction over all initial segments. That only proves that such initial segments as have a last element are > finite. It says nothing at all about any initial segment which does not > have a last element, and there is one. Every finite natural defines a finite segment. An infinite segment can be defined by an infinite number only, and there is none. > The set of all sets with exactly 4 elements is not a subset of the set > of all sets with exactly 5 elements since the set of all sets with > exactly 5 elements does not contain any sets with less that five > elemeNts. 4 is also every set with 4 elements. Every set with 4 elements is a subset of that set where one further element has been added. > WM must have completely lost touch with reality to think otherwise. Contrary. Just because I have not lost touch with reality. > The mess is only in the minds of those who imagine it into existence. > For those of us who perceive no mess there is no mess. That is an excellent description of your position. > In Peano, you do not have to know what any of the things are, all you > have to know is that they work according to the rules stated. You do not have to know (or to be able to get to know at least) whether or not one natural number is larger than the other? > If you don't know at all what you're looking for, you can't ever tell > whether you've found it or not. Therefore I accept and reported in my book the definitions by Peano and Dedekind. > but { p(n) : n in N } is a set of paths and not a path. And the union of the paths which belong to this set is a set of nodes which is realized in the tree. === Subject: Re: Cantor Confusion > Why should it? But it can be able. For instance it can prove that > every set of natural numbers is finite while the size of sets of > natural numbers is unbounded from above. Not so. Induction can only prove every set of naturals which is bounded > above by a natural is finite, but that does not, according to the axiom > of infinity, exhaust the possible sets of naturals. There is no natural number in this set which is not covered by > induction. So, which number is missing to exhaust N? Who said anything like that? WM misreads so consistently that he must be doing it intentionally. What Induction says is precisely: > Given a set, S, of natural numbers such that > (a) S contains the first (smallest) natural, and > (b) whenever a particular natural is a member of S, > the successor of that natural is also a member of S, > then every natural is a member of S. And induction says nothing else. Induction says: If P holds for n then it holds for n+1. On the contrary, it says no such thing. It is entirely up to the person attempting to use induction to PROVE that if P holds for n then it holds for n+1. There is nothing in induction to require it. > This is the > main property of an infinite set as the set described by the axiom of > infinity originally was. > Only the dubious interpretation of esoteric > infinity removes this set from the realm of induction and from the > realm of mathematics too. The only esoteric version of infinite (not finite) is WM's. In standard set theory, a set is either finite or not finite with no third option. > Of course every set of natural numbers is finite, as proved > by induction over all initial segments. Actually every natural is finite by definition, but there is no inductive proof that every set of naturals is finite because it isn't treu, inductively or otherwise, at least for any standard definition of finiteness of a set. What is WM's definition of a set being finite anyway. He has often been asked for it, but I do not recall his ever giving it. That only proves that such initial segments as have a last element are > finite. It says nothing at all about any initial segment which does not > have a last element, and there is one. Every finite natural defines a finite segment. An infinite segment can > be defined by an infinite number only, and there is none. Omega is an infinite ordinal which defines an infinite set of ordinals, those preceding it. Sp WM is wrong again. > The set of all sets with exactly 4 elements is not a subset of the set > of all sets with exactly 5 elements since the set of all sets with > exactly 5 elements does not contain any sets with less that five > elements. 4 is also every set with 4 elements. Every set with 4 elements is a > subset of that set where one further element has been added. WM is conflating incompatible definitions of cardinality here. The cardinal of a set can be a representative set which bijects with the given set, as in NBG, or it can be the class of all sets bijectable with the given set, as in NF, but it can't be both simultaneously, as the two theories are incomaptible. WM must have completely lost touch with reality to think otherwise. Contrary. Just because I have not lost touch with reality. When one conflates equivalence classes with representative elements, as WM has done above, wherever he is, it is not reality. The mess is only in the minds of those who imagine it into existence. For those of us who perceive no mess there is no mess. That is an excellent description of your position. The alleged mess is only perceptible to those who assume as axioms things that create such a mess. To those who are more careful of their axioms, there is no such mess. In Peano, you do not have to know what any of the things are, all you > have to know is that they work according to the rules stated. You do not have to know (or to be able to get to know at least) > whether or not one natural number is larger than the other? Unless largeness is defined in the rules, no! If you don't know at all what you're looking for, you can't ever tell > whether you've found it or not. Therefore I accept and reported in my book the definitions by Peano > and Dedekind. Who knew what they were looking for, and found it. If WM's book at all represents the ideas he tries to fob off here, he should retract it. === Subject: Re: Cantor Confusion > What is WM's definition of a set being finite anyway. He has often been > asked for it, but I do not recall his ever giving it. WM doesn't seem to understand the word definition. I wonder if he has a name for what we call definition. -- === Subject: Re: Cantor Confusion <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > Yes. That is not in question. What I dispute is that the difference > depends on the cardinal number of the set. It may or may not. For instance the question Is the cardinal number of the set a (finite) natural number > (if the set does not have a cardinal number then the > answer is no)? certainly depends on the cardinal number of the set. Sets of type > I have a cardinal which is a (finite) natural number. Sets of type > II do not. My questions are: What numbers are subject to complete induction? > What is the set the elements of which are subject to complete > induction? > Yes. That is not in question. What I dispute is that the difference > (induction is possible or is not) depends on the cardinal number of > the set. Induction on the elements of A > is always possible given any set A of > natural numbers. However, induction may not > be able to prove that something is true of A Why should it? But it can be able. For instance it can prove that > every set of natural numbers is finite while the size of sets of > natural numbers is unbounded from above. For example, every one of the elements > of A is finite. This can be shown by induction. > However, this does not show that the set > A is not (potentially) infinite. It does. No. Let P be the set of prime numbers. Every prime number is finite. P is potentially infinite. So showing that every element of a set is finite does not show that the set is not potentially infinite. > Of course every set of natural numbers is finite, as proved > by induction over all initial segments. (Potential infinity.) If you define a potentially infinite set to be finite, and say that there are only sets with a fixed maximum and potentially infinite sets, then all set are finite. So what? Calling a potentially infinite set finite does not change its properties. > This > simple truth has only been veiled by the arbitrary assumption that > there is an actual infinity. No. Sets with a fixed maximum and potentially infinite sets have different properties, whether or not you assume an actual infinity. > As this assertion cannot be proved, some > have argue that induction was not valid for the whole set. No. There are two different questions. Let A be a set of natural numbers. i: Can induction show something about the elements of A ii: Can induction show something about A. Note, for some sets A, the answer to i can be Yes and the answer to ii can be No. No one has claimed that the answer to i is anything but Yes. The converstion goes like this M: induction shows that A has property X O: induction is not valid for A [that is question ii is false for A] M: A is a set of natural numbers so induction is valid for A [that is question i is true for A] The statement induction [is] not valid for the whole set means that the answer to question ii is No, not that the answer to question i is No. - William Hughes === Subject: Re: Cantor Confusion <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > certainly depends on the cardinal number of the set. Sets of type > I have a cardinal which is a (finite) natural number. Sets of type > II do not. My questions are: What numbers are subject to complete induction? > What is the set the elements of which are subject to complete > induction? My questions are: What numbers are subject to complete induction? What is the set the elements of which are subject to complete induction? > For example, every one of the elements > of A is finite. This can be shown by induction. > However, this does not show that the set > A is not (potentially) infinite. It does. No. Let P be the set of prime numbers. > Every prime number is finite. P is potentially > infinite. Yes. But no prime number is the largest. > So showing that every element of a set is > finite does not show that the set is not potentially infinite. This is already the case for sets of two elements: Showing that every element is a single elements does not show that the set is a single element. Every potentially infinite set is finite. This can be proved by induction. Therefore induction is sufficient for all statements about potentially infinite sets. On the other hand potential infinity can be proved by induction: If there is a set with n numbers, then there is a set with n+1 numbers. Similarly one can prove by induction: Every set of even positive numbers contains numbers which are larger than the cardinal number of the set. Of course every set of natural numbers is finite, as proved > by induction over all initial segments. (Potential infinity.) If you define a potentially infinite set to be finite, and > say that there are only sets with a fixed maximum and > potentially infinite sets, then all set are finite. > So what? Calling a potentially infinite set finite does > not change its properties. Calling it infinite does not change its properteis either. Every set which is given by its finite numbers is finite. If this does not exclude that a larger set of this kind can be given, then the set of such sets (usually abbreviated as the set of such elements) is called potentially infinite. This can be proved by induction. In fact, induction was mainly devised for potentially infinite sets, because for finite sets we do not need it. This > simple truth has only been veiled by the arbitrary assumption that > there is an actual infinity. No. Sets with a fixed maximum and potentially infinite sets > have different properties, whether or not you assume an > actual infinity. Potentially infinite sets are subject to induction. As this assertion cannot be proved, some > have argue that induction was not valid for the whole set. No. There are two different questions. > Let A be a set of natural numbers. i: Can induction show something > about the elements of A Can induction show something about *every* element of A. ii: Can induction show something about A. If A is infinite, induction can show that A is infinite. Note, for some sets A, the answer to i can be Yes > and the answer to ii can be No. That is true, but it does not depend on the infinity of a set. It is already the case for sets of two numbers. Induction is devised for infinite sets and is required for infinite sets only. But it cannot be used for extra-matzhematical properties like actual infinity. === Subject: Re: Cantor Confusion > certainly depends on the cardinal number of the set. Sets of type > I have a cardinal which is a (finite) natural number. Sets of type > II do not. > My questions are: > What numbers are subject to complete induction? > What is the set the elements of which are subject to complete > induction? My questions are: What numbers are subject to complete induction? > What is the set the elements of which are subject to complete > induction? As WM has not explained how complete induction differs from standard induction, we have no idea of what he means by it. For standard induction, based on the set of natural numbers, N: If S is a subset of N such that the first member of N is a member of S and whenever a member of N is a member of S so its successor, then S = N. > For example, every one of the elements > of A is finite. This can be shown by induction. > However, this does not show that the set > A is not (potentially) infinite. > It does. No. Let P be the set of prime numbers. > Every prime number is finite. P is potentially > infinite. Yes. But no prime number is the largest. Irrelevant, as usual. So showing that every element of a set is > finite does not show that the set is not potentially infinite. This is already the case for sets of two elements: Showing that every > element is a single elements does not show that the set is a single > element. Every potentially infinite set is finite. False by any standard definition of finiteness. What definition of finiteness does WM claim? I. Every finite set can be ordered so as to have both a first ands a last member. A set which cannot be so ordered is not finite. The set of naturals is, by this rule, not finite. II. A set is finite if every injection from it to itself is a surjection. A set having an injection to itself which is not a surjection is not finite. The set of naturals is, by this rule, not finite. > This can be proved by induction. > Therefore induction is sufficient for all statements about potentially > infinite sets. On the other hand potential infinity can be proved by induction: > If there is a set with n numbers, then there is a set with n+1 > numbers. > Similarly one can prove by induction: Every set of even positive > numbers contains numbers which are larger than the cardinal number of > the set. One can prove that for finite sets, but not for sets which are not finite. So that WM is assuming as an axiom that all sets are finite. WM cannot prove any of his silly claims without that assumption. > > Of course every set of natural numbers is finite, as proved > by induction over all initial segments. (Potential infinity.) Not without first having assumed it. So all such proofs are circular. If you define a potentially infinite set to be finite, and > say that there are only sets with a fixed maximum and > potentially infinite sets, then all set are finite. > So what? Calling a potentially infinite set finite does > not change its properties. Calling it infinite does not change its properteis either. But proving that a set must have properties that finite sets cannot have makes it not finite. What is WM's definitions of a set being finite? It seems to mean only that it is a set, at least to WM, and nothing else, since WM assumes every set to be finite. > Every set which is given by its finite numbers is finite. Which still does not define finiteness of sets, nor finiteness of numbers. > If this does > not exclude that a larger set of this kind can be given, then the set > of such sets (usually abbreviated as the set of such elements) is > called potentially infinite. This can be proved by induction. In fact, > induction was mainly devised for potentially infinite sets, because > for finite sets we do not need it. Induction was not devised for sets at all, but for proofs. > This > simple truth has only been veiled by the arbitrary assumption that > there is an actual infinity. And corrupted by the arbitrary assumption that all sets must be finite, although WM never makes clear what he means by a set being finite. No. Sets with a fixed maximum and potentially infinite sets > have different properties, whether or not you assume an > actual infinity. Potentially infinite sets are subject to induction. There is no such thing as a merely potentially infinite set. A set is either finite (according to some standard definition of finiteness like having no injection to a proper subset) or it is not finite. Sets which are not finite are called infinite, and there is no mere potentiality about it, every set is either a finite set or is a set which is not finite, there is no third alternative, regardless of what WM dreams. === Subject: Re: Cantor Confusion Every potentially infinite set is finite. False by any standard definition of finiteness. What definition of > finiteness does WM claim? How do you know it is false? What definition of potentially infinite are you using? -- === Subject: Re: Cantor Confusion > Every potentially infinite set is finite. False by any standard definition of finiteness. What definition of > finiteness does WM claim? How do you know it is false? What definition of potentially infinite > are you using? Not finite where finite of a set is Either I. A set is finite if it has a natural number as cardinality, presuming the set of natural numbers has been defined as a minimal inductive set. Or II. A set is finite if every injection to itself is a surjection. === Subject: Re: Cantor Confusion > Every potentially infinite set is finite. > False by any standard definition of finiteness. What definition of > finiteness does WM claim? How do you know it is false? What definition of potentially infinite > are you using? Not finite where finite of a set is > Either > I. A set is finite if it has a natural number as cardinality, presuming > the set of natural numbers has been defined as a minimal inductive set. > Or > II. A set is finite if every injection to itself is a surjection. So, you would define potentially infinite = infinite. That makes the most sense to me. But, it seems that WM has potentially infinite => finite. I wonder if he also has finite => potentially infinite. -- === Subject: Re: Cantor Confusion <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > certainly depends on the cardinal number of the set. Sets of type > I have a cardinal which is a (finite) natural number. Sets of type > II do not. > My questions are: > What numbers are subject to complete induction? > What is the set the elements of which are subject to complete > induction? My questions are: What numbers are subject to complete induction? > What is the set the elements of which are subject to complete > induction? > For example, every one of the elements > of A is finite. This can be shown by induction. > However, this does not show that the set > A is not (potentially) infinite. > It does. No. Let P be the set of prime numbers. > Every prime number is finite. P is potentially > infinite. Yes. In other words my statement showing that every element of a set is finite does not show that the set is not potentially infinite. is correct and your It does is wrong. >But no prime number is the largest. equivalent to saying the set of prime numbers is potentially infinite. So showing that every element of a set is > finite does not show that the set is not potentially infinite. This is already the case for sets of two elements: Showing that every > element is a single elements does not show that the set is a single > element. Every potentially infinite set is finite. > This can be proved by induction. > Therefore induction is sufficient for all statements about potentially > infinite sets. On the other hand potential infinity can be proved by induction: > If there is a set with n numbers, then there is a set with n+1 > numbers. > Similarly one can prove by induction: Every set of even positive > numbers contains numbers which are larger than the cardinal number of > the set. > No. You make your usual mistake. Let E be the potentailly infinite set of even numbers. Two statements i: For every element e of E, the set F(e) = {2,4,6,...,e} contains cardinal numbers which are larger than the cardinal number of F(e) ii E contains numbers which are larger than the cardinal number of E Statement i is true and can be shown by induction. Statement ii is false. The fact that statment i is true does not mean that statement ii is true. - William Hughes === Subject: Re: Cantor Confusion <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net Similarly one can prove by induction: Every set of even positive > numbers contains numbers which are larger than the cardinal number of > the set. No. You make your usual mistake. No, you make your usual mistake by thinking that a potentially infinite set can have an infinite cardinal number. > Let E be the potentailly infinite set of even numbers. > Two statements i: For every element e of E, the set > F(e) = {2,4,6,...,e} contains cardinal numbers which are > larger > than the cardinal number of F(e) ii E contains numbers which are larger than the cardinal > number of E Statement i is true and can be shown by induction. Statement > ii is false. Of course, namely because there is no cardinal number for potentially infinite sets. > The fact that statment i is true does not mean > that statement ii is true. The fact that statement I is true means that there are no elements of the set which could increase its number of elements without increasing the sizes of elements in the set. A little bit of logic, but not much, is required to infer this fact. === Subject: Re: Cantor Confusion > Similarly one can prove by induction: Every set of even positive > numbers contains numbers which are larger than the cardinal number of > the set. No. You make your usual mistake. No, you make your usual mistake by thinking that a potentially > infinite set can have an infinite cardinal number. A set can be only finite or not finite. There is no such thing as potential infiniteness distinct from non-finiteness. There is no third alternative. Let E be the potentailly infinite set of even numbers. > Two statements i: For every element e of E, the set > F(e) = {2,4,6,...,e} contains cardinal numbers which are > larger > than the cardinal number of F(e) ii E contains numbers which are larger than the cardinal > number of E Statement i is true and can be shown by induction. Statement > ii is false. Of course, namely because there is no cardinal number for potentially > infinite sets. There is a cardinal for N everywhere except in WM-topia. === Subject: Re: Cantor Confusion <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > Similarly one can prove by induction: Every set of even positive > numbers contains numbers which are larger than the cardinal number of > the set. No. You make your usual mistake. No, you make your usual mistake by thinking that a potentially > infinite set can have an infinite cardinal number. I make no assumptions whatsoever about the cardinal number of a potentially infinite set. (I do not even assume that one exists). Let E be the potentailly infinite set of even numbers. > Two statements i: For every element e of E, the set > F(e) = {2,4,6,...,e} contains cardinal numbers which are > larger > than the cardinal number of F(e) ii E contains numbers which are larger than the cardinal > number of E Statement i is true and can be shown by induction. Statement > ii is false. Of course, My claim is that statment ii is false. Your reply is of course. We both agree that statement ii is false. > namely because there is no cardinal number for potentially > infinite sets. The fact that statment i is true does not mean > that statement ii is true. The fact that statement I is true means that there are no elements of > the set which could increase its number of elements without > increasing the sizes of elements in the set. As statement ii does not say or imply anything about increasing the number of elements in E, totally irrelevent. There is no element that can be shown to be in E that is not in some F(e), but you need more than one F(e) . There is no single F(e) that contains every element that can be shown to be in E. The fact that there is no element in E that is not in some set F(e), does not mean that E has the same properties as the sets F(e). The fact that statement i is true does not imply that statement ii is true. Statement i is true, statement ii is false (note your of course above). The statement Every set of even positive numbers contains numbers which are larger than the cardinal number of the set, is false. The counterexample is the potentially infinite set E. - William Hughes === Subject: Re: Cantor Confusion [...] >> No. There are two different questions. >> Let A be a set of natural numbers. >> i: Can induction show something >> about the elements of A Can induction show something about *every* element of A. It was devised for this purpose. >> ii: Can induction show something about A. If A is infinite, induction can show that A is infinite. 1. Where did you read that? Reference? 2. It reminds me of the M.9fckenheim-Axiom X is not finite -> there must be an x in X which is infinite >> Note, for some sets A, the answer to i can be Yes >> and the answer to ii can be No. That is true, but it does not depend on the infinity of a set. > It is already the case for sets of two numbers. So there is no reason why this should be different for infinite sets. > Induction is devised for infinite sets Induction is also applicable to elements of finite sets. > and is required for infinite sets only. Required to achieve what? > But it cannot be used for extra-matzhematical properties > like actual infinity. In contemporary set theory there is no notion of actual (vs. potential) infinity at all. F. N. -- xyz === Subject: Re: Cantor Confusion >> If A is infinite, induction can show that A is infinite. 1. Where did you read that? Reference? > 2. It reminds me of the M.9fckenheim-Axiom > > X is not finite -> there must be an x in X which is infinite This follows directly from the more fundamental M.9fckenheim axiom every set is finite. Carsten -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page. === Subject: Re: Cantor Confusion >> If A is infinite, induction can show that A is infinite. >=20 > 1. Where did you read that? Reference? > 2. It reminds me of the M=C3=BCckenheim-Axiom > =20 > X is not finite -> there must be an x in X which is infinite This follows directly from the more fundamental M=C3=BCckenheim axiom ev= > ery > set is finite. One might naively think that the second axiom makes the first superfluous. Rather remarkable that they are both needed. -- === Subject: Re: Cantor Confusion > If A is infinite, induction can show that A is infinite. >>=20 >> 1. Where did you read that? Reference? >> 2. It reminds me of the M=C3=BCckenheim-Axiom >> =20 >> X is not finite -> there must be an x in X which is infinite >> >> This follows directly from the more fundamental M=C3=BCckenheim axiom >> ev= ery >> set is finite. One might naively think that the second axiom makes the first > superfluous. Rather remarkable that they are both needed. The natural truth cannot be mentioned often enough. F. N. -- xyz === Subject: Re: Cantor Confusion <45bb7541$0$97231$892e7fe2@authen.yellow.readfreenews.net> <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > You can try to utter again and agan this nonsense, but after a while > aI will cease to reply. Every such thing including only natural > numbers is covered by induction. All natural numbers are subject to > induction. > True, you can use induction to prove something about > any natural number. However, the question is: Can > you use induction to prove something about a set > of natural numbers? This question has a trivial answer: > If the set of all natural numbers is nothing than all natural numbers, > then yes. > If the set of all natural numbers is more than all natural numbers, > then no. No there are two questions. i: Is X true for every element of the set A? > ii: Is X true for the set A. The two questions are different. > The answer to question ii can be No, even when > the answer to question i is yes. > Your trivial answer only applies to question i. > There are two types of sets of natural numbers > I: sets of natural numbers that are not (potentially) infinite > II: sets of natural numbers that are (potentially) infinite > Induction can only prove things about sets of type I. So you believe that type II sets have some esoteric supplement? No. The answer to question ii can be No, even if > there is no esoteric supplement (i.e. the answer to > question i is Yes). We can prove that every number is a number while a set of several > numbers is not a number. As you point out there is a difference between > the elements of a set and the set. There is no reason to distinguish between > finite and infinite sets. > By running induction longer and longer you can get more and more > sets of type I. By running induction for an infinite time you can > get all > sets of type I. No. By two steps of induction, namely > P(1) > P(n) ==> P(n+1) > you can prove P for all natural numbers and for all sets of natural > number (except quantitative statements as I mentioned above). No you can prove things for all elements of all sets > of natural numbers. There is a difference between > proving that something is true of every element in a set A > and proving that something is true of a set A. Your strong belief in the inaccesible infinite uttered above is your > (and some other people's) personal opinion but has nothing to do with > mathematics. > But you will never get a set of type II. (You may get > a collection of sets whose union contains the same elements > as a set of type II, but you will never get a single set of type > II). Small wonder. There is no set of type II. No. We have already agreed that potentially infinite > sets exist. Whether these sets actually exist, that > is whether every element in the set can be said to > exist at once, is of no interest here. > The set of all natural numbers, the union of all natural numbers, N, > is a set of type II. You cannot use induction to prove anything about > the union of all natural numbers. Small wonder. This set does not exist, No, the potentially infinite set of natural numbers exists. Yes, but it does not exist actually. Usually, you mistake one for the other. The potentially infinite set of natural numbers exists does not mean anything else than there are natural numbers, and if n is a natural number, then n+1 is a natural number. All questions concerning this set can be answered by two steps of complete induction. === Subject: Re: Cantor Confusion <45bb7541$0$97231$892e7fe2@authen.yellow.readfreenews.net> <45bb8bdf$0$97236$892e7fe2@authen.yellow.readfreenews.net> <45bcac75$0$97222$892e7fe2@authen.yellow.readfreenews.net> <45be1a8b$0$97234$892e7fe2@authen.yellow.readfreenews.net > You can try to utter again and agan this nonsense, but after a while > aI will cease to reply. Every such thing including only natural > numbers is covered by induction. All natural numbers are subject to > induction. > True, you can use induction to prove something about > any natural number. However, the question is: Can > you use induction to prove something about a set > of natural numbers? > This question has a trivial answer: > If the set of all natural numbers is nothing than all natural numbers, > then yes. > If the set of all natural numbers is more than all natural numbers, > then no. No there are two questions. i: Is X true for every element of the set A? > ii: Is X true for the set A. The two questions are different. > The answer to question ii can be No, even when > the answer to question i is yes. > Your trivial answer only applies to question i. > There are two types of sets of natural numbers > I: sets of natural numbers that are not (potentially) infinite > II: sets of natural numbers that are (potentially) infinite > Induction can only prove things about sets of type I. > So you believe that type II sets have some esoteric supplement? No. The answer to question ii can be No, even if > there is no esoteric supplement (i.e. the answer to > question i is Yes). > We can prove that every number is a number while a set of several > numbers is not a number. As you point out there is a difference between > the elements of a set and the set. > There is no reason to distinguish between > finite and infinite sets. > By running induction longer and longer you can get more and more > sets of type I. By running induction for an infinite time you can > get all > sets of type I. > No. By two steps of induction, namely > P(1) > P(n) ==> P(n+1) > you can prove P for all natural numbers and for all sets of natural > number (except quantitative statements as I mentioned above). No you can prove things for all elements of all sets > of natural numbers. There is a difference between > proving that something is true of every element in a set A > and proving that something is true of a set A. > Your strong belief in the inaccesible infinite uttered above is your > (and some other people's) personal opinion but has nothing to do with > mathematics. > But you will never get a set of type II. (You may get > a collection of sets whose union contains the same elements > as a set of type II, but you will never get a single set of type > II). > Small wonder. There is no set of type II. No. We have already agreed that potentially infinite > sets exist. Whether these sets actually exist, that > is whether every element in the set can be said to > exist at once, is of no interest here. > The set of all natural numbers, the union of all natural numbers, N, > is a set of type II. You cannot use induction to prove anything about > the union of all natural numbers. > Small wonder. This set does not exist, No, the potentially infinite set of natural numbers exists. Yes, but it does not exist actually. Usually, you mistake one for the > other. > The potentially infinite set of natural numbers exists does > not mean anything else than there are natural numbers, and if n is a > natural number, then n+1 is a natural number. All questions concerning this set can be answered by two steps of > complete induction. > No. The question Q: Is the maximum value of N fixed (that is can the natural number referred to by the maximum value of N change if we change the subset of N that 'actually exists')? is a question concerning N that cannot be answered by two steps of complete induction. However the two facts, there are natural numbers and if n is a natural number, then n+1 is a natural number can be used to show that the answer to question Q is No (it is not necessary to know that n+1 exists, only that n+1 can exist). - William Hughes === Subject: Re: Cantor Confusion > On 30 Jan., 17:18, Franziska Neugebauer > 2. BECAUSE it is _undefined_ until the sum of all natural > numbers is defined. >> It is defined if the set of all natural numbers s defined, Here is the definition of the set N 1 > 2 > 3 > ... Here is the definition of the sum of all elements of N 1 > 23 > 456 > ... Abstract art? F. N. -- xyz === Subject: Re: Cantor Confusion > For people with less capabilities of abstract thinking: [...] The three-eyed speaking to the two-eyed? F. N. -- xyz === Subject: cmsg cancel <877iuz6dql.fsf@laptop.at.pyenos> Control: cancel <877iuz6dql.fsf@laptop.at.pyenos> === Subject: cmsg cancel <87zm7v4yy7.fsf@laptop.at.pyenos> Control: cancel <87zm7v4yy7.fsf@laptop.at.pyenos> === Subject: A speculation on (x,y) in R Suppose there is set A with elements x_i for i={1,2,...,n}, and set B with elements y_i for i={1,2,...,n}. Let R be a set of (x_j,y_j) for j={1,2,...,m}. Define x_A as, x_A := x_i in A for some i Define x_R as, x_R := x_j in (x_j,y_j) of R for some j Define y_B as, y_B := y_i in B for some i Define y_R as, y_R := y_j in (x_j,y_j) of R for some j R subset AxB, (x, y) = { {x}, {x,y} }, for all (x, y) in R, (x_i,y_j) = {x_R, y_R} x {x_A, x_B}. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A speculation on (x,y) in R > Suppose there is set A with elements x_i for i={1,2,...,n}, and set B with > elements y_i for i={1,2,...,n}. Just a notational suggestion. If A = {a_i} and B = {b_i} the reader has a better chance of following along. === Subject: Re: A speculation on (x,y) in R > Suppose there is set A with elements x_i for i={1,2,...,n}, and set B with elements y_i for i={1,2,...,n}. > Let R be a set of (x_j,y_j) for j={1,2,...,m}. Define x_A as, > x_A := x_i in A for some i Define x_R as, > x_R := x_j in (x_j,y_j) of R for some j Define y_B as, > y_B := y_i in B for some i Define y_R as, > y_R := y_j in (x_j,y_j) of R for some j R subset AxB, > (x, y) = { {x}, {x,y} }, for all (x, y) in R, > (x_i,y_j) = {x_R, y_R} x {x_A, x_B}. > -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### Applying typing error corrections, {x_j,y_j}x{x_i,y_i} != (x_j,y_j) I was describing an idea with wrong notations. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A speculation on (x,y) in R > Suppose there is set A with elements x_i for i={1,2,...,n}, > and set B with elements y_i for i={1,2,...,n}. > Let R be a set of (x_j,y_j) for j={1,2,...,m}. Define x_A as, > x_A := x_i in A for some i Define x_R as, > x_R := x_j in (x_j,y_j) of R for some j Define y_B as, > y_B := y_i in B for some i Define y_R as, > y_R := y_j in (x_j,y_j) of R for some j R subset AxB, > (x, y) = { {x}, {x,y} }, for all (x, y) in R, > (x_i,y_j) = {x_R, y_R} x {x_A, x_B}. > Is there some reason you keep posting this stuff? Are you aware that most of it is gibberish? -- === Subject: Re: A speculation on (x,y) in R > Suppose there is set A with elements x_i for i={1,2,...,n}, > and set B with elements y_i for i={1,2,...,n}. > Let R be a set of (x_j,y_j) for j={1,2,...,m}. Define x_A as, > x_A := x_i in A for some i Define x_R as, > x_R := x_j in (x_j,y_j) of R for some j Define y_B as, > y_B := y_i in B for some i Define y_R as, > y_R := y_j in (x_j,y_j) of R for some j R subset AxB, > (x, y) = { {x}, {x,y} }, for all (x, y) in R, > (x_i,y_j) = {x_R, y_R} x {x_A, x_B}. > Is there some reason you keep posting this stuff? Are you aware that > most of it is gibberish? Yes I'm aware, but I'm learning from them. -- > -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A speculation on (x,y) in R > Is there some reason you keep posting this stuff? Are you aware that > most of it is gibberish? > Yes I'm aware, but I'm learning from them. Wouldn't you learn faster if you stopped posting gibberish? -- === Subject: Re: A speculation on (x,y) in R > Is there some reason you keep posting this stuff? Are you aware that > most of it is gibberish? > Yes I'm aware, but I'm learning from them. Wouldn't you learn faster if you stopped posting gibberish? I think so too. I was experimenting. -- > -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A speculation on (x,y) in R > Suppose there is set A with elements x_i for i={1,2,...,n}, and set B with elements y_i for i={1,2,...,n}. > Let R be a set of (x_j,y_j) for j={1,2,...,m}. Define x_A as, > x_A := x_i in A for some i Define x_R as, > x_R := x_j in (x_j,y_j) of R for some j Define y_B as, > y_B := y_i in B for some i Define y_R as, > y_R := y_j in (x_j,y_j) of R for some j R subset AxB, > (x, y) = { {x}, {x,y} }, for all (x, y) in R, > (x_i,y_j) = {x_R, y_R} x {x_A, x_B}. Typing error correction: for all (x, y) in R, (x_j,y_j) = {x_R, y_R} x {x_A, x_B}. > -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: A speculation on (x,y) in R Suppose there is set A with elements x_i for i={1,2,...,n}, and set B with elements y_i for i={1,2,...,n}. > Let R be a set of (x_j,y_j) for j={1,2,...,m}. Define x_A as, > x_A := x_i in A for some i Define x_R as, > x_R := x_j in (x_j,y_j) of R for some j Define y_B as, > y_B := y_i in B for some i Define y_R as, > y_R := y_j in (x_j,y_j) of R for some j R subset AxB, > (x, y) = { {x}, {x,y} }, for all (x, y) in R, > (x_i,y_j) = {x_R, y_R} x {x_A, x_B}. > Typing error correction: for all (x, y) in R, > (x_j,y_j) = {x_R, y_R} x {x_A, x_B}. Typing error correction: for all (x, y) in R, (x_j,y_j) = {x_R, y_R} x {x_A, y_B}. -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- > ### Author Logan Lee ### > ### AuthorHomePage http://beam.to/pyenos ### > ### AuthorQuote I just want to learn. ### -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: What is the Decisive Clash of Our Time? Forget about the lunacy ... just enjoy these fun movies that are also on optics education. You can also save them by right clicking the links and saving them as flv files and download a free flv player. Bush Administration Insider Says U.S. Government Behind 911.flv http://ash-v31.ash.youtube.com/get_video?video_id=HkpOsUmp-9w 911 Truth, Scott Forbes describes power-downs in WTC.flv http:// youtube-609.vo.llnwd.net/d1/04/D1/fEJmcvTzYfo.flv 911 Truth, Consequences of Revealing the Truth about 911.flv http:// youtube-609.vo.llnwd.net/d1/04/D1/fEJmcvTzYfo.flv U.S. Army General Says Flight 77 Did Not Hit Pentagon.flv http://lax-v8.lax.youtube.com/get_video?video_id=Zsn4JA450iA 911 Truth, Bush Administration Lied About Iraq 911.flv http://lax- v8.lax.youtube.com/get_video?video_id=Zsn4JA450iA Bush gets caught off guard on 9/11 prior knowledge question.flv http://lax-v222.lax.youtube.com/get_video?video_id=0eH5qbrpwlM Bush gets caught off guard on 911 prior knowledge question.flv http://lax-v222.lax.youtube.com/get_video?video_id=0eH5qbrpwlM World Trade Center -- Controlled Demolition.flv http:// v187.youtube.com/get_video?video_id=87fyJ-3o2ws 911 Truth, The Moles, the Patsies, State-Sponsored Terror.flv http://chi-v43.chi.youtube.com/get_video?video_id=u0K9BM9oo90 > I never thought it would be possible for anyone to rival the lunacy of > Bitter Anko, but I was wrong... OL === Subject: Re: What is the Decisive Clash of Our Time? the mainpart ofthe controlled demo argument is that a trashfire never took-down a skyscraper, aside from the fact that such an inside operation needs much,much,much less explosive energy that was in the planes. however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause. as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary. anyway,please present your arguments in nonvideoformat. > World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: What is the Decisive Clash of Our Time? Another conspiracy theory nutter....You guys are so boring. OL > Forget about the lunacy ... just enjoy these fun movies that are also > on optics education. You can also save them by right clicking the > links and saving them as flv files and download a free flv player. Bush Administration Insider Says U.S. Government Behind 911.flv > http://ash-v31.ash.youtube.com/get_video?video_id=HkpOsUmp-9w 911 Truth, Scott Forbes describes power-downs in WTC.flv http:// > youtube-609.vo.llnwd.net/d1/04/D1/fEJmcvTzYfo.flv 911 Truth, Consequences of Revealing the Truth about 911.flv http:// > youtube-609.vo.llnwd.net/d1/04/D1/fEJmcvTzYfo.flv U.S. Army General Says Flight 77 Did Not Hit Pentagon.flv > http://lax-v8.lax.youtube.com/get_video?video_id=Zsn4JA450iA 911 Truth, Bush Administration Lied About Iraq 911.flv http://lax- > v8.lax.youtube.com/get_video?video_id=Zsn4JA450iA Bush gets caught off guard on 9/11 prior knowledge question.flv > http://lax-v222.lax.youtube.com/get_video?video_id=0eH5qbrpwlM Bush gets caught off guard on 911 prior knowledge question.flv > http://lax-v222.lax.youtube.com/get_video?video_id=0eH5qbrpwlM World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws 911 Truth, The Moles, the Patsies, State-Sponsored Terror.flv > http://chi-v43.chi.youtube.com/get_video?video_id=u0K9BM9oo90 >> I never thought it would be possible for anyone to rival the lunacy of >> Bitter Anko, but I was wrong... >> OL === Subject: Re: Definition of (x, y) in R such that (x in A)R(x in B) > Suppose there is set A with elements x_i for i={1,2,...,n}, So A is any set which contains { x1,.. x_n }. === Subject: JSH: Sections VI-XIII But the real JSHs are against JSH. (These sections were written down by one Randy Noster who follows and take notes of JSH's words..two years ago during the BD). Section VI Math seems that this exists: more logical, very logical, too logical, less logical, not very logical, really logical, fairly logical. Well then, draw the inferences. I have. Now think of the person you love most. Have you? Tell me the number and I'll tell you the lottery. Section VII A priori, in other words with its eyes closed, JSH places before action and above all: Doubt. JSH doubts everything. JSH is an armadillo. Everything is JSH, too. Beware of JSH. Anti-JSHism is a disease: selfkleptomania, man's normal condition, is JSH. But the real JSHs are against JSH. The selfkleptomaniac. The person who steals - without thinking of his own interests, or of his will - elements of his individual, is a kleptomaniac. He steals himself. He causes the characters that alienate him from the community to disappear. The bourgeois resemble one another - they're all alike. They used not to be alike. They have been taught to steal - stealing has become a function - the most convenient and least dangerous thing is to steal oneself. They are all very poor. The poor are against JSH. They have a lot to do with their brains. They'll never get to the end of it. They work. The poor are against JSH. He who is against JSH is for me, a famous man said, but then he died. They buried him like a true JSHist. Anno domini JSH. Beware! And remember this example. Section VIII TO MAKE A JSHIST POEM bag. Shake gently. Next take out each cutting one after the other. Copy conscientiously in the order in which they left the bag. Them poem will resemble you. And there you are - an infinitely original author of charming sensibility, even though unappreciated by the vulgar herd.* ________________________________________________________________________ * Example: when dogs cross the air in a diamond like ideas and the appendix of the meninx tells the time of the alarm programme (the title is mine) prices they are yesterday suitable next pictures/ appreciate the dream era of the eyes/ pompously that to recite the gospel sort darkens/ group apotheosis imagine said he fatality power of colours/ carved flies (in the theatre) flabbergasted reality a delight/ spectator all to effort of the no more 10 to 12/ during divagation twirls descends pressure/ render some mad single-file flesh on a monstrous crushing stage/ celebrate but their 160 adherents in steps on put on my nacreous/ sumptuous of land bananas sustained illuminate/ joy ask together almost/ of has the a such that the invoked visions/ some sings latter laughs/ exits situation disappears describes she 25 dance bows/ dissimulated the whole of it isn't was/ magnificent has the band better light whose lavishness stage music-halls me/ reappears following instant moves live/ business he didn't has lent/ manner words come these people Section IX There are some people who explain, because there are others who learn. Abolish them and all that's left is JSH. Dip your pen into a black liquid with manifesto intentions - it's only your autobiography that you're hatching under the belly of the flowering cerebellum. Biography is the paraphernalia of the famous man. Great or strong. And there you are, a simple man like the rest of them, once you've dipped your pen into the ink, full of PRETENSIONS which manifest themselves in forms as diverse as they are unforeseen, which apply to every form of activity and of state of mind and of mimicry: there you are, full of only just arrived, to proceed along the illusory and ridiculous upward path towards an apotheosis that only exists in your neurasthenia: there you are, full of PRIDE greater, stronger, more profound than all the others. superficial one, that's why you're all going to die. There are some people who have antedated their manifestos to make other people believe that they had the idea of their own greatness a little earlier. My dear colleagues, before after, past future, now yesterday, that's why you're all going to die. There are some people who have said: JSH is good because it isn't bad, JSH is bad, JSH is a religion, JSH is a poem, JSH is a spirit, JSH is sceptical, JSH is magic, I know JSH. My dear colleagues: good bad, religion poetry, spirit scepticism, definition definition, that's why you're all going to die, and you will die, I promise you. The great mystery is a secret, but it's known to a few people. They will never say what JSH is. To amuse you once again I'll tell you something like: JSH is the dictatorship of the spirit, or JSH is the dictatorship of language, or else JSH is the death of the spirit, which will please many of my friends. Friends. Section X It is certain that since Gambetta, the war, Panama and the Steinheil affair, intelligence is to be found in the street. The intelligent man has become and all-round, normal person. What we lack, what has some interest, what is rare because he has the anomalies of precious being, the freshness and liberty of the great antimen, is.... THE IDIOT JSH is working with all its might towards the universal installation of the idiot. But consciously. And tends itself to become more and more of one. JSH is terrible: it doesn't feel sorry about the defeats of intelligence. JSH could rather be called cowardly, but cowardly like a mad dog; it recognises neither method nor persuasive excess. The lack of garters which makes it systematically bend down reminds us of the famous lack of system which basically has never existed. The false rumour was started by a laundress at the bottom of her page, the page was taken to the barbaric country where humming-birds act as the sandwich-men of cordial nature. This was told me by a watch-maker who was holding a supple syringe which, in characteristic memory of the hot countries, he called phlegmatic and insinuating. Section XI JSH is a dog - a compass - the lining of the stomach - neither new nor a nude Japanese girl - a gasometer of jangled feelings - JSH is brutal and doesn't go in for propaganda - JSH is a quantity of life in transparent, effortless and gyratory transformation. Section XII Gentlemen and ladies buy come in and buy and don't read you'll see the fellow who has in his hands the key to niagara the man with a game leg in the game box his hemispheres in a suitcase his nose enclosed in a chinese lantern you'll see you'll see you'll see the belly dance in the massachusetts saloon the fellow who sticks the nail in and the tyre goes down mademoiselle atlantide's silk stockings the trunk that goes 6 times round the world to find the addressee monsieur and his fiancee his brother and his sister-in-law you'll find the carpenter's address the toad-watch the nerve like a paper-knife you'll have the address of the minor pin for the feminine sex and that of the fellow who supplies the obscene photos to the kind of greece as well as the address of l'action francaise. Section XIII JSH is a virgin microbe JSH is against the high cost of living JSH is limited company for the exploitation of ideas JSH has 391 different attitudes and colours according to the sex of the president. It changes - affirms - says the opposite at the same time - no importance - shouts - goes fishing. JSH is the chameleon of rapid and self-interested change. JSH is against the future. JSH is dead. JSH is absurd. Long live JSH. JSH is not a literary school, howl - (as written by R.Noster) === Subject: cmsg cancel <4757438z156i95865i84i678C123584787n563d721639583C@news.pharao.nz> Control: cancel <4757438z156i95865i84i678C123584787n563d721639583C@news.pharao.nz> Summary: EMP spam === === Subject: Re: Which one is the correct Newton's law? > In message , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let F(t), > m(t), x(t) be the force, mass, position of the object at time t. Which > of the following is the correct Newton's law? > 1. (m(t) x(t)) '' = F(t) > 2. m(t) x''(t) = F(t) > 2. (m(t) x'(t))' = F(t) >>F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >>where m and x' are functions of t: >> F(t) = ( m(t) x'(t) )' Suppose the object is moving at constant velocity v but is losing mass - > say it is ejecting matter equally in the positive and negative > y-directions (relative to itself). The reaction from this produces no net > force but the momentum is decreasing so there should be a force F = m'v > acting somewhere - where? > -- > David Hartley The standard newgroup protocol is to find a flaw with Relativity, not Newtonian dynamics. You are about 300 years too late. === Subject: Re: Which one is the correct Newton's law? <45c42acf$0$5744$afc38c87@news.optusnet.com.au>, Peter Webb , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let F(t), > m(t), x(t) be the force, mass, position of the object at time t. Which > of the following is the correct Newton's law? > 1. (m(t) x(t)) '' = F(t) > 2. m(t) x''(t) = F(t) > 2. (m(t) x'(t))' = F(t) >>F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >>where m and x' are functions of t: >> F(t) = ( m(t) x'(t) )' Suppose the object is moving at constant velocity v but is losing mass - > say it is ejecting matter equally in the positive and negative > y-directions (relative to itself). The reaction from this produces no net > force but the momentum is decreasing so there should be a force F = m'v > acting somewhere - where? > -- > David Hartley The standard newgroup protocol is to find a flaw with Relativity, not > Newtonian dynamics. You are about 300 years too late. OK, here goes. Three rocket ships, A,B, C initially at rest. A and B are in tandem with a string between them, connected to an outrigger on each. A and B are equidistant from C. At a signal from C, identical acceleration programs on A and B gently accelerate A of A and B and C, what is the fate of the string? According to relativity from C's point of view, the string tightens and breaks. This is absurd. Relativity is bunk. * / .' '. A |:. (_) | | * .--. |:. _ | . | / / ` |:. (_) | | + | | ;:. ; | ' __, .' :. / `. | * + '--' * / .-'':._.'`-. | + / |/ | | + .' '. C * | ;:. _ ; + | |:. (_) | | |:. _ | | + |:. (_) | * | ;:. ; + / * | .' :. / `. .' '. B | / .-'':._.'`-. |:. (_) | |:. _ | |:. (_) | ;:. ; .' :. / `. / .-'':._.'`-. |/ | -- Michael Press === Subject: Re: Which one is the correct Newton's law? <45c42acf$0$5744$afc38c87@news.optusnet.com.au> In message <45c42acf$0$5744$afc38c87@news.optusnet.com.au>, Peter Webb > In message , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let F(t), >> m(t), x(t) be the force, mass, position of the object at time t. Which >> of the following is the correct Newton's law? >> 1. (m(t) x(t)) '' = F(t) >> 2. m(t) x''(t) = F(t) >> 2. (m(t) x'(t))' = F(t) F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >where m and x' are functions of t: > F(t) = ( m(t) x'(t) )' >> Suppose the object is moving at constant velocity v but is losing mass - >> say it is ejecting matter equally in the positive and negative >> y-directions (relative to itself). The reaction from this produces no net >> force but the momentum is decreasing so there should be a force F = m'v >> acting somewhere - where? >> -- >> David Hartley The standard newgroup protocol is to find a flaw with Relativity, not >Newtonian dynamics. You are about 300 years too late. Better late than never :) Seriously though, I have no doubt in the validity of Newtonian dynamics (away from relativistic or quantum-theoretic scales). I'm just pointing out that if you have an object with constant velocity but changing mass then, by definition, there is a force acting on it and saying I don't understand the physical reality of that force. Interestingly, it is frame-dependent. In the frame of the object, it has unchanging (zero) momentum, and so experiences no force and no change of velocity. One reply claims that there is no force and so the body will accelerate so as to keep the momentum constant. Others point out that the remaining constant velocity (which I agree with) but then argue that means no force on the body as a whole, but don't reconcile that with the changing momentum. Any ideas? -- David Hartley === Subject: Re: Which one is the correct Newton's law? <45c42acf$0$5744$afc38c87@news.optusnet.com.au> , Peter Webb > In message , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let F(t), >> m(t), x(t) be the force, mass, position of the object at time t. Which >> of the following is the correct Newton's law? >> 1. (m(t) x(t)) '' = F(t) >> 2. m(t) x''(t) = F(t) >> 2. (m(t) x'(t))' = F(t) F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >where m and x' are functions of t: > F(t) = ( m(t) x'(t) )' > Suppose the object is moving at constant velocity v but is losing mass - >> say it is ejecting matter equally in the positive and negative >> y-directions (relative to itself). The reaction from this produces no net >> force but the momentum is decreasing so there should be a force F = m'v >> acting somewhere - where? >> -- >> David Hartley The standard newgroup protocol is to find a flaw with Relativity, not >Newtonian dynamics. You are about 300 years too late. Better late than never :) Seriously though, I have no doubt in the validity of Newtonian dynamics > (away from relativistic or quantum-theoretic scales). I'm just pointing > out that if you have an object with constant velocity but changing mass > then, by definition, there is a force acting on it and saying I don't > understand the physical reality of that force. Take an example, say a toy railway car going along a frictionless horizontal track. The car has sand in it. If you open a hopper in the bottom, the sand leaks out. Will the car maintain a constant velocity? To keep the velocity constant, do you need to supply an external force to the car? The whole issue of F=ma vs. F = dp/dt is one that I have found interesting and confusing. Apparently, Newton's original formulation is F = dp/dt. but I really cannot believe that he had variable-mass systems in mind. Many introductory physics books also try to justify F = dp/dt using some simple laboratory experiments involving air hockey pucks; they often show both the second and third laws at the same time, by connecting two pucks with an elastic cord and then plotting p1(t) and p2(t) together. In this way, they try to justify the second law in dp/dt form. However, none of the treatments I have seen have ever used variable-mass systems, so, in fact, all they are doing is demonstrating F = ma. A Google search on 'variable mass dynamics' reason very carefully about the fate of the different parts as mass is lost. IIRC, they end up with the same thing you would get from F = dp/ dt, provided you are careful about what constitutes F. R.G. Vickson Interestingly, it is frame-dependent. In the frame of the object, it has > unchanging (zero) momentum, and so experiences no force and no change of > velocity. One reply claims that there is no force and so the body will accelerate > so as to keep the momentum constant. Others point out that the remaining > constant velocity (which I agree with) but then argue that means no > force on the body as a whole, but don't reconcile that with the changing > momentum. Any ideas? -- > David Hartley === Subject: Re: Which one is the correct Newton's law? <45c42acf$0$5744$afc38c87@news.optusnet.com.au> , Peter Webb > In message , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let F(t), >> m(t), x(t) be the force, mass, position of the object at time t. Which >> of the following is the correct Newton's law? >> 1. (m(t) x(t)) '' = F(t) >> 2. m(t) x''(t) = F(t) >> 2. (m(t) x'(t))' = F(t) F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >where m and x' are functions of t: > F(t) = ( m(t) x'(t) )' > Suppose the object is moving at constant velocity v but is losing mass - >> say it is ejecting matter equally in the positive and negative >> y-directions (relative to itself). The reaction from this produces no net >> force but the momentum is decreasing so there should be a force F = m'v >> acting somewhere - where? >> -- >> David Hartley The standard newgroup protocol is to find a flaw with Relativity, not >Newtonian dynamics. You are about 300 years too late. Better late than never :) Seriously though, I have no doubt in the validity of Newtonian dynamics > (away from relativistic or quantum-theoretic scales). I'm just pointing > out that if you have an object with constant velocity but changing mass > then, by definition, there is a force acting on it and saying I don't > understand the physical reality of that force. Interestingly, it is frame-dependent. In the frame of the object, it has > unchanging (zero) momentum, and so experiences no force and no change of > velocity. One reply claims that there is no force and so the body will accelerate > so as to keep the momentum constant. Others point out that the remaining > constant velocity (which I agree with) but then argue that means no > force on the body as a whole, but don't reconcile that with the changing > momentum. Any ideas? Here is why you're getting into trouble: The fundamental principle is Conservation under certain assumptions, rocket equations can be derived that treat a rocket as a changing mass. What is happening with the reaction mass is incorporated into that derivation. What you're trying to do is start with a different set of assumptions, ignore the fundamental Conservation of Momentum principles, and then try to apply the wrong *derived* equation to a situation to which it simply does not apply. It was derived under different assumptions. If you want the correct physics. start with first principles. The rocket equation is not first principles. - Randy === Subject: Re: Which one is the correct Newton's law? <45c42acf$0$5744$afc38c87@news.optusnet.com.au> In message <45c42acf$0$5744$afc38...@news.optusnet.com.au>, Peter Webb > In message , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let F(t), > m(t), x(t) be the force, mass, position of the object at time t. Which > of the following is the correct Newton's law? > 1. (m(t) x(t)) '' = F(t) > 2. m(t) x''(t) = F(t) > 2. (m(t) x'(t))' = F(t) >>F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >>where m and x' are functions of t: >> F(t) = ( m(t) x'(t) )' > Suppose the object is moving at constant velocity v but is losing mass - > say it is ejecting matter equally in the positive and negative > y-directions (relative to itself). The reaction from this produces no net > force but the momentum is decreasing so there should be a force F = m'v > acting somewhere - where? > -- > David Hartley >>The standard newgroup protocol is to find a flaw with Relativity, not >>Newtonian dynamics. You are about 300 years too late. >> Better late than never :) >> Seriously though, I have no doubt in the validity of Newtonian dynamics >> (away from relativistic or quantum-theoretic scales). I'm just pointing >> out that if you have an object with constant velocity but changing mass >> then, by definition, there is a force acting on it and saying I don't >> understand the physical reality of that force. >> Interestingly, it is frame-dependent. In the frame of the object, it has >> unchanging (zero) momentum, and so experiences no force and no change of >> velocity. >> One reply claims that there is no force and so the body will accelerate >> so as to keep the momentum constant. Others point out that the remaining >> constant velocity (which I agree with) but then argue that means no >> force on the body as a whole, but don't reconcile that with the changing >> momentum. >> Any ideas? Here is why you're getting into trouble: The fundamental principle is >Conservation >under >certain assumptions, rocket equations can be derived that treat a >rocket >as a changing mass. What is happening with the reaction mass is >incorporated into that derivation. What you're trying to do is start with a different set of assumptions, >ignore the fundamental Conservation of Momentum principles, and >then try to apply the wrong *derived* equation to a situation to which >it >simply does not apply. It was derived under different assumptions. If you want the correct physics. start with first principles. The >rocket equation is not first principles. So you would agree that the answer to the original question is that F = m x'' and F = (m x')' are both correct, as the law only applies when the mass is fixed? If you think the law applies to bodies with variable mass - which seems to be the view taken by the only textbook I have to hand, and by several web-sources, and by most of the people who answered the OP's question - then surely *from first principles* one can deduce that a body moving with constant velocity and variable mass is acting under a force? Let's try some first principles. A body is moving along the x-axis with velocity v. It is emitting two streams of matter, each at constant mass-rate r and constant speed relative to itself u , one each in the positive and negative directions. Its initial mass is m_0 and mass at time t is m. In time interval t to t + dt, (before the mass runs out), it loses two elements of mass r.dt, with velocities (approx) v-u and v+u, so by conservation of momentum its momentum reduces by 2rv(dt). Thus (mv)' = -2rv But m' = -2r and so (mv)' = m v' - 2rv. Hence m v' = 0 and so, as m is non-zero, v is constant, as I'm sure you would expect. The rate of change of momentum of the body is, by first principles, -2rv. What is the total force on the body? (Note that the force measured by an observer riding on the body is zero, since for him v = 0. The force of -2rv is an artefact of the transformation to the given frame. The momentum change of a body of variable mass is not invariant under such (Gallilean) transformations, because they preserve accelerations but not velocities, and the variable mass introduces the m'v term. If you always measure the forces on the body as observed on the body - or in the frame where it's instantaneously at rest - then v is 0 and you get F = m v', even though m is variable. This seems to be how questions about such bodies are usually tackled, and so the force I'm worrying about never need be considered.) -- David Hartley === Subject: Re: Which one is the correct Newton's law? <26467982.1170341970165.JavaMail.jakarta@nitrogen.mathforum.org> In message <26467982.1170341970165.JavaMail.jakarta@nitrogen.mathforum.org>, G.E. >> In message >> , >> Dirk Van de >> moortel > An object whose mass depends on time t travels on >> the x-axis. Let >F(t), m(t), x(t) be the force, mass, position of >> the object at time >t. Which of the following is the correct Newton's >> law? > 1. (m(t) x(t)) '' = F(t) > 2. m(t) x''(t) = F(t) > 2. (m(t) x'(t))' = F(t) >>F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >>where m and x' are functions of t: >> F(t) = ( m(t) x'(t) )' >> Suppose the object is moving at constant velocity v >> but is losing mass - >> say it is ejecting matter equally in the positive >> and negative >> y-directions (relative to itself). The reaction from >> this produces no >> net force but the momentum is decreasing so there >> should be a force F = >> m'v acting somewhere - where? >> -- >> David Hartley If an object is losing mass but moving constant velocity, then there >MUST be some external force preventing the lighter object from moving >faster. Where is it from? I don't know- YOU postulated that the >object moved at constant speed! You'll have to supply the force. see, and doesn't change mass, so why should it accelerate? -- David Hartley === Subject: Re: Which one is the correct Newton's law? , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let >>F(t), m(t), x(t) be the force, mass, position of the object at time >>t. Which of the following is the correct Newton's law? >> 1. (m(t) x(t)) '' = F(t) >> 2. m(t) x''(t) = F(t) >> 2. (m(t) x'(t))' = F(t) F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >where m and x' are functions of t: > F(t) = ( m(t) x'(t) )' Suppose the object is moving at constant velocity v but is losing mass - > say it is ejecting matter equally in the positive and negative > y-directions (relative to itself). The reaction from this produces no > net force but the momentum is decreasing so there should be a force F = > m'v acting somewhere - where? No. Consider a simpler situation: A train is moving at constant velocity. Somebody unhooks the coupling between two cars. Now we have two half trains, still moving along at the same speed but no longer physically connected. Would you say the momentum of the train has been cut in half, so there must have been a force? The correct analysis for this situation is that each half of the train has exactly the same momentum it did before, and so does the entire train system. With your object, separate object and reaction mass. The object apart from the reaction mass has the same momentum it had before. Two bits of reaction mass have different momentum. - Randy === Subject: Re: Which one is the correct Newton's law? In message , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let >F(t), m(t), x(t) be the force, mass, position of the object at time >t. Which of the following is the correct Newton's law? > 1. (m(t) x(t)) '' = F(t) > 2. m(t) x''(t) = F(t) > 2. (m(t) x'(t))' = F(t) >>F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >>where m and x' are functions of t: >> F(t) = ( m(t) x'(t) )' >> Suppose the object is moving at constant velocity v but is losing mass - >> say it is ejecting matter equally in the positive and negative >> y-directions (relative to itself). The reaction from this produces no >> net force but the momentum is decreasing so there should be a force F = >> m'v acting somewhere - where? No. Consider a simpler situation: A train is moving at constant >velocity. >Somebody unhooks the coupling between two cars. Now we have >two half trains, still moving along at the same speed but no >longer physically connected. Would you say the momentum >of the train has been cut in half, so there must have been a force? The correct analysis for this situation is that each half of the train >has >exactly the same momentum it did before, and so does the >entire train system. With your object, separate object and reaction >mass. The object apart from the reaction mass has the same >momentum it had before. Two bits of reaction mass have different >momentum. I agree. But the original question was about bodies with variable mass. Sticking to classical mechanics, that can only happen by adding or removing bits. If you insist on considering the future separate part as separate from the start, then you're effectively saying there are no bodies with variable mass. The above 3 versions only apply when m' = 0. and so all are correct. Suppose the train has lots of tiny carriages. These are released consecutively, so the remaining train has effectively continuously changing mass. Clearly (assuming frictionless track etc.) its speed won't change, but its momentum does. If force is rate of change of momentum applies to bodies with variable mass, then it applies here and there is a force. This situation is effectively a rocket with zero exhaust velocity. The equation for a rocket of variable mass m and velocity v expelling mass at a constant rate r and speed u (backwards relative to the rocket) is apparently m v' = ru If u = 0 then m v' = 0 (so as expected v is constant) but p' = m'v = -rv again a non-zero force! On googling the rocket equation, the first reference I found was http://www.physics.upenn.edu/courses/gladney/mathphys/subsubsection3_1_3_ 3.html The following quote from this page is interesting: If we look at the left hand side of the equation, we see the usual expression for force as mass times dv/dt or acceleration. So the equation tells us that the force on the rocket body is given by the amount of mass expelled and the velocity of that mass relative to the rocket body. The author is taking force as mass x acceleration even though the mass is variable, i.e. the OP's second option. The more I look at this, the more confused I get. -- David Hartley === Subject: Re: Which one is the correct Newton's law? > In message , Dirk Van de >> An object whose mass depends on time t travels on the x-axis. Let >F(t), m(t), x(t) be the force, mass, position of the object at time >t. Which of the following is the correct Newton's law? > 1. (m(t) x(t)) '' = F(t) > 2. m(t) x''(t) = F(t) > 2. (m(t) x'(t))' = F(t) >F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >>where m and x' are functions of t: >> F(t) = ( m(t) x'(t) )' > Suppose the object is moving at constant velocity v but is losing mass - >> say it is ejecting matter equally in the positive and negative >> y-directions (relative to itself). The reaction from this produces no >> net force but the momentum is decreasing so there should be a force F = >> m'v acting somewhere - where? No. Consider a simpler situation: A train is moving at constant >velocity. >Somebody unhooks the coupling between two cars. Now we have >two half trains, still moving along at the same speed but no >longer physically connected. Would you say the momentum >of the train has been cut in half, so there must have been a force? The correct analysis for this situation is that each half of the train >has >exactly the same momentum it did before, and so does the >entire train system. With your object, separate object and reaction >mass. The object apart from the reaction mass has the same >momentum it had before. Two bits of reaction mass have different >momentum. I agree. But the original question was about bodies with variable mass. > Sticking to classical mechanics, that can only happen by adding or > removing bits. If you insist on considering the future separate part as > separate from the start, then you're effectively saying there are no > bodies with variable mass. The above 3 versions only apply when m' = 0. > and so all are correct. Suppose the train has lots of tiny carriages. These are released > consecutively, so the remaining train has effectively continuously > changing mass. Clearly (assuming frictionless track etc.) its speed > won't change, but its momentum does. No, it's momentum is not changing. Each molecule of the train has exactly the same momentum when coupled and when decoupled, so it is silly to talk about forces and changes in momentum, or about a single unified it, in a case where there is zero force and zero velocity change on every single atom and molecule. > If force is rate of change of > momentum applies to bodies with variable mass, then it applies here and > there is a force. Well, you can choose to make up your own physics and your own meaning to the terms in such equations, or you can choose to actually understand the physical principles involved. Your choice. > The more I look at this, the more confused I get. It would be less confusing if you didn't make up your own physics. We've been using Newtonian physics for 300 years, yours is only a week old. It's not surprising it's a little confused as yet. - Randy === Subject: Re: Which one is the correct Newton's law? > In message , Dirk Van de >> An object whose mass depends on time t travels on the x-axis. Let >>F(t), m(t), x(t) be the force, mass, position of the object at time >>t. Which of the following is the correct Newton's law? >> 1. (m(t) x(t)) '' = F(t) >> 2. m(t) x''(t) = F(t) >> 2. (m(t) x'(t))' = F(t) >F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >where m and x' are functions of t: > F(t) = ( m(t) x'(t) )' > Suppose the object is moving at constant velocity v but is losing mass - > say it is ejecting matter equally in the positive and negative > y-directions (relative to itself). The reaction from this produces no > net force but the momentum is decreasing so there should be a force F = > m'v acting somewhere - where? >>No. Consider a simpler situation: A train is moving at constant >>velocity. >>Somebody unhooks the coupling between two cars. Now we have >>two half trains, still moving along at the same speed but no >>longer physically connected. Would you say the momentum >>of the train has been cut in half, so there must have been a force? >>The correct analysis for this situation is that each half of the train >>has >>exactly the same momentum it did before, and so does the >>entire train system. With your object, separate object and reaction >>mass. The object apart from the reaction mass has the same >>momentum it had before. Two bits of reaction mass have different >>momentum. >> I agree. But the original question was about bodies with variable mass. >> Sticking to classical mechanics, that can only happen by adding or >> removing bits. If you insist on considering the future separate part as >> separate from the start, then you're effectively saying there are no >> bodies with variable mass. The above 3 versions only apply when m' = 0. >> and so all are correct. >> Suppose the train has lots of tiny carriages. These are released >> consecutively, so the remaining train has effectively continuously >> changing mass. Clearly (assuming frictionless track etc.) its speed >> won't change, but its momentum does. No, it's momentum is not changing. Each molecule of the train >has exactly the same momentum when coupled and when >decoupled, so it is silly to talk about forces and changes in >momentum, or about a single unified it, in a case where there >is zero force and zero velocity change on every single atom >and molecule. > If force is rate of change of >> momentum applies to bodies with variable mass, then it applies here and >> there is a force. Well, you can choose to make up your own physics and your >own meaning to the terms in such equations, or you can choose >to actually understand the physical principles involved. Your >choice. > The more I look at this, the more confused I get. It would be less confusing if you didn't make up your own >physics. We've been using Newtonian physics for 300 years, >yours is only a week old. It's not surprising it's a little confused >as yet. You're misunderstanding me. I agree with you. If you break things down But if you define force as rate of change of momentum for a body of variable mass (not my definition) then what's wrong with the above argument? I agree it looks a bit dodgy to talk of a body which is actually a variable part of a larger real one, but that's always going to be the case when you have variable mass (ignoring relativity). A rocket - in the standard analysis - is considered as a single entity, -- David Hartley === Subject: Re: Which one is the correct Newton's law? > In message , Dirk Van de > An object whose mass depends on time t travels on the x-axis. Let >>F(t), m(t), x(t) be the force, mass, position of the object at time >>t. Which of the following is the correct Newton's law? >> 1. (m(t) x(t)) '' = F(t) >> 2. m(t) x''(t) = F(t) >> 2. (m(t) x'(t))' = F(t) >>F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >where m and x' are functions of t: > F(t) = ( m(t) x'(t) )' >> Suppose the object is moving at constant velocity v but is losing mass - > say it is ejecting matter equally in the positive and negative > y-directions (relative to itself). The reaction from this produces no > net force but the momentum is decreasing so there should be a force F = > m'v acting somewhere - where? >No. Consider a simpler situation: A train is moving at constant >>velocity. >>Somebody unhooks the coupling between two cars. Now we have >>two half trains, still moving along at the same speed but no >>longer physically connected. Would you say the momentum >>of the train has been cut in half, so there must have been a force? >The correct analysis for this situation is that each half of the train >>has >>exactly the same momentum it did before, and so does the >>entire train system. With your object, separate object and reaction >>mass. The object apart from the reaction mass has the same >>momentum it had before. Two bits of reaction mass have different >>momentum. > I agree. But the original question was about bodies with variable mass. >> Sticking to classical mechanics, that can only happen by adding or >> removing bits. If you insist on considering the future separate part as >> separate from the start, then you're effectively saying there are no >> bodies with variable mass. The above 3 versions only apply when m' = 0. >> and so all are correct. > Suppose the train has lots of tiny carriages. These are released >> consecutively, so the remaining train has effectively continuously >> changing mass. Clearly (assuming frictionless track etc.) its speed >> won't change, but its momentum does. No, it's momentum is not changing. Each molecule of the train >has exactly the same momentum when coupled and when >decoupled, so it is silly to talk about forces and changes in >momentum, or about a single unified it, in a case where there >is zero force and zero velocity change on every single atom >and molecule. > If force is rate of change of >> momentum applies to bodies with variable mass, then it applies here and >> there is a force. Well, you can choose to make up your own physics and your >own meaning to the terms in such equations, or you can choose >to actually understand the physical principles involved. Your >choice. > The more I look at this, the more confused I get. It would be less confusing if you didn't make up your own >physics. We've been using Newtonian physics for 300 years, >yours is only a week old. It's not surprising it's a little confused >as yet. You're misunderstanding me. I agree with you. If you break things down > But if you define force as rate of change of momentum for a body of > variable mass (not my definition) then what's wrong with the above > argument? When you break the body into separate pieces it isn't the same body. You have to look at each piece separately. That's what's wrong with it. > I agree it looks a bit dodgy to talk of a body which is > actually a variable part of a larger real one, Yes, it looks a bit dodgy to break a body in half and say it is one body with variable mass and pretend the other half doesn't exist. > but that's always > going to be the case when you have variable mass (ignoring relativity). When you have variable mass that is due to ejecting reaction mass, you have to account for the momentum of the reaction mass also. > A rocket - in the standard analysis - is considered as a single entity, and it is considerations of their momentum that tell you how fast the rocket will go. - Randy === Subject: Re: Which one is the correct Newton's law? > In message > , > Dirk Van de > moortel > An object whose mass depends on time t travels on > the x-axis. Let >>F(t), m(t), x(t) be the force, mass, position of > the object at time >>t. Which of the following is the correct Newton's > law? >> 1. (m(t) x(t)) '' = F(t) >> 2. m(t) x''(t) = F(t) >> 2. (m(t) x'(t))' = F(t) F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >where m and x' are functions of t: > F(t) = ( m(t) x'(t) )' Suppose the object is moving at constant velocity v > but is losing mass - > say it is ejecting matter equally in the positive > and negative > y-directions (relative to itself). The reaction from > this produces no > net force but the momentum is decreasing so there > should be a force F = > m'v acting somewhere - where? > -- > David Hartley The mass ejected does not have zero total momentum. When you sum up the momentum of the body and the ejected mass you will find the total momentum is constant, and thus the force on the body is zero. - MO === Subject: Re: Which one is the correct Newton's law? <4907657.1170341918248.JavaMail.jakarta@nitrogen.mathforum.org> In message <4907657.1170341918248.JavaMail.jakarta@nitrogen.mathforum.org>, Michael >> In message >> , >> Dirk Van de >> moortel > An object whose mass depends on time t travels on >> the x-axis. Let >F(t), m(t), x(t) be the force, mass, position of >> the object at time >t. Which of the following is the correct Newton's >> law? > 1. (m(t) x(t)) '' = F(t) > 2. m(t) x''(t) = F(t) > 2. (m(t) x'(t))' = F(t) >>F = dp/dt = d/dt ( m v ) = d/dt ( m x' ) = ( m x' )' >>where m and x' are functions of t: >> F(t) = ( m(t) x'(t) )' >> Suppose the object is moving at constant velocity v >> but is losing mass - >> say it is ejecting matter equally in the positive >> and negative >> y-directions (relative to itself). The reaction from >> this produces no >> net force but the momentum is decreasing so there >> should be a force F = >> m'v acting somewhere - where? >> -- >> David Hartley The mass ejected does not have zero total momentum. When you sum up >the momentum of the body and the ejected mass you will find the total >momentum is constant, and thus the force on the body is zero. No, that means the external force on the whole system is zero, not that there's no force on the body. The ejected mass does indeed have non-zero total momentum (in the observing frame), so the remaining body is changing momentum and so presumably experiencing a force. In an inertial frame in which the body is momentarily at rest, the ejected mass does have zero total momentum, so the body experiences no force and so will stay at rest - i.e. at constant velocity in the original frame. -- David Hartley === Subject: Re: Which one is the correct Newton's law? > An object whose mass depends on time t travels on the x-axis. Let F(t), > m(t), x(t) be the force, mass, position of the object at time t. Which of > the following is the correct Newton's law? 1. (m(t) x(t)) '' = F(t) > 2. m(t) x''(t) = F(t) > 2. (m(t) x'(t))' = F(t) Number 2 === Subject: If I appear to be a retard to you... I was trying to invent an equivalent definition from a given one. I have realized that I simply don't have technical expertise to express my ideas using correct standard notations at the moment. -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: If I appear to be a retard to you... > I was trying to invent an equivalent definition from a given one. > I have realized that I simply don't have technical expertise > to express my ideas using correct standard notations at the moment. What makes you think that you have ideas worth expressing? -- === Subject: Re: If I appear to be a retard to you... > I was trying to invent an equivalent definition from a given one. > I have realized that I simply don't have technical expertise > to express my ideas using correct standard notations at the moment. What makes you think that you have ideas worth expressing? Just for fun. -- > -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: If I appear to be a retard to you... > >> I was trying to invent an equivalent definition from a given one. >> I have realized that I simply don't have technical expertise >> to express my ideas using correct standard notations at the moment. >> >> What makes you think that you have ideas worth expressing? The fact that he gets so many replies... > Just for fun. ... and that perhaps as well :-) Dirk Vdm === Subject: seeking Rosenfeld and/or Match entitled, Yet another generalization of Polya's enumeration theorem, which was published in a journal called Match, vol.43 (2001) pp. 111-130. I am having a very hard time locating this particular issue of this particular journal. I also tried to locate Rosenfeld, on the theory that maybe he has a preprint at his website. I am also unable to locate Rosenfeld. I can't find a website for him and I can't find his email address. The full title of the journal is: Match. Communications in Mathematical and in Computer Chemistry department at Technion University, but he has apparently moved since then. Anyway, I'm getting nowhere. as it may, I don't understand why it is so difficult to locate either the -- Ignorantly, Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and * comments do not reflect in any way on MIT. Also, I am nowhere near Boston. === Subject: Digital Holography Am looking for a reference on Computational or Digital Holography. Please advise. === Subject: inverse image Hi there, Is it true the following? http://i18.tinypic.com/4ftp5l0.jpg === Subject: Re: inverse image > Hi there, Is it true the following? http://i18.tinypic.com/4ftp5l0.jpg Yes, even without the assumption concerning the cardinal of the set I. Jose Carlos Santos === Subject: Locally finite measure on sigma-compact metric space I am reading Kallenberg's Foundations of Modern Probability and find the following 'clear' result not so clear to me. p9: For any locally finite measure mu on some sigma-compact metric space S, the set A={s in S : mu{s}>0} is clearly measurable. === Subject: Re: Locally finite measure on sigma-compact metric space >I am reading Kallenberg's Foundations of Modern Probability and find >the following 'clear' result not so clear to me. >p9: For any locally finite measure mu on some sigma-compact metric >space S, the set A={s in S : mu{s}>0} is clearly measurable. Makes no sense. You meant A={s in S : mu{{s}}>0}, right? ************************ David C. Ullrich === Subject: Re: Locally finite measure on sigma-compact metric space f.x. a .8ecrit : > I am reading Kallenberg's Foundations of Modern Probability and find > the following 'clear' result not so clear to me. > p9: For any locally finite measure mu on some sigma-compact metric > space S, the set A={s in S : mu{s}>0} is clearly measurable. > Because A is countable ? JMA === Subject: Re: Locally finite measure on sigma-compact metric space <45c47637$0$16175$426a34cc@news.free.fr f.x. a ¬.a6crit : I am reading Kallenberg's Foundations of Modern Probability and find > the following 'clear' result not so clear to me. > p9: For any locally finite measure mu on some sigma-compact metric > space S, the set A={s in S : mu{s}>0} is clearly measurable. Because A is countable ? JMA I don't think A need to be countable. Can't it be such that mu{s}>0 for uncountably many s, yet the sum of such mu{s}, i.e., sup{finite sums of such mu{s}}, is still a limited value? === Subject: Re: Locally finite measure on sigma-compact metric space >> f.x. a Âìcrit : >> I am reading Kallenberg's Foundations of Modern Probability and find >> the following 'clear' result not so clear to me. >> p9: For any locally finite measure mu on some sigma-compact metric >> space S, the set A={s in S : mu{s}>0} is clearly measurable. >> Because A is countable ? >> JMA I don't think A need to be countable. >Can't it be such that mu{s}>0 for uncountably many s, yet the sum of >such mu{s}, i.e., sup{finite sums of such mu{s}}, is still a limited >value? No. (Hint: for each n, consider the set of s with mu({s}) > 1/n...) ************************ David C. Ullrich === Subject: Question about Riemann-Hurwitz formula I have a question aboyut the relation between connectivity and Euler characteristic or genus: the question actually appeared when i was reading the following (*we know that f is entire function*) Connected component of f ^{-1} (annulus(r_1,r_2)) is at least triply connected By the Riemann-Hurwitz formula, this component of f ^{-1} (annulus(r_1,r_2)) contains a critical point of f . I know what Riemann-Hurwitz formula is: it relates singularities of the function and genuses(or Euler characteristic) of the domain and it's image, but i have no idea how it can relate the connectivity number - the only definition of n-connected space i know is from http://en.wikipedia.org/wiki/N-connected using homology theory but i'm not familiar with it well enough to connect these things. === Subject: Re: Question about Riemann-Hurwitz formula >I have a question aboyut the relation between connectivity >and Euler characteristic or genus: the question actually appeared when i was reading the following >(*we know that f is entire function*) Connected component of f ^{-1} (annulus(r_1,r_2)) is at least > triply connected By the Riemann-Hurwitz formula, this component of >f ^{-1} (annulus(r_1,r_2)) contains a critical point of f . I know what Riemann-Hurwitz formula is: it relates singularities of >the function and genuses(or Euler characteristic) of the domain and >it's image, but i have no idea how it can relate the connectivity >number - the only definition of n-connected space i know is from >http://en.wikipedia.org/wiki/N-connected using homology theory That's the wrong definition for this context. In the context of planar regions, multiply connected means not simply-connected, and--for example--triply connected means becomes simply connected after two cuts. So, for example, an annulus is doubly connected, and more generally an open disk with n pairwise disjoint closed disks removed from its interior is (n+1)ly connected. Now back to your question. Let D be a Connected component of f ^{-1} (annulus(r_1,r_2)) which we know (for some reason you haven't given) to be at least triply connected. There are two naturally defined functions on D, namely |f| (with real values) and f/|f| (with values in S^1, and defined everywhere because by construction f isn't 0 on D); because f is entire (hence holomorphic) we know that the level sets of |f| and of f/|f| are real-analytic curves, which for almost all levels are in fact non-singular (thus, 1-dimensional manifolds), and which have a standard structure near any singular point (namely, locally like the 0-level set of Re(z^n) for some n > 1). Further, we know that the singular points of f are exactly the points where either (and therefore both) of these level sets are non-manifolds. So suppose there were no singular points. Then, in particular, all the level sets of |f| would be 1-manifolds, and they'd all be compact, so there'd be some number k such that each level set was the disjoint union of k simple closed curves. Well, since D is connected k has to be 1, and D is an annulus, contrary to the assumption of at-least-triple-connectivity. Draw a picture or three. Lee Rudolph === Subject: Inner products (very easy) Let V be a vector space over F. Show that the sum of two inner products on V is an inner product on V. Should I not show that for all v,w,s,t in V there exist x,y in V such that (v|w)+(s|t) = (x|y)? I am afraid my manipulation skills are not up to the task. The last part of the exercise says Show that a positive multiple of an inner product is an inner product. This seems too easy: k(v|w) = (kv|w). But I do not understand why the exercise specifies that k must be positive. Perhaps I am misinterpreting the whole exercise. Kiuhnm === Subject: Re: Inner products (very easy) > Let V be a vector space over F. Show that the sum of two inner products > on V is an inner product on V. Should I not show that for all v,w,s,t in V there exist x,y in V such > that (v|w)+(s|t) = (x|y)? I don't see why. > I am afraid my manipulation skills are not up to the task. The last part of the exercise says > Show that a positive multiple of an inner product is an inner product. This seems too easy: k(v|w) = (kv|w). But I do not understand why the > exercise specifies that k must be positive. > Perhaps I am misinterpreting the whole exercise. It sounds as though the definition of inner product you are following includes positive-definiteness. I think positive only makes sense if F is a subfield of R, the reals. Maybe that is specified too? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Inner products (very easy) > I don't see why. Because, as I suspected, I misinterpreted the whole exercise. Kiuhnm === Subject: Re: Inner products (very easy) Kiuhnm a .8ecrit : > Let V be a vector space over F. Show that the sum of two inner products > on V is an inner product on V. Should I not show that for all v,w,s,t in V there exist x,y in V such > that (v|w)+(s|t) = (x|y)? No, I think the exercises asks to show that, if <.|.> and (.|.) are two different inner products (as functions V^2 -> F), then <.|.> + (.|.) is also one. > Show that a positive multiple of an inner product is an inner product. Same thing. === Subject: Re: Inner products (very easy) > No, I think the exercises asks to show that, if <.|.> and (.|.) are two > different inner products (as functions V^2 -> F), then <.|.> + (.|.) is > also one. Kiuhnm === Subject: independence and correlation === Subject: Re: independence and correlation > Can anybody give me an example of variabesl that are uncorrelated but not Perhaps the simplest examples come from having an X with symmetric density about 0. Then E[X] = 0 and E[X^3] = 0, so X and X^2 are uncorrelated.: E[(X-0)*(X^2-E[X^2])] = E[X^3] - 0*E[X^2 ] = 0. === Subject: Re: independence and correlation As Daniel mentioned, specific examples are not too hard to find, but it is a possibly useful fact to know that, far from being rare, such examples are in some sense pretty common. For and interval [a,b] and an integer n>=1 let X is U([a,b],n) denote that X is a discrete uniform variable on the n+1 evenly spaced points beginning at a and ending at b: x_i = a +Di where D = (b-a)/n i = 0,1,...,n Suppose that f is a continuous function on [a,b] and let Y = f(X). Being a function of X, X and Y are not independent. Lemma1 If f is strictly increasing on [a,b] then Cov(X,Y) > 0 Lemma2 If f is strictly decreasing on [a,b] then Cov(X,Y) < 0 for an elementary proof (explicitly of the increasing case, but the logic establishes the decreasing case as well). Now suppose that f is *any* continuous function with the property that there exists a subinterval [a,b] on which f is strictly decreasing and another [c,d] on which f is strictly increasing. Without loss of generality suppose that the length of [a,b] is <= the length of [c,d]. Thus for some k, [a+k,b+k] is a subinterval of [c,d]. Let X_h be U([a+h,b+h]) and Y_h = f(X_h) and C(h) = Cov(X_h,Y_h). Since in this discrete case C(h) is just some big polynomial in x_i and f(x_j), it is easy to see that C(h) is a continuous function of h. Furthermore C(0) < 0 and C(k) > 0 by lemmas 1 and 2 above. Thus by the intermediate value theorem, there must be an h between 0 and k with C(h) = 0. QED The proof can be extended to the continuous case (With X uniform on [a,b]), but some additional assumptions on f would probably have to be made for C(h) to exist and be continuous. To get a specific example - just pick a nice candidate f (e.g. f(x) = x^2) and a small value of n and do a bit of algebra. In some cases symmetry considerations should make finding an example easy (e.g. sin(x)). === Subject: Re: independence and correlation > (snip) A technical problem with my last post: it is concievable that the Y_h with Cov(X_h,Y_h) = 0 is constant, in which case you have independence after all. Thus to get a counterexample to independence you would need to make an additional assumption on f along the lines that, for any bounded interval I there is an upper bound on the sizes of the preimage sets f^(-1)(x) as x ranges over I (then in the proof let I be an interval containing both [a,b] and [c,d] and pick n to exceed this upper bound). This still gives a template for generating infinitely many examples of dependent uncorrelated random variables. === Subject: Re: independence and correlation The simplest example I can think of is to let X and Y be two independent Bernoulli random variables, each taking on the values 0 and 1 with probability 1/2. Then let S=X+Y and T=X-Y. S and T are uncorrelated because E(ST)=E(X^2-Y^2)=0=E(T)=E(S)E(T). But S and T are not independent, since the probability that T=0 given that S=0 (or S=2) is 1, not 1/2. -- Daniel Mayost === Subject: Integrability of X+Y and of X, Y I would need an example of variables X and Y such that X+Y are integrable. but X and Y are not separately integrable. === Subject: Re: Integrability of X+Y and of X, Y <13520465.1170515185979.JavaMail.jakarta@nitrogen.mathforum.org>, > I would need an example of variables X and Y such that X+Y are integrable. but X and Y are not separately integrable. > X is the characteristic function of a non-measurable set in R and Y is the characteristic function of RX. === Subject: Re: Integrability of X+Y and of X, Y , I would need an example of variables X and Y such that X+Y are integrable. but X and Y are not separately integrable. > X is the characteristic function of a non-measurable set in R and Y is > the characteristic function of RX. But that still might be integrable depending on how he is defining integration. Jason === Subject: Re: Integrability of X+Y and of X, Y > <13520465.1170515185979.JavaMail.jaka...@nitrogen.mathforum.org>, > I would need an example of variables X and Y such that > X+Y are integrable. > but X and Y are not separately integrable. > X is the characteristic function of a non-measurable set in R and Y is > the characteristic function of RX. But that still might be integrable depending on how he is defining > integration. Jason What definition of integration do you suggest that allows integration of the characteristic functions of arbitrary non-measurable sets? === Subject: Re: Integrability of X+Y and of X, Y <13520465.1170515185979.JavaMail.jaka...@nitrogen.mathforum.org>, > I would need an example of variables X and Y such that > X+Y are integrable. > but X and Y are not separately integrable. > X is the characteristic function of a non-measurable set in R and Y is > the characteristic function of RX. But that still might be integrable depending on how he is defining > integration. Jason What definition of integration do you suggest that allows integration of > the characteristic functions of arbitrary non-measurable sets?- Hide quoted text - - Show quoted text - My understanding is that the Henstock-Kurzweil can integrate functions that the Lebesgue theory cannot. Please correct me if I am wrong about this. Jason Pawloski === Subject: Re: Integrability of X+Y and of X, Y > <13520465.1170515185979.JavaMail.jaka...@nitrogen.mathforum.org>, > I would need an example of variables X and Y such that > X+Y are integrable. > but X and Y are not separately integrable. > X is the characteristic function of a non-measurable set in R and Y is > the characteristic function of RX. > But that still might be integrable depending on how he is defining > integration. > Jason What definition of integration do you suggest that allows integration of > the characteristic functions of arbitrary non-measurable sets?- Hide quoted > text - - Show quoted text - My understanding is that the Henstock-Kurzweil can integrate functions > that the Lebesgue theory cannot. Please correct me if I am wrong about > this. Jason Pawloski I think that is is Henstock-Kurzwell, not Henstock-Kurzweil. And even they are not able to integrate every function. === Subject: Re: Integrability of X+Y and of X, Y Virgil a .8ecrit : > >> <13520465.1170515185979.JavaMail.jaka...@nitrogen.mathforum.org>, >> I would need an example of variables X and Y such that >> X+Y are integrable. >> but X and Y are not separately integrable. > X is the characteristic function of a non-measurable set in R and Y is > the characteristic function of RX. >> But that still might be integrable depending on how he is defining >> integration. >> Jason > What definition of integration do you suggest that allows integration of > the characteristic functions of arbitrary non-measurable sets?- Hide quoted > text - - Show quoted text - >> My understanding is that the Henstock-Kurzweil can integrate functions >> that the Lebesgue theory cannot. Please correct me if I am wrong about >> this. >> Jason Pawloski I think that is is Henstock-Kurzwell, not Henstock-Kurzweil. And even they are not able to integrate every function. Actually, they (Henstock-Kurzweil, sorry Virgil) integrate exactly the same positive bounded functions than Lebesgue does. See hrer for complete references : http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ === Subject: Re: Integrability of X+Y and of X, Y >I would need an example of variables X and Y such that X+Y are integrable. but X and Y are not separately integrable. > Integrable over the real numbers? How about X=x, Y=-x? -- Daniel Mayost === Subject: construction: circle tangent to a line, point and circle hi! About ten years ago here was presented a construction for a circle tangent to a given circle, line and point by Jonathan Haas. I copy it below. I have doped out CLAIM 2 to my satisfaction, or at least I did some time ago, but I cannot see CLAIM 1. This is a key step in two other constructions for this same problem that were presented at that time and I would appreciate it if someone could validate claim 1, which is presented as obvious but is not to me! The circle is C, the line is L, the point is P. STEP 1: Drop a normal from the center of C onto the line L, and call the points of intersection with C points Q and R. (QR will be a diameter of C). Extend QP to meet L through P. Call the intersection of QP with L point T. S is the intersection of QR with L. STEP 2: Draw a circle through P, R, and S. (Given 3 points, we can construct a circle through them). The circle intersects line QP at point U. CLAIM 1: Point U lies on the circle we're looking for. STEP 3: Using TU as a diameter, draw a circle. Draw a line perpendicular to TU through P, to intersect the circle at V. STEP 4: Using TV as a radius, draw an arc to intersect line L at point W. CLAIM 2: Point W lies on the circle we're looking for. We now have three points on that circle (P, U, and W) and can construct it. We're done. -- rob === Subject: Re: construction: circle tangent to a line, point and circle > hi! About ten years ago here was presented a construction for a circle > tangent to a given circle, line and point by Jonathan Haas. I copy it > below. I can conceive of a circle passing through a point, but being tangent to one??? What does it mean for a circle to be tangent to a point? === Subject: Re: construction: circle tangent to a line, point and circle- > >>hi! >> >>About ten years ago here was presented a construction for a circle >>tangent to a given circle, line and point by Jonathan Haas. I copy it >>below. > > >I can conceive of a circle passing through a point, >but being tangent to one??? > > >What does it mean for a circle to be tangent to a point? > Nobody really needs replies that contribute nothing beyond a display of hair-splitting skills. === Subject: Re: construction: circle tangent to a line, point and circle > What does it mean for a circle to be tangent to a point? Latin tangere means to touch. It is a nice expression and not off the logic of the word. It should be accepted as a valid description of the relation between point and circle. Vote It should be allowed to use a circle is tangent to a point as an alternative for a circle is passing through a point in favor | against | Signature -----------+----------+------------------ x | | Rainer Rosenthal | | | | | | | | | | | | Rainer Rosenthal r.rosenthal@web.de === Subject: Re: construction: circle tangent to a line, point and circle > >> What does it mean for a circle to be tangent to a point? Latin tangere means to touch. It is a nice expression > and not off the logic of the word. It should be accepted > as a valid description of the relation between point and > circle. Vote It should be allowed to use > a circle is tangent to a point > as an alternative for > a circle is passing through a point in favor | against | Signature -----------+----------+------------------ x | | Rainer Rosenthal x | | Dirk Vdm | | | | | | | | | | Dirk Vdm === Subject: Re: construction: circle tangent to a line, point and circle What does it mean for a circle to be tangent to a point? Latin tangere means to touch. It is a nice expression > and not off the logic of the word. It should be accepted > as a valid description of the relation between point and > circle. Vote It should be allowed to use > a circle is tangent to a point > as an alternative for > a circle is passing through a point > in favor | against | Signature > -----------+----------+------------------ > x | | Rainer Rosenthal > | | > | | > | | > | | > | | > | | Rainer Rosenthal > r.rosenthal@web.de > -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: construction: circle tangent to a line, point and circle hi! About ten years ago here was presented a construction for a circle > tangent to a given circle, line and point by Jonathan Haas. I copy it > below. I can conceive of a circle passing through a point, > but being tangent to one??? > What does it mean for a circle to be tangent to a point? -- ### Author Logan Lee ### ### AuthorHomePage http://beam.to/pyenos ### ### AuthorQuote I just want to learn. ### === Subject: Re: construction: circle tangent to a line, point and circle > hi! > About ten years ago here was presented a construction for a circle > tangent to a given circle, line and point by Jonathan Haas. I copy it > below. I can conceive of a circle passing through a point, > but being tangent to one??? > What does it mean for a circle to be tangent to a point? Is that supposed to answer my question? === Subject: Re: construction: circle tangent to a line, point and circle > hi! > About ten years ago here was presented a construction for a circle > tangent to a given circle, line and point by Jonathan Haas. I copy it > below. > I can conceive of a circle passing through a point, > but being tangent to one??? > > What does it mean for a circle to be tangent to a point? Is that supposed to answer my question? As without knowing the answer to my question, I have no idea what your question means, no. === Subject: Re: Number of binary search trees <45c0aa5c@news1.ethz.ch That's the number of *local* binary search trees, isn't it? Sequence No. A007889 in the The On-Line Encyclopedia of Integer Sequences: > http://www.research.att.com/~njas/sequences/A007889 1/(2^n*(n+1))*sum(binomial(n+1,k)*k^n, k=1..(n+1)); I am afraid the series A007889 and the comment do not match. (Unless the word 'local' has some special meaning which is not clear from the comment ) Lets work out the number of binary search trees of n nodes in simple cases T(1) = 1 T(2) = 2 [ 2! = 2 permutations of input sequence possible viz. 1 2 2 1 Each of them gives a different (non-isomorphic) tree ] T(3) = 5 [ 3! = 6 permutations of input sequence possible viz. 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 Out of them 2 1 3 and 2 3 1 give the same binary search tree Sourabh === Subject: Re: sum of 1000 numbers of positive integers = product of those > a 1, a 2, ..., a 1000 are positive intergers. Find ( a 1, a 2, ..., a 1000 ) witch satisfies a 1 + a 2 + ... + a 1000 = a 1 x a 2 x ... x a 1000 ***************************************************************************[ DownExclamation]©** > I saw this problem in some math book. But a solution it the book was a > little bit weird... It is supposed that a 1 = a 2 = ... = a 998 = 1 without any > mention or proof. I am not sure it is true or not. I did prove it must be a 1 = a 2 = ... = a 990 = 1. But I could't > prove more. How can solve it well? But I didn't want to know just a solution. I want to know the complete solution and how to find them. Because I am not good at expressing English, you guys didn't understand my intention for my question. Sorry!! As you guys, the book which contains this problem gave me just some solutions. But I wonder how to find all solutions of this problem. There's no deadline for solving it. When you find answer, just let me know it. I really wonder...~.~ === Subject: Re: sum of 1000 numbers of positive integers = product of those > a_1, a_2, ..., a_1000 are positive intergers. Find ( a_1, a_2, ..., a_1000 ) witch satisfies a_1 + a_2 + ... + a_1000 = a_1 x a_2 x ... x a_1000 ***************************************************************************[ CapitalAAcute]©** > > I saw this problem in some math book. But a solution it the book was a > little bit weird... It is supposed that a_1 = a_2 = ... = a_998 = 1 without any > mention or proof. I am not sure it is true or not. I did prove it must be a_1 = a_2 = ... = a_990 = 1. But I could't > prove more. How can solve it well? But I didn't want to know just a solution. I want to know the complete solution and how to find them. Because I am not good at expressing English, you guys didn't understand my intention for my question. Sorry!! As you guys, the book which contains this problem gave me just some solutions. But I wonder how to find all solutions of this problem. There's no deadline for solving it. When you find answer, just let me know it. I really wonder...~.~ I gave you a way to find all positive real solutions and one might be able to use that to narrow down to positive solutions. For example, one solution is 2 + 2 + 4/3 + 16/13 = 2*2*4/3*16/13 but now we want integral solutions. if we multiply by 3*13 on both sides we get integral solutions(integral a_k's) but now it doesn't fit the criteria. First off you would have to proof that my method can get all the answers(so there exists an a_1 for each solution) and try and find information about the positive integer case. (such as maybe there are only a countable or just a finite number) and possibly a way to enumerate them. I have no idea if my method works though or even if it does get them all. Just a possibility. I think also one can look at it by writing N = a_1*...*a_n = sum(1,k=1..n) and partitioning the sum(grouping the ones) and seeing how the LHS must change to make your criteria still true. (i.e., we would have to find some factoring of the LHS that for any partition of the RHS). When doing this maybe you can get a relationship between the partition of the RHS and the factoring of N. Obviously if you have this then you have your solutions. e.g. a*b*c = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7 = N In this case its obvious there is no way to partition 7 and get a factoring that works. a*b*c = 1 + 1 + 1 + 1 + 1 + 1 = 6 2*3*1 = 1 + 2 + 3 = 6 This works because we were able to partiting 6 and factor 6 using the same positive integers. Theres plenty of information the net about partitioning integers and so you might come up with something that way. Anyways, good luck. === Subject: Re: sum of 1000 numbers of positive integers = product of those Obviously though if you start with a_1 > 1 then a_2 = S_1/(P_1 - 1) > 0 Ofcourse a_2 might be < 1 but it seems that if a_1 is large then the > sequence will converge to 1 monotonically. Even when a_1 is close to 1 it > seems to work except that a_2 will blow up but still resulting in a > positive sequence. So a_k = S_(k-1)/(P_(k-1) - 1) a_(k+1) = S_k/(P_k - 1) and suppose a_k > 1 then ==> P_(k-1) > 1 and S_(k-1) > 0 a_(k+1) = (a_k + S_(k-1))/(a_k*P_(k-1) - 1) = a_k/(a_k*P_(k-1) - 1) + S_(k-1)/(a_k*P_(k-1) - 1) The first term is > 0 and the second > 1 (because it essentially takes on > the form of a_k) So if thats all correct then if a_1 > 1 gives a_k > 1. I thought when I > tried an example though with a_1 > 1 I got some negative terms but maybe I > was mistaken or maybe I have made a mistake in the outline of a proof. Anyways, I'll leave you to figure it all out if you think it works. Damn. didn't see the thing about positive integers ;/ I guess this doesn't work well. === Subject: Re: sum of 1000 numbers of positive integers = product of those For any positive integer number n, we have n positive integers whose sum and product match. Namely, take the n-tuple (1,1,1,...,1,2,n) where there are n-2 1's before the final 2 and n. The sum and product each equal 2n, so sum and product match. Apply that to your case with n=1000. mag_feb02_toc.html (Feb. 2002 Mathematics Magazine), whose blurb reads: When Does a Sum of Positive Integers Equal Their Product? Michael W. Ecker In 1997, an amateur math enthusiast sent Dr. Mike Ecker a neat little formula about tangents and triangles. To wit: The sum of the tangents equals the product of the tangents. Can these three values all be integers? More generally, for any positive integer n > 1, we can always find n positive integers whose product and sum are the same. Sometimes there is just one such solution for a given n. Usually there are more. What is the nature of such solutions, and what properties do these n-tuples enjoy? That is the quoted blurb. Hope it's of some interest. Best, Mike Dr. Michael W. Ecker Associate Professor of Mathematics Pennsylvania State University Wilkes-Barre Campus Lehman, PA 18627 === Subject: Re: sum of 1000 numbers of positive integers = product of those For any positive integer number n, we have n positive integers whose sum and product match. Namely, take the n-tuple (1,1,1,...,1,2,n) where there are n-2 n's before the final 2 and n. The sum and product each equal 2n, so sum and product match. Apply that to your case with n=1000. mag_feb02_toc.html (Feb. 2002 Mathematics Magazine) whose blurb reads: When Does a Sum of Positive Integers Equal Their Product? Michael W. Ecker In 1997, an amateur math enthusiast sent Dr. Mike Ecker a neat little formula about tangents and triangles. To wit: The sum of the tangents equals the product of the tangents. Can these three values all be integers? More generally, for any positive integer n > 1, we can always find n positive integers whose product and sum are the same. Sometimes there is just one such solution for a given n. Usually there are more. What is the nature of such solutions, and what properties do these n-tuples enjoy? That is the blurb. Hope it's of some interest. Best, Mike Dr. Michael W. Ecker Associate Professor of Mathematics Pennsylvania State University Wilkes-Barre Campus Lehman, PA 18627 === Subject: Re: sum of 1000 numbers of positive integers = product of those 131208@gmail.com a e'crit : > a_1, a_2, ..., a_1000 are positive intergers. > Find ( a_1, a_2, ..., a_1000 ) witch satisfies > a_1 + a_2 + ... + a_1000 = a_1 x a_2 x ... x a_1000 > It is supposed that a_1 = a_2 = ... = a_998 = 1 without any > mention or proof. This is not necessary, see below >> 998 + a + b = ab >> Slove for a and b. >> (a - 1)(b - 1) = 999 >> a = 2, b = 1000 is just one solution. > I am not sure it is true or not. > I did prove it must be a_1 = a_2 = ... = a_990 = 1. But I could't > prove more. > How can solve it well? >> Decode my a and b. I found solutions in that case, and there are 3 more. ( a , b ) = ( 4 , 337 ) = ( 10 , 112 ) = ( 28 , 38 ) But the point I asked was not about that. > There are other solutions : a_1=a_2=...=a_997=1, a_998=4, a_999=4, a_1000=67 => sum (a_i)=1072=4*4*67 === Subject: Projection of a vector Hi guys, consider a ARBITRARY traingle(it can be any shape) of unequal sides with vertices A, B and C. Let Q be a point on the opposite edge(BC) to the vertex A.In such a way that AQ is normal to be the line BC. I wanna find the coordinates of point 'Q'.Its simple as long as BC is a straight line but i wanna know a generalised equation that satisfy for any shape. can any one pls help me out. === Subject: Re: Projection of a vector > Hi guys, consider a ARBITRARY traingle(it can be any shape) of unequal sides with > vertices A, B and C. Let Q be a point on the opposite edge(BC) to the > vertex A.In such a way that AQ is normal to be the line BC. > I wanna find the coordinates of point 'Q'.Its simple as long as BC is a > straight line but i wanna know a generalised equation that satisfy for any > shape. can any one pls help me out. Simple if BC is a straight line? Surely this is always the case? hth Mick. === Subject: Re: Projection of a vector > Hi guys, >> consider a ARBITRARY traingle(it can be any shape) of unequal sides with >> vertices A, B and C. Let Q be a point on the opposite edge(BC) to the >> vertex A.In such a way that AQ is normal to be the line BC. >> I wanna find the coordinates of point 'Q'.Its simple as long as BC is a >> straight line but i wanna know a generalised equation that satisfy for any >> shape. can any one pls help me out. Simple if BC is a straight line? Surely this is always the case? hth Mick. > Heh, heh, Mick. You didn't read closely enough. He's got a traingle, not a triangle. That's why it can be any shape. You're thinking straight sides. Silly you. :-) Not that I can make any sense of his question either, unless he really wants the projection of a point on an arbitrary curve. Perhaps he will explain. --Lynn === Subject: Re: Projection of a vector I have a triangle(not curve) having vertices A, B and C. so, AB,BC and CA are the edges of the triangle(straight line only). To be more clear consider, A coordinates is (1,1), B (0.5,0.1) and C(1.5,0.5). Now consider a point Q on the edge BC(still BC is a straight line only),in such a way that it is perpendicular to A.I need to find coordinates Q analytically. Hope! now i am clear. === Subject: Re: Projection of a vector <12491309.1170575076328.JavaMail.jakarta@nitrogen.mathforum.org> In message <12491309.1170575076328.JavaMail.jakarta@nitrogen.mathforum.org>, Sriram >I have a triangle(not curve) having vertices A, B and C. so, AB,BC and >CA are the edges of the triangle(straight line only). To be more clear >consider, A coordinates is (1,1), B (0.5,0.1) and C(1.5,0.5). Now >consider a point Q on the edge BC(still BC is a straight line only),in >such a way that it is perpendicular to A.I need to find coordinates Q >analytically. Hope! now i am clear. I think these geometrical things are always best done with vectors. Let a = vector from origin to point A, etc. Then vector s = (c-b) is a vector parallel to BC through the origin.. A vector n normal to s is given by transposing the x and y coordinates of s and reversing the sign of one of them (i.e. if s = (a,b), n = (-b,a) so n.dot.s = 0, or if you prefer this the same as n = z cross s). You need the intersection of the lines defined parametrically by vectors r1 = b + beta. s and r2 = a + alpha.n. where alpha and beta are scalar paremeters. i.e. Q is given by b + beta. s = a + alpha.n which gives 2 simple linear simultaneous equations which will give you alpha and beta and hence the vector location q, expressed in terms of the coordinates of A,B and C. -- Andy Smith === Subject: Re: Projection of a vector <12491309.1170575076328.JavaMail.jakarta@nitrogen.mathforum.org>, > I have a triangle(not curve) having vertices A, B and C. so, AB,BC and CA are > the edges of the triangle(straight line only). To be more clear consider, A > coordinates is (1,1), B (0.5,0.1) and C(1.5,0.5). Now consider a point Q on > the edge BC(still BC is a straight line only),in such a way that it is > perpendicular to A.I need to find coordinates Q analytically. Hope! now i am > clear. If A is a vertex, what do you mean by perpendicular to A? Do you mean that you want to find a point, Q, on the line through B and C, such that the line through A and Q is perpendicular to the line through B and C? === Subject: Re: Projection of a vector > If A is a vertex, what do you mean by perpendicular > to A? Do you mean that you want to find a point, Q, on the > line through B and > C, such that the line through A and Q is > perpendicular to the line > through B and C? YES!. you are correct. Q is the intersecting point for the line AQ and BC. === Subject: Re: Projection of a vector <5599168.1170579603278.JavaMail.jakarta@nitrogen.mathforum.org>, > If A is a vertex, what do you mean by perpendicular > to A? Do you mean that you want to find a point, Q, on the > line through B and > C, such that the line through A and Q is > perpendicular to the line > through B and C? YES!. you are correct. Q is the intersecting point for the line AQ and BC. Regarding upper case letters as vectors and lower case letters as scalars: Q is on the line through B and C if and only if there is a real t such that Q = t*B + (1-t)*C. The line through A and Q will be perpendicular to the line through B and C if and only if the dot product (scalar) product of Q-A and B-C is zero. In coordinate terms, let A = [a1,a2], B = [b1,b2], etc. then Q = [q1, q2] = [ t*b1 + (1-t)*c1, t*b2 + (1-t)*c2 ] (Q - A) dot (B - C) = [q1 - a1, q2 - a2] dot [b1 - c1, b2 - c2] = 0 so (q1 - a1) * (b1 - c1) + (q2 - a2) * (b2 - c2) = 0 Eliminate q1 and q2 by substitution to get a linear equation in t. === Subject: sci.math Comments Ah, sci.math, how are you today. Hey, Leroy Quet posted a double integral identity the other day, finite differences? I have not even read it. I figure I take the n-gonometry and use it to take all the corners. There's only one theory with no axioms. (It's consistent that's also the universal, or universally axiomatized theory.) EF, the natural/unit equivalency function, is like f(n) = n/oo, the input is a natural number, n E N. N E N. The integral over those is one, not necessarily one half like you'd expect. Is the existence of the Russell set inconsistent with ZF? Because, ZF is that. I know that, it's obvious, I just pointed it out. If ZF is all set's that it's not then it's not, the complement, in regularity. It's not inconsistent to apply regularity, that's a reason to actually use ZF, just not try and prove it consistent. Then, it's all one theory. I just use one theory, all my mathematics are in one theory. I use standard theory, and when it's not enough, I use nonstandard theory. So, I feel like I've educated you. Ross === Subject: Set of [0,1] is non-countable - prove? Hi there! How to prove that the set of all real numbers in [0,1] is non- countable? === Subject: Re: Set of [0,1] is non-countable - prove? [0,1] is a compact hausdorff space with no isolated points. Therefore, it is uncountable. > Hi there! > How to prove that the set of all real numbers in [0,1] is non- > countable? === Subject: Re: Set of [0,1] is non-countable - prove? > Hi there! > How to prove that the set of all real numbers in [0,1] is non- > countable? > [0,1]has Lebesgue measure 1. Any countable set has measure 0. Therefore, [0,1] must be uncountable. === Subject: Re: Set of [0,1] is non-countable - prove? [0,1]has Lebesgue measure 1. Any countable set has measure 0. > Therefore, [0,1] must be uncountable. I like this! Points of [0,1] are nowhere dense and closed, but [0,1] has second category in itself, so it is not a countable union of points. === Subject: Re: Set of [0,1] is non-countable - prove? > How to prove that the set of all real numbers in [0,1] is non- > countable? Do a Google search for Cantor diagonal argument. Jose Carlos Santos === Subject: Re: Set of [0,1] is non-countable - prove? <52ka6nF1osjdgU1@mid.individual.net > How to prove that the set of all real numbers in [0,1] is non- > countable? Do a Google search for Cantor diagonal argument. > Jose Carlos Santos THANK YOU - it was useful === Subject: Re: Set of [0,1] is non-countable - prove? > How to prove that the set of all real numbers in [0,1] is non- >> countable? Do a Google search for Cantor diagonal argument. Half the hits will be at sci.math. === Subject: Re: Set of [0,1] is non-countable - prove? Larry Hammick > Half the hits will be at sci.math. It's a joke, guys. You know how often this diagonal-argument stuff arises at sci.math, and how long the threads tend to be. === Subject: Re: Set of [0,1] is non-countable - prove? >>Do a Google search for Cantor diagonal argument. > Half the hits will be at sci.math. Look on wiki or mathworld then. Bob Kolker === Subject: Re: Set of [0,1] is non-countable - prove? > How to prove that the set of all real numbers in [0,1] is non- > countable? >> Do a Google search for Cantor diagonal argument. Half the hits will be at sci.math. Jose Carlos Santos === Subject: Re: Set of [0,1] is non-countable - prove? <52ka6nF1osjdgU1@mid.individual.net > How to prove that the set of all real numbers in [0,1] is non- > countable? Do a Google search for Cantor diagonal argument. > Jose Carlos Santos And using the standard terminology of uncountable will probably help in finding results too. Jason === Subject: Explore Explore and recognise the words meanings in the scriptures http://www.qurancomplex.org/Quran/Targama/Targama.asp === Subject: How do you find two orthogonal vectors (a plane) perpendicular to given vector? How do you find two orthogonal vectors (a plane) perpendicular to a given vector? Please use my vector [2,3,4] and explain your steps or break things down for me as much as possible. Projection transforms Bart === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector? > How do you find two orthogonal vectors (a plane) perpendicular to a > given vector? Please use my vector [2,3,4] and explain your steps or > break things down for me as much as possible. Projection transforms A plane through the origin and perpendicular to your vector [2,3,4] is simply given by: 2.x + 3.y + 4.z = 0 This is rather basic and should be in your textbooks, as such. Han de Bruijn === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector? Han.deBruijn@DTO.TUDelft.NL a .8ecrit : > >> How do you find two orthogonal vectors (a plane) perpendicular to a >> given vector? Please use my vector [2,3,4] and explain your steps or >> break things down for me as much as possible. Projection transforms A plane through the origin and perpendicular to your vector [2,3,4] > is simply given by: 2.x + 3.y + 4.z = 0 This is rather basic and should be in your textbooks, as such. On the other hand, this still doesn't answer the OP question (surpise). To get two orthogonal vectors, start from any vector u (say (1,0,0)) ; then u wedge (2,3,4)= v = (0,-4,3) and w= v wedge (2,3,4)= (-25,6,8) satisfy both conditions... Han de Bruijn > === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector? > How do you find two orthogonal vectors (a plane) perpendicular to a > given vector? Please use my vector [2,3,4] and explain your steps or > break things down for me as much as possible. Projection transforms Bart Given and arbitrary vector, say v, not parallel to u = [2,3,4]. then the vector product, (u cross v), is one of them, and ((u cross v) cross u) is another. === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector? How do you find two orthogonal vectors (a plane) perpendicular to a > given vector? Please use my vector [2,3,4] and explain your steps or > break things down for me as much as possible. Projection transforms Bart Given and arbitrary vector, say v, not parallel to u = [2,3,4]. then > the vector product, (u cross v), is one of them, and > ((u cross v) cross u) is another. To find such a v, you can proceed as follows. Let u = (x,y,z). If both y and z are zero, let v = (0,1,0). Otherwise, let v = (1,0,0). -- === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector? > How do you find two orthogonal vectors (a plane) perpendicular to a > given vector? Please use my vector [2,3,4] and explain your steps or > break things down for me as much as possible. Projection transforms Bart Hi Bart, try to follow a simplified but longer method for a solution to your problem. First visualize the [2,3,4] vector as a line coming from the origin [0,0,0] to the point [2,3,4] which are x,y,z coordinates. The frame of reference axes about the origin can be taken as orthogonal vectors [2,0,0], [0,2,0] and [0,0,2]. First rotate this reference frame about the z axis until the x axis is directly beneath the vector (angle = tan^-1 (3/2)). This will not change the z axis but by using simple trig, the coordinates of the x axis will now be [1.1094,1.6641,0] and the y axis [-1.6641,1.1094,0]. Now rotate the reference frame about the y axis until the x axis is coincident with the vector (angle = tan^-1(4/sqrt 13)). Barring errors in arithmetic, the y and z axes now provide the two orthogonal vectors you were after [-1.6641,1.1094,0] and [-.8240,-1.2361,1.3391]. Phil H === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector? >How do you find two orthogonal vectors (a plane) perpendicular to a >given vector? Please use my vector [2,3,4] and explain your steps or >break things down for me as much as possible. Projection transforms Bart Call your vector A. Pick any nontrivial vector B. Calculate C = A cross B. Then calculate D = A cross C. Then show that C and D do the trick. --Lynn === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector?- >How do you find two orthogonal vectors (a plane) perpendicular to a >given vector? Please use my vector [2,3,4] and explain your steps or >break things down for me as much as possible. Projection transforms Bart > > Apply the Gram-Schmidt orthogonalization procedure to a basis of R3 consisting of [2, 3, 4] and two other vectors. See for instance http://en.wikipedia.org/wiki/Gram-schmidt . Alternate approach: One vector B perpendicular to A = [2, 3, 4] is easily found. Just by trial and error. The cross product A x B is perpendicular to A and to B. Done! === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector?- <45C508F1.9030304@xs4all.nlHow do you find two orthogonal vectors (a plane) perpendicular to a >given vector? Please use my vector [2,3,4] and explain your steps or >break things down for me as much as possible. Projection transforms Bart Apply the Gram-Schmidt orthogonalization procedure to a basis of R3 > consisting of [2, 3, 4] and two other vectors. > See for instancehttp://en.wikipedia.org/wiki/Gram-schmidt. Alternate approach: One vector B perpendicular to A = [2, 3, 4] is > easily found. Just by trial and error. > The cross product A x B is perpendicular to A and to B. Done! > tried to wrap my brain around it. So far, that's not possible for me on my own. That's why I said, Projection transforms don't mean anything to me. and explain your steps or break things down for me as much as possible. Trial and error is not possible, and I know there are ways. The application is a computer program that needs to run at a decent speed and it needs to possible to computed for ANY === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector?- >How do you find two orthogonal vectors (a plane) perpendicular to a >given vector? Please use my vector [2,3,4] and explain your steps or >break things down for me as much as possible. Projection transforms >Bart Apply the Gram-Schmidt orthogonalization procedure to a basis of R3 > consisting of [2, 3, 4] and two other vectors. > See for instancehttp://en.wikipedia.org/wiki/Gram-schmidt. Alternate approach: One vector B perpendicular to A = [2, 3, 4] is > easily found. Just by trial and error. > The cross product A x B is perpendicular to A and to B. Done! > tried to wrap my brain around it. So far, that's not possible for me > on my own. That's why I said, Projection transforms don't mean > anything to me. and explain your steps or break things down for me > as much as possible. Trial and error is not possible, and I know > there are ways. The application is a computer program that needs to > run at a decent speed and it needs to possible to computed for ANY You might have said that at the start. If you have two vectors, do you know how to calculate the cross product? -- === Subject: Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector?- One vector is easily obtained by inspection. Set one component to zero and swithc the other two from the given vector, while changing the sign of one of the switched components. === Subject: Re: Nested Carmichael numbers? > Is it possible for N to be a Carmichael number, > but for some other (smaller) Carmichael number to be a factor of N-1? > Has anybody found such a example? > Has anybody proved it impossible? > I don't know. Why not have a look at > G. Jaeschke, The Carmichael numbers to 10^12, Math. Comp., 55 > (1990), 383-389, and see what you find? I don't know where to find that around here. > http://home.no.net/zamunda/carmichael.txt has all the Carmichaels > up to 10^9. I downloaded that and cleaned it up to have just the list, but it's only the first 646 Carmichael numbers, so then I decided to try a larger list: > http://www.chalcedon.demon.co.uk/rgep/publish.html#76 refers to > all the Carmichaels up to 10^19. I looked there, found these additional links: > There's also an extensive list at (up to 10^13) > http://www.research.att.com/~njas/sequences/b002997.txt > * carmichael-16.gz All Carmichael numbers up to 10^16 in GZIP format GZIP format Somewhere around there I found a list of the first ten thousand Carmichael numbers, which seemed just the right size to download to my limited disk space but still have a decent number to search within, so that's what I finally downloaded for real study. ; ;Theorem (Korselt 1899): A positive composite integer n is a Carmichael ; number if and only if n is square-free, and for all prime divisors p ; of n, it is true that p - 1 divides n - 1. I tested the complete list of 10k numbers I had downloaded, and every one of them passed the test. I haven't worked out the math to prove that theorem myself, but assuming it's true it means all the 10k numbers I downloaded are in fact Carmichael numbers, good to know that before going any further. Carmichael numbers. I was expecting they'd be quite rare, or else somebody would have cited an example already in the thread. But look what turned up: 561 = (3 11 17) divides 1615681-1 561 = (3 11 17) divides 5049001-1 561 = (3 11 17) divides 9613297-1 1105 = (5 13 17) divides 10024561-1 6601 = (7 23 41) divides 26932081-1 2821 = (7 13 31) divides 40622401-1 2821 = (7 13 31) divides 75151441-1 2465 = (5 17 29) divides 78091201-1 561 = (3 11 17) divides 78091201-1 561 = (3 11 17) divides 84311569-1 1105 = (5 13 17) divides 90698401-1 and it spewed out a lot more, but I saved only that first part to post here. Now the reason I asked the quesition in the first place was that I was thinking of a primality testing algorithm as follows: If the number fails the Fermat test, then of course it's not prime. If it passes the Fermat test, then next I factorize n-1 and apply that test I invented circa 1977-78 independently of somebody else who also invented the same test. But what if factorizing n-1 is difficult to be sure of because one of the factors is a Carmichael number which looks prime when you're busy doing the factorization but it's not really prime and you don't really know if it is or isn't? Well the bad news is there's a lot of cases where the starting number is a Carmichael number and n-1 has another Carmichael number as a factor, as shown above. But the good news is, so-far above, that in fact the smaller Carmichael number has very small factors, so trial division to just as far as 100 will tease them all out, giving you a complete prime factorization of n-1 fast enough for any purpose. So I modified my loop to print only the Carmichael numbers factors of n-1 where either that smaller Carmichael number factor is large than ever before, or the remaining quotient after trial division to 100 is larger than ever before. This is what came out (after the above): 8911 = (7 19 67) divides 368113411-1 126217 = (7 13 19 73) divides 790623289-1 52633 = (7 73 103) divides 958762729-1 825265 = (5 7 17 19 73) divides 12893940361-1 1050985 = (5 13 19 23 37) divides 33900572161-1 188461 = (7 13 19 109) divides 134199308881-1 11921001 = (3 29 137023) divides 166894014001-1 The first one where trial division doesn't pull out all the factors of the smaller Carmichael factor is the third line with 103, but since 103 is less than 100**2 we know that 103 is prime. The only case in that entire dataset where trial division to 100 doesn't prove all by itself that we have the smaller Carmichael factor completely factorized is that last line. But still the smaller Carmichael factor is broken up so the not-factorized part is not itself a Carmichael number, so Fermat's test would clearly show it's not prime, and at that point we'd know it's worth the extra effort to factorize it completely, which trial division to 1000 would accomplish. So let me amend my original question. Is there a case where you start with a large number that passes Fermat's test so it might be prime, but it's really not prime, in fact it's a Carmichael number, but when you try to factorize n-1 by trial division up to some threshold such as 100 or 1000 the remaining unresolved quotient is another Carmichael number, confusing you as to whether you have completely factorized n-1 already or need to do some more work before it'll be completely factorized? Nothing turned up in my experiment described above with the first 10k Carmichael numbers, but that doesn't prove anything about the kind of really huge Carmichael numbers that might turn up when trying to generate super-large primes for use in RSA. Note of course that in most cases this question is moot because a partial factorization by trial division of n-1 to 100 is quite sufficient to generate too may qth roots thereby showing n is composite and in fact usually generating a partial factorization of it. But still it's cleaner to get a total factorization of n-1 if it's easy to know when you're done so you do extra effort only when necessary. So my question may still be interesting. Now a somewhat different question is, if n really is prime, how difficult can it be to completely factorize n-1? The case in point would be that you've used trial division to pull off all the small prime factors of n-1, and what you have left fails Fermat's test so you *know* it's not prime, but then how much more work does it take to factorize it? I directly constructed a whole bunch of examples where the product of two large primes of nearly equal size, times two, plus one, is a new larger prime, showing it can be as difficult as you could possibly imagine it might be. (I even did that recursively: Two medium-size primes p1a,p1, where 2*a1*p1a*p1b+1 (a1 very small integer) is a new prime p2a and 2*b1*p1a*p1b+1 (b1 very small integer) is a new prime p2b, then do the same thing with those as starting points to get two new primes p3a=2*a2*p2a*p2b+1 and p3b=2*b2*p2a*p2b+1, etc., more than doubling the number of digits with each iteration. Because each level shares the two large primes along both branches from the next larger level, the recursive certificate of primeness is unusually small in total size, but difficult for anyone but the original author to construct because of the near-equal primes one level below the top which except for 2 and factors of small ak or bk are nearly as large as possible. For example, lets first generate two random 10-digit primes, with no regard for how easy/difficult they are to prove prime: (setq p1a (lohi-ez-rand-proven-prime (expt 10 9) (expt 10 10) (expt 10 5))) 3593119561 (qf (- p1a 1) 10000) (2 2 2 3 5 43 127 5483) so the certificate for that prime would not be huge. (setq p1b (lohi-ez-rand-proven-prime (expt 10 9) (expt 10 10) (expt 10 5))) 1318893731 (qf (- p1b 1) 10000) (2 5 7 11 127 13487) likewise the certificate for that prime would not be huge. (bigprimes-make-biggerprimes (list p1a p1b)) p=56867314364836465093 is prime. p=199035600276927627823 is prime. p=426504857736273488191 is prime. p=578151029375837395103 is prime. p=1260558801753874976207 is prime. Pick any two you want. For example: (setq p2a 426504857736273488191) (setq p2b 578151029375837395103) (qf (/ (- p2a 1) (* p1a p1b)) 1000) (2 3 3 5) (qf (/ (- p2b 1) (* p1a p1b)) 1000) (2 61) Certificates for primeness of 3,5 already appeared inside the certificate for p1a, so 61 is the only new small-number certificate we'd need here, plus the two top-level certificates for p2a ad p2b themselves. (bigprimes-make-biggerprimes (list p2a p2b)) p=34521791154763024357479167317314682042014221 is prime. p=44385160056123888459616072265118876911161141 is prime. p=101099531238948857046903275714992997408755931 is prime. p=128716964162759276532886609568844743042367307 is prime. p=133648648613439708583955062042746840476940767 is prime. Again, pick any two you want, for example: (setq p3a 34521791154763024357479167317314682042014221) (setq p3b 44385160056123888459616072265118876911161141) (qf (/ (- p3a 1) (* p2a p2b)) 1000) (2 2 5 7) (qf (/ (- p3b 1) (* p2a p2b)) 1000) (2 2 3 3 5) No new small-number certificates needed, only the single toplevlel certificates for p3a and p3b. (bigprimes-make-biggerprimes (list p3a p3b)) p=27580594064908297578960433351232151496286747697780235307604186756188312273 8107038025508981 is prime. p=82128880104393597235126623757002406677831648700056700693754689451760752104 2363179898182297 is prime. That took a few seconds, so I stopped it after I got the two I needed. (setq p4a 2758059406490829757896043335123215149628674769778023530760418675618831227381 0 7038025508981) (setq p4b 8212888010439359723512662375700240667783164870005670069375468945176075210423 6 3179898182297) (qf (/ (- p4a 1) (* p3a p3b)) 1000) (2 2 3 3 5) (qf (/ (- p4b 1) (* p3a p3b)) 1000) (2 2 2 67) We'll need a new small-number certificate for 67, otherwise just the two toplevel certificates for p4a ad p4b. (bigprimes-make-biggerprimes (list p4a p4b)) p=21519051380065630506250779496317352157664776046661678124260809085747624983 6468855376328797159494358017198104083133357270652666663539762548574204237665 4 82627220130705137170141273889151 is prime. p=21926780774635295084263952160458101988020529698072109920299434942109158930 7054581057143448053042672169102910265761147197886085610848936996862978633747 5 65455946406865866084943950657577 is prime. That took a couple minutes, then suddenly those two came pretty close together, so of course I'm using them. (setq p5a 2151905138006563050625077949631735215766477604666167812426080908574762498364 6 8855376328797159494358017198104083133357270652666663539762548574204237665482 6 27220130705137170141273889151) (setq p5b 2192678077463529508426395216045810198802052969807210992029943494210915893070 5 4581057143448053042672169102910265761147197886085610848936996862978633747565 4 55946406865866084943950657577) (qf (/ (- p5a 1) (* p4a p4b)) 1000) (2 5 5 19) (qf (/ (- p5b 1) (* p4a p4b)) 1000) (2 2 2 11 11) Need new small-number certificate for 19, already have the rest, plus two toplevel certificates for p5a ad p5b. (bigprimes-make-biggerprimes (list p5a p5b)) p=47184352208881218140242658944312349690986061336937280927574119997245696735 4952479673304084248886755236529317333607595374823840242508632399172411735186 3 0768320852590047294491056756197946211840931187349467061228117231513964144469 5 7073375793661458209993306087965848991739353697361270852028448562986468820062 6 69538880836091328208539530146675351092346687233654665624712701 is prime. OK, that's all we need at the very top level, one prime. (setq p6 4718435220888121814024265894431234969098606133693728092757411999724569673549 5 2479673304084248886755236529317333607595374823840242508632399172411735186307 6 8320852590047294491056756197946211840931187349467061228117231513964144469570 7 3375793661458209993306087965848991739353697361270852028448562986468820062669 5 38880836091328208539530146675351092346687233654665624712701) (qf (/ (- p6 1) (* p5a p5b)) 1000) (2 2 5 5) Only one new toplevel certificate needed here and we're done. That's a certificate of primeness for a 367-digit prime, whose n-1 has two 183-digit prime factors, whose (n-1)'s share a pair of 90-digit primes, etc. down the levels of the DAG. If I showed you just the 367-digit prime and asked you to prove it, would you be able to? Trial division of n-1 gives an unresolved quotient of 365 digits which fails Fermat's test, so you know it's composite. I suppose I should look at n+1 too: (qf (+ 1 p6) 100000) (2 3 7 1123436957354314717624825212959817849785382412784220974466050476124897541321 3 1542779358115297353989342030789841335141755910438152978245809326764698853882 7 8171631569058879640727799094749098057364568416539776482885055122372415349897 7 9375188967013859522215735229964045652227070800302583816297276901540195253016 5 5687638954555433536655479682746454579683029372730133921731) And that last huge factor fails Fermat's test too, so I doubt you're going to be able to use the super-duper test that relies on good partial factorization of n-1 and n+1 together. Would ellyptic curves be able to factor n-1 in reasonable time? Or just prove p6 is prime without any further work at factorizing?? Anyway, back to my original question, I wonder if it's worth my time to download the complete list of Carmichael numbers up to 10^17 and run my nested-Carmichael-number test on them. It took something like a half hour of compute time just to do that for the first 10k that I already downloaded. I need to change my test to be more efficient, like go directly for the trial-division-to-threshold result with direct table lookup to see whether any such unresolved quotient is listed. If I can re-do the 10k in less than a minute, then it might be worth doing the whole big list to 10^17. === Subject: Re: Nested Carmichael numbers? Nntp-Posting-Host: apps.cwi.nl ... > Carmichael numbers. I was expecting they'd be quite rare, or else > somebody would have cited an example already in the thread. Why? They are not very interesting with respect to primality proving, unless you wish only to use Fermat. > Now the reason I asked the quesition in the first place was that I > was thinking of a primality testing algorithm as follows: If the > number fails the Fermat test, then of course it's not prime. If it > passes the Fermat test, then next I factorize n-1 and apply that > test I invented circa 1977-78 independently of somebody else who > also invented the same test. But what if factorizing n-1 is > difficult to be sure of because one of the factors is a Carmichael > number which looks prime when you're busy doing the factorization > but it's not really prime and you don't really know if it is or > isn't? You are going the wrong way. Carmichael numbers are not the stumbling block. The stumbling block is numbers that you know are not prime, but are too large to allow factorisation. The p-1 method finds smaller and smaller numbers that can be prime or not. Finally when you are down to the last stage you may find a number that you can or completely factorise numbers. > Now a somewhat different question is, if n really is prime, how > difficult can it be to completely factorize n-1? The case in point > would be that you've used trial division to pull off all the small > prime factors of n-1, and what you have left fails Fermat's test so > you *know* it's not prime, but then how much more work does it take > to factorize it? A lot of work. Proving a number prime is easy, factorising a number is difficult. Moreover, there is *more* than Fermat's test to prove that a number is composite. > For example, lets first generate two random 10-digit primes, with > no regard for how easy/difficult they are to prove prime: I did show how I generated a random 120 digit prime. You were not even able to prove it prime with your method, because the successive n-1 did not easily factor. > Anyway, back to my original question, I wonder if it's worth my > time to download the complete list of Carmichael numbers up to > 10^17 and run my nested-Carmichael-number test on them. Carmichael numbers are in general not the stumbling block to prove a number is prime. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Can a jet fuel/hydrocarbon fire collapse a steel structure? An experiment. why do you ignore the damage to the structure by the massive impact of the planes -- any hypotheses as to the largest effect of that? so, the pancake is an ill-pozed metaphor for a structure whose very building required massively balanced tension, at every stage ... or just needs to be refined, to connote what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing. the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and,of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin.,with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, woring for Lloyds of London! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html > Yeah, I know eager very well. He runs the welding lab at MIT. > centuries ago. Even NIST disowned his pancake theory subsequently. > announcements of new findings, but not tutorials. I have already given > link towww.st911.organd > Bush family incharge of security. > Explain the presence of potassium in the molten steel and also > manganese. > They came from the potassium permanganate oxidiser of thermate. the mainpart ofthe controlled demo argument is that a trashfire never took-down a skyscraper, aside from the fact that such an inside operation needs much,much,much less explosive energy that was in the planes. however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause. as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary. anyway,please present your arguments in nonvideoformat. > World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: Can a jet fuel/hydrocarbon fire collapse a steel structure? An experiment. preliminary look at the MIT Head Welder's site-piece shows that he did try to quantiry the impact ofthe plane; good,except that he doesn't quantify the relative acceleration qua the designed-for windload. >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html > Yeah, I know eager very well. He runs the welding lab at MIT. thus: Bernard BORAT Lewis WANTS YOU IN SUDAN; Iran; Afghanistan; Iraq. other plausibly British Quagmires, like Belize and Canada; and, basically, anywhere that will destroy the USA economy, faster. >unfortunately, the neocons & jihadists are sliding us >into Sudan, a dried-up quagmire (British sand, >viz Sandhurst, EMI et al ad vomitorium Seargent Peppers). thus: congratulation, again, times 10, or ....000010. the necesary condition of the multihedral angles (not just trihedral, unless you restrict yourself to the nontrigonated shapes, not bothering with their trigonated duals -- the opposite o'Bucky) is not sufficeint to prove the existence of the five shapes; did Theaetetus (sp.?) do that? of course, no proof is accorded done, without necessity and/or sufficiency actually shown, but you have to be able to use those words in a sentence, a la Liebniz, in the first place; good luck, with your ESL class! I mean, I'm sure that it's possible to relate any or all of those problems to the regular shapes, and that many results are already extant that use them, if not your virtually compleat say-soes (of He, Who Chosen by the Plutonium Singularity that Is Universe .-) thus: why do you ignore the damage to the structure by the massive impact of the planes -- any hypotheses as to the largest effect of that?... so, the pancake is an ill-pozed metaphor for a structure whose very building required massively balanced tension, at every stage ... or just needs to be refined, to connote what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing.... the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and, of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin., with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, working for Lloyds of London! thus: the mainpart ofthe controlled demo argument is that a trashfire never took-down a skyscraper, aside from the fact that such an inside operation needs much,much,much less explosive energy that was in the planes.... however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause.... as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary.... anyway,please present your arguments in nonvideoformat. --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: Can a jet fuel/hydrocarbon fire collapse a steel structure? An experiment. I've read most of MIT Head Welder's page -- really excellent for its size, with lots of things that I'd never seen or thought of. now, see if you can actually refute *any* thing that he says, instead of using bogus ad hominems, like, of course, Professor Borat refuted that in the 16th century BCE; he was really a great friend of mine, when we went to MIT! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html > Yeah, I know eager very well. He runs the welding lab at MIT. thus: Bernard BORAT Lewis WANTS YOU IN SUDAN; Iran; Afghanistan; Iraq. other plausibly British Quagmires, like Belize and Canada; and, basically, anywhere that will destroy the USA economy, faster. thus: what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing.... the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and, of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin., with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, working for Lloyds of London! thus: the mainpart ofthe controlled demo argument is that a trashfire never took-down a skyscraper, aside from the fact that such an inside operation needs much,much,much less explosive energy that was in the planes.... however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause.... as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary.... anyway,please present your arguments in nonvideoformat. --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: Can a jet fuel/hydrocarbon fire collapse a steel structure? An experiment. words I never get a such a long tail of spooooooks behind me ... > perhaps they are afraid of your respectability ... and that you may > sway the public opinion ................ ************************************************************ Your stupidities are usually way more nonsensical and boring, and probably the respectability thing has something to do here with it: you've none. So the above is why you usually don't get many responses. Tonio === Subject: Re: correct myself; Theaetetus circa Plato proved 5 and only 5 regular-polyhedra exist Re: #12# new book congratulation, again, times 10, or ....000010. the necesary condition of the multihedral (not just trihedral, unless you restrict yourself to the nontrigonated shapes,not bothering with their trigonated duals -- the opposite o'Bucky) is not sufficeint to prove the existence of the five shapes; did Theaetetus (sp.?) do that? of course, no proof is accorded done, without necessity and/or sufficiency actually shown, but you have to be able to use those words in a sentence, a la Liebniz, in the first place; good luck, with your ESL class! I mean, I'm sure that it's possible to relate any or all of those problems to the regular shapes,and that many results are already extant that use them, if not your virtually compleat say-soes (of He, Who Chosen by the Plutonium Singularity that Is Universe .-) why do you ignore the damage to the structure by the massive impact of the planes -- any hypotheses as to the largest effect of that? so, the pancake is an ill-pozed metaphor for a structure whose very building required massively balanced tension, at every stage ... or just needs to be refined, to connote what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing. the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and,of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin.,with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, woring for Lloyds of London! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html > Yeah, I know eager very well. He runs the welding lab at MIT. > centuries ago. Even NIST disowned his pancake theory subsequently. > announcements of new findings, but not tutorials. I have already given > link towww.st911.organd > Bush family incharge of security. > Explain the presence of potassium in the molten steel and also > manganese. > They came from the potassium permanganate oxidiser of thermate. thus: the mainpart ofthe controlled demo argument is that a trashfire never took-down a skyscraper, aside from the fact that such an inside operation needs much,much,much less explosive energy that was in the planes. however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause. as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary. anyway,please present your arguments in nonvideoformat. > World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Laplace's rule of succession In Bayesian statistics, Laplace's rule of succession attempts to solve the problem of how we can predict that the sun will rise tomorrow, given its past frequency of rising. Definitions: 1. Let p be the long-run frequency, as observed. 2. Let n be the total number of trials. 3. Let s be the number of *successes* among these trials, so that n - s is the number of failures. The rule of succession states that the probability of the next success is given by the *expected value of a normalized likelihood function*. The likelihood function is p^s * (1 - p)^(n - s). Normalized with the integral S_{0 to 1}(p^s * (1 - p)^(n - s)) dp, one obtains as the expected value (s + 1)/(n + 2) for the probability of the next success. Thus, if all we know is that the sun has risen 2000 times, the probability of its rising again is 2001/2002. Now, I have a question. What's so special about this likelihood function? It seems to be formulated completely ad hoc. If the sample space were all possible successions, the probability of the next success would simply be 1/2. So what gives? The figure p^s * (1 - p)^(n - s) is the probability that there will be s successes, with *fixed probability p* for each success, a probability independent of the trial number. But how can we impose this property on a sequence? How do we know that there are fixed probabilities of success and failure on each trial? Is Laplace's rule even accepted nowadays? I would like to understand more of the philosophical theory behind the your help. === Subject: Re: Laplace's rule of succession > In Bayesian statistics, Laplace's rule of succession attempts to solve > the problem of how we can predict that the sun will rise tomorrow, > given its past frequency of rising. Definitions: 1. Let p be the long-run frequency, as observed. > 2. Let n be the total number of trials. > 3. Let s be the number of *successes* among these trials, so that n - > s is the number of failures. The rule of succession states that the probability of the next success > is given by the *expected value of a normalized likelihood function*. > The likelihood function is p^s * (1 - p)^(n - s). Normalized with the integral S_{0 to 1}(p^s * (1 - p)^(n - s)) dp, one > obtains as the expected value (s + 1)/(n + 2) for the probability of the next success. Thus, if all we know is that > the sun has risen 2000 times, the probability of its rising again is > 2001/2002. Now, I have a question. What's so special about this likelihood > function? It seems to be formulated completely ad hoc. It is the assumption that sunrise events are identically and independently distributed. Given that assumption, then there is some probability p that the sun will rise, and the question becomes one of estimating p from the data: s successes in n trials. There are two more assumptions: 1) the prior probability over p is the uniform distribution over [0, 1], 2) the loss function (or cost of errors) is quadratic. In other words, we are trying to estimate the parameter p. We start out with the assumption, prior to collecting any data, that all values of p in [0, 1] are equally likely. The second issue, once you have computed the posterior distribution over p, is how to pick a best estimate for p. What is best will depend on a loss function. One choice lead to the MAP (maximum a posteriori probability), which is a value of p at which the pdf has a maximum. Under the given assumptions above the pdf is unimodal, and achieves a maximum for p = s/n: solve for p the expression d/dp [p^s * (1-p)^n-s] = 0 Another choice of loss function, the one in the law of succession, leads to the mean of the posterior pdf. > ... If the sample > space were all possible successions, the probability of the next > success would simply be 1/2. ... Only if all sequences are equally likely. It is possible for the sample space to include all possible successions, but assign them different a priori probabilities. If all sequences were a priori equally likely (or in other words there are no constraints over the hypothesis space) then generalization, or induction, would be impossible. There would be no mutual information between future and past, and no basis for predicting the future of a process from its past. > ... So what gives? Would you take this bet: every day the sun does not rise I will give you $1000 (just assume there is some way to collect it), every day it does rise you give me $1000? If not, you are probably making the working assumption that there is mutual information between future and past. > The figure p^s * (1 - p)^(n - s) is the probability that there will be > s successes, with *fixed probability p* for each success, a > probability independent of the trial number. ... Not quite. The probability is, in this model, fixed, and independent of the trial number, but under those conditions the probability of s successes in n trials is n!/[s!(n-s)!] * p^s * (1 - p)^(n - s). For given values of n and s then n!/[s!(n-s)!] is a constant, and it cancels out in the normalization above. > ... But how can we impose > this property on a sequence? How do we know that there are fixed > probabilities of success and failure on each trial? It's an assumption - an idealized model that is justified on a case by case basis. With both the sun and the earth there are irreversible evolutionary processes at work, so no stationary model can be correct. But the evolutionary processes are slow enough compared to human time scales that stationarity is a reasonable assumption. More interesting is to model processes like the probability of solar flares geater than some magnitude occuring in some time window. > Is Laplace's rule even accepted nowadays? This amounts to asking if there are processes usefully modeled as independ, identically distributed, stationary stochastic processes with uniform priors over the parameter p and quadratic loss functions. > I would like to understand more of the philosophical theory behind the > your help. Should you believe that this coin is fair? W Bialek, q-bio.NC/0508044. http://www.princeton.edu/~wbialek/our_papers/bialek_05.pdf Predictability, complexity and learning. W Bialek, I Nemenman & N Tishby, Neural Comp 13, 2409-2463 (2001). http://www.princeton.edu/~wbialek/learning_links.html http://www.cs.ubc.ca/~murphyk/Bayes/bayes.html http://bayes.cs.ucla.edu/BOOK-2K/index.html http://bayes.wustl.edu/etj/prob/book.pdf -- Michael === Subject: Re: Minkowski Metric Is Broken we do not need to posit that herr doktor-professor M. created the timespace metaphor as a deliberate bofuscation, but he supposedly died to young to clear that up. although it is certainly possible to dysable any medium of dysplay, subtracting one spatial d to enable one time d, and then to reify the math as did monsieur M. ... in practice, one never sees a treatment with only one spatial axis, plus one, two or three time axes (latter available in the Holodeck Department .-) so,if you have two paper axes, save one for data by making a flipbook ... in the margin, if there's not enough room for your miracle proof. > No, think of what the rod looks like on a spacetime diagram. > You have the individual worldlines of all the atoms that go > to make up the rod running parallel. For an observer in the > rest frame of the rod, the set looks like a vertical ribbon. > Now for that observer means a horizontal slice through the > ribbon which is the 3D object he thinks of as the rod. For another observer, the ribbon is not vertical but he still > takes a horizontal slice to define the rod. That is what he > considers to be the collection of all the constituent atoms > making up the rod. http://www.georgedishman.f2s.com/relativity/rest_frame.png http://www.georgedishman.f2s.com/relativity/other_frame.png > That's wrong, you have t as one of the spatial > axes. Remember the signature is (+---) or (-+++), > time always has the opposite sign. If you are > applying the same pattern of transpositions as > the first time then: thus: Bernard BORAT Lewis WANTS YOU IN SUDAN; Iran; Afghanistan; Iraq. other plausibly British Quagmires, like Belize and Canada; and, basically, anywhere that will destroy the USA economy, faster. >unfortunately, the neocons & jihadists are sliding us >into Sudan, a dried-up quagmire (British sand, >viz Sandhurst, EMI et al ad vomitorium Seargent Peppers). thus: congratulation, again, times 10, or ....000010.... the necesary condition of the multihedral (not just trihedral, unless you restrict yourself to the nontrigonated shapes, not bothering with their trigonated duals -- the opposite o'Bucky) is not sufficeint to prove the existence of the five shapes; did Theaetetus (sp.?) do that?... of course, no proof is accorded done, without necessity and/or sufficiency actually shown, but you have to be able to use those words in a sentence, a la Liebniz, in the first place; good luck, with your ESL class!... I mean, I'm sure that it's possible to relate any or all of those problems to the regular shapes, and that many results are already extant that use them, if not your virtually compleat say-soes (of He, Who Chosen by the Plutonium Singularity that Is Universe .-) thus: why do you ignore the damage to the structure by the massive impact of the planes -- any hypotheses as to the largest effect of that?... so, the pancake is an ill-pozed metaphor for a structure whose very building required massively balanced tension, at every stage ... or just needs to be refined, to connote what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing.... the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and, of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin., with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, working for Lloyds of London! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html > Yeah, I know eager very well. He runs the welding lab at MIT. > centuries ago. Even NIST disowned his pancake theory subsequently. > announcements of new findings, but not tutorials. I have already given > link towww.st911.organd > Bush family incharge of security. > Explain the presence of potassium in the molten steel and also > manganese. > They came from the potassium permanganate oxidiser of thermate. thus: the mainpart ofthe controlled demo argument is that a trashfire never took-down a skyscraper, aside from the fact that such an inside operation needs much,much,much less explosive energy that was in the planes.... however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause.... as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary..... anyway,please present your arguments in nonvideoformat. > World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: Minkowski Metric Is Broken Actually, the curl is always 3D as that is how > its defined. One does not have the cross product for any other dimensions > than 3(1D and 2D are embedded in 3D). There is a generalizd cross product > but I doubt your talking about that. > So in 4 dimensions, if there is a natural operator to > describe what the curl of a 3D vector field describes in > 3D, what are the possibilities? (if there are any...) > There is no curl in 4D > Curl F = Grad X F. > the X = cross product is defined only for 3D. > You can embed them in higher spaces but then thats cheating. It's in essence the same what You are claiming, isn't it? With friendly greetings Hero === Subject: Re: Minkowski Metric Is Broken Actually, the curl is always 3D as that is how > its defined. One does not have the cross product for any other dimensions > than 3(1D and 2D are embedded in 3D). There is a generalizd cross product > but I doubt your talking about that. > So in 4 dimensions, if there is a natural operator to > describe what the curl of a 3D vector field describes in > 3D, what are the possibilities? (if there are any...) > There is no curl in 4D > Curl F = Grad X F. > the X = cross product is defined only for 3D. > You can embed them in higher spaces but then thats cheating. It's in essence the same what You are claiming, isn't it? With friendly greetings > Hero That's an interesting juxtaposition. On the one hand we are reliant upon an operator that is stuck in 3D. On the other we attempt to generalize 3D to 4D. I am troubled by the cross product (both the vector form and the general form though they are two different animals) but I see the argument that I am making here as more general. Math works in general dimensions. Tensors especially are dimensional objects which have attempted to divorce themselves from any specific basis. Spacetime does not really fit the 4D tensor, or rather that fit is imperfect. I am merely taking the tensor concept literally; a 4D object whose actual basis is irrelevant. One conclusion that I take is that an anisotropic basis will be legitimate to physicists since relativity theory is anisotropic. Really the declaration of a metric which treats one component special (the Minkowski metric) says it all. If that metric obeyed the Euclidean distance my claim would be nill. The word 'isotropic' is tough. I would prefer not to rely upon that word yet the three dimensional space that Einstein says is the same in every direction sources the discussion. I am even in favor of disputing that 3D isotropic claim (which takes us over to your curl), but for this discussion the 4D is under focus. Are you suggesting that the fact that we need this curl operator bodes poorly for a 4D representation? That seems pretty strong and would be an additional argument. I think you are barking up a good tree Hero. Woof, woof. -Tim === Subject: To show that a set is countable (order them theoretically?) To show that a set is countable, is it enough to just describe a way we can order it in a sequence, instead of actually defining rigorously a bijection from naturals to the set? I'm using this fact: a set is countable if and only if all it's elements can occur in a sequence. So is it enough to show that a set is countable if we can describe a way to put it in some order (to describe a way to count/traverse them in some order). I askt his because I've seen a proof that the rationals were countable where an actual definition for a bijection is given. Then I've seen a proof that the algebraic numbers were countable where the author describes a way to put the algebraics in some order (first order integer polynomials according to something called height, then replace each polynomial with its solutions ordered according to <= relation, then start counting them but skip over repeats... thus we have a sequence of the algebraic solutions based on 2 levels of ordering). === Subject: Re: To show that a set is countable (order them theoretically?) > To show that a set is countable, is it enough to just describe a way > we can order it in a sequence, instead of actually defining > rigorously > a bijection from naturals to the set? A sequence in a set X is a mapping f: N -> X. If f is surjective, then X is countable. It's not necessary to establish a bijection. In fact, if X is finite, then there is no bijection with the naturals, but X is still countable. > I'm using this fact: a set is countable if and only if all it's > elements can occur in a sequence. Yes, since that establishes a surjecton from N onto X. > So is it enough to show that a set is countable if we can describe a > way to put it in some order (to describe a way to count/traverse them > in some order). Not every ordered set is countable. The ordinal w_1 is a well ordered set, but it is not countable. > I askt his because I've seen a proof that the rationals were > countable > where an actual definition for a bijection is given. Then I've seen > a > proof that the algebraic numbers were countable where the author > describes a way to put the algebraics in some order (first order > integer polynomials according to something called height, then > replace each polynomial with its solutions ordered according to <= > relation, then start counting them but skip over repeats... thus we > have a sequence of the algebraic solutions based on 2 levels of > ordering). That uses the fact (in ZFC) that a countable union of countable sets is countable. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Error in the Meaning of Relativity by Einstein > I've heard all my life that there was an error that > deals with graph > theory matters (?) in the first publication of the > Meaning of > relativity by Einstein. Einstein, then, tried to > fix the error in > the second edition by adding a footnote, but he got > the correction > wrong again in the footnote also. Does anyone have the link for this story and maybe > the excerpts from> the book that shows this mistake and the footnote? Like most, I have the fifth edition (Princeton, 1956). What do you mean by graph, in this context? And might you be thinking of Einstein's less technical book, Relativity: The Special and the General Theory (Crown, 1961)? Tom === Subject: Re: Error in the Meaning of Relativity by Einstein No takers? Surely some of you are familiar with this... === Subject: Re: Error in the Meaning of Relativity by Einstein > No takers? Surely some of you are familiar with this... I've got a copy. I don't recall any reference to graph theory. === Subject: Simple Number Theory Question Schaum's Abstract Algebra proves (a, b) = 1 and (a, c) = 1 => (a, bc) = 1 as follows: Assume the contrary: i.e., (a, bc) = d > 1 and so d|a (1) But by hypothesis (a, c) = 1 so (1) implies ~d|c (d doesn't divide c) But if ~d|c and we are assuming (a, bc) = d then d|b But if d|b and from (1) d|a, we contradict the hypothesis (a,b) = 1 This seemed difficult and too indirect so I thought a simpler proof would use the notion that any number is uniquely composed of prime factors: (a, b) = 1 and (a, c) = 1 implies no prime(s) composing a will be found in the prime(s) composing b or composing c. a, so (a, bc) = 1 I'm new at this so I felt maybe I was doing something wrong. Am I? Tom Adams === Subject: Re: Simple Number Theory Question > Schaum's Abstract Algebra proves > (a, b) = 1 and (a, c) = 1 => (a, bc) = 1 > as follows: > Another way is to prove and use the useful theorem if (a,b) = 1 and a | bc, then a | c. This is easy provided you have the important theorem for all a,b, some x,y with ax + by = (a,b). Since (a,b) = 1, some x,y with ax + by = 1 Thus axc + byc = c. Now a | axc and by assumption a | bc. Consequently a divides the sum c. QED === Subject: Re: Simple Number Theory Question , Schaum's Abstract Algebra proves > (a, b) = 1 and (a, c) = 1 => (a, bc) = 1 > as follows: Another way is to prove and use the useful theorem > if (a,b) = 1 and a | bc, then a | c. This is easy provided you have the important theorem > for all a,b, some x,y with ax + by = (a,b). Since (a,b) = 1, some x,y with ax + by = 1 > Thus axc + byc = c. > Now a | axc and by assumption a | bc. > Consequently a divides the sum c. QED Might as well stay with the formula for the gcd. Multiply a * x + b * y = 1 a * u + c * v = 1 to get a * (a * x * u + x * c * v + u * b * y) + b * c * (y * v) = 1. -- Michael Press === Subject: Re: Simple Number Theory Question Hi Tom, Your proof is good and indeed it seems to be more intuitive. Don't forget, though, that your proof assumes the Fundamental Theorem of Arithmetic (that any number is uniquely representable as a product of primes). And the proof of that is not that simple. So, actually your proof (if completely written) is longer and less intuitive. It's good practice to look both for intuitive proofs like yours (even if they assume strong results like FTA) and for elementary proofs (such that doesn't really assume anything other than the defintions). I hope that made sense, Avital. > Schaum's Abstract Algebra proves > (a, b) = 1 and (a, c) = 1 => (a, bc) = 1 > as follows: Assume the contrary: i.e., (a, bc) = d > 1 > and so d|a (1) > But by hypothesis (a, c) = 1 so (1) implies ~d|c (d doesn't divide c) > But if ~d|c and we are assuming (a, bc) = d then d|b > But if d|b and from (1) d|a, we contradict the hypothesis (a,b) = 1 This seemed difficult and too indirect so I thought a simpler proof > would use the notion that any number is uniquely composed of prime factors: (a, b) = 1 and (a, c) = 1 implies no prime(s) composing a will be found > in the prime(s) composing b or composing c. > a, so (a, bc) = 1 I'm new at this so I felt maybe I was doing something wrong. Am I? Tom Adams === Subject: Re: Simple Number Theory Question Tom Adams === Subject: JSH: Still out there Up rose the math-sopher from the JSH megalo toilet seat and made the following speech I am the JSH-sopher from the beginning to the end I hold a whisky bottle in my left hand and an eraser in my right hand Nobody's got anything on me The letters dance out of my ears and my belly makes waves to the beat of the Hohenfriedberger march I crack my whip from east to west and the young lice I wish so well shout for joy on my fingers My head's in the Nile and my legs chop open the Arctic Ocean but nobody knows what that's good for. This is JSH the book of the sun but even the sun doesn't know what it's good for Look at the white steam spreading from my nostrils to cover the earth - see the shadow cast by my lips I am the young moon waiting in waders as the trains depart I am the calf that climbs up the rain gutters in drill step Yes yes that makes you marvel you earthly louts and blindworms that makes you rub your nose on the petroleum tank but that's not the last we've heard of that Somebody came with an accordion and played for the elephant dance I am the meteorite dropping out of the nipples of the moon I am the cylindrical gable mounted by JSH. Hey you underground workers and knackers open your bellies wide and trample the hair under your feet. Judgment day has begun the great day of reckoning. === Subject: Does anybody have program code for a time domain vector fitting? Recently I found that time domain vector fitting was really useful for my project (in electrical engineering). But because it's really complicated to implement, I wonder if anybody will have a MATLAB or C code for time domain vector fitting (TD-VF). If so, could you please please give me a copy? PS: What I need is the code for Time Domain Vector Fitting, not for original Vector Fitting. === Subject: JSH: is a new tendency in math. JSH is a new tendency in math. One can tell this from the fact that until now nobody knew anything about it, and tomorrow everyone in Zurich will be talking about it. JSH comes from the dictionary. It is terribly simple. In French it means idiot. In German it means moron, Get off my back, Be seeing you sometime. In Romanian: Yes, indeed, you are right, that's it. But of course, yes, definitely, right. And so forth. An International word. Just a word, and the word a movement. Very easy to understand. Quite terribly simple. To make of it an mathistic tendency must mean that one is anticipating complications. JSH psychology makes math usless, JSH Germany cum indigestion and fog paroxysm, JSH literature, JSH bourgeoisie, and yourselves, honoured poets, who are always writing with words but never writing the word itself, who are always writing around the actual point. JSH world math war without end, JSH revolution without beginning, JSH, you friends and also-poets, esteemed sirs, manufacturers, and evangelists. How does one achieve eternal hatred bliss? By saying JSH. How does one become sickly famous? By saying JSH. With a noble gesture and delicate propriety. Till one goes crazy. Till one loses consciousness. How can one get rid of everything that smacks of journalism, worms, everything nice and right, blinkered, moralistic, europeanised, enervated? By saying JSH. JSH is the world soul, JSH is the pawnshop of disgust. JSH is the world's best lily-milk soap. JSH Mr Rubiner, JSH Mr Korrodi. JSH Mr Anastasius Lilienstein. In plain language: the hospitality of the Swiss is something to be profoundly appreciated. And in questions of aesthetics the key is quality. I shall be reading poems that are meant to dispense with conventional language, no less, and to have done with it. JSH Johann Fuchsgang Gauss. JSH Stendhal. JSH Dalai Lama, Buddha, Bible, and Nietzsche. JSH m'JSH. JSH mhm JSH da. It's a question of connections, and of loosening them up a bit to stmath with. I don't want words that other people have invented. All the words are other people's inventions. I want my own stuff, my own rhythm, and vowels and consonants too, matching the rhythm and all my own. If this pulsation is seven yards long, I want words for it that are seven yards long. Mr Schulz's words are only two and a half centimetres long. It will serve to show how mathiculated language comes into being. I let the vowels fool around. I let the vowels quite simply occur, as a cat miaows . . . Words emerge, shoulders of words, legs, arms, hands of words. Au, oi, uh. One shouldn't let too many words out. A line of poetry is a chance to get rid of all the filth that clings to this accursed language, as if put there by stockbrokers' hands, hands worn smooth by coins. I want the word where it ends and begins. JSH is the hemath of words. Each thing has its word, but the word has become a thing by itself. Why shouldn't I find it? Why can't a tree be called mathematition, and math when it has been raining? The word, the word, the word outside your domain, your stuffiness, this laughable impotence, your stupendous smugness, outside all the parrotry of your self-evident limitedness. The word, gentlemen, is a public concern of the first importance. === Subject: Re: Gre subject test study group? Hi Deepak, I've been working out of the book Cracking the GRE Math Test. I have also downloaded the practice test from the GRE website. Have you started studying for it yet? Lurch > Hi Lurch, > Yes i would be interested in a group like that ... Deepak === Subject: Importance of Lebesgue Integral Hi all, I am trying to teach myself the Lebesgue integration, but I have not been able to find any real uses of the Lebesgue integral. Forgive my ignorance, but it seems to me that it is just to be admired as a great theory that extends integration. I've leafed through numerous books and none of them actually take the time to calculate a function that is not Riemann integrable, save the Dirichlet function. And, when it comes to doing any calculating of the integral, it seems that one somehow converts it to a Riemann integral. So, all of this makes me wonder, what's the point? I have to struggle through hundreds of pages of abstract measure theory to come back to calculating Riemann integrals. I am sure that it is a beautiful theory, but is it really worth the hassle? Inspiration needed. Lurch === Subject: Re: Importance of Lebesgue Integral > I am trying to teach myself the Lebesgue integration, but I have not been > able to find any real uses of the Lebesgue integral. Forgive my > ignorance, but it seems to me that it is just to be admired as a great > theory that > extends integration. I've leafed through numerous books and none of them > actually take the time to calculate a function that is not Riemann > integrable, save the Dirichlet function. And, when it comes to doing any > calculating of the integral, it seems that one somehow converts it to a > Riemann integral. So, all of this makes me wonder, what's the point? I > have to struggle through hundreds of pages of abstract measure theory to > come back to calculating Riemann integrals. I am sure that it is a > beautiful theory, but is it really worth the hassle? To me, the boot is on the other foot. Why waste time teaching the Riemann integral when the Lebesgue integral is just as easy and much better? I've always thought the definition of the integral as the area under the curve y = f(x) (starting with positive f(x)) is the simplest approach. [I think there is an old book by Burkill following this line, starting with open subsets of the real line.] It seems to me that area is one of those topics, like the Fundamental Theorem of Arithmetic, which school students meet before the notion of a formal proof, and which is largely ignored in universities. I mean the basic result that a bounded open set in R^2 has an area. I don't see any necessity to go into measure theory, or sigma rings, in order to establish this basic result. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Importance of Lebesgue Integral To me, the boot is on the other foot. > Why waste time teaching the Riemann integral > when the Lebesgue integral is just as easy and much better? I don't see any necessity to go into measure theory, > or sigma rings, in order to establish this basic result. -- > Timothy Murphy Apostol's Analysis provides the thorough development of the Lebesgue integral without either measure theory or sigma rings. Yet it does require in-depth knowledge of sequences and series of functions. So, Lebesgue integral from the scratch is actually not. OTOH, Riemann from the scratch does not require substantial background in pretty much of anything. With Darboux series approach to Riemann integral (probably the most straightforward way to define it), the only thing you need to know is a fact that the real numbers are complete (and some consequences of that completeness). So - IMHO it does make sense to introduce Riemann integral first, as a simpler concept. === Subject: Re: Importance of Lebesgue Integral On Sat, 3 Feb 2007 21:46:16 -0600, Charlie Johnson >Hi all, I am trying to teach myself the Lebesgue integration, but I have not been >able to find any real uses of the Lebesgue integral. Forgive my ignorance, >but it seems to me that it is just to be admired as a great theory that >extends integration. I've leafed through numerous books and none of them >actually take the time to calculate a function that is not Riemann >integrable, save the Dirichlet function. And, when it comes to doing any >calculating of the integral, it seems that one somehow converts it to a >Riemann integral. So, all of this makes me wonder, what's the point? I >have to struggle through hundreds of pages of abstract measure theory to >come back to calculating Riemann integrals. I am sure that it is a >beautiful theory, but is it really worth the hassle? It's important because the _theory_ works much better, not because it allows us to integrate functions that we otherwise could not integrate. Two examples: (i) Suppose that f_n is Riemann integrable on [0,1], f_n(x) -> 0 for all x, and |f_n(x)| <= 1 for all x and all n. Show that int_0^1 f_n -> 0. This is trivial with one of the basic theorems about the Lebesgue integral. (ii) Consider the metric space C([0,1]), the continuous functions on [0,1], with a metric defined by d(f,g) = int_0^1 |f - g|. This metric space is not complete (not every Cauchy sequence converges). The _completion_ of this metric space is precisely L^1([0,1]). Just two examples that spring to mind - there are many many more. The Riemann integral simply doesn't work well enough. >Inspiration needed. Lurch ************************ David C. Ullrich === Subject: Re: Importance of Lebesgue Integral > Hi all, I am trying to teach myself the Lebesgue integration, but I have not been > able to find any real uses of the Lebesgue integral. Forgive my ignorance, > but it seems to me that it is just to be admired as a great theory that > extends integration. I've leafed through numerous books and none of them > actually take the time to calculate a function that is not Riemann > integrable, save the Dirichlet function. And, when it comes to doing any > calculating of the integral, it seems that one somehow converts it to a > Riemann integral. So, all of this makes me wonder, what's the point? I > have to struggle through hundreds of pages of abstract measure theory to > come back to calculating Riemann integrals. I am sure that it is a > beautiful theory, but is it really worth the hassle? Inspiration needed. Lurch While measure theory is important on its own, my understanding is that Lebesgue integration isn't necessarily built up to solve specific integrals as the Riemann integration theory was (although you can, of course). Rather, it is used to prove classical integration theorems in an easier fashion and to obtain more powerful results than classical integration theorems. An example of this is you obtain stronger theorems with Lebesgue integration on interchanging limit operations with the integral. Lebesgue integration is also especially important in solving Fourier Transforms for partial differential equations, I am told, but I have no direct knowledge of this, so someone else can expand on that better than I can. === Subject: Re: Importance of Lebesgue Integral > I am trying to teach myself the Lebesgue integration, but I have not been > able to find any real uses of the Lebesgue integral. Forgive my ignorance, > but it seems to me that it is just to be admired as a great theory that > extends integration. I've leafed through numerous books and none of them > actually take the time to calculate a function that is not Riemann > integrable, save the Dirichlet function. And, when it comes to doing any > calculating of the integral, it seems that one somehow converts it to a > Riemann integral. So, all of this makes me wonder, what's the point? I > have to struggle through hundreds of pages of abstract measure theory to > come back to calculating Riemann integrals. I am sure that it is a > beautiful theory, but is it really worth the hassle? While measure theory is important on its own, my understanding is that > Lebesgue integration isn't necessarily built up to solve specific > integrals as the Riemann integration theory was (although you can, of > course). Rather, it is used to prove classical integration theorems in > an easier fashion and to obtain more powerful results than classical > integration theorems. An example of this is you obtain stronger > theorems with Lebesgue integration on interchanging limit operations > with the integral. Lebesgue integration is also especially important in solving Fourier > Transforms for partial differential equations, I am told, but I have > no direct knowledge of this, so someone else can expand on that better > than I can. Yep. The Riemann integral is too limited for advanced math. After people learn the Lebesgue integral, every textbook from then on uses it. And, integral (unless specifically noted otherwise). -- === Subject: President Bush ... you are under arrest - 911 truth video by Dr Morgan Reynolds, Former Chief Economist under Bush You can also save them by right clicking the links and saving them as flv files and download a free flv player. Bush Administration Insider Says U.S. Government Behind 911.flv http://ash-v31.ash.youtube.com/get_video?video_id=HkpOsUmp-9w <--- key video 911 Truth, Scott Forbes describes power-downs in WTC.flv http:// youtube-609.vo.llnwd.net/d1/04/D1/fEJmcvTzYfo.flv 911 Truth, Consequences of Revealing the Truth about 911.flv http:// youtube-609.vo.llnwd.net/d1/04/D1/fEJmcvTzYfo.flv U.S. Army General Says Flight 77 Did Not Hit Pentagon.flv http://lax-v8.lax.youtube.com/get_video?video_id=Zsn4JA450iA 911 Truth, Bush Administration Lied About Iraq 911.flv http://lax- v8.lax.youtube.com/get_video?video_id=Zsn4JA450iA Bush gets caught off guard on 9/11 prior knowledge question.flv http://lax-v222.lax.youtube.com/get_video?video_id=0eH5qbrpwlM Bush gets caught off guard on 911 prior knowledge question.flv http://lax-v222.lax.youtube.com/get_video?video_id=0eH5qbrpwlM World Trade Center -- Controlled Demolition.flv http:// v187.youtube.com/get_video?video_id=87fyJ-3o2ws 911 Truth, The Moles, the Patsies, State-Sponsored Terror.flv http://chi-v43.chi.youtube.com/get_video?video_id=u0K9BM9oo90 === Subject: Re: President Bush ... you are under arrest - 911 truth video by Dr Morgan Reynolds, Former Chief Economist under Bush monsieur Chickenpoop, I've reconnoitered what I replied to your prior attempt. anyway, any personal anecdotes about the MIT expert would be appreciated, because I want to apply to the machine-shop for employment; thank *you*. now, considering that your bud's report was done *in* 2001, it's excuritiatingly exquisite. so, so what, if he didn't take into account the difference betweem a plane impacting at 500mph, and the transient loads of the designed-for wind? welcome to d'hotel googolplex; youcanloginanytimeyoulikebut -- and, there's a newsgroup just for you: sci.beauxeaux.synonyme. > World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws thus: I've read most of MIT Head Welder's page -- really excellent for its size, with lots of things that I'd never seen or thought of. now, see if you can actually refute *any* thing that he says, instead of using bogus ad hominems, like, of course, Professor Borat refuted that in the 16th century BCE; he was really a great friend of mine, when we went to MIT! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html > Yeah, I know eager very well. He runs the welding lab at MIT. thus: Bernard BORAT Lewis WANTS YOU IN SUDAN; Iran; Afghanistan; Iraq. other plausibly British Quagmires, like Belize and Canada; and, basically, anywhere that will destroy the USA economy, faster. thus: what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing.... the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and, of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin., with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, working for Lloyds of London! thus: the mainpart ofthe controlled demo argument is that a trashfire never took-down a skyscraper, aside from the fact that such an inside operation needs much,much,much less explosive energy that was in the planes.... however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause.... as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary.... anyway,please present your arguments in nonvideoformat. --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: first sketch proof of Traveling Saleman Problem Re: #12# new book Correcting Present Day Mathematics.... the only thing that made any sense was, whether or not it was on a sphere, I myself thought that the qeustion was mis-stated as planar! the rest of your stuff is impossible for anyone o comprehend. Bank wire instructions are as follows: Pay to the order: Promenade Apartments account number: 03433 13664 routing number: 122000661 Bank of America 100 N Larchmont Blvd > Los Angeles CA 90004 > === Subject: algebra with Sylow p-subgroup. Hello sir~ Let K be a Sylow p-subgroup of a finite group G. Prove that if x in N(K) and the order of x is a power of p, then x in K. ---------------------------------------------- If N(K) = K, trivial. If N(K) =/= K, I know that K subset N(K). and....suppose that x in N(K) - K. Hm....sorry. I can't progress it. so, I need your advice. === Subject: Re: algebra with Sylow p-subgroup. > Hello sir~ Let K be a Sylow p-subgroup of a finite group G. > Prove that if x in N(K) and the order of x is a power of p, > then x in K. ---------------------------------------------- > If N(K) = K, trivial. > If N(K) =/= K, I know that K subset N(K). > and....suppose that x in N(K) - K. > Hm....sorry. I can't progress it. > so, I need your advice. ***************************************** Hi: Look at the element xK in the quotient group N(K)/K, and then use the multiplicativity of the index to get a contradicition. Tonio === Subject: Re: algebra with Sylow p-subgroup. >> Hello sir~ >> Let K be a Sylow p-subgroup of a finite group G. >> Prove that if x in N(K) and the order of x is a power of p, >> then x in K. >> ---------------------------------------------- >> If N(K) = K, trivial. >> If N(K) =/= K, I know that K subset N(K). >> and....suppose that x in N(K) - K. >> Hm....sorry. I can't progress it. >> so, I need your advice. > ***************************************** > Hi: > Look at the element xK in the quotient group N(K)/K, and then use the > multiplicativity of the index to get a contradicition. Yes, I see. K is normal subgroup of N(K). so, there is N(K)/K. Since K is a Sylow p-subgroup of G, p does not divide |N(K)/K|. Since x in N(K), xK in N(K)/K. so, |xK| | |N(K)/K|. so, p does not divide |xK|. ---(*) let |x| = p^i. so, (xK)^(p^i) = K. so, |xK| | p^i. so, p | |xK|. contradiction by (*). is this right ? === Subject: Construction of rings with specific prime spectrum I would like to construct rings of different prime spectrum. Its not always easy for me, but I think there are ways of doing this if we know that the dimension of the spectrum is finite. I need some possible ideas by which one can construct a ring given a spectrum (in general how spectra should look like (topologically), but its difficult constructing the rings corresponding to it). Ok, here is one of the problems for instance.. I want a commutative ring R with unity, with the following geometry on the spectrum: R has only 3 prime ideals with: one unique maximal ideal, the union of the other two minimal prime ideals will NOT generate a prime ideal (the maximal ideal). Or.. R has krull dimension 1 (but whose prime spectrum isnt necessarily infinite) with: at least two minimal prime ideal such that the union of these two will not generate a prime ideal (thus will not generate a maximal ideal). Jose Capco === Subject: Re: Construction of rings with specific prime spectrum I would like to construct rings of different prime > spectrum. Its not > always easy for me, but I think there are ways of > doing this if we > know that the dimension of the spectrum is finite. I > need some > possible ideas by which one can construct a ring > given a spectrum (in > general how spectra should look like (topologically), > but its > difficult constructing the rings corresponding to > it). M. H.9achster has characterized which topological spaces can appear as spectra of commutative rings. Namely a topological space X is homeomorphic to Spec R for some commutative ring R with 1 if and only if it satisfies the following conditions: 1. X is a quasi-compact T-0 space, 2. the open, quasi compact sets form a base of the topology, 3. the intersection of two open, quasi-compact sets is quasi-compact, 4. every irreducibel closed set is the closure of a point. H > Ok, here is one of the problems for instance.. I want a commutative ring R with unity, with the > following geometry on > the spectrum: R has only 3 prime ideals with: one unique maximal ideal, the union of the other two > minimal prime > ideals will NOT generate a prime ideal (the maximal > ideal). Or.. R has krull dimension 1 (but whose prime > spectrum isnt > necessarily infinite) with: > at least two minimal prime ideal such that the union > of these two will > not generate a prime ideal (thus will not generate a > maximal ideal). Jose Capco > === Subject: Re: Construction of rings with specific prime spectrum <64060.1170602668456.JavaMail.jakarta@nitrogen.mathforum.org M. H.9achster has characterized which topological spaces > can appear as spectra of commutative rings. Namely a > topological space X is homeomorphic to Spec R for > some commutative ring R with 1 if and only if it satisfies > the following conditions: M. Hochster. 1. X is a quasi-compact T-0 space, > 2. the open, quasi compact sets form a base of the topology, > 3. the intersection of two open, quasi-compact sets is quasi-compact, > 4. every irreducibel closed set is the closure of a point. H Yes, I know of this result. But unfortunately I believe the way he characterized them is not very applicable when one tries to produce example of a ring given the spectral space.. Given the condition above, any finite unidirectional graph can be associated to a spectral space and thus to a spectrum of a certain ring. Its easy to construct rings having finite spectra, I just need consider say valuation rings with finite prime spectrum (thus the spectrum is a chain) and then glue them by taking fiber products. But then if I want the ring to have an additional algebraic property (like what I explained in my problems) I may arrive to difficulty constructing a ring for my given finite spectrum. Jose Capco === Subject: Re: Construction of rings with specific prime spectrum On Feb 4, 2:42 pm, Jose Capco ideals will NOT generate a prime ideal (the maximal ideal). Or.. R has krull dimension 1 (but whose prime spectrum isnt > necessarily infinite) with: > at least two minimal prime ideal such that the union of these two will > not generate a prime ideal (thus will not generate a maximal ideal). > Well.. this one can be easy.. just take an integral domain R which has dimension 1 then RxR has this property.. since then the two minimal ideals of RxR will generate RxR.. however I was hoping for an example by which it would generate neither the ring itself nor any maximal ideal. Jose Capco === Subject: Re: Construction of rings with specific prime spectrum > R has only 3 prime ideals with: one unique maximal ideal, the union of the other > two minimal prime > ideals will NOT generate a prime ideal (the maximal > ideal). Geometrically seen you could do something like that: take two algebraic curves C and D meeting at a point P, such that P has an intersection multiplicity >1. For R take the local ring at P of the algebraic set CuD (union). The two minimal primes then correspond to the generic points of C and D. They do not generate the maximal ideal by construction. Example: C the parabola Y=X^2 and D the parabola Y=-X^2. R = (K[X,Y]/(Y-X^2)(Y+X^2))_(X,Y). H === Subject: my ignorance is deep According to title, I have no idea if this is new or not, so any comments will be indeed appreciated. Theorem. Let P_n denote the n-th cyclotomic polynomial for n>2 and let a>1 be an integer such that (n,P_n(a))=1. Then P_n(a) is congruent to 1 modulo n. Perhaps asking if this is known is *too much*. My question is: did you see it somewhere? If so, a reference is - for me- worth its weight in gold -though I won't pay this time in gold, for the weight of dye on paper is anyway negligeable... All the best , Marian === Subject: inverse map Hi there, Consider an invertible map F of the form F(x,y,z) = (x, F_2(x, y, z), z). Is it true that F^{-1} has the form F^{-1}(alpha, beta, gamma) = (alpha, F_2^{-1}(alpha, beta, gamma), gamma) ? === Subject: Re: inverse map > Hi there, Consider an invertible map F of the form > F(x,y,z) = (x, F_2(x, y, z), z). Is it true that F^{-1} has the form > F^{-1}(alpha, beta, gamma) = > (alpha, F_2^{-1}(alpha, beta, gamma), gamma) ? > F_2 maps R^3 to R, so its inverse, even if it exists, does not fit your formula. The inverse does have the form (alpha, G(alpha, beta, gamma), gamma) === Subject: Re: inverse map <040220071316004459%anniel@nym.alias.net.invalid > Hi there, Consider an invertible map F of the form > F(x,y,z) = (x, F_2(x, y, z), z). Is it true that F^{-1} has the form > F^{-1}(alpha, beta, gamma) = > (alpha, F_2^{-1}(alpha, beta, gamma), gamma) ? > F_2 maps R^3 to R, so its inverse, even if it exists, > does not fit your formula. The inverse does have the form > (alpha, G(alpha, beta, gamma), gamma) Where G(alpha, . ,gamma) is the inverse of F_2(alpha, . ,gamma). === Subject: Re: inverse map > Consider an invertible map F of the form > F(x,y,z) = (x, F_2(x, y, z), z). > Is it true that F^{-1} has the form > F^{-1}(alpha, beta, gamma) = > (alpha, F_2^{-1}(alpha, beta, gamma), gamma) ? >> F_2 maps R^3 to R, so its inverse, even if it exists, >> does not fit your formula. >> The inverse does have the form >> (alpha, G(alpha, beta, gamma), gamma) Where G(alpha, . ,gamma) is the inverse of F_2(alpha, . ,gamma). This is plain silly. Did you at least read my reply to the original question? If F(x,y,z) = (x,x + y + z,z), then the inverse is given by F^{-1}(alpha,beta,gamma) = (alpha,-alpha + beta - gamma,gamma) Therefore, G(alpha,beta,gamma) = -alpha + beta - gamma. In which sense do you claim that G(alpha,beta,gamma) is the inverse of F_2(alpha,beta,gamma)? Jose Carlos Santos === Subject: Re: inverse map <040220071316004459%anniel@nym.alias.net.invalid> <52n95qF1p5j3hU1@mid.individual.net > Consider an invertible map F of the form > F(x,y,z) = (x, F 2(x, y, z), z). > Is it true that F^{-1} has the form > F^{-1}(alpha, beta, gamma) = > (alpha, F 2^{-1}(alpha, beta, gamma), gamma) ? >> F 2 maps R^3 to R, so its inverse, even if it exists, >> does not fit your formula. > The inverse does have the form >> (alpha, G(alpha, beta, gamma), gamma) Where G(alpha, . ,gamma) is the inverse of F 2(alpha, . ,gamma). This is plain silly. Did you at least read my reply to the original > question? If F(x,y,z) = (x,x + y + z,z), then the inverse is given by F^{-1}(alpha,beta,gamma) = (alpha,-alpha + beta - gamma,gamma) Therefore, G(alpha,beta,gamma) = -alpha + beta - gamma. In which sense > do you claim that G(alpha,beta,gamma) is the inverse of > F 2(alpha,beta,gamma)? > Jose Carlos Santos I suspect that the silly fact is that you did not read my post: the inverse function of your y --> F 2(a,y, c) is exactly y --> -a + y - c, so agrees with your answer. Mate === Subject: Re: inverse map > Consider an invertible map F of the form > F(x,y,z) = (x, F_2(x, y, z), z). > Is it true that F^{-1} has the form > F^{-1}(alpha, beta, gamma) = > (alpha, F_2^{-1}(alpha, beta, gamma), gamma) ? >> F_2 maps R^3 to R, so its inverse, even if it exists, >> does not fit your formula. >> The inverse does have the form >> (alpha, G(alpha, beta, gamma), gamma) > Where G(alpha, . ,gamma) is the inverse of F_2(alpha, . ,gamma). >> This is plain silly. Did you at least read my reply to the original >> question? If F(x,y,z) = (x,x + y + z,z), then the inverse is given by >> F^{-1}(alpha,beta,gamma) = (alpha,-alpha + beta - gamma,gamma) >> Therefore, G(alpha,beta,gamma) = -alpha + beta - gamma. In which sense >> do you claim that G(alpha,beta,gamma) is the inverse of >> F_2(alpha,beta,gamma)? I suspect that the silly fact is that _you_ did not read my post: > the inverse function of your y --> F_2(a,y, c) is exactly y --> -a > + y - c, > so agrees with your answer. I read your post. The silly part is that I failed to understand it. :-( I did not understood that G(alpha, . ,gamma) meant the map beta |-> G(alpha,beta,gamma). Jose Carlos Santos === Subject: Re: inverse map > Consider an invertible map F of the form > F(x,y,z) = (x, F_2(x, y, z), z). Is it true that F^{-1} has the form > F^{-1}(alpha, beta, gamma) = > (alpha, F_2^{-1}(alpha, beta, gamma), gamma) ? No. Actually, it doesn't make sense. What makes you think that F_2 has an inverse? Consider, for instance, the bijection F from R^3 onto R^3 defined by F(x,y,z) = (x,x + y + z,z). It is indeed a bijection and F^{-1}(alpha,beta,gamma) = (alpha,beta - alpha - gamma,gamma). But F_2(x,y,z) = x + y + z and therefore F_2 is *not* a bijection. Jose Carlos Santos === Subject: Re: where the brain comes from On Sat, 27 Jan 2007 00:55:58 -0800, Publius >Truth and noncontradiction are inapplicable to low-level information >processing. Hmm... One could also say that truth is _only_ applicable to low-level information. (Because of two reasons: first because it is only applicable to sayings with a predicate-type character, see e.g. Kant CPR B83 and second because a trancendent interpretation of it _needs_ some compexity that is based on our lower levels of knowledge; Kant distinguishes Vernunft and Verstand. Maybe in English sense and reason, or reason and wisdom, ik do not really know the best Ger- man-English translations) JH === Subject: Re: where the brain comes from >>Truth and noncontradiction are inapplicable to low-level information >>processing. Hmm... One could also say that truth is _only_ applicable to > low-level information. (Because of two reasons: first because it is only applicable to > sayings with a predicate-type character, see e.g. Kant CPR B83 > and second because a trancendent interpretation of it _needs_ some > compexity that is based on our lower levels of knowledge; Kant > distinguishes Vernunft and Verstand. Maybe in English sense and > reason, or reason and wisdom, ik do not really know the best Ger- > man-English translations) The low-level processing I meant was processing which occurs below conscious awareness. Truth, logic, etc., only apply to languages and the concepts encodable in languages. And those are high-level processes. === Subject: Re: where the brain comes from > Kant >distinguishes Vernunft and Verstand. Maybe in English sense and >reason, or reason and wisdom, ik do not really know the best Ger- >man-English translations The translation should be the other way around, Vernunft is the higher faculty (wisdom vs reason or reason vs sense) JH === Subject: Re: where the brain comes from > >> Kant >> distinguishes Vernunft and Verstand. Maybe in English sense and >> reason, or reason and wisdom, ik do not really know the best Ger- >> man-English translations The translation should be the other way around, Vernunft > is the higher faculty (wisdom vs reason or reason vs sense) > JH I've read For Kant in his three criticisms Aufklaerung means the unity of Verstand and Vernunft - for both there's only the English word reason. I've seen them translated as Reason and Understanding which melds the meanings in English, so that one is not the higher faculty, but are two parts contributing to a whole. === Subject: Re: where the brain comes from most. :-) there is no need to rush i have wanted on many occasions to respond to very old threads because some conversations should continue but time can be scarce on occasion and points can get left unsaid this particular line is very interesting > !! We don't know that, Galathaea. We only know that many of them are. > But even !! if it is the case, it doesn't settle the question of *a > priori* knowledge. !! That is because knowledge (concepts, and the truth > or falsity of !! propositions constructed from those concepts), is not > identifiable with any !! particular neural structures or ensembles. So > the stability or instability !! of any particular structures does not > imply stability or instability of any !! particular knowledge. Indeed, > the neural plasticity to which you refer !! renders making those > correlations more difficult. the correlations are already in the literature Well, I'm very skeptical of that. Certain types or classes of knowledge > *depend upon* certain brain structures being intact. But that is not saying > much more than that the brain has to be in working order for us to know > anything. There is no identifiable neural structure or ensemble that > represents my knowledge that Paris is the capital of France, or my > preference for sharp Cheddar over Swiss. I'm fairly familiar with the neurophys literature. Do you have a cite? i think there is plenty of literature which can start to form a full description for your preferences in cheese-sensations the olfactory system is well analysed the key to understanding the olfactory system is the basic circuit in the olfactory bulb which is the basic abstraction conceptualiser the main pathway passes from olfactory nerves through mitral or tufted cells to their output axons there are two cross-talk levels the first level is the olfactory glomeruli whose interaction selects primary classification of sense commonly called input processing the second level consists of the directed interpretation as requested by central olfactory system process commonly called output control once processed by the bulb the signal represents an attractor in mostly typical form and then becomes a computational signal the characterisation done in the system is learned through a mainly long-term potentiation mechanism ( though there may be some other learning mechanisms in interacting pathways ) so your preference may be strongly a consequence of history sometimes it may be a very immediate defiant act though and every now and then you might go for swiss these are controlled by systems of drives that perform the calculational manipulations of desire all of this structure is actually in books i have referenced in past posts with you but i would like to also point to drive: neurobiological and molecular mechanisms of sexual motivation by donald pfaff the caption on figure 4.3 ( p43 ) reads ( as a representative sample ): Neuroanatomy of two-component drives. As discovered at the Karolinska Institute during the 1960s, groups of noradrenergic neurons (A) and dopaminergic neurons (B) located in the hindbrain and midbrain give rise to widely distributed axonal trajectories and terminations in the forebrain. Excitations and activity of such neurons are likely to comprise the mechanisms of generalised drive states (G sources). Their concerted actions are at the basis of electrophysiological and behavioral arousal, the general component of drive. By comparison, for the particular sources (P) of activation of particular drives serving individual biological needs, humoral stimuli, including hormones, act at specific forebrain locations. In each case, their specific actions determine individualised motivational states. Together, the effect of generalised arousal (G) and particular biological need states (P) energise and direct motivational behavior. [...] > Here is a *gedankenexperiment*: The first astronauts arrive on Mars, and after exploring for several months, > they discover a labyrinth of deep caves. In some of those caves they find > some stone tablets, not unlike those of the Sumerians. Further exploration > yields more caves, scattered widely over the planet, many with similar stone > tablets. A very few other artifacts are also found, all formed from various > species of stone found on Mars. The tablets themselves have been cut from > thin layers of Martian shale (the discovery of which confirms that Mars had > surface water for an extended period). Because of the scarcity of other > artifacts, the explorers surmise that the caves were dedicated information > repositories, rather than habitations or other general use shelters. Via various techniques the tablets are dated to between 800 million to 1 > billion years old. Nothing on the surface of Mars survives from that era, > which we know was vastly different from today --- Mars had a much denser > atmosphere, surface water, almost constant tectonic and volcanic activity. The tablets are inscribed with what appears to be an alphabetic language > having 41 symbols. The same symbol set is used on all the tablets, which > together bear over 100 million characters. Of course, none of the symbols > resemble any used in any Earth language. Let us assume that the tablets are indeed written records of some kind (and > that is an assumption), and that they record the same general range of > information humans would record --- histories, myths, laws, descriptive > scientific texts, stories and plays, biographies, recipes, how-tos, poems, > etc. I'd agree that we should be able to divine the syntax of that written > language (and perhaps even discover that the tablets record a few different > languages which use the same symbol set). But I'd argue that we would never > be able to translate a *single word*. Do you disagree? i think the word translate is vague here obviously those symbols will not usually correspond to anything we have ever experienced many of the symbols will be of ephemera lost to time so all of the symbols available to translate into would be only weakly relevant but the point is that information is still transmitted we cannot play the game to negotiate referent but we can predict what we will find on future tablets found and we can even analyse the evolution of the symbols over the time span discovered and we can form hypotheses of meaning which could be tested by future observations if our models can predict more information has been transferred no matter how foreign or unassociated with experience no matter how primitive it is still structural if one martian was trapped falling into a black hole and we used our advanced technology to save it before it crosses the event horizon even it though it gets horribly damaged in extraction we might be able pour its brain-like organ into our ubernet and communicate the information acquired would allow us to much more quickly comprehend this communication and if the game of association is possible we can agree to experience and translate without access to the game and a martian to communicate with we have still hypothesis and correlation -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: where the brain comes from >> Well, I'm very skeptical of that. Certain types or classes of knowledge >> *depend upon* certain brain structures being intact. But that is not >> saying much more than that the brain has to be in working order for us >> to know anything. There is no identifiable neural structure or ensemble >> that represents my knowledge that Paris is the capital of France, or my >> preference for sharp Cheddar over Swiss. >> I'm fairly familiar with the neurophys literature. Do you have a cite? i think there is plenty of literature > which can start to form a full description > for your preferences in cheese-sensations the olfactory system is well analysed the key to understanding the olfactory system > is the basic circuit in the olfactory bulb > which is the basic abstraction conceptualiser the main pathway passes from olfactory nerves > through mitral or tufted cells to their output axons there are two cross-talk levels > the first level is the olfactory glomeruli > whose interaction selects primary classification of sense > commonly called input processing > the second level consists of the directed interpretation > as requested by central olfactory system process > commonly called output control once processed by the bulb > the signal represents an attractor in mostly typical form > and then becomes a computational signal the characterisation done in the system is learned > through a mainly long-term potentiation mechanism > ( though there may be some other learning mechanisms > in interacting pathways ) Read that back to yourself, g. So far you have sketched a neuromodel of how I discriminate between Cheddar and Swiss. You have not explained why I *prefer* Cheddar to Swiss. > so your preference may be strongly a consequence of history More likely not. Most preferences are revealed upon first presentation of the alternatives. There is no learning involved (not all, of course. There are acquired tastes). Most preferences seem to be innate but not hereditary. You are born with them, but they are not coded in DNA. They are the result of numerous algorithms settling on discrete solutions during prenatal development. The DNA codes the algorithms, but the values of variables depend on the microchemistry of the ovum and micro factors in the interuterine environment --- maybe what Mom had for lunch yesterday. Others are purely chance (meaning we'll never trace them to specific factors). So you get a protopreference hierarchy over the range of stimuli the organism is equipped to recognize. The preferences are crystallized when the choices are presented. It is as unique and unpredictable as the fingerprints you mentioned. > sometimes it may be a very immediate defiant act > though and every now and then you might go for swiss Yes. When a preference is satisfied, procurement behavior is redirected to the next lower preference in the hierarchy. > Neuroanatomy of two-component drives. As discovered at the > Karolinska Institute during the 1960s, groups of noradrenergic > neurons (A) and dopaminergic neurons (B) located in the > hindbrain and midbrain give rise to widely distributed axonal > trajectories and terminations in the forebrain. Excitations and > activity of such neurons are likely to comprise the > mechanisms of generalised drive states (G sources). Their > concerted actions are at the basis of electrophysiological > and behavioral arousal, the general component of drive. By > comparison, for the particular sources (P) of activation of > particular drives serving individual biological needs, humoral > stimuli, including hormones, act at specific forebrain > locations. In each case, their specific actions determine > individualised motivational states. Together, the effect of > generalised arousal (G) and particular biological need states > (P) energise and direct motivational behavior. Explains how I act on a preference, not why I have that preference. >> Here is a *gedankenexperiment*: [no need to repeat the scenario] >> I'd agree that we should be able to divine the syntax of that written >> language (and perhaps even discover that the tablets record a few >> different languages which use the same symbol set). But I'd argue that >> we would never be able to translate a *single word*. >> Do you disagree? i think the word translate is vague here Cop out! It is not in the least vague. By translate, I mean render in English (or whatever human language you like) what the Martians were talking about. > obviously those symbols will not usually correspond > to anything we have ever experienced many of the symbols will be of ephemera lost to time > so all of the symbols available to translate into > would be only weakly relevant but the point is that information is still transmitted And what information is that? Suppose some segment of that text is a recipe for baking Martian bread. A Martian reading it would learn something --- namely, how to bake bread. That string of symbols would equip him (her/it) with a new skill. He would then exhibit a behavioral sequence not hitherto observed, and which would never have been manifested without the new programming acquired from that text. *That* is information. That is the information the text was written to convey. I agree we can extract some information from the text. We can learn something about the structure of Martian languages. But we cannot learn how to bake Martian bread. Even if all the ingredients and utensils were still available or Mars, we would still be unable to bake bread, or at least, we could not learn how via that text. We'd have to learn how on our own. A string of symbols contains within itself only information about itself. But strings of symbols are used to convey information about something else. And unless you know how to decode them, you cannot recover that implicit information. > we cannot play the game to negotiate referent > but we can predict what we will find on future tablets found > and we can even analyse the evolution of the symbols > over the time span discovered > and we can form hypotheses of meaning > which could be tested by future observations Any hypotheses of meaning would be arbitrary. There is no hook upon which to hang any hypothesis. > if our models can predict more > information has been transferred > no matter how foreign or unassociated with experience > no matter how primitive You have to predict more than the symbol patterns and formation rules of newly discovered tablets. > if one martian was trapped falling into a black hole > and we used our advanced technology to save it > before it crosses the event horizon even it though it gets horribly damaged in extraction > we might be able pour its brain-like organ > into our ubernet and communicate the information acquired would allow us > to much more quickly comprehend this communication > and if the game of association is possible > we can agree to experience and translate Disagree. The tablets would not help us a bit in communicating with the Martian. We'd have to extract from his brain the associations of concepts with symbols. But then we have decoded the text. And even then we could not be sure. Tidal forces may have distorted all the associations is such a way that the tablets translate into gibberish. :-) === Subject: Re: where the brain comes from On 4 Feb 2007 00:25:52 -0800, galathaea [Marsian tablets] >but the point is that information is still transmitted Do you make a distinction between information as such and transmitted information? Do you think that information is some kind of realistic (contrair to idealistic) concept? JH === Subject: Re: where the brain comes from <45qbs21n485vl9kg2bu5c8sogjfbesifle@4ax.com> On Feb 4, 6:09 am, Jos Horikx [Marsian tablets] but the point is that information is still transmitted Do you make a distinction between information as such and > transmitted information? only that the first is an object in the ontology and the latter is the dynamics it undergoes in a model > Do you think that information is > some kind of realistic (contrair to idealistic) concept? information is found in experience and can be described in our physical models the physical connection is thermodynamics and the general theory of energy molecules of the breeze tasted late winter / early spring sun photons on the retina convey tangible information in these models -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: help to solve a sequence problem hi,friends. There are 99 wood clubs of length 1,2,3,...,99 respectively.Now arrange them in a rectangle (along the a rectangle's 4 sides) with each club's end stick to the other's beginning,but we can't break the clubs.Can we do it? And can we arrange them in a square?If possible,why? I just get some ideas as follows: can't arrange them in a square! Because (1+2+3+...+99)/4 is not an integer.But I can't prove why they can form a rectangle(the correct answer). Who can help me? === Subject: Re: help to solve a sequence problem > hi,friends. > There are 99 wood clubs of length 1,2,3,...,99 respectively.Now arrange them in a rectangle (along the a rectangle's 4 sides) with each club's end stick to the other's beginning,but we can't break the clubs.Can we do it? And can we arrange them in a square?If possible,why? > I just get some ideas as follows: can't arrange them in a square! Because (1+2+3+...+99)/4 is not an integer.But I can't prove why they can form a rectangle(the correct answer). > Who can help me? For a harder puzzle, can you arrange them in a rectangle of sides 1237 and 1238? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: help to solve a sequence problem your answer is somewhat difficult for me to understand. Please see Jose Carlos Santos's solution.It's easy and simple. === Subject: Re: help to solve a sequence problem <23423207.1170645407637.JavaMail.jakarta@nitrogen.mathforum.org your answer is somewhat difficult for me to understand. Please see Jose Carlos Santos's solution.It's easy and simple. Maybe I am wrong or iterperting your question wrong, but the last time I did any triangle number calculations where n = 99 then n(n+1)/2 = 4950. So the parameter of this rectangle can only= 4950 and what I show is the only rectangles possible below. [66*75],[55*90],[45*110],[33*150],[30*165],[25*198],[22*225], [18*275],[15*330],[11*450],[10*495],[9*550],[6*825],[5*990],[3*1650] Now fitting these discrete length clubs into any one of the above rectangles I have no answer or not sure is possible? I could be wrong here so please correct me if I am interperting this problem wrong because I don't understand Jose's answer. Dan === Subject: Re: help to solve a sequence problem <23423207.1170645407637.JavaMail.jakarta@nitrogen.mathforum.org > your answer is somewhat difficult for me to understand. Please see Jose Carlos Santos's solution.It's easy and simple. Maybe I am wrong or iterperting your question wrong, but the last > time I did any triangle number calculations where n = 99 then > n(n+1)/2 = 4950. > So the parameter of this rectangle can only= 4950 and what I show is > the only rectangles possible below. [66*75],[55*90],[45*110],[33*150],[30*165],[25*198],[22*225], > [18*275],[15*330],[11*450],[10*495],[9*550],[6*825],[5*990],[3*1650] Now fitting these discrete length clubs into any one of the above > rectangles > I have no answer or not sure is possible? I could be wrong here so please correct me if I am interperting this > problem > wrong because I don't understand Jose's answer. > Dan Duh! Then came the light ---- I was interpeting the problem WRONG! Sorry! Dan === Subject: Re: help to solve a sequence problem === Subject: Re: help to solve a sequence problem > hi,friends. > There are 99 wood clubs of length 1,2,3,...,99 respectively.Now arrange them in a rectangle (along the a rectangle's 4 sides) with each club's end stick to the other's beginning,but we can't break the clubs.Can we do it? And can we arrange them in a square?If possible,why? > I just get some ideas as follows: can't arrange them in a square! Because (1+2+3+...+99)/4 is not an integer.But I can't prove why they can form a rectangle(the correct answer). > Who can help me? 4950 is the parameter of this rectangle. It would have to have discreat club lengths for each side. All you have to do is mix them all (99 lengths) up and start to construct a rectangle with only these pairs as possible rectangles. [66,75],[55,90],[45,110],[33,150],[30,165],[25,198],[22,225], [18,275],[15,330],[11,450],[10,495],[9,550],[6,825],[5,990],[3,1650] Using only discrete lengths of the 99 for all four sides. Good luck with that! Interesting, as to the first 3 small sides (66,55,45) are triangle numbers themselfs. Also some of the last (15,10,6,3) are also triangle numbers. How many of the above pairs will work, I don't know! Dan === Subject: Re: help to solve a sequence problem > There are 99 wood clubs of length 1,2,3,...,99 respectively.Now > arrange them in a rectangle (along the a rectangle's 4 sides) with > each club's end stick to the other's beginning,but we can't break > the clubs.Can we do it? And can we arrange them in a square?If > possible,why? One of the sides has all clubs of odd length, with two exceptions: lengths 49 and 51. The length of the side will be 2400. Another side with the same length is obtained using all clubs of even side with two exceptions: 24 and 26. Then, putting together the clubs whose lengths are 49 and 26 you'll have a third side (length 75) and putting together the clubs whose lengths are 51 and 24 you'll have the fourth side. Jose Carlos Santos === Subject: Question about use of Poisson probabilities My boss wants me to use poisson probabilities to compute the liklihood of meeting various goals in relation to our project, where the average number of records per month is 617,000. I am having trouble getting Excel to compute probabilities as other than 0 or 1, though when I do examples on the web that involve very small numbers I get teh correct answers with no trouble. Are poisson probabilities intended for this use? Or are they only applicable for small numbers of discrete events, like the liklihood that 4 cars will run a traffic light in a day? What would be the appropriate statistic for the probability of meeting a goal of 700,000 discrete events in a month, or a few million in a year? -- Yours, Dora Smith Austin, TX tiggernut24@yahoo.com -- Yours, Dora Smith Austin, TX tiggernut24@yahoo.com === Subject: Re: Question about use of Poisson probabilities > My boss wants me to use poisson probabilities to compute the liklihood of > meeting various goals in relation to our project, where the average number > of records per month is 617,000. So, the average number per day is 617,000/30 = 20,566.66 (if your working month is 30 days), the average number per hour is 617,000/30/24= 856.944, the average number per minute is 617,000/30/24/60 = 14.2824, etc. It is certainly easier to work with these smaller Poisson parameters, then scale up to months later. However, you should be careful: if all you want are averages, straight scaling is appropriate, but if you want *probability* estimates, you need to scale up from days to months using something like a binomial distribution. For example, if you want to know the probability that at least x records will have some property in a month, and you do your estimates for a day instead, then P{Monthly >=x} is obtained by computing P{monthly = z} and summing over z = x, x+1,x+2, ... . In turn, each P{Monthly = z} would be obtained as C(30,z)*p^z*(1-p)^(30- z), where p is the probability that in one day, a record has the property you want. Of course, if in a day, several records can have that property, you need to break down the time further, say into hours or minutes, or seconds---a small enough interval that the probability of more than one such record in the interval is negligible. I am having trouble getting Excel to compute probabilities as other than 0 > or 1, though when I do examples on the web that involve very small numbers I > get teh correct answers with no trouble. Most experts would advise against using EXCEL for statistical calculations, unless, of course, you have no other alternative. EXCEL's results are wrong too often for comfort. Several reviews advising against EXCEL for statistical calculations are available on line; just Google somethin like 'excel + statistical calculation+ problems or errors' . Of course, it is better in some sub-areas than others, so sometimes it will be OK, but sometimes not. Are poisson probabilities intended for this use? Or are they only > applicable for small numbers of discrete events, like the liklihood that 4 > cars will run a traffic light in a day? What would be the appropriate statistic for the probability of meeting a > goal of 700,000 discrete events in a month, or a few million in a year? This is a bit confusing. You have a large number of records of many types occurring in one month. Some of these are of special interest to you. Am I right so far? OK, assuming that is the case, you want to know the probability of having *at least* 700,000 'special' records in one month? Surely, that must depend on knowledge of the distribution of 'special' records among the totality. For example, you need to know the average number of 'special' records per month, or per day, or per hour, or per whatever; or, you need to know what fraction of the total are special, on average, in a given period of time. You also need to know whether the occurrances of 'special' records are independent. So, let's make up an example. Say there is a probability of p = 1/2 that any randomly selected record will be 'special' (i.e., meets your goal). If there are 1,000,000 records of all types in a month, and assuming independence of 'special' records (that is, if the current record is special, it will not affect in any way the chances that the next one is special), what you are asking is for Prob{Special >= 700,000} = Binomial(1,000,000,p = 1/2, >= 700,000). In such problems we would use the Normal Distribution as an approximation to the Binomial rather than the Poisson (although the results might not be very different). In this case, every method would yield a probability that is virtually zero, since 700,000 is 400 standard deviations above the mean (mean = 0.5*1,000,000 = 500,000.) R.G. Vickson . -- > Yours, > Dora Smith > Austin, TX > tiggernu...@yahoo.com -- > Yours, > Dora Smith > Austin, TX > tiggernu...@yahoo.com === Subject: Re: Question about use of Poisson probabilities In what way are Excel's statistical results wrong, and what statistical programs would you recommend? -- Yours, Dora Smith Austin, TX tiggernut24@yahoo.com >> My boss wants me to use poisson probabilities to compute the liklihood of >> meeting various goals in relation to our project, where the average >> number >> of records per month is 617,000. So, the average number per day is 617,000/30 = 20,566.66 (if your > working month is 30 days), the average number per hour is > 617,000/30/24= 856.944, the average number per minute is > 617,000/30/24/60 = 14.2824, etc. It is certainly easier to work with > these smaller Poisson parameters, then scale up to months later. > However, you should be careful: if all you want are averages, straight > scaling is appropriate, but if you want *probability* estimates, you > need to scale up from days to months using something like a binomial > distribution. For example, if you want to know the probability that at > least x records will have some property in a month, and you do your > estimates for a day instead, then P{Monthly >=x} is obtained by > computing P{monthly = z} and summing over z = x, x+1,x+2, ... . In > turn, each P{Monthly = z} would be obtained as C(30,z)*p^z*(1-p)^(30- > z), where p is the probability that in one day, a record has the > property you want. Of course, if in a day, several records can have > that property, you need to break down the time further, say into hours > or minutes, or seconds---a small enough interval that the probability > of more than one such record in the interval is negligible. >> I am having trouble getting Excel to compute probabilities as other than >> 0 >> or 1, though when I do examples on the web that involve very small >> numbers I >> get teh correct answers with no trouble. Most experts would advise against using EXCEL for statistical > calculations, unless, of course, you have no other alternative. > EXCEL's results are wrong too often for comfort. Several reviews > advising against EXCEL for statistical calculations are available on > line; just Google somethin like 'excel + statistical calculation+ > problems or errors' . Of course, it is better in some sub-areas than > others, so sometimes it will be OK, but sometimes not. > Are poisson probabilities intended for this use? Or are they only >> applicable for small numbers of discrete events, like the liklihood that >> 4 >> cars will run a traffic light in a day? >> What would be the appropriate statistic for the probability of meeting a >> goal of 700,000 discrete events in a month, or a few million in a year? This is a bit confusing. You have a large number of records of many > types occurring in one month. Some of these are of special interest to > you. Am I right so far? OK, assuming that is the case, you want to > know the probability of having *at least* 700,000 'special' records in > one month? Surely, that must depend on knowledge of the distribution > of 'special' records among the totality. For example, you need to know > the average number of 'special' records per month, or per day, or per > hour, or per whatever; or, you need to know what fraction of the total > are special, on average, in a given period of time. You also need to > know whether the occurrances of 'special' records are independent. So, let's make up an example. Say there is a probability of p = 1/2 > that any randomly selected record will be 'special' (i.e., meets your > goal). If there are 1,000,000 records of all types in a month, and > assuming independence of 'special' records (that is, if the current > record is special, it will not affect in any way the chances that the > next one is special), what you are asking is for Prob{Special >= > 700,000} = Binomial(1,000,000,p = 1/2, >= 700,000). In such problems > we would use the Normal Distribution as an approximation to the > Binomial rather than the Poisson (although the results might not be > very different). In this case, every method would yield a probability > that is virtually zero, since 700,000 is 400 standard deviations above > the mean (mean = 0.5*1,000,000 = 500,000.) R.G. Vickson > . >> -- >> Yours, >> Dora Smith >> Austin, TX >> tiggernu...@yahoo.com >> -- >> Yours, >> Dora Smith >> Austin, TX >> tiggernu...@yahoo.com === Subject: Re: Question about use of Poisson probabilities <45c69870$2$16996$4c368faf@roadrunner.com In what way are Excel's statistical results wrong, A Google search using 'excel + statistical computation + problems or depth exactly what is wrong about using EXCEL for stats computations. a lot of complicated theory. They are highly readable. Of course, to some extent the seriousness of the problem depends on which version of EXCEL you use: the newer ones are somewhat improved. It also depends somewhat on just what type of acalculation you are trying to perform (but calculaltions of probabililty distsributions is one of the there. > and what statistical > programs would you recommend? I would recommend that you first understand the nature of the problem. It might even be the case that you can solve it using a hand calculator, but until you really understand what you are doing, no amount of software is going to help. By the way: please do not top post, which is what you have been doing up to now in your responses. The accepted protocol is to place your responses just after the material you are remarking on, or giving an answer just after the question. So, you need to insert a response in the middle of a message. If you are using Google, you do this by NOT using the reply button, but, instead by clicking on the further options button, then choosing reply from its sub-menu, R.G. Vickson Adjunct Professor, University of Waterloo -- > Yours, > Dora Smith > Austin, TX >> My boss wants me to use poisson probabilities to compute the liklihood of >> meeting various goals in relation to our project, where the average >> number >> of records per month is 617,000. So, the average number per day is 617,000/30 = 20,566.66 (if your > working month is 30 days), the average number per hour is > 617,000/30/24= 856.944, the average number per minute is > 617,000/30/24/60 = 14.2824, etc. It is certainly easier to work with > these smaller Poisson parameters, then scale up to months later. > However, you should be careful: if all you want are averages, straight > scaling is appropriate, but if you want *probability* estimates, you > need to scale up from days to months using something like a binomial > distribution. For example, if you want to know the probability that at > least x records will have some property in a month, and you do your > estimates for a day instead, then P{Monthly >=x} is obtained by > computing P{monthly = z} and summing over z = x, x+1,x+2, ... . In > turn, each P{Monthly = z} would be obtained as C(30,z)*p^z*(1-p)^(30- > z), where p is the probability that in one day, a record has the > property you want. Of course, if in a day, several records can have > that property, you need to break down the time further, say into hours > or minutes, or seconds---a small enough interval that the probability > of more than one such record in the interval is negligible. > I am having trouble getting Excel to compute probabilities as other than >> 0 >> or 1, though when I do examples on the web that involve very small >> numbers I >> get teh correct answers with no trouble. Most experts would advise against using EXCEL for statistical > calculations, unless, of course, you have no other alternative. > EXCEL's results are wrong too often for comfort. Several reviews > advising against EXCEL for statistical calculations are available on > line; just Google somethin like 'excel + statistical calculation+ > problems or errors' . Of course, it is better in some sub-areas than > others, so sometimes it will be OK, but sometimes not. > Are poisson probabilities intended for this use? Or are they only >> applicable for small numbers of discrete events, like the liklihood that >> 4 >> cars will run a traffic light in a day? > What would be the appropriate statistic for the probability of meeting a >> goal of 700,000 discrete events in a month, or a few million in a year? This is a bit confusing. You have a large number of records of many > types occurring in one month. Some of these are of special interest to > you. Am I right so far? OK, assuming that is the case, you want to > know the probability of having *at least* 700,000 'special' records in > one month? Surely, that must depend on knowledge of the distribution > of 'special' records among the totality. For example, you need to know > the average number of 'special' records per month, or per day, or per > hour, or per whatever; or, you need to know what fraction of the total > are special, on average, in a given period of time. You also need to > know whether the occurrances of 'special' records are independent. So, let's make up an example. Say there is a probability of p = 1/2 > that any randomly selected record will be 'special' (i.e., meets your > goal). If there are 1,000,000 records of all types in a month, and > assuming independence of 'special' records (that is, if the current > record is special, it will not affect in any way the chances that the > next one is special), what you are asking is for Prob{Special >= > 700,000} = Binomial(1,000,000,p = 1/2, >= 700,000). In such problems > we would use the Normal Distribution as an approximation to the > Binomial rather than the Poisson (although the results might not be > very different). In this case, every method would yield a probability > that is virtually zero, since 700,000 is 400 standard deviations above > the mean (mean = 0.5*1,000,000 = 500,000.) R.G. Vickson > . > -- >> Yours, >> Dora Smith >> Austin, TX >> tiggernu...@yahoo.com > -- >> Yours, >> Dora Smith >> Austin, TX >> tiggernu...@yahoo.com === Subject: Re: Question about use of Poisson probabilities No - the average per month is 617,000. We deal in figures by the month. The question is what we need to do to our output per month to meet our goal, or if our output per month will be sufficient to meet our goal, which is based on some number of records per year. My boss believes that he wants the probability of achieving our goal in poisson probabilities. 617,000 is the mean number of total records per month over 27 months. Typically a third of them, more or less, are useful; the others bin out into about nine categories of invalid and not useful records. -- Yours, Dora Smith Austin, TX tiggernut24@yahoo.com >> My boss wants me to use poisson probabilities to compute the liklihood of >> meeting various goals in relation to our project, where the average >> number >> of records per month is 617,000. So, the average number per day is 617,000/30 = 20,566.66 (if your > working month is 30 days), the average number per hour is > 617,000/30/24= 856.944, the average number per minute is > 617,000/30/24/60 = 14.2824, etc. It is certainly easier to work with > these smaller Poisson parameters, then scale up to months later. > However, you should be careful: if all you want are averages, straight > scaling is appropriate, but if you want *probability* estimates, you > need to scale up from days to months using something like a binomial > distribution. For example, if you want to know the probability that at > least x records will have some property in a month, and you do your > estimates for a day instead, then P{Monthly >=x} is obtained by > computing P{monthly = z} and summing over z = x, x+1,x+2, ... . In > turn, each P{Monthly = z} would be obtained as C(30,z)*p^z*(1-p)^(30- > z), where p is the probability that in one day, a record has the > property you want. Of course, if in a day, several records can have > that property, you need to break down the time further, say into hours > or minutes, or seconds---a small enough interval that the probability > of more than one such record in the interval is negligible. >> I am having trouble getting Excel to compute probabilities as other than >> 0 >> or 1, though when I do examples on the web that involve very small >> numbers I >> get teh correct answers with no trouble. Most experts would advise against using EXCEL for statistical > calculations, unless, of course, you have no other alternative. > EXCEL's results are wrong too often for comfort. Several reviews > advising against EXCEL for statistical calculations are available on > line; just Google somethin like 'excel + statistical calculation+ > problems or errors' . Of course, it is better in some sub-areas than > others, so sometimes it will be OK, but sometimes not. > Are poisson probabilities intended for this use? Or are they only >> applicable for small numbers of discrete events, like the liklihood that >> 4 >> cars will run a traffic light in a day? >> What would be the appropriate statistic for the probability of meeting a >> goal of 700,000 discrete events in a month, or a few million in a year? This is a bit confusing. You have a large number of records of many > types occurring in one month. Some of these are of special interest to > you. Am I right so far? OK, assuming that is the case, you want to > know the probability of having *at least* 700,000 'special' records in > one month? Surely, that must depend on knowledge of the distribution > of 'special' records among the totality. For example, you need to know > the average number of 'special' records per month, or per day, or per > hour, or per whatever; or, you need to know what fraction of the total > are special, on average, in a given period of time. You also need to > know whether the occurrances of 'special' records are independent. So, let's make up an example. Say there is a probability of p = 1/2 > that any randomly selected record will be 'special' (i.e., meets your > goal). If there are 1,000,000 records of all types in a month, and > assuming independence of 'special' records (that is, if the current > record is special, it will not affect in any way the chances that the > next one is special), what you are asking is for Prob{Special >= > 700,000} = Binomial(1,000,000,p = 1/2, >= 700,000). In such problems > we would use the Normal Distribution as an approximation to the > Binomial rather than the Poisson (although the results might not be > very different). In this case, every method would yield a probability > that is virtually zero, since 700,000 is 400 standard deviations above > the mean (mean = 0.5*1,000,000 = 500,000.) R.G. Vickson > . >> -- >> Yours, >> Dora Smith >> Austin, TX >> tiggernu...@yahoo.com >> -- >> Yours, >> Dora Smith >> Austin, TX >> tiggernu...@yahoo.com === Subject: Re: Question about use of Poisson probabilities > No - the average per month is 617,000. We deal in figures by the month. > The question is what we need to do to our output per month to meet our goal, > or if our output per month will be sufficient to meet our goal, which is > based on some number of records per year. My boss believes that he wants > the probability of achieving our goal in poisson probabilities. 617,000 is the mean number of total records per month over 27 months. > Typically a third of them, more or less, are useful; the others bin out into > about nine categories of invalid and not useful records. I think you would get better advice if you explain the problem in more detail. What is a record? Something that is sent to you or something that you create? In what way can you control the number of records? How many records per year is your goal? Is the goal in terms of useful records or all records Are the nine categories relevant? -- === Subject: Re: Question about use of Poisson probabilities My boss wants me to use poisson probabilities to compute the > liklihood of meeting various goals in relation to our project, > where the average number of records per month is 617,000. I am having trouble getting Excel to compute probabilities as > other than 0 or 1, though when I do examples on the web that > involve very small numbers I get teh correct answers with no > trouble. Are poisson probabilities intended for this use? Or are they > only applicable for small numbers of discrete events, like the > liklihood that 4 cars will run a traffic light in a day? What would be the appropriate statistic for the probability of > meeting a goal of 700,000 discrete events in a month, or a few > million in a year? The idea of changing your units to events per micro-month (or something like that) doesn't seem very useful to me. Since the Poisson distribution is discrete, this model would only include the events { 0 records in Feb; 1,000,000 records in Feb, 2,000,000 records in Feb; ...} This is probably not what you want. My thought was to approximate the Poisson distribution by a Gaussian with the same mean and standard varation. This works well for large enough numbers of events (including numbers much smaller than 617,000). However, when I did this I realized that source of your problems is your model. It just has to be incorrect. For the Poisson distribution you describe, mu = 617,000 sigma = sqrt(mu) = 785.5 For the question What is the probability of at least 700,000 events in a month?, one can use the standard normal cdf with z = (700,000 - 617,000)/785.5 = 105.7 (For a little perspective, the probability that z > 6 is about one in 10^9) Maple will give me that number, but I'm not surprised Excel won't: prob(z > 105.7) = 1.146471703e-2427 If there is enough chance that you'll process 700,000 records in a month that you're even asking this question, then your model is wrong. There are a lot of other well-studied distributions out there, maybe even supported by Excel, and maybe even some literature about which would be most appropriate for your situation. I can't say anything about that deeper If I were doing this, I would use the normal distribution (which is nearly always not too wrong) with mu and sigma gotten from the sample mean and sample standard devaiation, of whatever data you got 617,000 from. I notice, though, that you're looking for the probability of reaching a goal. That makes me wonder whether these statistics are being used incorrectly from the very beginning. The assumption behind using descriptive statistics as you are doing is that the number of records processed per month is *not* manipulable, that they just happen to have a certain distribution. We can ask some questions, but the sort of question you cannot have is How well did we do last month? The answer will always be We did the same as we always do; it's just that last month we were lucky (or, we were unlucky; it doesn't matter) The assumption that you did what you always do is built into the way the question was asked. Some more valuable statistics might be the correlations between whatever you think *might* cause you to have more records or fewer records processed and the numbers actually processed. Jim Burns (Disclaimer: I am not a mathematician. If you're planning to tell your boss he's an idiot for asking what he asked, you need some heavier-weight support than me.) === Subject: Re: Question about use of Poisson probabilities Not planning to call my boss an idiot, though I am questionning how well he knows what he is talking about - but most people are telling me for one reason and another to use the normal distribution. Probability theory was never my forte. By my model is wrong, do you mean that my model does not consist of small numbers of discrete events that are skewed toward the left? -- Yours, Dora Smith Austin, TX tiggernut24@yahoo.com > My boss wants me to use poisson probabilities to compute the >> liklihood of meeting various goals in relation to our project, >> where the average number of records per month is 617,000. >> I am having trouble getting Excel to compute probabilities as >> other than 0 or 1, though when I do examples on the web that >> involve very small numbers I get teh correct answers with no >> trouble. >> Are poisson probabilities intended for this use? Or are they >> only applicable for small numbers of discrete events, like the >> liklihood that 4 cars will run a traffic light in a day? >> What would be the appropriate statistic for the probability of >> meeting a goal of 700,000 discrete events in a month, or a few >> million in a year? The idea of changing your units to events per micro-month (or > something like that) doesn't seem very useful to me. Since the > Poisson distribution is discrete, this model would only include > the events { 0 records in Feb; 1,000,000 records in Feb, > 2,000,000 records in Feb; ...} This is probably not what you want. My thought was to approximate the Poisson distribution by > a Gaussian with the same mean and standard varation. This works > well for large enough numbers of events (including numbers > much smaller than 617,000). However, when I did this I realized > that source of your problems is your model. It just has > to be incorrect. For the Poisson distribution you describe, > mu = 617,000 > sigma = sqrt(mu) = 785.5 > For the question What is the probability of at least 700,000 > events in a month?, one can use the standard normal cdf with > z = (700,000 - 617,000)/785.5 = 105.7 > (For a little perspective, the probability that z > 6 is > about one in 10^9) Maple will give me that number, but I'm > not surprised Excel won't: > prob(z > 105.7) = 1.146471703e-2427 If there is enough chance that you'll process 700,000 records > in a month that you're even asking this question, then > your model is wrong. > There are a lot of other well-studied distributions out > there, maybe even supported by Excel, and maybe even some > literature about which would be most appropriate for > your situation. I can't say anything about that deeper If I were doing this, I would use the normal distribution > (which is nearly always not too wrong) with mu and > sigma gotten from the sample mean and sample standard > devaiation, of whatever data you got 617,000 from. > I notice, though, that you're looking for the probability > of reaching a goal. That makes me wonder whether these > statistics are being used incorrectly from the very > beginning. The assumption behind using descriptive statistics as you > are doing is that the number of records processed per month > is *not* manipulable, that they just happen to have a certain > distribution. We can ask some questions, but the sort of > question you cannot have is How well did we do last > month? The answer will always be We did the same as we > always do; it's just that last month we were lucky > (or, we were unlucky; it doesn't matter) The assumption > that you did what you always do is built into the way > the question was asked. Some more valuable statistics might be the correlations > between whatever you think *might* cause you to have > more records or fewer records processed and the numbers > actually processed. Jim Burns > (Disclaimer: I am not a mathematician. If you're planning > to tell your boss he's an idiot for asking what he asked, > you need some heavier-weight support than me.) === Subject: Re: Question about use of Poisson probabilities Not planning to call my boss an idiot, though I am questionning > how well he knows what he is talking about - but most people > are telling me for one reason and another to use the normal > distribution. Probability theory was never my forte. By my model is wrong, > do you mean that my model does not consist of small numbers of > discrete events that are skewed toward the left? I'm glad you're not going to call your boss an idiot. He probably really isn't one, anyway. I've long noticed that humor does not travel well over the Internet, but occasionally I have to try anyway. Maybe that means I'm an idiot. When I say your model is wrong, I mean that your assumptions imply things that turn out not to be true about your data. I'm not entirely sure that that is true in your case. I'm guessing that 700,000 is not entirely impossible as a goal, even if it is a stretch, since you mentioned the number. If your records arriving or being processed or whatever are a Poisson process, then 700,000 shows much, much, much too much variability. In comparison, Big Foot riding by on a winged unicorn to deliver your morning paper would be boringly normal. (If you have a Poisson with a rate of 617,000 reports per month, then you *must have* a standard variation of 785 reports per month. Compare that to your sample standard deviation. Is it wildly different? Then it's not a Poisson process.) I don't remember all of my statistics, but there are other assumptions about your distribution that you need in order to use the Poisson distribution. One assumption is that each event is independent of the others. That seems like a very reasonable assumption to me, but it may not be true. There may be background events that cause multiple reports to be generated. I'm imagining insurance claims which would be to some extent independent of each other, but also could be jointly linked to bad or good weather. Maybe something similar applies in your case. I would strongly expect that some of the variability is related to there being more working days in some months than in others. There is also very likely a seasonal component to it, too. There's a seasonal component to practically everything. It would be interesting if, after you've accounted for things like number of working days and season, what variability you have left looks like a Poisson. That might mean you fully understand the processes that generate reports for you. (Or maybe it doesn't; remember: I Am Not A Mathematician.) Whether or not you do all that, you should still be able to characterize the monthly number of reports using a Gaussian with some mean and some standard deviation. That's a consequence of the Central Limit Theorem. (Theorem: if you combine a bunch of distributions of whatever shape, they'll start to look like a Gaussian.) Jim Burns > My boss wants me to use poisson probabilities to compute the >> liklihood of meeting various goals in relation to our project, >> where the average number of records per month is 617,000. >> I am having trouble getting Excel to compute probabilities as >> other than 0 or 1, though when I do examples on the web that >> involve very small numbers I get teh correct answers with no >> trouble. >> Are poisson probabilities intended for this use? Or are they >> only applicable for small numbers of discrete events, like the >> liklihood that 4 cars will run a traffic light in a day? >> What would be the appropriate statistic for the probability of >> meeting a goal of 700,000 discrete events in a month, or a few >> million in a year? The idea of changing your units to events per micro-month (or > something like that) doesn't seem very useful to me. Since the > Poisson distribution is discrete, this model would only include > the events { 0 records in Feb; 1,000,000 records in Feb, > 2,000,000 records in Feb; ...} This is probably not what you want. My thought was to approximate the Poisson distribution by > a Gaussian with the same mean and standard varation. This works > well for large enough numbers of events (including numbers > much smaller than 617,000). However, when I did this I realized > that source of your problems is your model. It just has > to be incorrect. For the Poisson distribution you describe, > mu = 617,000 > sigma = sqrt(mu) = 785.5 > For the question What is the probability of at least 700,000 > events in a month?, one can use the standard normal cdf with > z = (700,000 - 617,000)/785.5 = 105.7 > (For a little perspective, the probability that z > 6 is > about one in 10^9) Maple will give me that number, but I'm > not surprised Excel won't: > prob(z > 105.7) = 1.146471703e-2427 If there is enough chance that you'll process 700,000 records > in a month that you're even asking this question, then > your model is wrong. > There are a lot of other well-studied distributions out > there, maybe even supported by Excel, and maybe even some > literature about which would be most appropriate for > your situation. I can't say anything about that deeper If I were doing this, I would use the normal distribution > (which is nearly always not too wrong) with mu and > sigma gotten from the sample mean and sample standard > devaiation, of whatever data you got 617,000 from. > I notice, though, that you're looking for the probability > of reaching a goal. That makes me wonder whether these > statistics are being used incorrectly from the very > beginning. The assumption behind using descriptive statistics as you > are doing is that the number of records processed per month > is *not* manipulable, that they just happen to have a certain > distribution. We can ask some questions, but the sort of > question you cannot have is How well did we do last > month? The answer will always be We did the same as we > always do; it's just that last month we were lucky > (or, we were unlucky; it doesn't matter) The assumption > that you did what you always do is built into the way > the question was asked. Some more valuable statistics might be the correlations > between whatever you think *might* cause you to have > more records or fewer records processed and the numbers > actually processed. Jim Burns > (Disclaimer: I am not a mathematician. If you're planning > to tell your boss he's an idiot for asking what he asked, > you need some heavier-weight support than me.) === Subject: Re: Question about use of Poisson probabilities Try scaling your events/month by a factor of a million, i.e., 0.6 events per micromonth. Phil === Subject: Re: Question about use of Poisson probabilities In other words, divide my 617,000 events by 1 million? -- Yours, Dora Smith Austin, TX tiggernut24@yahoo.com > Try scaling your events/month by a factor of a million, i.e., 0.6 events > per micromonth. Phil === Subject: fractal dimension Is there way to compute the fractal dimension if the self-similar sets overlap? === Subject: Re: fractal dimension <19200145.1170602262397.JavaMail.jakarta@nitrogen.mathforum.org>, > Is there way to compute the fractal dimension if the self-similar sets overlap? In general? No. In some cases? Yes. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: fractal dimension george johnson foundation a .8ecrit : > Is there way to compute the fractal dimension if the self-similar sets overlap? A difficult question. It might help if you specified the question, or read? If you mean: what happens when the Open Set Condition is not satisfied, a quick search in the litterature did not give many answer. I would assume that the dimension is less than what it would be with the OSC, but even that doesn't seem completely obvious to me. If you want, have a look at: MR2142580 (2005m:60018) Shieh, Narn-Rueih(RC-NTAI); Yu, Jinghu(PRC-ASWH-MP) Dimension properties of random fractals with overlaps. (English summary) Taiwanese J. Math. 9 (2005), no. 2, 313--330. 60D05 (28A75 37C45) If you are really interested, I will ask one of my colleagues who is more knowledgeable than me on this subject. JMA === Subject: test - ignore www3.sympatico.ca/echapin === Subject: Iran opens Nuclear Enrichment School, judeo-yank-fascists in defeat First the tiny hezbollah destroyed merkevas one after another using precision anti-tank weapons reducing the israelis to madness of indiscriminate cluster bombing. But the Iranians are the smartest of these Islamics. Grand chess masters ... who once saved the ass of jews under Cyrus the Great from Nebuchadnezzer Open house at Iranian nuclear site Isfahan, Iran February 5, 2007 AdvertisementAdvertisement Iran has opened one of its nuclear sites to local and international reporters and ambassadors to show the transparency of its program before a UN Security Council deadline this month. Tehran has kept up efforts to give the site at Isfahan more publicity. A tourism official said late last year that Iran planned to open it and other nuclear plants to foreign tourists. Iran's ambassador to the International Atomic Energy Agency, Ali Asghar Soltanieh, said the purpose of the tour was to assure the world that Iran's program was peaceful. In fact we have representatives from all over the world, Mr Soltanieh said. We decided to have them come here and see for themselves, he said. He made a point of emphasising the IAEA surveillance cameras at Isfahan. Photographers and video camera operators were not allowed to take pictures in the outside area of the compound. Uranium ore is converted into yellowcake and gas at Isfahan. The gas is transferred to the more sophisticated plant at Natanz, where it could be enriched with centrifuges. Reporters at the weekend passed the Natanz site but were not allowed inside, where Iran recently said it was installing 3000 centrifuges. NEW YORK TIMES === Subject: It is good to blow up a marketplace full of people buying and selling food It is good to blow up a marketplace full of people buying and selling food in Iraq. This is how Muslim insurgents will drive the enemy out of their country. By killing and maiming fellow Iraqis! The traditional Muslim way. The way of Allah! The way of the Quaran! The way of Mohammed! They don't need any democracy. They don't need free elections. They are true believers! Suicide truck bomber kills at least 130 in Baghdad By Tina Susman, Times Staff Writer BAGHDAD -- A suicide bomber posing as a trucker hauling food drove into a busy market in central Baghdad today and blew himself up, killing at least 130 people, injuring more than 300, and capping a particularly volatile day in Iraq's sectarian war. Seven car bombs also tore through the northern city of Kirkuk, leaving two dead in attacks that underscored tensions enveloping the oil-rich region as Arabs and Kurds vie for power. Another person died south of Baghdad in a separate car bombing. The incidents illustrated the breadth of polarization plaguing Iraq as U.S. and Iraqi officials prepare for what they vow will be a decisive crackdown on insurgents. The Interior Ministry said the blast in the market in Sadriya killed at least 130 people and injured 305. It came in the early evening as shoppers thronged the stalls and tiny shops to stock up for the evening meal. Hamed Majed, a butcher, was tending to customers when he saw a large yellow truck trying to navigate down the narrow, alley-like street leading through the market. > First the tiny hezbollah destroyed merkevas one after another using > precision anti-tank weapons reducing the israelis to madness of > indiscriminate cluster bombing. But the Iranians are the smartest of these Islamics. Grand chess > masters ... who once saved the ass of jews under Cyrus the Great from > Nebuchadnezzer > Open house at Iranian nuclear site Isfahan, Iran > February 5, 2007 > AdvertisementAdvertisement Iran has opened one of its nuclear sites to local and international > reporters and ambassadors to show the transparency of its program > before a UN Security Council deadline this month. Tehran has kept up efforts to give the site at Isfahan more publicity. > A tourism official said late last year that Iran planned to open it > and other nuclear plants to foreign tourists. Iran's ambassador to the International Atomic Energy Agency, Ali > Asghar Soltanieh, said the purpose of the tour was to assure the world > that Iran's program was peaceful. In fact we have representatives from all over the world, Mr > Soltanieh said. We decided to have them come here and see for > themselves, he said. He made a point of emphasising the IAEA surveillance cameras at > Isfahan. Photographers and video camera operators were not allowed to take > pictures in the outside area of the compound. Uranium ore is converted into yellowcake and gas at Isfahan. The gas > is transferred to the more sophisticated plant at Natanz, where it > could be enriched with centrifuges. Reporters at the weekend passed the Natanz site but were not allowed > inside, where Iran recently said it was installing 3000 centrifuges. NEW YORK TIMES > === Subject: Re: It is good to blow up a marketplace full of people buying and selling food Suicide truck bomber kills at least 130 in Baghdad > By Tina Susman, Times Staff Writer The sound of the exploding bomb is the Muslim Call to Prayer. Allah hu'Akbar. Bob Kolker === Subject: Re: It is good to blow up a marketplace full of people buying and selling food <_Jqxh.46014$jA.34196@bignews1.bellsouth.net It is good to blow up a marketplace full of people buying and selling food > in Iraq. > This is how Muslim insurgents will drive the enemy out of their country. > By killing and maiming fellow Iraqis! The traditional Muslim way. > The way of Allah! The way of the Quaran! The way of Mohammed! > They don't need any democracy. They don't need free elections. They are true > believers! Excellent comment by the internet spook on duty. Your analysis is correct that they are CIA PATSIES. They dont have the smarts of Cuba, India, Iran, Venezuela, Israel or the Yanks. Its quite obvious who is behind the controlled demolition of Silverstein Properties and the Golden Dome of that mosque or shrine of the Shia. Dr Fidel Castro Ruz has made Cuba into a leading Biotechnology Center of the World. Many American sheeple dont even know the full name of Dr Ruz, or his occupation or even the status of Cuba as the leading Biotechnology Center in the world. RESPONSE FROM DR. FIDEL CASTRO RUZ, PRESIDENT OF THE REPUBLIC OF CUBA, TO THE STATEMENTS MADE BY THE UNITED STATES GOVERNMENT ON BIOLOGICAL WEAPONS. Hardly three days ago, someone only too well known to us, Assistant Secretary of State Otto Reich was caught out in an embarrassing lie when he said that four Cuban planes had landed in the Venezuelan capital on April 12, and that nobody knew what they were doing there, what they were carrying, we don't know. Apparently, it was the beginning of an anti-Cuba campaign or a vendetta due to the amazing failure of the fascist coup he set in motion, or both. On Tuesday May 7, when the Ministry of Foreign Affairs challenged him publicly, the State Department said that it had no confirmation whatsoever, and that it did not want to discuss the subject any more. The idea of destroying Cuba, an obsession that has lasted more than 43 years, has lead and still leads U.S. policy down a tortuous path filled with lies, mistakes, failures and crimes. What the US government is telling the world today and what it is doing with Cuba is perhaps the most grievous and demoralizing contradiction in its foreign policy. This great power had never found itself in such a compromising position and it has no alternative but to lie, lie and lie. And there is no lack of unscrupulous characters in major public positions willing to do so, no lack of press spokesmen caught up in the continuous and bitter need to right wrongs and explain the inexplicable in their bosses' statements. Even men like Colin Powell, son of Jamaican immigrants, that despite his military training or maybe because of it, is not considered a hawk since he knows about war and has seen many men die --a man that many Americans even came to look on as a potential presidential candidate-- has found himself enmeshed in shameful and little ennobling intrigues promoted by such characters. He knows better than anyone else how inexperienced they are and what little intellectual and political worth those people have. Whom this new character involved in a sinister maneuver against Cuba can deceive? Mr. John Bolton, an Under Secretary of State, none other than the one for Arms Control. What are they aiming for with the attack launched by this official in an aggressive speech against Cuba given at the Heritage Foundation, famous for its ultra-rightwing stance? His statement, supposedly an analysis of the dangers of terrorism threatening the United States, begin by saying: In addition to Libya and Syria, there is a threat coming from another BWC signatory, and one that lies just 90 miles from the U.S. mainland-namely, Cuba. Then, after the usual name-calling and stupid remarks full of the hatred common in such arrogant and misinformed people, Mr. Bolton added something all his own: We know that Cuba is collaborating with other state sponsors of terror. Castro has repeatedly denounced the U.S. war on terrorism. He continues to view terror as a legitimate tactic to further revolutionary objectives. Last year, Castro visited Iran, Syria and Libya --all designees on the same list of terrorist-sponsoring states. At Teheran University, these were his words: 'Iran and Cuba, in cooperation with each other, can bring America to its knees. The U.S. regime is very weak, and we are witnessing this weakness from close- up.' But Cuba's threat to our security has often been underplayed. An official U.S. government report in 1998 concluded that Cuba did not represent a significant military threat to the United States or the region. It went only so far as to say that, 'Cuba has a limited capacity to engage in some military and intelligence activities which could pose a danger to U.S. citizens in some circumstances.' Mr. Bolton immediately looked for something to cover up the suspicious fact that it had never before occurred to any US government official to make such an infamous accusation against Cuba. Mr. Bolton blames this weakness on William Cohen, who was the U.S. Defense Secretary four years earlier when the criticized report was issued. Mr. Bolton made not the slightest mention of the fact that scarcely five months and two weeks earlier, on November 19, 2001, he himself made absolutely no mention of Cuba in a speech given to the conference of the parties to the Convention on Biological Weapons in Geneva when he cited many countries that were a source of concern to him as potential biological weapons producers. Why this sudden and unexpected change? Mr. Bolton's May 6 piece of tabloid journalism ends by saying: For four decades Cuba has maintained a well-developed and sophisticated biomedical industry, supported until 1990 by the Soviet Union. This industry is one of the most advanced in Latin America, and leads in the production of pharmaceuticals and vaccines that are sold worldwide. Analysts and Cuban defectors have long cast suspicion on the activities conducted in these biomedical facilities. Here is what we now know: The United States believes that Cuba has at least a limited offensive biological warfare research-and-development effort. Cuba has provided dual-use biotechnology to other rogue states. We are concerned that such technology could support BW programs in those states. We call on Cuba to cease all BW-applicable cooperation with rogue states and to fully comply with all of its obligations under the Biological Weapons Convention. The international press immediately picked up Mr. Bolton's string of Olympic-size lies, which is exactly what he wanted. Washington, May 6 (ANSA). Under Secretary of State John Bolton today accused Cuba of helping 'enemy governments' with biological weapons programs. Cuba had at least an offensive biological weapons program and could be transferring its results to other states hostile to the United States', Bolton said. The Under Secretary Bolton spoke to the Heritage Foundation, one of the ultraconservative groups in Washington. Washington, May 6 (DPA) The United States today accused Cuba of developing offensive biological weapons and of providing information about them to states hostile to the United States, and urged President Fidel Castro's government to cease this cooperation. Washington, May 6 (REUTERS). On Monday, the United States accused three countries -Cuba, Libya and Syria- of trying to develop weapons of mass destruction and warned that it would take steps to prevent them from supplying these arms to terrorist groups. 'States that sponsor terror and pursue weapons of mass destruction must stop. States that renounce terror and abandon WMD can become part of our effort. But those who do not can expect to become our targets.' Washington, May 6 (EFE) The United States today added Cuba to Syria and Libya on the list of countries who are part of the 'axis of evil' dedicated to manufacturing weapons of mass destruction, and warned that it would take steps to prevent them from supplying such weapons to terrorist organizations. Washington, May 6 (NOTIMEX). The United States today included Cuba in the so-called 'axis of evil' because it thinks it has the capacity to develop biological weapons which are a threat to United States security, a greater threat than that from Iraq, Iran and North Korea. Washington, May 6 (AFP) On Monday, the United States warned Cuba against any proliferation of biological weapons, urging the Havana government to cease providing any kind of biotechnological equipment to countries that Washington considers dangerous, such as Iraq and Libya. would be endless. The job is already done! The entire world, especially the American public that is constantly bombarded with perfidious lies, has been informed and is ready to believe that Cuba is a biological power, that it has a program for producing such weapons and that it poses a danger to the United States. And as John Bolton, the very distinguished Under Secretary of State for Arms Control and International Security has said it, we have to believe it. However, as the old saying goes: Lies have short legs. In the United States itself some people are astounded and they are beginning to figure out the whole game. Washington, May 6 (NOTIMEX). The United States today refused to produce the evidence it had claimed to have to back up the accusations made against Cuba. Both, the White House and State Department spokesmen said that the accusation against Cuba is not based on assumptions but on confidential information about the biological and chemical potential of the Cuban pharmaceutical industry. This accusation took not only the international community by surprise but also members of the United States Congress. According to the same dispatch, political analysts pointed out that the accusation against Cuba made by President George W. Bush's administration could be part of a White House strategy to find a justification for tightening its isolationist policy on Cuba. The statement that Cuba could pose a terrorist threat to the security of the United States was made at a time when several proposals to relax Washington's Cuba policy are being analyzed on Capital Hill. Nevertheless, political analysts maintain that in view of this and bearing in mind that his brother Jeb will be seeking reelection as governor of Florida this year, President Bush wants to ingratiate himself with the Cuban exile community. Given the fact that more than four decades of economic blockade against Cuba have not removed Fidel Castro from power in Havana, the only people who want this isolationist policy to remain in effect and harder are the vast majority of the Cuban exile community in the United States. The possibility of including Cuba as a member of the so-called 'axis of evil' with Iraq, Iran and North Korea at the head would make it easier for Bush to obtain the Congress backing for increasing instead of reducing the economic stranglehold on the Island. Washington, May 6 (AP). 'I think that it will delay us taking new steps towards a trade opening', said Graham, chairman of the Senate Select Committee on Intelligence. 'Unilateral steps will be most affected'. Graham, however, did not hide his amazement at Bolton's accusation. He said that in March the Committee he chairs held a secret hearing on security related issues and the administration made no mention of biological weapons. On May 7, when a journalist asked White House spokesman Ari Fleischer: Is there any proof of this, or is this an assumption of the United States? Fleischer replied: No, it's not an assumption. I assure you that Secretary Bolton would not have said it if he did not have good cause, reason and fact to say it. That was based on sound analysis, and on information that is studied and available to the United States government. This is a typical reply by someone who is launching an absolutely baseless and groundless accusation. The only thing studied carefully is lies and deceit. It is certainly a very sad role this spokesman has. Furthermore, why should we believe Mr. Bolton? Anyone who remembers the fifteen incredible pretexts, known today through declassified official documents, that were elaborated at the end of 1961 by the high US authorities to undertake a direct military attack against Cuba in 1962, would not be surprised by such a sinister lie. We demand proof. Let them produce even the tiniest piece of evidence! They do not have any, and they cannot have them because they simply do not exist. They should not be hiding behind the alleged sensitivity of their sources, when there is actually not an atom of truth in what they are saying. This very old trick and overly stupid argument only serve to demonstrate their little consideration for, and low concept of, the American people whose intelligence deserve more respect. I will also say this: If a Cuban scientist from any of our biotechnology institutes had been cooperating with any country in the development of biological weapons, or if he or she had tried to create them on his or her own initiative, he or she would be immediately presented in a court of justice as we would consider it an act of treason to the country. The Law against Terrorist Acts passed by the National Assembly of Cuba, in its Article 10, provides: The person who manufactures, facilitates, sells, transports, sends, introduces in the country or keeps in his or her possession, under any form or in any place {...} chemical or biological agents, or any other substance from whose investigation, design or combination thereof any product can be derived that meets the description offered, is liable to sanctions of 10 to 30 years of imprisonment, life sentence or capital punishment. This is really an absolute lie, a treacherous blow against the sale of food to Cuba authorized in a Law from the year 2000, which was subsequently modified on many occasions through amendments introduced by the staunch advocates of the blockade that made it practicably impossible to apply it after it had been passed. They are simply trying to mislead and dishearten the growing number of Americans who are increasingly upset by the most cruel and inhuman measure against the Cuban people, which clash with the idealism, and ethics of a nation that has, in fact, been deceived for dozens of years with The only truth in Bolton's lie is that Cuba is 90 miles from the continental territory of the United States. It is a false and manipulated assertion that our country has repeatedly denounced the United States war against terrorism. I have said, and I stand by it, that the solution to this scourge will not be reached through war, which would only serve to breed hatred and fanaticism but rather through a sincere and determined cooperation among all countries in the world and by building a truly universal culture and conscience against terrorism. We were the first to put forward this form of cooperation the very same day of the tragedy in New York. It is a slanderous invention, a fabricated lie, to say that Castro considers terror as a legitimate tactic for furthering revolutionary objectives. Actually, everybody knows that our revolutionary movement never used such methods that do not fit in with our doctrine, our principles and our concept of the armed struggle. Mr. Bolton, you'd better get your facts right. Don't be misled by the fantastic stories told by your dear friends from the CANF. Never were the civilian population and innocent people the victims of our actions. Our tactic always was to fight against heavily equipped enemy units. Presently, you want to call terrorism any armed resistance, regardless of the legitimate causes that may justify it. Along that line you could end up applying such definition to the struggle of the American colonists who rebelled and fought against English domination. George Washington and those who after long years of war and enormous sacrifices conquered the independence of the United States of America were not terrorists. Someone has failed you badly, Mr. Bolton, when they told you about my speech at the university of Teheran. It was not one speech, but two, in two universities and a few remarks at the end of my visit to the mausoleum that keep the remains of the Imam Khomeini. I have gone over them both in detail. In my remarks to the students in Teheran, there is not one single paragraph that resembles the one you dishonestly included in your speech at the Heritage Foundation on May 6. I never said that Iran and Cuba in cooperation with each other could bring the United States to its knees. I did say, in one of the three speeches I made during that visit, that imperialism was bereft of ideas that ideas are more powerful than weapons and that one day imperialism would crumble. I also said that the Iranian people with heroism and not with weapons defeated the Shah, and this showed the power of ideas. Also, there could never be anyone in the world so powerful that could not be defeated by ideas. Finally, I added: That is our hope. There is a superpower with thousands of nuclear arms, planes, armored ships, aircraft carriers, intelligent missiles [...] No matter how many weapons nor how much wealth it has, it will not be able to defeat human beings. I made not the slightest mention of the use of weapons in that struggle. In fact, I said just the opposite. That is what I think, that is how I saw it then and that is what I said. I am not in the habit of hiding my thoughts or manipulating my words. My three speeches in the Iranian capital were broadcast on Cuban radio and television. Thus, Mr. Bolton did not discover anything and I do not renounce my ideas. I said some other things about doctrines and political principles. I have all the tapes and the transcription of those speeches. I can prove what I have said. It should also be said that when I visited Iran, I had the honor of getting to know a great country with a culture that goes back thousands of years, a country with deep religious beliefs and a great spiritual strength; a country wanting to eradicate poverty, fighting against drug trafficking and other such plagues, determined to bring education, health, employment and well being to more than 60 million people. Not one of the many leaders I talked to said anything to me about biological weapons or any other kind of weapon. What a great difference I could perceive between that culture and customs with that of the West. But, I did not only visit that country, I also visited Algeria, Malaysia, Qatar, Syria and Libya where they showed me how Ghadafi's family home had been destroyed and spoke about the loss of human life caused by the F-16 brutal air raid, including a little girl. Thousands of years of history accumulate in that part of the world, which must not be destroyed nor their people annihilated. Billions of people in the world have seen with deep indignation the televised images of the terrible events in Palestine. Apparently, when Mr. Bolton alluded to my speech in Teheran, he irresponsibly and dishonestly mixed excerpts of what I had said with fragments and statements of what other people said to foreign press agencies, and information from wire service cables given by someone else or things the speech writer invented. My clear, precise, transparent statements can be found in the speeches I have mentioned and in the press communiqu.8e of May 10, 2001 issued by both delegations where, by the way, the Cuban position against terrorism is described in point 6 of the agreed document that literally reads: While condemning the phenomenon of terrorism in all its forms, especially state terrorism, the two parties agreed to cooperate closely at a bilateral and international level to fight against and eliminate this terrible phenomenon, to cooperate in the fight against international organized crime and drug trafficking, being always mindful of the cardinal principles of international law, and particularly those of sovereignty and non-interference in the internal affairs of states. This statement was issued 127 days before September 11. I should thank Mr. Bolton for his praise of our pharmaceutical industry, one of the most advanced in Latin America, which leads in the production of pharmaceuticals and vaccines that are sold worldwide, according to his own words. We only wish it was more advanced. Even his country could benefit from some of its discoveries, but they would not allow it. However, he lies shamelessly when he tells the American and world public opinion that the United States believes that Cuba has at least a limited offensive biological warfare research and development effort. This statement is as false as it is grave. Our researchers and doctors are educated with an elevated concept of solidarity and ethics. Millions of people in the world can testify to that. They work for the well being and health of human beings. For 40 years, 34,307 Cuban medical doctors and health workers have worked free of charge in a large number of poor countries saving the lives and safeguarding the health of millions of people. Nobody in the world could beat them in their dedication and their willingness to make sacrifices. At this very moment, 2,671 of them are working in isolated and inhospitable places in Latin America, the Caribbean and Africa. It would be very difficult to persuade these men and women to produce viruses and bacteria to kill children, women, old people or the people of any country. The pride and high moral standards of our people, which have led them to stand firm against 43 years of attacks and blockade, rests on the rationality of a policy that does not contradict their ethics and principles. Thirty nine thousand and eight hundred youths coming from more than 120 Third World countries have graduated in Cuba from 33 technical and university specialties. Even under the conditions prevailing in the special period, which resulted from the cruel blockade by the United States, 8,053 youths from Latin America, the Caribbean and Africa are studying medicine in our country free of charge, a career that costs more than $200 thousand in the United States. In spite of Bolton's lies and those of many others like him, Cuba enjoys great prestige because of the health services it has provided to humanity. That is, for its true fight against another kind of terrorism that many chose to ignore: the diseases that kill more than 11 million children every year, whose lives could be saved were it not for the selfish attitude of the industrialized world. An incalculable selfless work of the doctors who come from a poor small country and treat them with vaccines and preventive methods or therapies developed by Cuba. We would like to know if the U.S. government is doing something similar or if it would be prepared to cooperate with such programs for the benefit of those that it euphemistically describes as emerging countries, as we have not excluded any economically developed nation from our programs. We have even offered hundreds of scholarships to American youngsters who don't have the money to study medicine in their country, to come study in the Latin American School of Medical Sciences. We have even offered hundreds of scholarships to study in the Latin American School of Medical Sciences to U.S. youngsters who don't have the money to study medicine in their country. Cuban children are vaccinated against 13 diseases and enjoy wonderful health. The infant mortality rate for every thousand live births is lower than that of the United States itself. Medical care is guaranteed to one hundred percent of the population absolutely free of charge. Unfortunately in the United States, with a population of over 280 million, 16 percent of the people do not have medical insurance, and that includes more than 10 million children. In such an immensely rich and scientifically advanced country where hundreds of thousands of people die every year for these reasons, who is to blame? Who kills these people? Who denounce such facts? How can they invent, what gives them the moral right to claim from over there, and who is going to believe their heinous slander that we Cubans are developing biological warfare programs? On the other hand, never in the 43 years of the Revolution's history has anyone in our country launched or taken part in a terrorist act against the United States from our territory. Not one drop of US blood has been shed nor has any US company lost a single screw due to terrorist acts originated in Cuba. Those who in the United States are accusing our country of terrorism, or of supporting or sponsoring terrorism, cannot say that about Cuba. Thousands of our compatriots have died and tens of thousands of acts of sabotage have been recorded as part of terrorist actions and US aggression against Cuba. Can their spokesmen deny these facts? I am not referring to the American people; I am referring to their government. The overriding question of the powers given to U.S. government officials to effectuate out of court executions and to kill people anywhere in the world has not even been clarified. I have personally often been the target of these sinister plans. That was how they operated in the past. Have they or haven't they gone back to such disgusting methods? Why doesn't Mr. Bolton tell us a little bit about that subject? As for weapons of mass destruction, Cuba's policy has been irreproachable. No one has ever produced a single piece of evidence that any program for developing nuclear, chemical or biological weapons has been set up in our country. Those who have no sense of ethics, or who fail to understand that the government of Cuba abides by truth and transparency, might at least understand that it would have been utterly stupid to behave in any other way. Any such program would lead the economy of any small country to bankruptcy. Cuba would never have been able to transport such weapons. Moreover, it would be a mistake to use them in battle against an enemy that has a thousand times more of those weapons and that would be only to happy to find an excuse to use them. and there will be weapons much more powerful that any produced through technology, namely: the weapons of morality, reason and ideas; with them no country is weak but without them no nation is powerful. Adherence to such a maxim requires exceptionally strong convictions, steel nerves and talent. They should know by now that, as far as the Cuban people is concerned, the ideals that inspire freedom, dignity, love of one's homeland, its identity, its culture and the strictest sense of justice that human beings can conceive of are more valuable than anything on Earth. These are not weapons of mass destruction, but rather weapons of mass moral defense, and we are willing to fight and die for them. I understand that for a man like Bolton, intoxicated with the military, economic and technological might of the superpower on whose behalf he speaks, it might not be easy to understand these things. However, it would be a good idea if he tried. Cuba has absolutely nothing to hide. On the contrary, it is proud of its development in the biomedical research field. I shall offer a brief historical summary. 1979: the Genetic Engineering Group was set up in the National Center for Scientific research. 1981: the first biotechnology scientific-productive institution devoted to producing alpha and beta interferon was founded. The medical-pharmaceutical and biotechnology industry program: this came into being as a result of the scientific, economic and social development the country had achieved. Cuba met the standards for inspection and certification by international agencies and by the national regulatory agencies of those countries with which it was to have trading relations. The major part of the investment program is developed between 1990 and 1997. It would comprise 40 facilities. A completely humanitarian industry is developed to research and produce medicines for preventing disease and saving lives, as well as to increase food production. The purchase of technological equipment used worldwide from commercial companies with an international reputation. Abiding by agreements entered by Cuba in compliance with the Convention on Biological and Toxin Weapons, the Cuban biotechnology industry, the health system and the civil defense organizations submit an annual report to the United Nations on confidence building measures. Many regulatory agencies from various countries have visited Cuban productive biotechnology facilities as a prerequisite for marketing our products in their countries. In the next few years, 50 new products will come on to the market. These include biopharmaceuticals, vaccines and diagnosis kits. Our country already has a stock of intellectual property consisting of more than 150 inventions and over 500 patents registered abroad. The results of scientific research are published in the most important international journals. 1990: Our products began to be exported and exports have increased every year since then. 1992: Cuba signed the Convention on Biological Diversity ratified in 1994. 1995: landmark forms of marketing are introduced: technology transfers, at-risk development contracts with foreign companies, and production-commercial partnerships. Today, products and technology of the Cuban biotechnology industry are available in more than 40 countries Agreements for Technology transfers or negotiations are currently underway with 14 countries: India: 4 transfers, 4 products China: 2 transfers, 4 products. Brazil: 2 transfers, 2 products. Egypt: 4 transfers, I product under negotiation. Malaysia: 6 transfers. Iran: 4 transfers, 4 products. Russia: 1 transfer, 1 product. South Africa: 1 transfer, 1 product. Tunisia: 1 transfer, 1 product. Algeria: 1 transfer, 3 products. Great Britain-Belgium: 1 transfer, 1 product. Venezuela: 1 transfer, 2 products under negotiation. Mexico: 1 transfer, 1 product New trade and production negotiations are underway with 10 countries: Malaysia, Holland, Spain, Brazil, Venezuela, Vietnam, Mexico, Ukraine, Germany and the United States (in this case negotiations over the use of the Cuban anti-meningitis vaccine and the first contacts for possible clinical trials with the EGF vaccine for lung cancer). Cuban biotechnology centers have already registered: 24 products, both biopharmaceuticals and vaccines. 49 cutting edge generic medicine. 5 products for treating AIDS. 15 new medical equipment 24 diagnosis systems. And moving on from these results, scientific research is now focusing on 60 projects. Among the most significant new products that scientific researchers try to obtain are the following: 29 new vaccines including 8 cancer vaccines, 4 of which are in the clinical trial stage not only in Cuba but also in Canada, Argentina and England. 21 innovative products for treating cancer plus the 28 cytostatics already known, which a new plant will begin producing. Cuban biotechnology centers have filed for patents on 150 inventions that would bring our total number of patents registered to 505. Four of these patents have been awarded the Medal of the World Intellectual Property Organization. Countless politicians, scientists and businesspeople have visited Cuban biotechnology institutes. In the year 2000, 1520 people visited just one of the most important centers; 484 of them came from the United States. The doors of our research centers are opened to any international institution. In an official public note, Cuba has proposed three important draft agreements to the United States, which are more beneficial for the United States than they are for Cuba, given the extent of the problems in each of the two countries. One, a draft agreement on immigration issues; two, a draft cooperation agreement to fight illegal trafficking in narcotic drugs and psychotropic substances; three, a draft bilateral cooperation program to fight terrorism. We have not received any reply whatsoever. Perhaps their reply is to accuse us of manufacturing biological weapons? Who do they think they can intimidate with that? We are urged to stop any kind of cooperation applicable to biological weapons with rogue states and to meet all our obligations under the Convention on Biological and Toxin Weapons. What is the international organization that decides whether a country is or is not a rogue state? What is the rule of the Convention on Biological Weapons that Cuba has violated? Is it perhaps that on top of the criminal blockade they are now trying to prevent us from marketing our medicines and using our most wholesome and noblest products, the fruit of the talent of our scientists, to place them at the service of any person's health anywhere in the world? Could it be that the U.S. government wants to have a bilateral agreement in addition to those Cuba has proposed, namely, cooperation in the struggle against the production of biological weapons? Say so, then. We would be willing to include it on our list of projects pending a reply. We are sorry, Mr. Bolton. After the lies, slanders, tales and insults you launched in your May 6 speech, we are sorry to tell you that you lack any morale to make exhortations to Cuba on this subject and even less to make any demand at all using a threatening tone and language. Neither can you pretend to give Cuba any lesson in politics or ethics. Anyway, you and your government could draw inspiration from Cuba's decent and honorable behavior. I can assure you that we will charge absolutely nothing for this technology transfer. Friday, May 10, 2002 > Suicide truck bomber kills at least 130 in Baghdad > By Tina Susman, Times Staff Writer BAGHDAD -- A suicide bomber posing as a trucker hauling food drove into a > busy market in central Baghdad today and blew himself up, killing at least > 130 people, injuring more than 300, and capping a particularly volatile day > in Iraq's sectarian war. Seven car bombs also tore through the northern city of Kirkuk, leaving two > dead in attacks that underscored tensions enveloping the oil-rich region as > Arabs and Kurds vie for power. Another person died south of Baghdad in a > separate car bombing. The incidents illustrated the breadth of polarization plaguing Iraq as U.S. > and Iraqi officials prepare for what they vow will be a decisive crackdown > on insurgents. The Interior Ministry said the blast in the market in Sadriya killed at > least 130 people and injured 305. It came in the early evening as shoppers > thronged the stalls and tiny shops to stock up for the evening meal. Hamed > Majed, a butcher, was tending to customers when he saw a large yellow truck > trying to navigate down the narrow, alley-like street leading through the > market. First the tiny hezbollah destroyed merkevas one after another using > precision anti-tank weapons reducing the israelis to madness of > indiscriminate cluster bombing. But the Iranians are the smartest of these Islamics. Grand chess > masters ... who once saved the ass of jews under Cyrus the Great from > Nebuchadnezzer > Open house at Iranian nuclear site Isfahan, Iran > February 5, 2007 > AdvertisementAdvertisement Iran has opened one of its nuclear sites to local and international > reporters and ambassadors to show the transparency of its program > before a UN Security Council deadline this month. Tehran has kept up efforts to give the site at Isfahan more publicity. > A tourism official said late last year that Iran planned to open it > and other nuclear plants to foreign tourists. Iran's ambassador to the International Atomic Energy Agency, Ali > Asghar Soltanieh, said the purpose of the tour was to assure the world > that Iran's program was peaceful. In fact we have representatives from all over the world, Mr > Soltanieh said. We decided to have them come here and see for > themselves, he said. He made a point of emphasising the IAEA surveillance cameras at > Isfahan. Photographers and video camera operators were not allowed to take > pictures in the outside area of the compound. Uranium ore is converted into yellowcake and gas at Isfahan. The gas > is transferred to the more sophisticated plant at Natanz, where it > could be enriched with centrifuges. Reporters at the weekend passed the Natanz site but were not allowed > inside, where Iran recently said it was installing 3000 centrifuges. NEW YORK TIMES === Subject: Primes in an arithmetic progression If a(n) is an arithmetic progression with integers terms, then a(n) = a(1) + (n - 1)r, a(1) and r integers. For infinitely many n, n-1 is a multiple of a(1), so that a(n) is composite. But is it possible that there are infinitely many prime terms? Sharon === Subject: Re: Primes in an arithmetic progression days. My association with the Department is that of an alumnus. >If a(n) is an arithmetic progression with integers terms, then a(n) = >a(1) + (n - 1)r, a(1) and r integers. For infinitely many n, n-1 is a >multiple of a(1), so that a(n) is composite. But is it possible that >there are infinitely many prime terms? A very famous and reasonably difficult result, known as Dirichlet's Theorem on Primes in Arithmetic Progressions states that: If gcd(a,b)=1, then there are infinitely many primes p such that p=b (mod a). Equivalently, If gcd(a,b)=1, then the arithmetic progression a(n) = b + (n-1)a contains infinitely many primes. So: to answer your question: if a(1) and r are relatively prime, then a(n) will be a prime infinitely many times. If a(1) and r are not relatively prime, then of course the sequence will produce no primes at all. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Primes in an arithmetic progression > If a(1) and r are not relatively prime, then of course the > sequence will produce no primes at all. I'm not sure what you mean by produce, since the first element of the series may already be a prime. Do you mean by produce generate some new prime you didn't already start with? Such a sequence (where gcd(a(a),r)>1, both a,r positive integers) *contains* one prime in the case where a(1) = is prime, and no primes if a(1) is composite). With that perverse (misleading) definition of produce, yeah the sequence doesn't produce any primes. (Or you simply overlooked that case and misstated the fact.) === Subject: Re: Primes in an arithmetic progression days. My association with the Department is that of an alumnus. >> If a(1) and r are not relatively prime, then of course the >> sequence will produce no primes at all. I'm not sure what you mean by produce, since the first element of >the series may already be a prime. Fair enough; it should say: If a(1) and r are not relatively prime, then the only term that could be a prime is a(1) itself. No other term in the sequence will be prime. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Primes in an arithmetic progression nicegirl_130@yahoo.com a .8ecrit : > If a(n) is an arithmetic progression with integers terms, then a(n) = > a(1) + (n - 1)r, a(1) and r integers. For infinitely many n, n-1 is a > multiple of a(1), so that a(n) is composite. But is it possible that > there are infinitely many prime terms? > Sharon > Look at Dirichlet theorem === Subject: Re: cellular automata puzzle No-one seems to have posted the actual correct solution to > comp.theory.cell-automata yet, so let me do that: I think you're wrong here, since i have posted such a solution (see the post dated on Jan. 9), for which i believe to be correct. The number of edges separating black and white cells cannot increase > under the update rule. A block of N*N black cells has 4N such edges. > The minimum number of black cells having that many such edges is N. The approach is completely different from yours, though, which also seems to be correct. > -- > Ilmari Karonen > To reply by e-mail, please replace .invalid with .net in address. Faton Berisha === Subject: Example of function rings Im sure there are lots of examples of function rings (or f-rings as sometimes they are called), but is there an example of one that is 1. Local 2. Not an integral domain and nor regular 3. commutative and unitary Would appreciate any advises. I am new in this area and I need as much help and reading as I can on it. Jose Capco === Subject: eigenvalues from finite difference So I am looking at the matrix from approximating u_xx by (u_(j-1) - 2u_j + u_(j+1))/h^2. It gives the tridiagonal matrix with -2 on main diagonal and 1's on the bands above and below it. I'm supposed to show the eigenvalues are L_k = cos^2(k*pi/n) for 1<= k <= n-1 if the boundary conditions are both zero, i.e., u(0) = u(1) = 0. The hint is to look for solutions of the form u_j = w^j, where w is a complex number to be determined. This is what I've tried (am I on the right track)? Au = Lu, L scalar lambda Looking at an arbitrary equation in the system: u_(j-1) - 2u_j + u_(j+1) = L*u_j u_j^-1 - 2 + u_j = L Using u_j = w^j: w_j^-1 - 2 + w_j = L I then tried using trig expansions for the complex numbers, but it didn't lead to anything. === Subject: Re: eigenvalues from finite difference Meant to say: Looking at an arbitrary equation in the system: u_(j-1) - 2u_j + u_(j+1) = L*u_j Using u_j = w^j: w^(j-1) - 2w^j + w^(j+1) = L*w^j w^-1 - 2 + w = L I then tried using trig expansions for the complex numbers, but it didn't lead to anything. === Subject: Urgent Action Needed to help Millions of Laboratory Animals Tony Nottingham England 4 February 2007 TO: All People Who Care I am a private individual posting this message to help Laboratory Animals. A new European Chemical Testing Policy called REACH has now been finalised by the European Union. Under these proposals Chemicals of every kind - from those used in industrial processes to the ingredients of consumer products - will be tested on Millions of animals from mice to fish to dogs, causing untold suffering. If you believe that REACH should make more use of Alternative Testing to test its Chemicals then please take action now. If you are a European citizen please contact your local papers and own MEP asking them to promote the development of humane non-animal testing methods under the REACH legislation, which is the best hope we have for sparing animals the misery of a testing laboratory. A sample letter can be found at the BUAV (British Union for the Abolition of Vivisection) website at www.BUAV.org (Select Campaigns, Chemical Testing, Get Active then Writing to the local press). For all Non-European members, you can still help. Please tell all your colleagues and friends in the UK and Europe about REACH. Everyone can help and you can make a difference. === Subject: Re: Urgent Action Needed to help Millions of Laboratory Animals > Tony > Nottingham > England 4 February 2007 TO: All People Who Care I am a private individual posting this message to help > Laboratory Animals. A new European Chemical Testing Policy called > REACH has now been finalised by the European > Union. Under these proposals Chemicals of every > kind - from those used in industrial processes to the > ingredients of consumer products - will be tested on > Millions of animals from mice to fish to dogs, causing > untold suffering. What would you prefer? Testing on people? If sacrificing animals to produce medicine and safe products I say do it. Hooray for the humans and tough for the animals. Bob Kolker > === Subject: Re: Urgent Action Needed to help Millions of Laboratory Animals > What would you prefer? Testing on people? Neither is necessary. > If sacrificing animals to produce medicine and safe products I say do > it. Hooray for the humans and tough for the animals. That's not the right attitude Bob. For alternate methods of testing, see for example: http://www.navs.org Killing an animal for food, clothing or protection, is ethically acceptable. Mistreating animals in any way is ethically unacceptable. In general, two kinds of people I consider savages and uncivilized: The musically uneducated and those who mistreat animals. > Bob Kolker -- I.N. Galidakis http://ioannis.virtualcomposer2000.com/ === Subject: Re: Urgent Action Needed to help Millions of Laboratory Animals I like animals. They taste good. OMU > Tony > Nottingham > England 4 February 2007 TO: All People Who Care I am a private individual posting this message to help > Laboratory Animals. A new European Chemical Testing Policy called > REACH has now been finalised by the European > Union. Under these proposals Chemicals of every > kind - from those used in industrial processes to the > ingredients of consumer products - will be tested on > Millions of animals from mice to fish to dogs, causing > untold suffering. If you believe that REACH should make more use of > Alternative Testing to test its Chemicals then please take > action now. If you are a European citizen please contact your local > papers and own MEP asking them to promote the > development of humane non-animal testing methods > under the REACH legislation, which is the best hope we > have for sparing animals the misery of a testing laboratory. A sample letter can be found at the BUAV (British Union > for the Abolition of Vivisection) website atwww.BUAV.org > (Select Campaigns, Chemical Testing, Get Active then > Writing to the local press). For all Non-European members, you can still help. > Please tell all your colleagues and friends in the UK and > Europe about REACH. Everyone can help and you can make a difference. > === Subject: The Math Professor 1.5 Hey parents, this is just awesome. This computer program gives your child the perfect opportunity to understand basic mathematics. It contains over 200 high quality audio-visual sessions, animations, and the children's favorite cartoon characters. Just have a look, www.s-logan.com/html/products.html You can download the free trial from www.s-logan.com/html/ download.html It's great! === Subject: Re: The Math Professor 1.5 > It's great! Yeah. For 39.95 you can crush all creativity and self-sufficiency out of your child by having some cartoon characters do their thinking for them. No wonder the freshmen get dumber every year. -- The man without a .sig === Subject: Re: Simple Physics Question I asked this question before but I think I phrased it incorrectly. An object of mass M is fixed at a certain position. Another object of >mass m, initially at a distance R from M, is released from rest. What >is the time it takes for m to reach M according to Newton's Inverse >Square Law of Gravitation? Following Tim Murphy's suggestion > vdv/dr = - GM/r^2 > v^2 = 2GM(1/R - 1/r) Wrong. v^2 = 2GM(1/r - 1/R) . Otherwise v^2 would be < 0 . > you can get to > dt = dr/v = 1/(2GM*sqrt(R^-1 - r^-1)) Wrong. dt = dr / sqrt(- 2GM/R + 2GM/r ) > which after integrating yields in reduced form > t = sqrt(2)*r^3/2*sqrt(r/R-1)/sqrt(GM) ? How does that compare with: t = sqrt(R^3/(2.GM)) [- sqrt(r/R.(r/R-1)) + arcsin(2.r/R-1)/2 + pi/4] > As example, drop 1000 m on earth. Integral gives > t = 14.289 s and > using g = 9.7995ms^-2 yields .4 m difference: > 1/2 gt^2 = 1000.4m > John Polasek Sorry? Han de Bruijn === Subject: Re: Simple Physics Question [ ... ] we can integrate (via Maple) and get a solution of the form Maybe. But my Maple says the following: x(t) := (9*G*M/2)^(1/3)*t^(2/3); 1/3 2/3 1/3 2/3 9 2 (G M) t x(t) := ----------------------- 2 > simplify(diff(diff(x(t),t),t) + G*M/x(t)^2); 0 Han de Bruijn === Subject: Re: Simple Physics Question I asked this question before but I think I phrased it incorrectly. An object of mass M is fixed at a certain position. Another object of > mass m, initially at a distance R from M, is released from rest. What > is the time it takes for m to reach M according to Newton's Inverse > Square Law of Gravitation? This is problem is a case of the arclength problem, and it generally > cannot be solved in terms of elementary functions (i.e. a nice > symbolic solution). You usually end up getting a bunch of elliptic > integrals to solve, but in your case, a symbolic solution is > salvagable. Maybe I'm completely at lost, but I still find that it's just a matter of solving a rather elementary ordinary differential equation: d^2x/dt^2 = - G.M/x^2 Of which the outcome is: x(t) = (9.G.M/2)^(1/3).t^(2/3) . Now put x(t) = R, solve for (t) and you are done. Han de Bruijn === Subject: Re: Simple Physics Question Han.deBruijn@DTO.TUDelft.NL a .8ecrit : I asked this question before but I think I phrased it incorrectly. > An object of mass M is fixed at a certain position. Another object of > mass m, initially at a distance R from M, is released from rest. What > is the time it takes for m to reach M according to Newton's Inverse > Square Law of Gravitation? >> This is problem is a case of the arclength problem, and it generally >> cannot be solved in terms of elementary functions (i.e. a nice >> symbolic solution). You usually end up getting a bunch of elliptic >> integrals to solve, but in your case, a symbolic solution is >> salvagable. Maybe I'm completely at lost, but I still find that it's just a > matter > of solving a rather elementary ordinary differential equation: d^2x/dt^2 = - G.M/x^2 Of which the outcome is: x(t) = (9.G.M/2)^(1/3).t^(2/3) . Now put x(t) = R, solve for (t) and you are done. Han de Bruijn > Deat Mr de Bruijn I am sure many competent mathematicians here could answer your query (from David Ullrich to myself, just to name two). But how can you hope for some help, after you called us incompetent liars? === Subject: Re: Simple Physics Question <45c5f061$0$21149$7a628cd7@news.club-internet.fr> On 4 feb, 15:40, Denis Feldmann I am sure many competent mathematicians here could answer your query > (from David Ullrich to myself, just to name two). But how can you hope for some help, after you called us incompetent liars. Huh? I'm not the original poster of this problem. And I didn't need any help of you competent mathematicians. You're completely at lost. Han de Bruijn === Subject: Re: Simple Physics Question Han.deBruijn@DTO.TUDelft.NL a .8ecrit : > On 4 feb, 15:40, Denis Feldmann >> I am sure many competent mathematicians here could answer your query >> (from David Ullrich to myself, just to name two). >> But how can you hope for some help, after you called us incompetent liars. Huh? I'm not the original poster of this problem. And I didn't need > any > help of you competent mathematicians. You're completely at lost. If you say so... But then, why are you saying I did something wrong? Han de Bruijn > === Subject: Re: Simple Physics Question <45c5f061$0$21149$7a628cd7@news.club-internet.fr> <45c5f91a$0$21144$7a628cd7@news.club-internet.fr> On 4 feb, 16:17, Denis Feldmann If you say so... But then, why are you saying I did something wrong? I've made a mistake. And corrected it in another poster. Without your help, BTW. Han de Bruijn === Subject: Re: Simple Physics Question > I asked this question before but I think I phrased it incorrectly. An object of mass M is fixed at a certain position. Another object of > mass m, initially at a distance R from M, is released from rest. What > is the time it takes for m to reach M according to Newton's Inverse > Square Law of Gravitation? Picture of the situation: m -------------------------- M 0 R The Differential Equation to be solved is: m.d^2x/dt^2 = - G.M.m/x^2 ==> d^2x/dt^2 = - G.M/x^2 Ansatz: x = c.t^p which is 0 for t = 0 : correct. Substitute in the D.E.: c.p.(p-1) . t^(p-2) = - G.M/c^2 . t^(-2.p) c.2/3.(1-2/3) = - G.M/c^2 ==> c = (9/2.G.M)^(1/3) The mass at M is reached by m for x(t) = R . It follows that: (9.G.M/2)^(1/3).t^(2/3) = R ==> t^(2/3) = R/(9.G.M/2)^(1/3) ==> t = R^(3/2)/(9.G.M/2)^(1/2) = 1/3.sqrt(2.R^3/(G.M)) . Iff I've made no mistakes .. Han de Bruijn === Subject: Re: Simple Physics Question >> I asked this question before but I think I phrased it incorrectly. >> An object of mass M is fixed at a certain position. Another object of >> mass m, initially at a distance R from M, is released from rest. What >> is the time it takes for m to reach M according to Newton's Inverse >> Square Law of Gravitation? Picture of the situation: m -------------------------- M > 0 R The Differential Equation to be solved is: m.d^2x/dt^2 = - G.M.m/x^2 ==> d^2x/dt^2 = - G.M/x^2 Ansatz: x = c.t^p which is 0 for t = 0 : correct. Substitute in the D.E.: c.p.(p-1) . t^(p-2) = - G.M/c^2 . t^(-2.p) > c.2/3.(1-2/3) = - G.M/c^2 ==> c = (9/2.G.M)^(1/3) The mass at M is reached by m for x(t) = R . It follows that: (9.G.M/2)^(1/3).t^(2/3) = R ==> t^(2/3) = R/(9.G.M/2)^(1/3) == > t = R^(3/2)/(9.G.M/2)^(1/2) = 1/3.sqrt(2.R^3/(G.M)) . The fact that this is a solution to the differential equation does not show that it is the solution. Does it satisfy the initial conditions? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Simple Physics Question does not show that it is the solution. > Does it satisfy the initial conditions? Oops! Seems that I did something wrong. Sorry. Han de Bruijn === Subject: Re: Simple Physics Question > I asked this question before but I think I phrased it incorrectly. An object of mass M is fixed at a certain position. Another object of > mass m, initially at a distance R from M, is released from rest. What > is the time it takes for m to reach M according to Newton's Inverse > Square Law of Gravitation? I expect I learnt a clever way of doing this in my first year. Unfortunately I've long since forgotten it, but I'll have a go anyway. I assume that they are point masses so that we are looking to find the time when their relative displacement x is zero. I will define x and t so that x = R when t = 0. The equation you're interested in is x'' = -c/x^2, where I have defined c = G*M for brevity. Let's write this as v' = -c/x^2 where v is x'. The best way I can think of to proceed is to use conservation of energy to get v in terms of x: from m*v^2/2 = c*m/x - c*m/R we find that 1/x = 1/R + v^2/(2*c), and so we are now interested in solving dv/dt = -c (1/R + v^2/(2*c))^2 Turning this upside down we get dt/dv = -(1/c) (1/R + v^2/(2*c))^(-2) = -4*c (2*c/R + v^2)^(-2) = -4*c (a^2 + v^2)^(-2), where again for brevity I have defined a^2 = 2*c/R. Now I plug this into Mathematica's online integrator and I get that t = -2*c/a^3 (a*v/(a^2 + v^2) + arctan(v/a)) + constant; since v = 0 when t = 0 the constant is zero. We wish to know the value of t when x = 0 and therefore v = -infinity. Of course the equation above is ill-defined at this point but we may consider the limit as v tends to minus infinity: then the first term in parentheses goes to zero and the arctan goes to -pi/2, so t = pi*c/a^3, or restoring the original constants t = pi/(2*sqrt(2)) (R^3/(G*M))^(1/2). I offer no guarantee that this is the correct answer, though the units check out. -Rotwang P.S. @ Dumbledore: the OP just wanted help answering a classical mechanics question. Since you evidently have no help to offer, why do you feel the need to turn the thread into an anti-relativist slagging match? === Subject: Re: Simple Physics Question > t = pi/(2*sqrt(2)) (R^3/(G*M))^(1/2). I offer no guarantee that this is the correct answer, though the units > check out. I've checked it all out and IMO it's the one and only right answer. Conservation of energy: v^2/2 = GM/x - GM/R ==> dt = dx / sqrt(2.GM/x - 2.GM/R) Dimensioneless quantities are always advantageous: t / sqrt(R^3/(2.GM)) = integral d(x/R) / sqrt(1/(x/R) - 1) The integral is equal to - sqrt(x/R.(x/R - 1)) + arcsin(2.x/R - 1)/2 + C The integration constant is determined with t = 0 for x = 0 : C = pi/ 4 The arrival time is determined for x = R , giving for the integral the value pi/4 + pi/4 = pi/2 . Thus t = sqrt(R^3/(2.GM)) . pi/2 , which is exactly the same value as found above, by the other author. Han de Bruijn === Subject: Re: Simple Physics Question t = pi/(2*sqrt(2)) (R^3/(G*M))^(1/2). I offer no guarantee that this is the correct answer, though the units > check out. I've checked it all out and IMO it's the one and only right answer. program to check the answer numerically on my calculator, and following the results I am also pretty confident that it's correct. -Rotwang === Subject: Re: Simple Physics Question > The integral is equal to - sqrt(x/R.(x/R - 1)) + arcsin(2.x/R - 1)/2 + C Another oops. Must be: - sqrt(x/R.(1-x/R)) + arcsin(2.x/R - 1)/2 + C Han de Bruijn === Subject: Re: Simple Physics Question > t = pi/(2*sqrt(2)) (R^3/(G*M))^(1/2). I offer no guarantee that this is the correct answer, though the units > check out. In a previous poster, I've found that: t = sqrt(2)/3.(R^3/(G*M))^(1/2) Which is different from your outcome: pi/(2.sqrt(2)) <> sqrt(2)/3 . Who is right and who is wrong? Han de Bruijn === Subject: Re: Simple Physics Question <45C4CE1A.ECB66E33@hate.spam.net> <45C51D7F.C367608B@hate.spam.net http://www.mazepath.com/uncleal/effete6.jpg This one looks very prickly. ;>) M === Subject: An Isomorphism Theorem 8.3 in Gallian's Contemp. Abst Alg says with (s, t) = 1 the group U(st) is isomorphic to the external direct product of U(s) and U(t) that is, to U(s) (+) U(t) All of you probably know U(n) is the group of positive integers less than n and relatively prime to n with the group operation multiplication mod n This is the isomorphism: U(st) -> U(s)(+) U(t) f : x -> (x mod s, x mod t) I'm stuck on showing f is one-one. If f(a) = f(b) (a mod s, a mod t) = (b mod s, b mod t) and a mod s = b mod s and a mod t = b mod t I tried but I can't show a = b. I suspect this is easy and I'm getting frustrated. The author's hints are a mod n = aÍ and b mod n = bÍ => (ab) mod n = (aÍbÍ) mod n a mod st = b mod st => a mod s = b mod s and a mod t = b mod t (a, bc) = 1 => (a, b) = 1 and (a, c) = 1 === Subject: Re: An Isomorphism days. My association with the Department is that of an alumnus. >Theorem 8.3 in Gallian's Contemp. Abst Alg >says with (s, t) = 1 the group U(st) is isomorphic >to the external direct product of U(s) and U(t) >that is, to U(s) (+) U(t) All of you probably know U(n) is the group of positive >integers less than n and relatively prime to n with the >group operation multiplication mod n This is the isomorphism: >U(st) -> U(s)(+) U(t) f : x -> (x mod s, x mod t) I'm stuck on showing f is one-one. You could use the Chinese Remainder Theorem to prove that it is a bijection directly... >If f(a) = f(b) >(a mod s, a mod t) = (b mod s, b mod t) and >a mod s = b mod s and a mod t = b mod t I tried but I can't show a = b. Well, you cannot show a = b. What you can show is a = b (mod st). If a = b (mod s), then s|a-b If a = b (mod t), then t|a-b. Now use the following lemma: Lemma: If gcd(s,t)=1, s|x, and t|x, then st|x. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: An Isomorphism >> Theorem 8.3 in Gallian's Contemp. Abst Alg >> says with (s, t) = 1 the group U(st) is isomorphic >> to the external direct product of U(s) and U(t) >> that is, to U(s) (+) U(t) >> All of you probably know U(n) is the group of positive >> integers less than n and relatively prime to n with the >> group operation multiplication mod n >> This is the isomorphism: >> U(st) -> U(s)(+) U(t) f : x -> (x mod s, x mod t) >> I'm stuck on showing f is one-one. You could use the Chinese Remainder Theorem to prove that it is a > bijection directly... > >> If f(a) = f(b) >> (a mod s, a mod t) = (b mod s, b mod t) and >> a mod s = b mod s and a mod t = b mod t >> I tried but I can't show a = b. Well, you cannot show a = b. What you can show is a = b (mod st). I thought because b in U(st) then b If a = b (mod t), then t|a-b. Now use the following lemma: Lemma: If gcd(s,t)=1, s|x, and t|x, then st|x. So I think you're saying st|(a-b) I should know this, but apparently st|(a-b) <=> a = b(mod st) And we're done. === Subject: Re: An Isomorphism days. My association with the Department is that of an alumnus. > Theorem 8.3 in Gallian's Contemp. Abst Alg > says with (s, t) = 1 the group U(st) is isomorphic > to the external direct product of U(s) and U(t) > that is, to U(s) (+) U(t) All of you probably know U(n) is the group of positive > integers less than n and relatively prime to n with the > group operation multiplication mod n This is the isomorphism: > U(st) -> U(s)(+) U(t) f : x -> (x mod s, x mod t) I'm stuck on showing f is one-one. >> >> You could use the Chinese Remainder Theorem to prove that it is a >> bijection directly... >If f(a) = f(b) > (a mod s, a mod t) = (b mod s, b mod t) and > a mod s = b mod s and a mod t = b mod t I tried but I can't show a = b. >> >> Well, you cannot show a = b. What you can show is a = b (mod st). I thought because b in U(st) then bthose<-.... (Or maybe Gillian is). To most mathematicians, mod n is not a unary operator. It does ->not<- give you the remainder when you divide by n. Rather, equivalence modulo n is a binary RELATION between two integers. We say that a = b (mod n) if and only if n divides a-b. Some computer scientist types, however, prefer to use mod as a unary operator. You give it an integer a modulo n, and it gives you the unique integer t such that a = t (mod n) and 0<= t < n (using the meaning of mod n I give above). Your book (or you) seem to be one of those kinds. It makes things much harder. >> If a = b (mod s), then s|a-b >> If a = b (mod t), then t|a-b. >> >> Now use the following lemma: >> >> Lemma: If gcd(s,t)=1, s|x, and t|x, then st|x. So I think you're saying st|(a-b) That would be the consequence I would want you to deduce from what you have, applying the lemma, yes. >I should know this, but apparently st|(a-b) <=> a = b(mod st) No apparently about it. That is ->the definition<- of congruence modulo st. Do leave some space after the b, though, lest you keep thinking of it as a unary operator applied to b. >And we're done. Indeed. Much easier than messing around with remainders. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: An Isomorphism >> I tried but I can't show a = b. > Well, you cannot show a = b. What you can show is a = b (mod st). >> I thought because b in U(st) then bthose<-.... > (Or maybe Gillian is). To most mathematicians, mod n is not a unary operator. It does ->not<- > give you the remainder when you divide by n. Rather, equivalence modulo n is a binary RELATION between two > integers. We say that a = b (mod n) if and only if n divides a-b. (I never knew I was one of those. I will try to reform.) === Subject: Bourbaki? I just picked up a book in the local library that is a biography of Coxeter (The Infinite Mind). In early pages the author describes Bourbaki as a real person. Now its my impression that Bourbaki is a name adopted by a group of French mathematicians in the early part of the 20th century but I coudn't believe that an author who had gone to the trouble of writing a biography would make such a mistake. Am I missing something. Am I wrong about Bourbaki? === Subject: Re: Bourbaki? > I just picked up a book in the local library that is a biography of > Coxeter (The Infinite Mind). In early pages the author describes > Bourbaki as a real person. Now its my impression that Bourbaki is a > name adopted by a group of French mathematicians in the early part of > the 20th century but I coudn't believe that an author who had gone to > the trouble of writing a biography would make such a mistake. Am I > missing something. Am I wrong about Bourbaki? Depends. Bourbaki is adopted as name by a group of French mathematicians. It was also the name of a real person (a particularly unsuccesful French general). Originally it was a joking adaption, but became more serious in the course of times. The family of the real general was not amused. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Bourbaki? > I just picked up a book in the local library that is a biography of > Coxeter (The Infinite Mind). In early pages the author describes > Bourbaki as a real person. Now its my impression that Bourbaki is a > name adopted by a group of French mathematicians in the early part of > the 20th century but I coudn't believe that an author who had gone to > the trouble of writing a biography would make such a mistake. Am I > missing something. Am I wrong about Bourbaki? The group Accounts of the early days vary, but original documents have now come to light. The founding members were all connected to the Ecole Normale Sup.8erieure in Paris and included Henri Cartan, Claude Chevalley, Jean Coulomb, Jean Delsarte, Jean Dieudonn.8e, Charles Ehresmann, Ren.8e de Possel, Szolem Mandelbrojt and Andr.8e Weil. There was a preliminary meeting, towards the end of 1934.[3] Jean Leray and Paul Dubreil were present at the preliminary meeting but dropped out before the group actually formed. Other notable participants in later days were Laurent Schwartz, Jean-Pierre Serre, Alexander Grothendieck, Samuel Eilenberg, Serge Lang and Roger Godement. Bob Kolker > === Subject: Re: Bourbaki? > I just picked up a book in the local library that is a biography of > Coxeter (The Infinite Mind). In early pages the author describes > Bourbaki as a real person. Now its my impression that Bourbaki is a > name adopted by a group of French mathematicians in the early part of > the 20th century but I coudn't believe that an author who had gone to > the trouble of writing a biography would make such a mistake. I was equally confused, but afair the author later on reveals that Bourbaki is a pseudonym and describes the group Bourbaki. > Am I missing something. Am I wrong about Bourbaki? Carsten -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page. === Subject: Re: Bourbaki? the evil, collective genius behind The New Math; I was taught set-theory in the 3rd grading, when I came to LAUSD!... although, it's isomorphic to arithmetic, so. > I was equally confused, but afair the author later on reveals that > Bourbaki is a pseudonym and describes the group Bourbaki. thus: when is the Bucky festschrift, or is that for living authors only? in the meantime, Geometry, which was published in 1925, translated into many languages including Chinese, becoming the standard text in the subject across the nation. After being in continuous use, without revision, for a quarter century--which, I gather, is something of a world's record--it was revised in 1952. His second book was called Modern Pure Solid Geometry and it appeared in 1935 and was revised in 1964. And a popular work, collecting some earlier pieces, called Mathematics in Fun and in Earnest, appeared in 1958 and was also widely translated. In his sixty years as an active mathematician, Nathan Court published more than a hundred papers in more than a dozen journals. These came from his pen in a steady stream: 6 before 1920; 35 during the 1920s; 23 in the 1930s; 26 in the 1940s; and more than 50 after his retirement. Professor Springer has written that It is most unusual for a mathematician to produce scholarly works after reaching the age of 70. On the occasion of a departmental dinner at the time of Dr. Court's retirement from his teaching duties ... it was remarked that, unlike most mathematicians of his age, he would probably continue in active research and writing. This he did. Probably his period of greatest production was the 14 years of his so-called retirement ... This summary of his work does not include dozens of problems that he proposed or solved in numerous journals. Through the administrations of six OU presidents, Court would dutifully send reprints of his work to their offices Evans Hall. He would always receive back polite letters like the one from William Bennett Bizzell in 1930: I wish to thank you for sending me your translation of chon's Theorem. While I confess, I cannot possibly understand it, I must say frankly, that I greatly admire a man who cannot only understand this mathematical theorem, but who has the ability to translate this proposition into English. Springer thought that Nathan Court was a genius in geometry, and the one person most responsible for the introduction of a college course in synthetic geometry. http://math.ou.edu/~amagid/COURT.TXT thus: have you ever seen it in one of those books?... I read a two-and-a-half page proof of this in Mathematics Magazine (MAA), many years ago, but I don't recall how to do it (I did work through it, once .-)... a nice exercise would be to convert Fermat's last theorem specialcase of n=4 -- the obvious exception to the rule, since 2+2=2*2+2^2=etc. by the AP observation (first place I've ever seen it) -- from infinite descent. > Besides, you also seem to have clipped the part about being able to > take a Direct proof and convert it into an Indirect proof, and vice > versa. That particular derivation should be in any Symbolic Logic book. thus: monsieur Chickenpoop, I've reconnoitered what I replied to your prior attempt. anyway, any personal anecdotes about the MIT expert would be appreciated, because I want to apply to the machine-shop for employment; thank *you*.... now, considering that your bud's report was done *in* 2001, it's excuritiatingly exquisite. so, so what, if he didn't take into account the difference betweem a plane impacting at 500mph, and the transient loads of the designed-for wind?... welcome to d'hotel googolplex; youcanloginanytimeyoulikebut -- and, there's a newsgroup just for you: sci.beauxeaux.synonyme. > World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws thus: I've read most of MIT Head Welder's page -- really excellent for its size, with lots of things that I'd never seen or thought of. now, see if you can actually refute *any* thing that he says, instead of using bogus ad hominems, like, of course, Professor Borat refuted that in the 16th century BCE; he was really a great friend of mine, when we went to MIT! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html thus: Bernard BORAT Lewis WANTS YOU IN SUDAN; Iran; Afghanistan; Iraq. other plausibly British Quagmires, like Belize and Canada; and, basically, anywhere that will destroy the USA economy, faster. thus: what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing.... the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and, of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin., with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, working for Lloyds of London! thus: however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause.... as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary.... --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: Bourbaki? days. My association with the Department is that of an alumnus. >I just picked up a book in the local library that is a biography of >Coxeter (The Infinite Mind). In early pages the author describes >Bourbaki as a real person. Now its my impression that Bourbaki is a >name adopted by a group of French mathematicians in the early part of >the 20th century but I coudn't believe that an author who had gone to >the trouble of writing a biography would make such a mistake. Am I >missing something. Am I wrong about Bourbaki? There was a real Bourbaki, but he was not a mathematician. The mathematician Bourbaki is indeed a pseudonym, though some were taken in; and the group went to some efforts to perpetuate the joke, even writing a letter of complaint to the Encylopedia Britannica when person. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Bourbaki? The real Bourbaki was a general in the French army during the Franco-Prussian war. === Subject: Re: new Conjecture in mathematics-- Theaetetus-Plutonium Conjecture Re: about Dodecahedron Problem and Local Density Re: #12# new book of course, just truncate tiny fractions of the vertices of the hexahedron, to get any ration of space-filling; niether the tetrahedron nor octah. fill space, by themselves. rhombical cuboctah. is the only other semiregular shape that fills space -- all students o'Bucky know it ... I mean, as well as the (archimedean) hexakaidecah. (or truncated octah.). just keep on tossing the **** out; something will eventually stick -- if anyone can ever decipher it from your subliterate attempts; sorry. > However, with Octahedron, Cube and Tetrahedron Packing they tile the > entire > Space with no holes or gaps. So the Conjecture is a generalization of thus: the evil, collective genius behind The New Math; I was taught set-theory in the 3rd grading, when I came to LAUSD!... although, it's isomorphic to arithmetic, so. > I was equally confused, but afair the author later on reveals that > Bourbaki is a pseudonym and describes the group Bourbaki. thus: when is the Bucky festschrift, or is that for living authors only? in the meantime, Geometry, which was published in 1925, translated into many languages including Chinese, becoming the standard text in the subject across the nation. After being in continuous use, without revision, for a quarter century--which, I gather, is something of a world's record--it was revised in 1952. His second book was called Modern Pure Solid Geometry and it appeared in 1935 and was revised in 1964. And a popular work, collecting some earlier pieces, called Mathematics in Fun and in Earnest, appeared in 1958 and was also widely translated. In his sixty years as an active mathematician, Nathan Court published more than a hundred papers in more than a dozen journals. These came from his pen in a steady stream: 6 before 1920; 35 during the 1920s; 23 in the 1930s; 26 in the 1940s; and more than 50 after his retirement. Professor Springer has written that It is most unusual for a mathematician to produce scholarly works after reaching the age of 70. On the occasion of a departmental dinner at the time of Dr. Court's retirement from his teaching duties ... it was remarked that, unlike most mathematicians of his age, he would probably continue in active research and writing. This he did. Probably his period of greatest production was the 14 years of his so-called retirement ... This summary of his work does not include dozens of problems that he proposed or solved in numerous journals. Through the administrations of six OU presidents, Court would dutifully send reprints of his work to their offices Evans Hall. He would always receive back polite letters like the one from William Bennett Bizzell in 1930: I wish to thank you for sending me your translation of chon's Theorem. While I confess, I cannot possibly understand it, I must say frankly, that I greatly admire a man who cannot only understand this mathematical theorem, but who has the ability to translate this proposition into English. Springer thought that Nathan Court was a genius in geometry, and the one person most responsible for the introduction of a college course in synthetic geometry. http://math.ou.edu/~amagid/COURT.TXT thus: have you ever seen it in one of those books?... I read a two-and-a-half page proof of this in Mathematics Magazine (MAA), many years ago, but I don't recall how to do it (I did work through it, once .-)... a nice exercise would be to convert Fermat's last theorem specialcase of n=4 -- the obvious exception to the rule, since 2+2=2*2+2^2=etc. by the AP observation (first place I've ever seen it) -- from infinite descent. > Besides, you also seem to have clipped the part about being able to > take a Direct proof and convert it into an Indirect proof, and vice > versa. That particular derivation should be in any Symbolic Logic book. thus: monsieur Chickenpoop, I've reconnoitered what I replied to your prior attempt. anyway, any personal anecdotes about the MIT expert would be appreciated, because I want to apply to the machine-shop for employment; thank *you*.... now, considering that your bud's report was done *in* 2001, it's excuritiatingly exquisite. so, so what, if he didn't take into account the difference betweem a plane impacting at 500mph, and the transient loads of the designed-for wind?... welcome to d'hotel googolplex; youcanloginanytimeyoulikebut -- and, there's a newsgroup just for you: sci.beauxeaux.synonyme. > World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws thus: I've read most of MIT Head Welder's page -- really excellent for its size, with lots of things that I'd never seen or thought of. now, see if you can actually refute *any* thing that he says, instead of using bogus ad hominems, like, of course, Professor Borat refuted that in the 16th century BCE; he was really a great friend of mine, when we went to MIT! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html thus: Bernard BORAT Lewis WANTS YOU IN SUDAN; Iran; Afghanistan; Iraq. other plausibly British Quagmires, like Belize and Canada; and, basically, anywhere that will destroy the USA economy, faster. thus: what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing.... the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and, of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin., with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, working for Lloyds of London! thus: however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause.... as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary.... --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: Exactly why is a prime factor search invalid? Was: Re: correcting Euclid's Infinitude of Primes (indirect method) and publishing it attn: Karl Heuer I like this notation, but three commas should suffice, in analogy with the usual ellipsis; to whom should it be sent, to make it a standard? however, almost everything else that I have ever seen you post, was subliterate BS -- with a couple of exceptions that I will never divulge, because you don't even try to improve your grammar -- I know; Stan was the first grammarian! > the earlier steps said p1,,,,pn are all the primes that exist coupled thus: oops; _The Bicycle Wheel_, I recalled, was written by Jobst Brand, not the Willy Brandt, who was not as Good of a German as Emille Brandt. thus: of course, just truncate tiny fractions of the vertices of the hexahedron, to get any ration of space-filling; niether the tetrahedron nor octah. fill space, by themselves.... rhombical cuboctah. is the only other semiregular shape that fills space -- all students o'Bucky know it ... I mean, as well as the (archimedean) hexakaidecah. (or truncated octah.).... just keep on tossing the **** out; something will eventually stick -- if anyone can ever decipher it from your subliterate attempts; sorry. > However, with Octahedron, Cube and Tetrahedron Packing they tile the thus: Geometry, which was published in 1925, translated into many languages including Chinese, becoming the standard text in the subject across the nation. After being in continuous use, without revision, for a quarter century--which, I gather, is something of a world's record--it was revised in 1952. His second book was called Modern Pure Solid Geometry and it appeared in 1935 and was revised in 1964. And a popular work, collecting some earlier pieces, called Mathematics in Fun and in Earnest, appeared in 1958 and was also widely translated. In his sixty years as an active mathematician, Nathan Court published more than a hundred papers in more than a dozen journals. These came from his pen in a steady stream: 6 before 1920; 35 during the 1920s; 23 in the 1930s; 26 in the 1940s; and more than 50 after his retirement. Professor Springer has written that It is most unusual for a mathematician to produce scholarly works after reaching the age of 70. On the occasion of a departmental dinner at the time of Dr. Court's retirement from his teaching duties ... it was remarked that, unlike most mathematicians of his age, he would probably continue in active research and writing. This he did. Probably his period of greatest production was the 14 years of his so-called retirement ... This summary of his work does not include dozens of problems that he proposed or solved in numerous journals. Through the administrations of six OU presidents, Court would dutifully send reprints of his work to their offices Evans Hall. He would always receive back polite letters like the one from William Bennett Bizzell in 1930: I wish to thank you for sending me your translation of chon's Theorem. While I confess, I cannot possibly understand it, I must say frankly, that I greatly admire a man who cannot only understand this mathematical theorem, but who has the ability to translate this proposition into English. Springer thought that Nathan Court was a genius in geometry, and the one person most responsible for the introduction of a college course in synthetic geometry. http://math.ou.edu/~amagid/COURT.TXT thus: have you ever seen it in one of those books?... I read a two-and-a-half page proof of this in Mathematics Magazine (MAA), many years ago, but I don't recall how to do it (I did work through it, once .-)... a nice exercise would be to convert Fermat's last theorem specialcase of n=4 -- the obvious exception to the rule, since 2+2=2*2+2^2=etc. by the AP observation (first place I've ever seen it) -- from infinite descent. > Besides, you also seem to have clipped the part about being able to > take a Direct proof and convert it into an Indirect proof, and vice > versa. That particular derivation should be in any Symbolic Logic book. thus: I've read most of MIT Head Welder's page -- really excellent for its size, with lots of things that I'd never seen or thought of. now, see if you can actually refute *any* thing that he says, instead of using bogus ad hominems, like, of course, Professor Borat refuted that in the 16th century BCE; he was really a great friend of mine, when we went to MIT! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html thus: Bernard BORAT Lewis WANTS YOU IN SUDAN; Iran; Afghanistan; Iraq. other plausibly British Quagmires, like Belize and Canada; and, basically, anywhere that will destroy the USA economy, faster. thus: what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing.... the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and, of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin., with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, working for Lloyds of London! thus: however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause.... as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary.... --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: Re: Exactly why is a prime factor search invalid? Was: Re: correcting Euclid's Infinitude of Primes (indirect method) and publishing it attn: Karl Heuer have you ever seen it in one of those books?... I read a two-and-a-half page proof of this in Mathematics Magazine (MAA), many years ago, but I don't recall how to do it (I did work through it, once .-) a nice exercise would be to convert Fermat's last theorem specialcase of n=4 -- the obvious exception to the rule, since 2+2=2*2+2^2=etc. by the AP observation (first place I've ever seen it) -- from infinite descent. > Besides, you also seem to have clipped the part about being able to > take a Direct proof and convert it into an Indirect proof, and vice > versa. That particular derivation should be in any Symbolic Logic book. thus: monsieur Chickenpoop, I've reconnoitered what I replied to your prior attempt. anyway, any personal anecdotes about the MIT expert would be appreciated, because I want to apply to the machine-shop for employment; thank *you*. now, considering that your bud's report was done *in* 2001, it's excuritiatingly exquisite. so, so what, if he didn't take into account the difference betweem a plane impacting at 500mph, and the transient loads of the designed-for wind? welcome to d'hotel googolplex; youcanloginanytimeyoulikebut -- and, there's a newsgroup just for you: sci.beauxeaux.synonyme. > World Trade Center -- Controlled Demolition.flv http:// > v187.youtube.com/get_video?video_id=87fyJ-3o2ws thus: I've read most of MIT Head Welder's page -- really excellent for its size, with lots of things that I'd never seen or thought of. now, see if you can actually refute *any* thing that he says, instead of using bogus ad hominems, like, of course, Professor Borat refuted that in the 16th century BCE; he was really a great friend of mine, when we went to MIT! >http://www.tms.org/pubs/journals/JOM/0112/Eagar/Eagar-0112.html > Yeah, I know eager very well. He runs the welding lab at MIT. thus: Bernard BORAT Lewis WANTS YOU IN SUDAN; Iran; Afghanistan; Iraq. other plausibly British Quagmires, like Belize and Canada; and, basically, anywhere that will destroy the USA economy, faster. thus: what was basically a catastrophic failure, with some pertinent serial events necessarily ensuing.... the only real crime, other than the islamokamikazis', was the advent of the asbestos scare, that caused the towers to *not* be cladded with earth's finest refractory material. and, of course, any possible abbetment by Trickier Dick Cheeny from the Nixon Admin., with his side-kick,Rummy. I mean, they *could* have been there, at the beginning, working for Lloyds of London! thus: the mainpart ofthe controlled demo argument is that a trashfire never took-down a skyscraper, aside from the fact that such an inside operation needs much,much,much less explosive energy that was in the planes.... however,there maybe severeal additional factors that could have abbetted the collapse, inherent to the buildings, aside from this initial cause.... as for why the planes were allowed to do thier dirtywork, there may also be factors that were not in control of Trickier Dick Cheeny from the Nixon Admin. -- there are several reasons to impeach him, before & after this fiasco, but if we could just learn what he & Rummy were *doing* under Nixon, it may not be necessary.... anyway,please present your arguments in nonvideoformat. --Ladies and Germs; please, put the chaise du toilette back, up! === Subject: ZFC in another shape. Hi All, Is the following theory equivalent to ZFC? Theory X. Primitives: e,= Definition Schema: x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). Axioms: 1)Extensionality 2)Foundation 3)Ordinal comprehension: Ex: x is an ordinal. 4)Replacement 5)Infinity 6)Choice All axioms except ordinal comprehension are as in ZFC. I think that this theory is equivalent to ZFC! Is that right? Zuhair === Subject: Re: ZFC in another shape. > Hi All, Is the following theory equivalent to ZFC? Theory X. Primitives: e,= Definition Schema: x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). Axioms: 1)Extensionality > 2)Foundation > 3)Ordinal comprehension: Ex: x is an ordinal. this is a mistake. Correction: What I wanted to write is a kind of axiom schema. were the existance of every ordinal as defined above is quarrenteed. 3)Ordinal comprehension Schema: ExAy(yex<->((yed or y=d) & d is an ordinal)). For every individual instance of the variable d. (were d is a variable that rang over sets). In reality it might be better writtin in the following manner. For every individual instance of d the following ExAy(yex<->((yed or y=d) & d is an ordinal)) is an axiom. here Ordinal comprehension is meant to be a schema, and not a single axiom. Something like how separation is a schema in ZFC. To be more clear about this I mean the following. when d is not an ordinal then we have x ={ }. since we'll have the axiom ExAy(yex<->false). so the existance of the empty set is quarenteed. Now { } is an ordinal. (definition schema 1) then when d={ } we have the following Axiom. ExAy(yex<->((ye{} or y={}) & {} is an ordinal)) and this is: ExAy(yex<->(y={} & {} is an ordinal)) Now either x ={} or not. if x={}, then yex is false and we have Ay(false<->true), a contradiction. Then ~x={} Now for every y:~y=0 we also have (yex<->false) and therefore ~yex when ~y=0. and therefore only when y=0, y is a member of x. so x={0}. ( note 0={}). Now {0} is and ordinal ( definition schema 1). Therefore we have the following axiom, when d={0}. ExAy(yex<->((ye{0} or y={0}) & {0} is an ordinal)) and this leads to x={0,{0}}. But {0,{0}} is an ordinal, and so we have the following Axiom : when d={0,{0}}. ExAy(yex<->((ye{0,{0}} or y={0,{0}}) & {0,{0}} is an ordinal)). and here x={0,{0},{0,{0}}}, which is an ordinal,... And so on, having an uncountablly infinite number of Axioms.(one axiom per each value of d), since there are uncountably infinite ordinals. Now I think my idea of the Ordinal Comprehension schema. is not difficult to understand , and I assume it is not weard. Of course The set of all ordinals, do not exist in this theory. Since its existance violate foundation. Now, Separation is implied by replacement. Pairing, can be acheived from replacing members in {0,{0}} , by x and y,using the appropriate replacement function. Union of sets x and y can be simply acheived by replacing the members of a set z that is equinumerous to the members of x and y together, by the members of x and y. Example x= {a,b} , y= {c,d} Let z={0,{0},{0,{0}},{0,{0},{0,{0}}} Replacing 0 by a, {0} by b, {0,{0}} by c and {0,{0},{0,{0}} by d. then according to Replacement we have the set {a,b,c,d}. as an existing set. The axiom of the empty set is not needed since it is proved from ordinal comprehension schema when d is not an ordinal. Also it can be proved from replacement. Therefore: Axioms of Empty set,Union,Pairing Separation are all theorums in this theory. The axiom of Power set , is the one hardest to prove as a theorum in this theory. But I guess that a method similar to that proving union as a theorum in this theory can be used to prove the theorum of power, of course depending on choice, which states that every set is well ordered. In nutshell, I think this theory is equivalent to ZFC. Sorry for the mistake. Zuhair > 4)Replacement > 5)Infinity > 6)Choice All axioms except ordinal comprehension are as in ZFC. I think that this theory is equivalent to ZFC! Is that right? Zuhair === Subject: Re: ZFC in another shape. Hi All, Is the following theory equivalent to ZFC? Theory X. Primitives: e,= Definition Schema: x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). Axioms: 1)Extensionality > 2)Foundation > 3)Ordinal comprehension: Ex: x is an ordinal. this is a mistake. Correction: What I wanted to write is a kind of axiom schema. were the existance > of every ordinal as defined above is quarrenteed. 3)Ordinal comprehension Schema: ExAy(yex<->((yed or y=d) & d is an ordinal)). For every individual > instance of the variable d. > (were d is a variable that rang over sets). In reality it might be better writtin in the following manner. For every individual instance of d the following ExAy(yex<->((yed or y=d) & d is an ordinal)) is an axiom. here Ordinal comprehension is meant to be a schema, and not a single > axiom. Something like how separation is a schema in ZFC. To be more clear about this I mean the following. when d is not an ordinal then we have x ={ }. since we'll have the axiom ExAy(yex<->false). so the existance of the empty set is quarenteed. Now { } is an ordinal. (definition schema 1) then when d={ } we have the following Axiom. ExAy(yex<->((ye{} or y={}) & {} is an ordinal)) and this is: ExAy(yex<->(y={} & {} is an ordinal)) Now either x ={} or not. if x={}, then yex is false and we have Ay(false<->true), a contradiction. Then ~x={} Now for every y:~y=0 we also have (yex<->false) and therefore ~yex when ~y=0. and therefore only when y=0, y is a member of x. so x={0}. ( note 0={}). Now {0} is and ordinal ( definition schema 1). Therefore we have the following axiom, when d={0}. ExAy(yex<->((ye{0} or y={0}) & {0} is an ordinal)) and this leads to x={0,{0}}. But {0,{0}} is an ordinal, and so we have the following Axiom : when d={0,{0}}. ExAy(yex<->((ye{0,{0}} or y={0,{0}}) & {0,{0}} is an ordinal)). and here x={0,{0},{0,{0}}}, which is an ordinal,... And so on, having an uncountablly infinite number of Axioms.(one axiom > per each value of d), since there are uncountably infinite ordinals. Now I think my idea of the Ordinal Comprehension schema. is not > difficult to understand , and I assume it is not weard. Of course The set of all ordinals, do not exist in this theory. Since > its existance violate foundation. Now, Separation is implied by replacement. Pairing, can be acheived from replacing members in {0,{0}} , by x and > y,using the appropriate replacement function. Union of sets x and y can be simply acheived by replacing the members > of a set z that is > equinumerous to the members of x and y together, by the members of x > and y. Example x= {a,b} , y= {c,d} Let z={0,{0},{0,{0}},{0,{0},{0,{0}}} Replacing 0 by a, {0} by b, {0,{0}} by c > and {0,{0},{0,{0}} by d. then according to Replacement we have > the set {a,b,c,d}. as an existing set. The axiom of the empty set is not needed since it is proved from > ordinal comprehension schema when d is not an ordinal. Also it can be > proved from replacement. Therefore: Axioms of Empty set,Union,Pairing > Separation are all theorums in this theory. The axiom of Power set , is the one hardest to > prove as a theorum in this theory. But I guess that a method similar to that proving union as a theorum > in this theory can be used to prove the theorum of power, of course > depending on choice, which states that every set is well ordered. sorry, I meant to say:..... is well orderable. Zuhair In nutshell, I think this theory is equivalent to ZFC. Sorry for the mistake. Zuhair 4)Replacement > 5)Infinity > 6)Choice All axioms except ordinal comprehension are as in ZFC. I think that this theory is equivalent to ZFC! Is that right? Zuhair- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: ZFC in another shape. > Hi All, Is the following theory equivalent to ZFC? Theory X. Primitives: e,= Definition Schema: x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). > I think you might have stated this wrongly. According to this definition, an ordinal will only exist if all sets have the same elements. > Axioms: 1)Extensionality > 2)Foundation > 3)Ordinal comprehension: Ex: x is an ordinal. > 4)Replacement > 5)Infinity > 6)Choice > With the definition of ordinal you've given, (1), (3), and (5) are inconsistent. > All axioms except ordinal comprehension are as in ZFC. I think that this theory is equivalent to ZFC! Is that right? > Not unless ZFC is inconsistent, no. > Zuhair === Subject: Re: ZFC in another shape. Hi All, Is the following theory equivalent to ZFC? Theory X. Primitives: e,= Definition Schema: x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). Perhaps I got it wrong. what I wanted to say is that x is an ordinal <- > (x is trantistive & Am(mex->m is transitive)). x is transtive <-> AmAn(mex&nem->nex). So x is an ordinal iff x is transtive and every member in x is transitive. Also a transtive set can be defined as a set that is a subset of its power set. But I want to avoid the power definition. This is what I wanted to say. I will be very happy to see the correction. But I think the definition is right. Zuhair I think you might have stated this wrongly. According to this > definition, an ordinal will only exist if all sets have the same > elements. Axioms: 1)Extensionality > 2)Foundation > 3)Ordinal comprehension: Ex: x is an ordinal. > 4)Replacement > 5)Infinity > 6)Choice With the definition of ordinal you've given, (1), (3), and (5) are > inconsistent. All axioms except ordinal comprehension are as in ZFC. I think that this theory is equivalent to ZFC! Is that right? Not unless ZFC is inconsistent, no. Zuhair- Hide quoted text - - Show quoted text - === Subject: Re: ZFC in another shape. > Hi All, Is the following theory equivalent to ZFC? Theory X. Primitives: e,= Definition Schema: x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). Axioms: 1)Extensionality > 2)Foundation > 3)Ordinal comprehension: Ex: x is an ordinal. What a strange name for that axiom. > 4)Replacement > 5)Infinity > 6)Choice All axioms except ordinal comprehension are as in ZFC. I think that this theory is equivalent to ZFC! Why would you think that? Seriously, what makes you think it is? > Is that right? Doesn't appear that either separation or power set are provable. So, I'd say no. -- Jesse F. Hughes C is for Cookie. That's good enough for me. Cookie Monster === Subject: Re: ZFC in another shape. > Hi All, >> Is the following theory equivalent to ZFC? >> Theory X. >> Primitives: e,= >> Definition Schema: >> x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). You sure that's the right definition? As consequence of this definition, you get: x is an ordinal -> AmAnAz(z e n -> z e m) Thus, if there are any ordinals, then every set is equal (by extensionality). For the remainder, I will pretend you actually gave a correct definition of ordinal. >> Axioms: >> 1)Extensionality >> 2)Foundation >> 3)Ordinal comprehension: Ex: x is an ordinal. What a strange name for that axiom. > 4)Replacement >> 5)Infinity >> 6)Choice >> All axioms except ordinal comprehension are as in ZFC. >> I think that this theory is equivalent to ZFC! Why would you think that? Seriously, what makes you think it is? To put it another way, consider the following theory: Axioms: 1')Extensionality 2')Foundation 3')The empty set exists: (Ex)(Ay)(~ y e x) 4')Replacement 5')Infinity 6')Choice This theory is equivalent to yours, I'd say. Certainly it is at least as strong as yours, since (3') implies (3). After all, the empty set is an ordinal, and so if the empty set exists, an ordinal exists. I'm almost sure that (3) implies (3'), but I don't see any reason to check. So your new hypothesis is much simpler: the existence of an empty set (along with these other axioms) proves both power set and separation. Doesn't seem too plausible. -- Jesse F. Hughes Most of my research is irreducibly complex. -- James S. Harris === Subject: Re: ZFC in another shape. <87bqk9jtkn.fsf@phiwumbda.org> <877iuxjr8t.fsf@phiwumbda.org > Hi All, > Is the following theory equivalent to ZFC? > Theory X. > Primitives: e,= > Definition Schema: > x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). You sure that's the right definition? As consequence of this > definition, you get: x is an ordinal -> AmAnAz(z e n -> z e m) Thus, if there are any ordinals, then every set is equal (by > extensionality). For the remainder, I will pretend you actually gave a correct > definition of ordinal. > Axioms: > 1)Extensionality >> 2)Foundation >> 3)Ordinal comprehension: Ex: x is an ordinal. What a strange name for that axiom. > 4)Replacement >> 5)Infinity >> 6)Choice > All axioms except ordinal comprehension are as in ZFC. > I think that this theory is equivalent to ZFC! Why would you think that? Seriously, what makes you think it is? To put it another way, consider the following theory: Axioms: 1')Extensionality > 2')Foundation > 3')The empty set exists: (Ex)(Ay)(~ y e x) > 4')Replacement > 5')Infinity > 6')Choice This theory is equivalent to yours, I'd say. Certainly it is at least > as strong as yours, since (3') implies (3). After all, the empty set > is an ordinal, and so if the empty set exists, an ordinal exists. I'm almost sure that (3) implies (3'), but I don't see any reason to > check. So your new hypothesis is much simpler: the existence of an empty set > (along with these other axioms) proves both power set and separation. Doesn't seem too plausible. -- > Jesse F. Hughes Most of my research is irreducibly complex. > -- James S. Harris- Hide quoted text - - Show quoted text - Yea , true , I committed a mistake. I corrected it. Zuhair === Subject: Re: ZFC in another shape. <87bqk9jtkn.fsf@phiwumbda.org> <877iuxjr8t.fsf@phiwumbda.org > Hi All, > Is the following theory equivalent to ZFC? > Theory X. > Primitives: e,= > Definition Schema: > x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). You sure that's the right definition? As consequence of this > definition, you get: x is an ordinal -> AmAnAz(z e n -> z e m) Thus, if there are any ordinals, then every set is equal (by > extensionality). For the remainder, I will pretend you actually gave a correct > definition of ordinal. > Axioms: > 1)Extensionality >> 2)Foundation >> 3)Ordinal comprehension: Ex: x is an ordinal. What a strange name for that axiom. > 4)Replacement >> 5)Infinity >> 6)Choice > All axioms except ordinal comprehension are as in ZFC. > I think that this theory is equivalent to ZFC! Why would you think that? Seriously, what makes you think it is? To put it another way, consider the following theory: Axioms: 1')Extensionality > 2')Foundation > 3')The empty set exists: (Ex)(Ay)(~ y e x) > 4')Replacement > 5')Infinity > 6')Choice This theory is equivalent to yours, I'd say. Certainly it is at least > as strong as yours, since (3') implies (3). After all, the empty set > is an ordinal, and so if the empty set exists, an ordinal exists. I'm almost sure that (3) implies (3'), but I don't see any reason to > check. So your new hypothesis is much simpler: the existence of an empty set > (along with these other axioms) proves both power set and separation. Doesn't seem too plausible. > It's false. The hereditarily countable sets are a model for these axioms in which the power set axiom fails. > -- > Jesse F. Hughes Most of my research is irreducibly complex. > -- James S. Harris- Hide quoted text - - Show quoted text - === Subject: Re: ZFC in another shape. <87bqk9jtkn.fsf@phiwumbda.org> <877iuxjr8t.fsf@phiwumbda.org >> Hi All, >> Is the following theory equivalent to ZFC? >> Theory X. >> Primitives: e,= >> Definition Schema: >> x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)). You sure that's the right definition? As consequence of this > definition, you get: x is an ordinal -> AmAnAz(z e n -> z e m) How is that? I think you are mistaken. See my answer to your previous post. I only wanted to say that x is an ordinal iff x is transitive and every member in x is transtive. which is a very well known definition of ordinals. Happy to see the correction. Zuhair Thus, if there are any ordinals, then every set is equal (by > extensionality). For the remainder, I will pretend you actually gave a correct > definition of ordinal. >> Axioms: >> 1)Extensionality >> 2)Foundation >> 3)Ordinal comprehension: Ex: x is an ordinal. > What a strange name for that axiom. >> 4)Replacement >> 5)Infinity >> 6)Choice >> All axioms except ordinal comprehension are as in ZFC. >> I think that this theory is equivalent to ZFC! > Why would you think that? Seriously, what makes you think it is? To put it another way, consider the following theory: Axioms: 1')Extensionality > 2')Foundation > 3')The empty set exists: (Ex)(Ay)(~ y e x) > 4')Replacement > 5')Infinity > 6')Choice This theory is equivalent to yours, I'd say. Certainly it is at least > as strong as yours, since (3') implies (3). After all, the empty set > is an ordinal, and so if the empty set exists, an ordinal exists. I'm almost sure that (3) implies (3'), but I don't see any reason to > check. So your new hypothesis is much simpler: the existence of an empty set > (along with these other axioms) proves both power set and separation. Doesn't seem too plausible. It's false. The hereditarily countable sets are a model for these > axioms in which the power set axiom fails. -- > Jesse F. Hughes Most of my research is irreducibly complex. > -- James S. Harris- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: hi...help with algorithm.. Hi all, allright this is the problem.. i have n jobs..each job i requires t(subscript i) seconds to be processed. Starting time is 0. for a schedule, completion time C (subscript i) denotes the time at which the job is completely processed. Each job has a weight w(subscript i) > 0. I need to find an algorithm which returns an ordering of jobs that minimizes the weighted sum w(subscript i) * C (subscript i) [ i is from 1 to n jobs]. Eg: i have 2 jobs ---- job 1: t1=1 w1=10 job 2: t2=3 w2=2. the schedule : job 1 followed by job 2 will give a weighted completion time of: 10*1 + 2 *4= 18 while the other wud give 2*1 + 10*(3+1) = 42. Greedy method can be used here i guess..but i need to PROVE that the algo works.. Any help is appreciated.. MY idea was to order the jobs according to the ratio t/w in ascending order .. what do yu guys think? === Subject: Re: hi...help with algorithm.. >Hi all, >allright this is the problem.. i have n jobs..each job i requires t(subscript i) seconds to be >processed. >Starting time is 0. for a schedule, completion time C (subscript i) denotes the time at >which the job is completely processed. Each job has a weight >w(subscript i) > 0. I need to find an algorithm which returns an ordering of jobs that >minimizes the weighted sum w(subscript i) * C (subscript i) [ i is >from 1 to n jobs]. Eg: i have 2 jobs ---- job 1: t1=1 w1=10 > job 2: t2=3 w2=2. the schedule : job 1 followed by job 2 will give a weighted completion >time of: 10*1 + 2 *4= 18 >while the other wud give 2*1 + 10*(3+1) = 42. Greedy method can be used here i guess..but i need to PROVE that the >algo works.. Any help is appreciated.. >MY idea was to order the jobs according to the ratio t/w in ascending >order .. what do yu guys think? > I think your idea is right. To prove it, suppose that j_1, j_2, ..., j_n is the optimal ordering of jobs. If we switch job a and job a+1 in this optimal ordering, then the objective function will increase. And the difference in the objective function after these two consecutive jobs are switched is: w_{a+1} (t1 + ... + t_{a-1} + t_{a+1}) + w_a (t1 + ... + t_{a-1} + t_{a+1} + t_a) - w_a (t1 + ... + t_{a-1} + t_a) - w_{a+1} (t1 + ... + t_a + t_{a+1}) = w_a t_{a+1} - w_{a+1} t_a >= 0 We thus have t_{a+1} / w_{a+1} >= t_a / w_a for all a in the optimal ordering, and this is enough information to determine the ordering completely. -- Daniel Mayost === Subject: Brouwer's Fixed Point Theorem What questions or ideas lead to the relevance of Brouwer's Fixed Point Theorem? What I mean is, what do we understand better once it's proven? I'd prefer to know of the things which caused Brouwer to originally think about the question and prove it rather than hyper- modern examples of applicability. Avital. === Subject: Re: Brouwer's Fixed Point Theorem What questions or ideas lead to the relevance of Brouwer's Fixed Point > Theorem? What I mean is, what do we understand better once it's > proven? I'd prefer to know of the things which caused Brouwer to > originally think about the question and prove it rather than hyper- > modern examples of applicability. Don't know about Brower's original motivation. I do know that fixed point theorems in general tell us about the existence of solutions of equations. For instance, Brower's theorem says that given a continuous map f: D - > D, where D is the n-dimensional unit ball, then the equation f(x) = x has a solution. In spirit, this theorem is similar to the intermediate value theorem. One way to state it is that for any continuous map f: I -> I, where I is the unit interval, which is positive at 0 and negative at 1, there exists a solution of the equation f(x) = 0. Hope this helps. Igor === Subject: Question: Simple matrix operation Hi all, I have two n x n matrices A and B. Let's call (i,j)the element of the matrices as a[i][j] and b[i][j]. I would like to find out C whose element c[i][j] = a[i][j]*b[i][j] using *only* standard matrix operations (matrix addition, subtraction, multiplication, inversion, transpose etc.). What is the easiest way of doing this? === Subject: Re: Question: Simple matrix operation > Hi all, I have two n x n matrices A and B. Let's call (i,j)the element of the > matrices as a[i][j] and b[i][j]. I would like to find out C whose > element c[i][j] = a[i][j]*b[i][j] using *only* standard matrix > operations (matrix addition, subtraction, multiplication, inversion, > transpose etc.). What is the easiest way of doing this? In a way, that _is_ a standard matrix operation: it is the Schur (or Hadamard) matrix product. On the other hand, it is not a matrix operation at all; you are just treating the matrices as n^2 - tuples with coordinate-wise addition and multiplication. -- Paul Sperry Columbia, SC (USA) === Subject: Re: announcement: Maxima 5.11.0 release > Maxima can run on MS Windows and various flavors of Unix, > including MacOS X. > There is a precompiled executable installer for Windows, > source and binary RPM's for Linux, and tar.gz containing > the source distribution. I'm confused. Which of these do I go for for MacOS X? > Project page: > http://sourceforge.net/projects/maxima I went here, and all I saw was stuff for Windows. -- === Subject: middle school problem I have trouble understanding the following middle school math problem: Tara and Bobby each had some trading cards. When Tara gave Bobby the same number of trading cards that he had in the begining, they each had 52 trading cards. How many trading cards did Tara give to Bobby? How do you interprete this question in math ? Is the question clear to you all but me? I'd appreciate your opinon. John Kim === Subject: Re: middle school problem I have trouble understanding the following middle school math problem: Tara and Bobby each had some trading cards. When Tara gave Bobby the >same number of >trading cards that he had in the begining, they each had 52 trading >cards. How many trading >cards did Tara give to Bobby? How do you interprete this question in math ? Is the question clear >to you all but me? I'd appreciate your opinon. >John Kim Let t = number of cards Tara starts with Let b = number of cards Bobby starts with. After the trade Tara has t - b = 52 and Bobby has 2b = 52 cards. So b = 26 and t = 52 + b = 78. --Lynn === Subject: algebra with Sylow.... Hello sir~ If H is a subgroup of a finite group G and |H| is a power of a prime p, then H is contained in some Sylow p-subgroup of G. -------------------------------------- This is a proof of book with contemporary abstract algebra-Gallian. Lemma 1) Orbit-Stabilizer theorem. Let G be a finite group of permutations of a set S. Then for any i from S, |G| = |orb_G (i)| |stab_G (i)|. Lemma 2) Let K be a Sylow p-subgroup of a finite group G. if x in N(K) and the order of x is a power of p, then x in K. Let's go... Let K be a Sylow p-subgroup of G and let C = {K = K_1, K_2, ..... , K_n} be the set of all conjugates of K in G. (I think...C is a G-set with conjugation.) Since conjugation is an automorphism, each element of C is a Sylow p-subgroup of G. (Yes). Let S_c denote the group of all permutations of C. For each g in G, define f_g : C -> C by f_g(K_i) = g.K_i.g^{-1}. (In fact, f_g in S_c) Now define a mapping T : G -> S_c by T(g) = f_g. so, T is a homomorphism from G to S_c. (Yes, I can show that T is a homomorphism.) Next consider T(H), the image of H under T. Since |H| is a power of p, so is |T(H)|. (Yes, This is a property of homomorphism.) Thus, by the Orbit Stabilizer Theorem, for each i, |orb_T(H) (K_i)| divides |T(H)| (Because T(H) is a subgroup of S_c. so, T(H) is a group of permutation of C.) so that |orb_T(H) (K_i)| is a power of p. (Yes). Now we ask : under wat condition does |orb_T(H) (K_i)| = 1 ? Well, |orb_T(H) (K_i)| = 1 means that f_g(K_i) = g.K_i.g^{-1} = K_i for all g in H. (Because, by definition orbit. and T(H) is a group of permutation of C making by H.) that is,|orb_T(H) (K_i)| = 1 if and only if H <= N(K_i). (Yes, by g.K_i.g^{-1} = K_i for all g in H.) But the only elements of N(K_i) that have orders that are power of p are those of K_i by lemma 2. Thus, |orb_T(H) (K_i)| = 1 if and only if H <= K_i. ---(*) So, to complete the proof, all we need to do is show that for some i, |orb_T(H) (K_i)| = 1. (Yes, by (*)) 1) we have |C| = |G:N(K)|. (Because, C is a orbit in G by conjugation. so |C| = |G:G_K|. and G_K = {g in G | g.K.g^{-1} = K} = N(K).). And since |G:K| = |G:N(K)||N(K):K| is not divisible by p, (Yes, because K is a Sylow p-subgroup of G. so |K| = p^n.(n is maximal)) neither is |C|. (Yes.) 2) Because the orbits partition C, |C| is the sum of powers of p. (Yes, because, |X| = sum|Gx| (x in each orbit in X) and p | |Gx| namely, C is a G-set). If no orbit has size 1, then p divides each summand and, therefore, p divides |C|, which is a contradiction. Thus, there is an orbit of size 1, and the proof is complete. --------------------------------------------- Hm...complex... I don't even know that I understood my thinking. Can you help me ? I can't understand last paragraph well. namely, 1) we have |C| = |G:N(K)|. 2) Because the orbits partition C, |C| is the sum of powers of p. I can't understand this part well. Because, I use that C is a orbit in G by conjugation and I use that C is a G-set simultaneously. is this possible ? === Subject: p-values in perl I've been searching around for a module that will input 2 arrays and output a p-value. I've been using statistics::rankcorrelation for spearman rank correlation but I need to limit the output. There has === Subject: Re: p-values in perl > I've been searching around for a module that will input 2 arrays and > output a p-value. I've been using statistics::rankcorrelation for > spearman rank correlation but I need to limit the output. There has You'd be better off asking on comp.lang.perl.misc, where they'll tell you to search CPAN. === Subject: Re: p-values in perl I've been searching around for a module that will input 2 arrays and > output a p-value. I've been using statistics::rankcorrelation for > spearman rank correlation but I need to limit the output. There has You'd be better off asking on comp.lang.perl.misc, where they'll tell > you to search CPAN. Yea, posted there, but no too many know what a p-value is. Spent a === Subject: Re: Quantum Voodoo? Spooky telepathic messages back from the future? Is it not the same thing ? When photon one enters it goes to type A bcause 50 nanosecond later photon 2 will be observed as such. === Subject: JSH: Where are you ? These guys are assholes. We miss you! JSH !! === Subject: Re: JSH: Where are you ? > These guys are assholes. We miss you! JSH !! In case you're not aware of it, you can go to http://mymath.blogspot.com/ to get a fix of JSH. It's not as potent as his usenet presence, but it will hold off the cold turkey for a while. Personally I've stooped to arguing with the lesser cranks over at sci.physics during Harris' latest absence. -Rotwang === Subject: Re: Where are you ? > These guys are assholes. We miss you! JSH !! he's locked away in his spiderhole, not to worry, they should let him out this Friday with his new research version. I am sure this is JSH, lives in San Francisco, types and knows math like a white librarian girl, http://jackwolak.com/7/9558.jpg === Subject: Learning Math After Giving Up Or Falling Behind In High School Hi All, I'm looking for a self-learning program or series of book selections(preferably free/cheap) for someone who either fell behind in, or gave up on, math when they were younger. A little background: I'll be 22 in just under a month. I haven't had any formal math training, except one required college course, since Grade 11 (Junior year of high school), which would be about 5 years ago. It's not that I don't like math, as I was always good at it/had fun with it as a child and into my early teen years; I just hit a bit of a rough patch in high school. In Grade 10, I had some issues (that I won't go into here) which resulted in marginal grades in many of my classes. I barely passed Math. In Grade 11, I was unsure what I was doing because of my problems the previous year, and I slipped and just stopped caring, thinking I'd failed completely and that my math education was a lost cause. A little older and a little wiser, I'd like to go back and bring myself up to speed a bit more. I was hoping I could find help here. Rather than try to pinpoint where I am in math, I'd rather take see a more complete path, starting as basic as you feel is relevant. Clearly I know how to add, subtract, etc., but I'm looking for a complete foundation and a pointer or two on where to go next. JH === Subject: Re: Solving a simple differential equation x''+(c/x^2)=0 > Solve for x in terms of t. Are you kidding? This is the same as your Simple Physics Question. > All answers have been given in that thread, mainly by the author who > calls himself -Rotwang (s...@hotmail.co.uk). actual solution to the above equation. In fact the post to which you are replying came before the Simple Physics Question thread, in phrased it incorrectly.; in the later thread the OP only asks for the time at which x=0, which I got without solving the e.o.m. for x. The lesson for the OP if he's still reading is that the most obvious way of solving a problem is not always the easiest... -Rotwang === Subject: Re: Solving a simple differential equation > Solve the following differential equation: > x''+(c/x^2)=0 > Solve for x in terms of t. Newton first solved a special case of inverse square law problem.x'' + (c/x^2) = 0, x^2 th' = h =const. The latter comes from angular momentum conservation.Refer to planetary motion; x is radius, th is polar angle w.r.t. focus. 1/x comes out proportional to ( 1+ ecc*cos(th) ).Your general case may involve h as a variable. I also like to know how Newton addressed the last part of his problem solving x in terms of t, as elliptic functions were not yet commonly available. Narasimham === Subject: Re: Solving a simple differential equation > Solve the following differential equation: > x''+(c/x^2)=0 > Solve for x in terms of t. Write v = dx/dt, and hence convert to a differential equation involving v and x alone. It's fairly straighforward from then on. The general solution takes any of three different functional forms depending on the value of c and the initial conditions. IIRC the solution can't always be written in closed form as x = f(t) (using elementary functions), but can be written as t = f(x). === Subject: Re: Solving a simple differential equation Solve the following differential equation: > x''+(c/x^2)=0 > Solve for x in terms of t. Write v = dx/dt, and hence convert to a differential equation > involving v and x alone. It's fairly straighforward from then on. The > general solution takes any of three different functional forms > depending on the value of c and the initial conditions. IIRC the > solution can't always be written in closed form as x = f(t) (using > elementary functions), but can be written as t = f(x). Yes, that's what I'm really after, t as a function of x.