mm-4089 === Subject: Re: Equalizing Golf to Baseball Re: B. Bonds test on distance of playing field Re: better to slow pitch in majors > (snipped) > Because Golf is similar to Baseball in that the ball tee-ed off is a ball at zero > speed and would simulate a self pitched baseball which is nearly zero speed. > So, all I would have to do is compute a factor of a golf club compared to a > baseball bat. Compute another factor of the aerodynamics of a golf ball > compared to a major league baseball. Then one more factor of the muscles > used in baseball compared to the muscles used in golf. > Plugging in all those factors in an equation. Can you imagine how far a golf ball would go if it was fired toward the tee at 200 MPH and you could hit the dang thing? My God, that would be spectacular! === Subject: Sum (1/(N logN) [was Re: is this a solid proof for series?] > ... a divergent series needn't be satisfy the > condition that a_n > k / n^p for some p in [0, 1] - as an example > consider a_n = 1 / (n log n). I was led to this series when trying to construct a series which diverged more slowly than the harmonic. On the face of it it seemed to diverge such that you needed to square the number of terms to increase the sum so far by ln2. I was however unable to prove it. Is there a simple proof? === Subject: Re: Sum (1/(N logN) [was Re: is this a solid proof for series?] > ... a divergent series needn't be satisfy the > condition that a_n > k / n^p for some p in [0, 1] - as an example > consider a_n = 1 / (n log n). > I was led to this series when trying to construct a series which diverged > more slowly than the harmonic. On the face of it it seemed to diverge such > that you needed to square the number of terms to increase the sum so far by > ln2. I was however unable to prove it. Is there a simple proof? There is a simple method that handles quite a few series like this. If the elements of the series a(n) are positive and decreasing as they are here, then the sum of a(n) converges/diverges if and only if the sum of 2^k * a(2^k) converges/diverges. That sum is in your example the sum over 2^k * (1 / 2^k / log 2^k) = 1 / (k log 2) which is divergent. === Subject: Re: Sum (1/(N logN) [was Re: is this a solid proof for series?] >>... a divergent series needn't be satisfy the >>condition that a_n > k / n^p for some p in [0, 1] - as an example >>consider a_n = 1 / (n log n). > I was led to this series when trying to construct a series which diverged > more slowly than the harmonic. On the face of it it seemed to diverge such > that you needed to square the number of terms to increase the sum so far by > ln2. I was however unable to prove it. Is there a simple proof? Well, my proof would be the integral test. Sum ( i = 3 to n ) 1 / (n log n) >= Sum ( i = 2 to n ) Integral ( x = i- 1 to i ) 1 / (x log x) dx = Integral (x = 2 to n) 1 / ( x log x) dx = log (log n ) - c. So if you square n then you do indeed add log 2 to (a lower bound for) the sum. I suppose there are probably more elementary proofs, but the integral test is generally nice and handy so why not use it? :) David === Subject: Re: Sum Of All Integers In sci.physics, J 891 Note subject change. Followups to sci.math only. [snip for brevity] >> Pedant point: I'm not sure if the sum of integers is well-defined. >> Or one can simply observe that the partial sums >> S_0 = 0 >> S_1 = 0 + 1 >> S_2 = 0 + 1 - 1 >> S_3 = 0 + 1 - 1 + 2 >> S_4 = 0 + 1 - 1 + 2 - 2 >> ... >> S_k = 0 + 1 - 1 + 2 - 2 ... + n >> where there are (k+1) terms, is 0 if k is odd but k/2 if k is even. > [Accuracy warning, I am working from memory of my university days long > ago.] > In any sequence, this series does not converge in the standard sense > at all. But even if it did, it is not surprising that you can > rearrange it and get different answers. > A series is said to be absolutely convergent if the sum of the > absolute values of the terms also converges (not necessarily to the > same value). If the terms of an absolutely convergent series are > rearranged then it will still converge and to the same value. If the > terms of a convergent series are all positive (e.g. the power series > of exp(x) for any positive x) then the series is obviously absolutely > convergent. But there are many series which are not all positive that > are also absolutely convergent. The power series exp(x) for negative > x or sin(x) and cos(x) for any x are examples. The convergence of > exp(x) can be used to easily prove the convergence of these series. > If a series converges but is not absolutely convergent, then it is > said to be conditionally convergent. These are odd beasts. You can > rearrange the terms to make them diverge, and more surprisingly, you > can rearrange the terms to arrive at any desired answer. > The series 1 - 1/2 + 1/3 - 1/4 + 1/5 etc is an example of a > conditionally convergent series. It is easy to prove that it > converges but if you make all the terms positive then it diverges. So > you can rearrange it to come to any answer you like. > For the meaning of the terms as I am using them and some more detail, > see http://mathworld.wolfram.com/AbsoluteConvergence.html and links on > that page. This is a great site. You are correct as far as *my* memory goes (I graduated in 1983) :-). In any event, good explanation; also, I've found mathworld.wolfram.com is chock full of stuff. -- #191, ewill3@earthlink.net It's still legal to go .sigless. electron-dot-cloud are galaxies === Subject: Re: B. Bonds test on distance of playing field Re: better to slow pitch > Too simple. In fact, you just flunked physics 101. With stuff like Try getting a real full name, so you flunked what, second grade. When you make a silly assertion twice on the Internet that a 100mph fastball is a disadvantage to hitters compared to a 0 speed tossup hit. You must then be called either a troll or a dunce or both. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === > Too simple. In fact, you just flunked physics 101. With stuff like > Try getting a real full name, so you flunked what, second grade. > When you make a silly assertion twice on the Internet that a 100mph > fastball is a disadvantage to hitters compared to a 0 speed tossup hit. You must > then be called either a troll or a dunce or both. > Archimedes Plutonium, a_plutonium@hotmail.com > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies Recommended reading: The Physics of Baseball - Momentum at: http://library.thinkquest.org/11902/physics/momentum.html Bat Weight, Swing Speed and Ball Velocity by Daniel A. Russell: http://www.kettering.edu/~drussell/bats-new/batw8.htm Force of a Bat on a Baseball: http://hypertextbook.com/facts/2000/AlbertKlyachko.shtml Speed of the Fastest Pitched Baseball: http://hypertextbook.com/facts/2000/LoriGrabel.shtml HTH HP === Subject: normal subgroup...... thank you....always let H : subgroup of group G let N = intersection aHa^(-1) [a in G] show that N is normal subgroup of G ------------------------------------ i think.......... any g in G, any n in N => gng^(-1) in N (i use) let n = ai*hj*ai^(-1) in N g*ai*hj*ai^(-1)*g^(-1) = (g*ai) hj (g*ai)^(-1) in N um.............i have a doubt final result. (g*ai) hj (g*ai)^(-1) in N how do you think about it?? === Subject: Re: normal subgroup...... > thank you....always > let H : subgroup of group G > let N = intersection aHa^(-1) [a in G] > show that N is normal subgroup of G > ------------------------------------ I think you are having problems working out a proof. Here are some tips that may help you write out a formal proof. First, start off by prefixing your proof with the phrase Proof: so that one can see where the statement of the theorem ends and your proof begins. Second, restate the hypotheses that are given in the theorem. Third, restate the claim that you are going to prove. Thus, you would have here in your case the following: Proof: Let G be a group. Let H be a subgroup of G. Let N = intersection over a in G of aHa^(-1). Claim: N is normal subgroup of G Note that your claim has two things to prove: 1) N is a subgroup of G. 2) N is normal in G. You are concentrating on just proving the second statement below. But, you must also give a reason why 1) is true. You can give a reason for why 1) is true by quoting a prior theorem. Thus, you add to your proof the following: N is a subgroup of G since . > i think.......... > any g in G, any n in N Here you introduce any. My suggestion is that you should not work with *any* g or *all* g. Instead, you should work with a particular (but arbitrary) g. I find that this makes the proof less abstract. There is justification for doing this if you are aware of mathematical logic. Thus, you continue in your proof as follows: Let g be an element of G. Let n be an element of N. Note that implicitly you are aiming at showing that gng^(-1) is an element in N, which will show N is normal in G by the definition of being normal. Thus, you will have proved the second statement of your claim. > => gng^(-1) in N (i use) Here you use the implication sign =>. I don't like having a string of implies in a proof since it makes the proof statements unwieldly. However, this proof is not the best example of showing how to avoid using =>. So, I will pass on going into more detail on that. > let n = ai*hj*ai^(-1) in N Here, you lost the thread of what needs to be proved. Recall that you want to show that gng^(-1) is an element in N for the particular g and n that I am currently working with. You want to look at the definition of N: Let N = intersection over a in G of aHa^(-1). One method is to replace N by its definition. Thus, you want to prove: gng^(-1) is an element of the intersection over a in G of aHa^(-1). Now, how do you prove an element is in the intersection of several sets? The standard way is to prove that the element is in each of the sets. That is, you are replacing working with all a in G by working with a particular (but arbitrary) a in G. This again makes it more concrete. Thus, you have the following: Let a be an element of G. Claim: gng^(-1) is an element of aHa^(-1). Note that now a, g, n are particular elements. > g*ai*hj*ai^(-1)*g^(-1) = (g*ai) hj (g*ai)^(-1) in N What is ai? What is hj? You haven't introduce them yet. Instead, you want to prove that gng^(-1) is an element of aHa^(-1). But, you want to use the information that you have stated above for the elements a, g, n. In particular, you want to use the information that n is an element of N and that N is the the intersection over a in G of aHa^(-1). That is, by definition of intersection, n is an element of aHa^(-1) for all a in G. Note that this last a is not the same a that we selected earlier, since it is quantified by all. That is, you are free to select whatever a you want in G and still have n is an element of aHa^(-1). Now, the heart of the proof is to choose an appropriate a in the statement For all a in G, n is in aHa^(-1) so that you can show that gng^(-1) is an element of aHa^(-1). Let me eliminate the ambiguity of the two a's in that last paragraph. That is, you have the following situation. The heart of the proof is to choose an appropriate b in the statement For all b in G, n is in bHb^(-1) so that you can show that gng^(-1) is an element of aHa^(-1). You can work backwards to see what is the required choice for b. > um.............i have a doubt final result. (g*ai) hj (g*ai)^(-1) in N > how do you think about it?? I realize that the above may be a little vague. But, I think it gives you some suggestions on how to approach a (formal) proof. -- Bill Hale === Subject: Re: normal subgroup...... Adjunct Assistant Professor at the University of Montana. >thank you....always I think you should spend more time on your own homework. You are asking the newsgroup for ->really easy<- problems. That leads me to believe that either you are not trying at all, (in which case, why should we do your homework for you?) or else you are trying and failing (in which case, you should not be in the class you are in; these are ->easy<- problems, if you cannot do them, then you are failing the class). >let H : subgroup of group G >let N = intersection aHa^(-1) [a in G] >show that N is normal subgroup of G >------------------------------------ >i think.......... >any g in G, any n in N => gng^(-1) in N (i use) >let n = ai*hj*ai^(-1) in N >g*ai*hj*ai^(-1)*g^(-1) = (g*ai) hj (g*ai)^(-1) in N >um.............i have a doubt final result. (g*ai) hj (g*ai)^(-1) in N >how do you think about it?? Show that g (intersection of aHa^{-1} [a in G] ) g^{-1} is the same as intersection of (gaHa^{-1}g^{-1} [a in G]). Then take it from there. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: normal subgroup...... >let H : subgroup of group G >let N = intersection aHa^(-1) [a in G] >show that N is normal subgroup of G > Show that > g (intersection of aHa^{-1} [a in G] ) g^{-1} > is the same as > intersection of (gaHa^{-1}g^{-1} [a in G]). Then take it from there. That leads to some general theorems. Arbitrary intersection of subgroups is a subgroup. Arbitrary intersection of normal subgroups is a normal subgroup. Now when H is subgroup, aHa^-1 is subgroup, thus N is subgroup. Now if H is normal subgroup, then aHa^-1 is normal subgroup. and N too would be normal. Indeed H = aHa^-1 = N. However H is just any subgroup... === Subject: Re: normal subgroup...... Adjunct Assistant Professor at the University of Montana. >>let H : subgroup of group G >>let N = intersection aHa^(-1) [a in G] >>show that N is normal subgroup of G >> Show that >> g (intersection of aHa^{-1} [a in G] ) g^{-1} >> is the same as >> intersection of (gaHa^{-1}g^{-1} [a in G]). Then take it from there. >That leads to some general theorems. >Arbitrary intersection of subgroups is a subgroup. >Arbitrary intersection of normal subgroups is a normal subgroup. >Now when H is subgroup, aHa^-1 is subgroup, thus N is subgroup. >Now if H is normal subgroup, then aHa^-1 is normal subgroup. >and N too would be normal. Indeed H = aHa^-1 = N. >However H is just any subgroup... However what? Here's a further hint, then, for you: show that if g is fixed, then as a ranges over all elements of G, ga also ranges over all elements of G. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Homegeneous relations... Hello Can anyone tell me if the following is a linear homogeneous recurrence relation? a_n = a_(n-1) / n My first reaction was to say not homogeneous because of the n. But does that rule apply in this case or only in the form similiar to: === Subject: Re: Homegeneous relations... > Can anyone tell me if the following is a linear homogeneous recurrence > relation? > a_n = a_(n-1) / n Yes, it is. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Homegeneous relations... > Hello > Can anyone tell me if the following is a linear homogeneous recurrence > relation? > a_n = a_(n-1) / n > My first reaction was to say not homogeneous because of the n. But does that > rule apply in this case or only in the form similiar to: Linear? Yes. Homogeneous? Yes. Constant coefficient? No. -- Paul Sperry Columbia, SC (USA) === Subject: Re: algebra problem.... Adjunct Assistant Professor at the University of Montana. >group G : not cyclic group such that |G| = 27 >1. show that G : not simple group >i don't know second problem. Clearly. Two hints: (a) Show that if g^a = e, then a divides |G|. (Easy) (b) Assume there is some g for which g^9 is not equal to e; what is the order of g then? >um................. >advice please.......my doctor.... One advice I would give you, given the number of requests for help you have thrown into the group in the last few weeks, is to ask your professor for more help or to drop the class. If I had a student who needs to ask as many questions on his or her homework as you do, and at such a level as your questions, I would definitely suggest to him or her to re-evaluate his/her commitment to the course, and possibly just drop out. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: algebra problem.... >group G : not cyclic group such that |G| = 27 >1. show that G : not simple group >i don't know second problem. > Clearly. > Two hints: > (a) Show that if g^a = e, then a divides |G|. (Easy) Sigh; this is wrong as written, of course. Here, it should say Show that if a is the ORDER of g, then a divides |G|. That is, a is assumed to be the smallest positive integer such that g^a=e. Arturo Magidin, sans .sig === Subject: Re: algebra problem.... >>group G : not cyclic group such that |G| = 27 >>1. show that G : not simple group >> I agree with everything that Arrturo says below. But I would be intrigued to know how you managed to prove that a group of order 27 is not simple by using Sylow's Theorem. Derek Holt. >>2. show that any g in G, g^9 = e (e =identity) >>i don't know second problem. >Clearly. >Two hints: >(a) Show that if g^a = e, then a divides |G|. (Easy) >(b) Assume there is some g for which g^9 is not equal to e; what is >the order of g then? >>um................. >>advice please.......my doctor.... >One advice I would give you, given the number of requests for help you >have thrown into the group in the last few weeks, is to ask your >professor for more help or to drop the class. If I had a student who >needs to ask as many questions on his or her homework as you do, and >at such a level as your questions, I would definitely suggest to him >or her to re-evaluate his/her commitment to the course, and possibly >just drop out. >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magidin@math.berkeley.edu === Subject: Re: Groups isomorphic? Adjunct Assistant Professor at the University of Montana. >in message : >> Consider, for example, that there are nonabelian groups of exponent p >> and order p^3, so that p^3-1 elements have order p and one element has >> order 1. >Nitpick: This is only true for p odd. Of course Arturo knows >this, but maybe it will save the OP from looking for a >non-abelian group of exponent 2 and order 8. You are absolutely right; I've been working in my research with groups of exponent p and class 2, p and odd prime, and I seem to have developed the bad habit of not specifying the prime is odd... It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Set Exponentation Adjunct Assistant Professor at the University of Montana. >My understanding of taking sets to powers of other sets is really bad. >My professor briefly mentioned something about the number of all >possible functions that exist between the two sets, but he is so >MIND-NUMBING that I can't stay awake in his class . . . >Let A and B both be infinitely countable sets. What does A^B mean? >Let A be finite and B be countable. What does A^B mean In general, if A and B are sets, then A^B is the set of all function f:B->A. That is, A^B is the collection of all subsets S of BxA such that: (a) For every b in B there exists a in A such that (b,a) is in S; and (b) For every b in B, if (b,a) and (b,a') are both in S, then a=a'. >For example, we have the set 2^N, whose cardinality is equivalent to >that of the set of real numbers. 2^N is the set of all functions from N to 2={0,1}; you can think of it as the set of all characteristic functions of subsets of N. > If 2^N is the set of all infinite >sequences of zeros and ones, Yes. > then I can see how it relates to Cantor's >Diagonalization. >But HOW can 2^N = {0,1}^N DESCRIBE the set of all infinite sequences >of zeros and ones? Because, by definition, a sequence is a map whose domain is N, the natural numbers. Here, you are specifying the image, {0,1}. So any sequence of zeros and ones is a map from N to {0,1}, hence an element of 2^N. >Which leads to more questions . . . >{0,1)^1 = ? I assume you mean {0,1}^1; since 1, as a set, is 1={0}, that means all functions from {0} to {0,1}. There are exactly two functions: the one mapping 0 to 0, and the one mapping 0 to 1. So {0,1}^1 = {0,1}^{0} = { {(0,0)}, {(0,1)} }. >{0,1}^{0,1} = ? This is 2^2; it is the collection of all functions from {0,1} to {0,1}. There are four such functions: everything to 0; everything to 1; 0 to 0 and 1 to 1; 0 to 1 and 1 to 0. {0,1}^{0,1} = { {(0,0), (1,0)}, {(0,1),(1,1)}, } { {(0,0), (1,1)}, {(0,1),(1,0)}. } >{0,1}^n = ? (where n belongs to N) The collection of all functions from n={0,1,2,...,n-1} to {0,1} >Let A = 2 = {0,1}. Let B = 3 = >{0,1,2} What does A^B mean? The set whose elements are all functions from {0,1,2} to {0,1}. There's eight of them. > What does B^A mean? The set whose elements are all maps from {0,1} to {0,1,2}; there are 9 of them. >I'm betting that A^B= B^A is false. That's correct. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Set Exponentation Xevious grava .88 la saucisse et au marteau: > Let A and B both be infinitely countable sets. What does A^B mean? The set of the applications from B to A. To each element of B can be assigned an element in A. Therefore, the cardinality of this set is Card(A)^Card(B). > Let A be finite and B be countable. What does A^B mean? Same thing. > For example, we have the set 2^N, whose cardinality is equivalent to > that of the set of real numbers. If 2^N is the set of all infinite > sequences of zeros and ones, then I can see how it relates to Cantor's > Diagonalization. But HOW can 2^N = {0,1}^N DESCRIBE the set of all > infinite sequences of zeros and ones? When I look at the text, 2^N > means nothing to me. Which leads to more questions . . . > {0,1)^1 = ? The set containing the applications 1-> 1 and 1-> 0 (not very interesting applications). > {0,1}^{0,1} = ? The set containing the four applications: f1: 0 -> 0 1 -> 0 f2: 0 -> 1 1 -> 0 f3: 0 -> 0 1 -> 1 f4: 0 -> 1 1 -> 1 > {0,1}^n = ? (where n belongs to N) I let you imagine what it is. > Let A = 2 = {0,1}. Let B = 3 = {0,1,2} > What does A^B mean? What does B^A mean? Idem. > I'm betting that A^B = B^A is false. Yes. The first set contains 8 applications and the second one 9. -- Genji === Subject: Re: Set Exponentation By the way, what Le Roux calls an application is usually called a function in English. His point is correct. Given two sets A and B, by definition, A^B = {f in P(B x A): f is a function with domain B}. When both A and B are finite, card(A^B) = card(A)^card(B). In fact, that is true when either A or B are inifinite as well. One fact of interest: If 1 < card(A) <= card(C), 1 < card(B) <= card(C), and C is infinite, then card(A^C) = card(B^C). I am pretty sure one can demonstrate this without the axiom of choice; I am sure someone here will correct me if I am wrong. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Set Exponentation grava .88 la saucisse et au marteau: > By the way, what Le Roux calls an application is usually called a > function in English. His point is correct. Given two sets A and B, Oops, sorry. That's because there is a difference between fonction and application in French, and I thought there was the same difference in English. -- Genji === Subject: Re: Set Exponentation > grava .88 la saucisse et au marteau: >> By the way, what Le Roux calls an application is usually called a >> function in English. His point is correct. Given two sets A and B, >Oops, sorry. That's because there is a difference between fonction and >application in French, and I thought there was the same difference in >English. That's interesting. What's the difference? I have not seen application used this way in English, but I am not a professional mathematician. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Set Exponentation >> grava ? la saucisse et au marteau: > By the way, what Le Roux calls an application is usually called a > function in English. His point is correct. Given two sets A and B, >Oops, sorry. That's because there is a difference between fonction and >>application in French, and I thought there was the same difference in >>English. > That's interesting. What's the difference? I have not seen > application used this way in English, but I am not a professional > mathematician. Could it be the same distinction as between function and map? Marc === Subject: Re: Set Exponentation > grava .88 la saucisse et au marteau: > By the way, what Le Roux calls an application is usually called a >> function in English. His point is correct. Given two sets A and B, >Oops, sorry. That's because there is a difference between fonction and >application in French, and I thought there was the same difference in >English. > That's interesting. What's the difference? I have not seen > application used this way in English, but I am not a professional > mathematician. The main difference is nobody in America uses Eniglish Set theorists definition of function. And we're the only ones to do math applications, it's just the same old story for France. They act like they discovered DNA of something. But, since it's just another case of Napoleon fever, the only thing we can tell them is the same thing we tell Canadian doctors: Call us in morning, after you get rid of those Ruskie snivels. === Subject: French vs. English (Was: Re: Set Exponentation) I do not think Nicolas will mind my sharing this e-mail: >>That's interesting. What's the difference? I have not seen >>application used this way in English, but I am not a professional >>mathematician. >An application f:E->F is a fonction defined over the whole subset E. >log:R->R is a fonction (but valid only on R+*) >log:R+*->R is an application *Very* interesting. In English, the concept of function has a certain asymmetry to it. A function is a special type of relation. One of the axioms of a function is that, for each element in its domain (but not the range), there is at least one ordered pair in the function which has the element as its first coordinate. Apparently, the French fonction drops this requirement, so that if f is a fonction from A to B, neither projection must be the entire superset. I find the definition of application confirmed in the 1977 notes, Logique et Theories Axiomatiques, by J.L. Krivine. (I received these when I was taking an undergraduate logic/set theory course with G. Sabbagh at Jussieu (Paris VII) in 1980.) Interestingly, Krivine never defines the word fonction, though he uses it in reference to an application and as a subheading for the section where he introduces the concept. This is another interesting linguistic contrast between French and English mathematics. The other one I am thinking of is field (literally, champs) vs. corps (literally, body): In a field (corps commutatif), multiplication is by definition commutative; not commutativity is an essential feature of multiplication for English speaker; not so for French speakers. To this we may compare the lack of agreement for the definition of ring even in English alone. Some authors (notably Isaac) require a ring to contain a multiplicative identity; others do not. (BTW, as I was composing this, I received the following: >>That's interesting. What's the difference? I have not seen >>application used this way in English, but I am not a professional >>mathematician. >After a Google research, I found that the translation of application >is map or mapping. >hth ) -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: French vs. English (Was: Re: Set Exponentation) >>log:R->R is a fonction (but valid only on R+*) >>log:R+*->R is an application function and total function. Do mathematicians say that too? Martin === Subject: Re: French vs. English (Was: Re: Set Exponentation) > This is another interesting linguistic contrast between French and > English mathematics. The other one I am thinking of is field > (literally, champs) vs. corps (literally, body): In a field > (corps commutatif), multiplication is by definition commutative; not > commutativity is an essential feature of multiplication for English > speaker; not so for French speakers. I don't think that follows at all. In so far as there is any reason for the difference in usage, I suspect that it reflects a difference in the feel or philosophy of the two languages. A person writing in English would feel uneasy at repeating commutative field 50 or 100 times -- the natural tendency would be to find a shorthand form. The French, on the other hand, are perfectly happy as far as I can see to repeat a compound phrase indefinitely. (Do the Germans actually prefer it?) -- 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: French vs. English >> This is another interesting linguistic contrast between French and >> English mathematics. The other one I am thinking of is field >> (literally, champs) vs. corps (literally, body): In a field >> (corps commutatif), multiplication is by definition commutative; not >> commutativity is an essential feature of multiplication for English >> speaker; not so for French speakers. > I don't think that follows at all. > In so far as there is any reason for the difference in usage, > I suspect that it reflects a difference in the feel > or philosophy of the two languages. > A person writing in English would feel uneasy > at repeating commutative field 50 or 100 times -- > the natural tendency would be to find a shorthand form. > The French, on the other hand, are perfectly happy as far as I can see > to repeat a compound phrase indefinitely. > (Do the Germans actually prefer it?) Depending on which is used more often, one may use Schiefkoerper and Koerper (cr skew-field) or Koerper and kommutativer Koerper As you indicated above, it is mostly a matter of convenience. Marc === Subject: Re: French vs. English > This is another interesting linguistic contrast between French and >> English mathematics. The other one I am thinking of is field >> (literally, champs) vs. corps (literally, body): In a field >> (corps commutatif), multiplication is by definition commutative; not >> commutativity is an essential feature of multiplication for English >> speaker; not so for French speakers. You can infer that, but most English writers don't even use the word commutative in the context of multiplication. If you don't use the phrase symmetric field, they take it for granted that you're a Russian spy, and shoot you. I don't think that follows at all. > In so far as there is any reason for the difference in usage, > I suspect that it reflects a difference in the feel > or philosophy of the two languages. A person writing in English would feel uneasy > at repeating commutative field 50 or 100 times -- > the natural tendency would be to find a shorthand form. The French, on the other hand, are perfectly happy as far as I can see > to repeat a compound phrase indefinitely. > (Do the Germans actually prefer it?) > Depending on which is used more often, one may use > Schiefkoerper and Koerper (cr skew-field) > or > Koerper and kommutativer Koerper > As you indicated above, it is mostly a matter of convenience. > Marc electron-dot-cloud are galaxies === Subject: Re: new way of describing ellipse for math > Consider rotating the circle about a diameter, (which becomes the > major axis of your ellipse), while a perpendicular diameter becomes > the minor axis as the circle is projected on a stationary, > (non-rotating), plane paralel to the major axis. Still cannot picture your above. Regardless it is too complicated. I need an immediate and simple method. > The corners will not give you a unique elipse, but if you connect the > midpoints of the oposite sides and use them as major and minor axes, > you've got one. Sounds like the ellipse is then inscribed inside the rectangle. I want the rectangle inscribed inside the circle and ellipse. I want to get away from axes. I want to connect a unique circle to a unique ellipse using a unique rectangle. Every circle has a unique square inscribed. By throwing in a circle into the construction, that circle ought to allow me to get away without needing any axes or foci. If the expression that a ellipse is a squashed circle has any truth to it then some such construction ought to exist. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: new way of describing ellipse for math > Sounds like the ellipse is then inscribed inside the rectangle. I want the > rectangle inscribed inside the circle and ellipse. > I want to get away from axes. I want to connect a unique circle to a unique > ellipse using a unique rectangle. Every circle has a unique square inscribed. > By throwing in a circle into the construction, that circle ought to allow me to > get away without needing any axes or foci. > If the expression that a ellipse is a squashed circle has any truth to it then > some such construction ought to exist. > Archimedes Plutonium, a_plutonium@hotmail.com > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies If you are looking for a unique elipse defined only by an inscribed rectangle, you are out of luck. Four points arranged as the corners of a rectangle define a whole family of elipses including one circle. An elipse is not a 'squashed` circle, it is a rotated one. (Conic Sections - remember?) Whether you like it or not, I believe that the relationship you are looking for does involve the rotation posted earlier. Consider the circle with its inscribed square. (You didn't say you had trouble with that.) If you rotate this figure about one side of the square, and project it, you get a rectangle, (the square with one set of opposite sides forshortened), and a unique circumscribed elipse, (the circle rotated). Hint: if you wanted to graph this elipse, you could multiply one set of ordinates from the graph of the circle with its center at the projected intersection of the diagonals of the square by the cosine of the angle of rotation. This is High School stuff. If you have trouble with these concepts, you are wasting your time in attempting to understand, much less modify, string theory. === Subject: Re: Factorial/Exponential Identity, Infinity Fish, or cut bait. Is the dis-parity discrepancy contradictory? Why or why not? Ross === Subject: Re: Math dependency logic REVISED > A sure sign of the decline of sci.math/sci.logic presided > over by David Ullrich and the Boyz is that these NGs attract > loathsome mutants like you. Is it lighten up time yet? One of the invariants of Usenet (like, for the past twenty years or so) is that, except on the relatively few groups that are moderated, *nobody* is in charge. Anybody who wants to can pipe up and say anything. > Here one may shill for whatsoever and whosoever one pleases-- > including that puke, Lyndon LaRouche--as long as one > fires off a few at JSH. I don't read everything on sci.math; but my choices are informed in part by morbid curiosity, so I am surprised to have missed the plugs for Lyndon LaRouche. > Welcome to sci.math/sci.logic, where loathsome mutants gyre > and gimble in good company. Cool. Way cool. But how do you tell them apart? (Assuming you want to.) -- Chris Henrich Moral indignation is jealousy with a halo. -- H. G. Wells === Subject: Re: Math dependency logic REVISED Adjunct Assistant Professor at the University of Montana. [.newsgroups trimmed.] >>[cut] > Pray give a *proof* that your > ring (which includes all possible values of a_1(x)/7 and a_2(x)/7 for > integer x) does not contain 1/7. >>I realize that you are trying to get Harris to see his error by >>asking him to prove that 1/7 is not in his ring. You are using >>this approach since Harris rejects concrete counterexamples, >>says valid definitions are flawed, and refuses to go into more >>detail when questioned on how one statement follows from another >>logically (instead, he gives metaphysical reasons why the >>questioned statement *must* be true). >>However, I feel that your approach gives a misleading idea of >>what constitutes a proof. Your approach seems to say that >>even after a proof has been presented, other things need also >>be considered and proved. My objection is that what if I failed >>to notice that I must give a proof that 1/7 is not in the ring? >>It doesn't appear to be needed in the proof given by Harris. >> You have a point, of sorts. The problem here is that James refuses to >> tell us what IS in his ring; he only tells us what is NOT in his ring. >Mathematics is about tracing out an argument from a truth using >logical steps to get to a conclusion which you then know MUST BE TRUE. >If you start with a truth, and use only logical steps to get to a >conclusion, it MUST BE TRUE. >I do that, which is how my work should be faced. Keep telling yourself that; someday maybe you'll believe it. >It amazes me how many of you seem to doubt that fact!!! I don't doubt it. I know for a fact it is not true that your work proceeds along only logical steps. When you finish with the case of x=0, you perform a logical fallacy called a non-sequitur, when you claim that the only way a factorization can occur is the way you claim it occurs, and when you claim that something should be an algebraic integer. Neither of those two steps are justified by the previous (mostly correct if also mostly irrelevant) work. >Instead posters try to find a backdoor. >> His definition is that the object ring is the ring of numbers such >> that the only integers which are units are 1 and -1. He claims that it >> is more inclusive than the algebraic integers, but refuses to present >> a single element of the object ring which is not an algebraic integer, >> and refuses to explain why the integers themselves, which satisfy his >> definition, are not the object ring. >> His ->only<- objection to Dik's use of James's argument in a different >> polynomial is that the argument clearly does not apply, since in Dik's >> case it is easily provable that the conclusion would yield that 1/7 is >> in the ring where all these factorizations take place. >The argument I use relies on numbers like 7 being NUMBERS and not >variable, as well as on the distributive property: a(b+c) = ab + ac. Nonsense. That is only what you tell yourself because you cannot understand the objections raised to your work. You are assigning quasi-magical properties to the value of a function at zero. You are so confused that you attack your own method when I use it. >Now I've pointed out before that what I have is short and basic, but >posters continually try to make endruns and obfuscate the facts. No, what you have is short and wrong. We've told you EXACTLY where you are wrong, but you continue to lie and claim nobody has every pointed out a line of your argument which is wrong. >> It is not unreasonable to request that James explain to us why this is >> also not the case in his argument, since he is introducing inverses of >> algebraic integer factors of 7 all over the place. >And I have explained it. To nobody's satisfaction but your own. That is not an explanation, that is a reply. And while every explanation and every answer is a reply, not every reply is an answer and not every reply is an explanation. You are very good at replying. But you have not provided answers or explanations. At least, not answers or explanations that anybody finds satisfying, except for you. >I noticed something simple, which challenges the assumptions of >mathematicians, and rather than do what they're supposed to do: focus >on the argument itself; they're trying to cheat. >It's intellectual dishonesty. You *focus on the argument presented*. I have noticed that I've explained several times, by giving EXPLICIT formulas, why your argument about what the constant terms imply for arbitrary values is incorrect, and why your claim that a factorization must be in some specific way is wrong. Rather than address those points, you either remove them and repeat your argument, or you attack me personally. You do not focus on the argument presented, which is IN DIRECT REPLY, AND USING THE METODS OF, the argument ->you<- present. I also notice that you do not consider this to be intellectual dishonesty on YOUR part. Why is that? Is it just because, as always, you are being a hypocrite? Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Math dependency logic REVISED Adjunct Assistant Professor at the University of Montana. [.newsgroups trimmed.] >In sci.physics, Arturo Magidin >: [.snip.] >> His definition is that the object ring is the ring of numbers such >> that the only integers which are units are 1 and -1. He claims that it >> is more inclusive than the algebraic integers, but refuses to present >> a single element of the object ring which is not an algebraic integer, >> and refuses to explain why the integers themselves, which satisfy his >> definition, are not the object ring. >Uh....very dumb question, but, if a ring A contains a subring B, >aren't all the units of subring B also units of A? Yes; but the converse does not hold. That is, it is possible for something to be (a) an element of B; and (b) a unit in A' and (c) not a unit in B. >Subring B in this case is the ring of algebraic integers, which >contains many units: n - sqrt(n-1) and n + sqrt(n-1) being >some. In fact, any equation >x^n + a_{n-1} * x^{n-1} + ... + a_1 * x 1 = 0 >(where n > 0 and the a_i are integers) will have unit roots, >as one can easily prove. >Therefore, an encompassing object ring of the algebraic >integers cannot have 1 and -1 as its only units, James specifies that the only INTEGERS which are units are 1 and -1; he does not say anything about any other elements which may be units are 1 and -1 [emphasis added]. [.rest deleted.] Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Math dependency logic REVISED In sci.math, Arturo Magidin : > [.newsgroups trimmed.] >>In sci.physics, Arturo Magidin >>: > [.snip.] > His definition is that the object ring is the ring of numbers such > that the only integers which are units are 1 and -1. He claims that it > is more inclusive than the algebraic integers, but refuses to present > a single element of the object ring which is not an algebraic integer, > and refuses to explain why the integers themselves, which satisfy his > definition, are not the object ring. >>Uh....very dumb question, but, if a ring A contains a subring B, >>aren't all the units of subring B also units of A? > Yes; but the converse does not hold. That is, it is possible for > something to be (a) an element of B; and (b) a unit in A' and (c) not > a unit in B. Correct, of course. :-) For example, the ring of integers is a subring of the algebraic integers, but there are an infinite number of units in the algebraic integers, whereas the integers have only two. >>Subring B in this case is the ring of algebraic integers, which >>contains many units: n - sqrt(n-1) and n + sqrt(n-1) being >>some. In fact, any equation >>x^n + a_{n-1} * x^{n-1} + ... + a_1 * x 1 = 0 >>(where n > 0 and the a_i are integers) will have unit roots, >>as one can easily prove. >>Therefore, an encompassing object ring of the algebraic >>integers cannot have 1 and -1 as its only units, > James specifies that the only INTEGERS which are units are 1 and -1; And he is quite correct in this case. But I think he's very confused regarding factorization of algebraic integers. For example, 7^(2/3) is an algebraic integer. (It's not a unit, either.) x^3 - 49 can be rewritten (correctly) as x^3 - 49 = (x - 7^(2/3) * v1) * (x - 7^(2/3) * v2) * (x - 7^(2/3) * v3) where v1, v2, and v3 are units. v1 = 1, v2 = cos 2pi/3 + i sin 2pi/3, v3 = cos 4pi/3 + i sin 4pi/3. However, one cannot conclude x^3 - 49 = (x - 7 * w1) * (x - 7 * w2) * (x - w3) where w1, w2, and w3 are algebraic integers. (Algebraic *numbers*, yes.) This is of course only a strawman, intended to be vaguely similar to JSH's proof, but it turns out his suffers from a similar problem, at least in the examples I've seen thus far. > he does not say anything about any other elements which may be > units are 1 and -1 [emphasis added]. I'll admit I am now curious as to what subrings of the algebraic integers are such that the only units of the subring are 1 and -1. Or is James' object ring a superset? My brain hurts. :-) [rest deleted] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Math dependency logic REVISED >Arturo Magidin aren't all the units of subring A also units of B? >> Yes; but the converse does not hold. That is, it is possible >> for something to be a nonunit of A but a unit in B > Correct, of course. :-) For example, the ring of integers is > a subring of the algebraic integers, but there are an infinite > number of units in the algebraic integers, whereas the integers > have only two. That doesn't exemplify Arturo's remark, which is that a nonunit in a ring A need not stay a nonunit in a ring B containing A. However, this is true in case A < B is an _integral_ ring extension; then any (non)unit in A remains the same in B. Integral extensions cannot alter whether an element is a unit (or not). This is a fundamental property of integral extensions. Indeed, when this property is generalized to the LYING-OVER (LO) or SURVIVAL property and further combined with the INCOMPARABILITY (INC) property, one obtains a characterization of integral ring extensions, namely THEOREM For commutative rings R < T the following are equivalent (1) R < T is an integral extension (2) A < B has INC and LO for all A,B with R < A < B < T (3) A < A[u] has INC and LO for all A,B with R < A < T, all u in T Recall that one defines a ring extension R < T to be LYING-OVER (LO) if for any prime ideal P in R there exists a prime ideal Q in T lying over P, i.e. Q / R = P. INCOMPARABLE (INC) if two different primes P, Q in T with the same contraction in R are incomparable: neither P < Q nor Q < P SURVIVAL if each proper (prime) ideal P of R survives in T, i.e. P doesn't explode to 1 in T: PT != T In particular: r is a nonunit of R <=> (r) < 1, so if R < T is survival then rT < 1 so r remains a nonunit in T. For details see my prior post [1], to which I've just posted a reply [2] containing much further detail, including (online) references. -Bill Dubuque === Subject: Re: Math dependency logic REVISED Adjunct Assistant Professor at the University of Montana. >In sci.math, Arturo Magidin >: >> [.newsgroups trimmed.] >In sci.physics, Arturo Magidin >: >> [.snip.] >> His definition is that the object ring is the ring of numbers such >> that the only integers which are units are 1 and -1. He claims that it >> is more inclusive than the algebraic integers, but refuses to present >> a single element of the object ring which is not an algebraic integer, >> and refuses to explain why the integers themselves, which satisfy his >> definition, are not the object ring. >Uh....very dumb question, but, if a ring A contains a subring B, >aren't all the units of subring B also units of A? >> Yes; but the converse does not hold. That is, it is possible for >> something to be (a) an element of B; and (b) a unit in A' and (c) not >> a unit in B. >Correct, of course. :-) For example, the ring of integers is >a subring of the algebraic integers, but there are an infinite >number of units in the algebraic integers, whereas the integers >have only two. Sorry, but that is a BAD example of the situation I mentioned. In this situation, B=integers; A=algebraic integers, but there is NO element x of A such that: (1) x is in B (i.e., is an integer) (2) x is a unit in A; and (3) x is not a unit in B. That is, the situation I mentionted does NOT occur in your example. What you gave was an example of a subring B of A, and element x of A, such that: (1) x is a unit in A; and (2) x is not in B (hence not a unit in B). A better example would be setting B=the integers, A = Z[1/2], the ring of all rationals that can be written as p/q with q a power of 2. Then, settting x=2, gives you an element of B which is a unit in A but not in B. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Math dependency logic REVISED Adjunct Assistant Professor at the University of Montana. >my question, or thing that always bugs me, is why i (and -i) are not >included as units, Not in cluded as units of ->what<-? James is specifying that no INTEGERS other than 1 and -1 should be units; he does not say anything about what other sorts of units may be included. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Math dependency logic REVISED In sci.math, Arturo Magidin : >>[cut] > Pray give a *proof* that your > ring (which includes all possible values of a_1(x)/7 and a_2(x)/7 for > integer x) does not contain 1/7. >>I realize that you are trying to get Harris to see his error by >>asking him to prove that 1/7 is not in his ring. You are using >>this approach since Harris rejects concrete counterexamples, >>says valid definitions are flawed, and refuses to go into more >>detail when questioned on how one statement follows from another >>logically (instead, he gives metaphysical reasons why the >>questioned statement *must* be true). >>However, I feel that your approach gives a misleading idea of >>what constitutes a proof. Your approach seems to say that >>even after a proof has been presented, other things need also >>be considered and proved. My objection is that what if I failed >>to notice that I must give a proof that 1/7 is not in the ring? >>It doesn't appear to be needed in the proof given by Harris. > There is some truth to what you say. > On the other hand, James claims that a_1(x) and a_2(x) are divisible > by 7 in the object ring. The only property he has given on the > object ring is that no integers other than 1 and -1 are units. If his > claim that a_1(x)/7 and a_2(x)/7 are in the object ring is true, > then it ... must be the case that a_1(x) and a_2(x) are multiples of 7? Or what? :-) [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Graph terminology Given a directed graph G = (V, U), where V is the set of vertices, and U is the set of arcs, I define H = (U, W), where the set of arcs U of G becomes the set of vertices of H, and W is the set of arcs of H such that u = (u1, u2) is in H if and only if u1 = (v1, v2) and u2 = (v2, v3). I would like to know if there is a standard name for graph H. And is there some symbolism (like H = G' for instance) that comes with that? Christophe === Subject: Re: Region of Convergence of 1/zeta(s) >there are half-planes of convergence and of absolute convergence, >>but they need not be the same; e.g. sum (-1)^n/n^s has convergence >>iff Re(s) > 0 and abolute convergence iff Re(s) > 1. Is it possible for the limit function to be analytic in a larger > right-half-plane than the half-plane of conditional convergence? > Yes. This function is entire. A few follow-up questions; If a Dirichlet series does not contain any poles, does it converge everywhere? Related question; for all Dirichlet series that do have poles, does analytic continuation ALWAYS imply that one or more poles are cancelled ? For example, my understanding of analytic continuation of the series sum n^(-s) is that we multiply by (1 - 2^(1-s)), which has a zero that cancels the pole at s = 1. This extends the ROC to re(s) > 0. We can then divide by the same factor (1 - 2^(1-s)) to get back the same function we started with. I am just a lowly electrical engineer dabbling in the world of math, so if I have made a terminology error, please don't crucify me! Bob Adams Bob Adams === Subject: Re: Region of Convergence of 1/zeta(s) > If a Dirichlet series does not contain any poles, does it converge > everywhere? I think you're asking the following: If the Dirichlet series sum(a_n/n^s) converges for sufficiently large real part to a function F, and F can be continued to an entire function, must the original series converge everywhere? An example (a_n = (-1)^n) has already been posted showing that the answer is no. You may be interested to know that *if* all the a_n are nonnegative, then the answer is yes. This follows from a theorem of Landau which says that a Dirichlet series with nonnegative coefficients cannot be continued to a region containing its real abscissa of convergence. For applications of this corollary, see Chapter VI of Newman's Analytic Number Theory, where it is employed in natural proofs of the nonvanishing of L-series on the line Re(s) = 1. Take care, Paul === Subject: Re: Region of Convergence of 1/zeta(s) > If a Dirichlet series does not contain any poles, does it converge > everywhere? Hello Liz. I'm not sure what you mean by does not contain any poles. As I pointed out sum_{n=1}^infinity (-1)^n/n^s is a Dirichlet series converging iff Re(s) > 0. But its limit is the restriction of an entire function (no poles anywhere is C). Does this Dirichlet series not contain any poles? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: How to find probability mass function of this? >>Suppose that we have two (fair) dice. Let >>X = the highest [sic - SJH] number on any of the dice, >>Y = the sum of both dice. >>What is the joint mass function here? >There are 6*6 ways of throwing two dice. X can be any integer from 1 >to 6. Given X, Y can be any integer from X+1 to 2X, but 2X only occurs >if both are the highest, while in the other cases either dice can be >the highest . So the formula becomes: >Prob(X=x,Y=y) > =2/36 if 1<=x<=6, x+1<=y<=2x-1 > =1/36 if 1<=x<=6, y=2x > =0/36 otherwise. Next exercise: Let U and V be independent random variables each with Uniform(0,1) distribution. Determine the joint density of X = maj(U,V) and Y= U+V. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: How to find probability mass function of this? > So the formula becomes: > Prob(X=x,Y=y) > =2/36 if 1<=x<=6, x+1<=y<=2x-1 > =1/36 if 1<=x<=6, y=2x > =0/36 otherwise. > Next exercise: > Let U and V be independent random variables each with Uniform(0,1) > distribution. Determine the joint density of X = maj(U,V) and Y= U+V. Assuming that is a serious question The natural way of extending the previous thinking would be (for some constant K) p(x,y) =2K if 01, or y2x and if the boundary needs a density then take account of the precise definition of the density of Uniform(0,1) at 0 and 1. === Subject: Re: How to find probability mass function of this? >So the formula becomes: >Prob(X=x,Y=y) > =2/36 if 1<=x<=6, x+1<=y<=2x-1 > =1/36 if 1<=x<=6, y=2x > =0/36 otherwise. >Next exercise: >>Let U and V be independent random variables each with Uniform(0,1) >>distribution. Determine the joint density of X = maj(U,V) and Y= U+V. >Assuming that is a serious question It was meant as an exercise for Konrad (and other interested beginning students of probability). >The natural way of extending the previous thinking would be (for some >constant K) >p(x,y) >=2K if 0=0 outside the boundary [...] Not at all rigorous. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: How to find probability mass function of this? > Determine the joint density of X = maj(U,V) and Y= U+V. What is maj? -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. 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: How to find probability mass function of this? >>Determine the joint density of X = maj(U,V) and Y= U+V. >What is maj? the larger -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Some people have gotten excited by their ability to find *different* factorizations than the ones I've been giving, where they end up with units other than 1 or -1, and maybe an example will help some of you, so here's this post. Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1) which I pick because it's valid in the ring of integers, but NOT always in the ring of algebraic integers. That, of course, is because every integer is either even or 1 away from being even, so it works. Now then, what if someone wanted to question that fact, so they came back at me with a *different* factorization that did something different. Why would that be relevant? It wouldn't. You MUST FOLLOW A PROOF, and not second guess the process. Now I've given a proof, and mathematicians need to behave like people who believe in what they're doing. And make no mistake, if you choose NOT to behave like mathematicians, then it seems to me that there's no reason for you to remain mathematicians, even if it is only in name. After all, fair is fair, and it's inhumane what you people are doing to me, basically punishing me for making discoveries. I have rights!!! James Harris === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) In sci.math, James Harris factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. > Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1) That's not a factorization, just a rewrite (although one could quibble over the '2'). In this case P(x) = (x - (-1/2 + i*(sqrt(7))/2)) * (x - (-1/2 - i*(sqrt(7))/2)) is a factorization. > which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. Erm, it's an abstract formalism/equation/rewrite; how can it not be valid? Unless 2 = 0, which is only the case in the field of integers mod 2, which isn't all that interesting. :-) > That, of course, is because every integer is either even or 1 away > from being even, so it works. This is true. > Now then, what if someone wanted to question that fact, so they came > back at me with a *different* factorization that did something > different. > Why would that be relevant? > It wouldn't. > You MUST FOLLOW A PROOF, and not second guess the process. I'm assuming there's a proof to follow somewhere on the Web; do you have a Web site? I for one want to see the whole proof, ideally with the corrections others have mentioned in this newsgroup. > Now I've given a proof, and mathematicians need to behave like people > who believe in what they're doing. You've given a sequence of what you purport are logical deductions. Some of them are logical enough. However, others are jumps over crevasses which I for one can't follow. I can only illustrate with one of my own strawman examples, however: x^3 - 49 is factorizable into (x - r1)(x - r2)(x - r3), but none of the roots is divisible by 7 -- x divisible by 7 over the algebraic integers, means that one can find an algebraic integer y such that x = 7 * y. Since all the r's are products of (-sqrt(3)/2 + i/2)^n * 7^(2/3) (n=0,1,2), it's clear that there is no such algebraic integer in this strawman case. Does your proof suffer from a similar flaw? I hope not. > And make no mistake, if you choose NOT to behave like mathematicians, > then it seems to me that there's no reason for you to remain > mathematicians, even if it is only in name. > After all, fair is fair, and it's inhumane what you people are doing > to me, basically punishing me for making discoveries. I have > rights!!! Personally, I prefer to critique your proof, not you. I can't say anything at all whether you smell good or bad, run around in your underwear (or not in your underwear), do weird things with a shishkebab, a chainsaw, and a Superball, attend church, give to charity, drive around the city with a large sign saying I [heart] MATH, etc. :-) Who really cares? We judge according to what one posts. It's all we have. :-) > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > In sci.math, James Harris > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1) > That's not a factorization, just a rewrite (although one could You're showing troubling ignorance here considering how often you've posted in my threads as if you know basic mathematics. In my example P(x) is factored into 2 and x(x+1)/2 + 1, and *both* are indeed factors. Possibly the symbols are confusing you, so here are some numbers: Let x=1, then P(1) = 4, and the factors are 2 and 2. It IS a factorization poster. You're just lost on basics. > quibble over the '2'). In this case > P(x) = (x - (-1/2 + i*(sqrt(7))/2)) * (x - (-1/2 - i*(sqrt(7))/2)) > is a factorization. Yes, it is a factorization, but it's not the only factorization poster. Ok, let's get one thing straight: the discussions here are over *advanced* topics, so if you don't even know what a factorization is, you might want to sit back and just read without posting. And yes, when I see such posts I downgrade a poster. If I downgrade a poster far enough, I just don't worry about reading their posts, unless I start seeing evidence that the newsgroup as a whole is convinced by them in their confusion. James Harris === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana. >Some people have gotten excited by their ability to find *different* >factorizations than the ones I've been giving, where they end up with >units other than 1 or -1, and maybe an example will help some of you, >so here's this post. >Consider P(x) = x^2 + x + 2, which factors nicely as >P(x) = 2(x(x+1)/2 + 1) That's not a factorization; that's a rewriting. >which I pick because it's valid in the ring of integers, but NOT >always in the ring of algebraic integers. Nope. For every algebraic integer value of x, P(x) is also an algebraic integer. I think what you mean is that not every partial calculation is an algebraic integer; that is, x(x+1)/2 is not an algebraic integer for every algebraic integer value of x; even though x(x+1)/2 is always an integer for every integer value of x. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana. >>Some people have gotten excited by their ability to find *different* >>factorizations than the ones I've been giving, where they end up with >>units other than 1 or -1, and maybe an example will help some of you, >>so here's this post. >>Consider P(x) = x^2 + x + 2, which factors nicely as >>P(x) = 2(x(x+1)/2 + 1) >That's not a factorization; that's a rewriting. Oh, I get it. You are factoring it as 2 times ((x)(x+1)/2) + 1. >>which I pick because it's valid in the ring of integers, but NOT >>always in the ring of algebraic integers. To be honest, I fail to see your point. Nobody is saying that you cannot factor your polynomial as P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7). What people are saying is that this factorization does not yield algebraic integer factors whenever P(x) is irreducible; you claim it SHOULD. Are you claiming that your factorization above, which is valid in algebraic integers when x=0, should also be valid for arbitrary algebraic integer values of x? And if so, why do you think so? Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) >[...] >After all, fair is fair, and it's inhumane what you people are doing >to me, basically punishing me for making discoveries. Huh??? The only punishment anyone's been giving you is to say you're wrong and explain why. A minute ago you asked me What the does agreement matter? This is a little puzzling, since you regard disagreement as punishment. >I have >rights!!! Yes. You have the right to say what you want. And we all have the right to say what we want about what you say. We'd even have the _right_ to say you were wrong if you were actually _right_! But you're not, and you _don't_ have the right to agreement from people about things you're wrong about, sorry. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > Some people have gotten excited by their ability to find *different* > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. > Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1) > which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. Can you give an example where x^2 + x + 2 is not divisible by 2 when x is an algebraic integer? -- 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: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > Some people have gotten excited by their ability to find *different* > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1) which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. > Can you give an example where x^2 + x + 2 is not divisible by 2 when > x is an algebraic integer? It's the FACTORIZATION Dik Winter, and if you can't grasp that point, you'll never get it. The factorization is valid in integers for any integer x, but not in the ring of algebraic integers for any algebraic integer x. My point is that you have to focus on the *factorization* and its validity in particular rings. Some factorizations will be valid in one ring, but not another. James Harris === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) ... > Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1) which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. Can you give an example where x^2 + x + 2 is not divisible by 2 when > x is an algebraic integer? > It's the FACTORIZATION Dik Winter, and if you can't grasp that point, > you'll never get it. > The factorization is valid in integers for any integer x, but not in > the ring of algebraic integers for any algebraic integer x. Oh, there are enough algebraic integers where it is valid in the ring of algebraic integers. For instance: x = sqrt(2). Arturo gave an example for which it is indeed not valid. > My point is that you have to focus on the *factorization* and its > validity in particular rings. > Some factorizations will be valid in one ring, but not another. Yes, and your factorisation P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22) is in general not valid in the ring of algebraic integers. So what are you trying to show? -- 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: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > ... > Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1) which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. Can you give an example where x^2 + x + 2 is not divisible by 2 when > x is an algebraic integer? It's the FACTORIZATION Dik Winter, and if you can't grasp that point, > you'll never get it. The factorization is valid in integers for any integer x, but not in > the ring of algebraic integers for any algebraic integer x. > Oh, there are enough algebraic integers where it is valid in the ring > of algebraic integers. For instance: x = sqrt(2). Arturo gave an > example for which it is indeed not valid. > My point is that you have to focus on the *factorization* and its > validity in particular rings. Some factorizations will be valid in one ring, but not another. > Yes, and your factorisation > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22) > is in general not valid in the ring of algebraic integers. So what > are you trying to show? That's the point. I *prove* that if you have coprimeness between 7 and 22 in the ring in which the factorization is valid, where 7 is NOT a unit (and neither is 22), then the constant terms of the factors that result from dividing P(x) by 49 *MUST* be coprime to 7. That result is then ring independent to the extent that it's not particular to a particular ring, but to a particular property *of* the ring. However, the ring of algebraic integers, which has the desired coprimeness property, does NOT always have a_1(x)/7 and a_2(x)/7 as members! It's a great result, which follows from some fascinatingly basic algebra. Get it yet Dik Winter? James Harris === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana. >> ... >> Consider P(x) = x^2 + x + 2, which factors nicely as >> P(x) = 2(x(x+1)/2 + 1) >> which I pick because it's valid in the ring of integers, but NOT >> always in the ring of algebraic integers. >> Can you give an example where x^2 + x + 2 is not divisible by 2 when >> x is an algebraic integer? >> It's the FACTORIZATION Dik Winter, and if you can't grasp that point, >> you'll never get it. >> The factorization is valid in integers for any integer x, but not in >> the ring of algebraic integers for any algebraic integer x. >> Oh, there are enough algebraic integers where it is valid in the ring >> of algebraic integers. For instance: x = sqrt(2). Arturo gave an >> example for which it is indeed not valid. >> My point is that you have to focus on the *factorization* and its >> validity in particular rings. >> Some factorizations will be valid in one ring, but not another. >> Yes, and your factorisation >> P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22) >> is in general not valid in the ring of algebraic integers. So what >> are you trying to show? >That's the point. I *prove* that if you have coprimeness between 7 >and 22 in the ring in which the factorization is valid, where 7 is NOT >a unit (and neither is 22), then the constant terms of the factors >that result from dividing P(x) by 49 *MUST* be coprime to 7. What you FAIL to realize, though, is that this DOES NOT MEAN that the factorization is P(x)/49 = (5a1/7 + 1)(5a2/7 + 1)(5b3 + 22). What you get is that 49 = w_1(x)*w_2(x)*w_3(x), with w_i varying with the values of x, and P(x)/49 = (( 5a1(x)+7)/w_1(x) - 1) + 1)* (((5a_2(x)+7)/w_2(x) - 1) + 1) * ((( 5b_3(x)+7)/w_3(x) - 22) + 22). That's IT. You are ASSUMING your conclusion when you claim that w_1(x)=7 for all x. Remember: Given ANY function f(x) and ANY constant r, f(x) can be written as f(x)=g(x) + r, where r is constant, and g(x) is a function. Given ANY function f(x), and any two values r and s, we can write f(x) = g(x) + c, where c is a constant, and g(r)=s: just let g(x) = f(x) - f(r) + s and let c = f(r)-s. Then g(x) + c = f(x)-f(r) + s + f(r) - s = f(x) and g(r) = f(r) - f(r) + s = s. So you can take the function 5a_1(x) + 7, divide it by ANY FUNCTION r(x) WHICH SATISFIES r(0)=7, and then you can write (5a_1(x)+7)/r(x) = h_1(x) + 1 where h_(1) = 0. Likewise, you can divide 5a_2(x)+7 by ANY FUNCTION s(x) which satisfies s(0)=7, and then you can write it as (5a_2(x)+7)/s(x) = h_2(x) + 1 And you can divide 5b_3(x) + 22 by ANY FUNCTION t(x) that satisfies t(0)=1, and then you can write the result as (5b_3(x)+22)/t(x) = h_3(x) + 22. So, if you make ANY choice for which r(x)*s(x)*t(x) = 49, the factorization will work; and if you make the choice in such a way that r(x)*s(x)*t(x) = 49, r(x) divides a_1(x) and 7, s(x) divides a_2(x) and 7, and t(x) divides a_3(x) and 7 (all in the ring of algebraic integers), then the factorization will work, and it will NOT NECESSARILY BE OF THE FORM YOU CLAIM IT ->MUST<- BE. You cannot conclude that, because you can ALWAYS write the function as h_1(x) + 1 with h_1(x) = 0, regardless of the value of w_1(x). Likewise for w_2 and for w_3. And if you choose them CORRECTLY, then you get that all of them have values in the algebraic integers. Get it, James S. Harris? Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana. > Some people have gotten excited by their ability to find *different* > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1) which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. >Can you give an example where x^2 + x + 2 is not divisible by 2 when >x is an algebraic integer? Ah, I think I see now what James is trying to claim; that we can write it as x(x+1)/2 + 1 times something... If x=i, then i^2+i+2 = -1+i+2 = 1+i; 1+i is a divisor of 2 and not a unit (in Z[i], its norm is 2, and if r is an algebraic integer unit, then it is a unit in the ring of integers of Q(r)); and it is a proper divisor of 2, since (1-i)(1+i) = 2 and 1-i is also not an algebraic integer unit. If it were a multiple of 2, then 1+i would be an associate of 2, which would make 1-i into a unit, which is impossible. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > Some people have gotten excited by their ability to find *different* > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1) which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. > Can you give an example where x^2 + x + 2 is not divisible by 2 when > x is an algebraic integer? I think James' point is that since x(x+1)/2 is not an algebraic integer for all algebraic integers x, writing the polynomial in that form implicitly involves 'going into a larger ring'. On the other hand, if x is an integer, x(x+1) is always even, so x(x+1)/2 makes sense within the integers. But then, as so often before, this is a factorization of *numbers* written as if it were one of *polynomials*. -- Dave Taylor When I want your opinion, I'll ... I'll never want your opinion [BtVS] === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > Some people have gotten excited by their ability to find *different* > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. > Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1) > which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. > Can you give an example where x^2 + x + 2 is not divisible by 2 when > x is an algebraic integer? That's impossible because if you do a little mathematical induction, you find that P(1)=4 and then evaluating for k+1, you get (k+1)^2+k+1+2 k^2+2k+1+k+1+2 k^2+3k+4 So P(x) is even for all x. David Moran > -- > 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: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > Some people have gotten excited by their ability to find *different* > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1) which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. Can you give an example where x^2 + x + 2 is not divisible by 2 when > x is an algebraic integer? > That's impossible because if you do a little mathematical induction, you > find that P(1)=4 and then evaluating for k+1, you get > (k+1)^2+k+1+2 > k^2+2k+1+k+1+2 > k^2+3k+4 (= (k^2 + k + 2) + 2(k + 1)) > So P(x) is even for all x. Yes, for the integers. Bit how about the algebraic integers? -- 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: limitations on mutual anti-correlation? After all that about how I use Google to cross-post, I mindlessly posted this just to sci.math. So this will appear twice on sci.math, sorry. morning. | |Interesting to see how many people do post from Google. I do it when I decide to cross-post something, which aol's own newsreader doesn't do. I sometimes think of getting a better ISP, but it hasn't been a priority-- and using Google for just when I cross-post is ok. Doing it this way has the useful feature of giving me a slight disincentive to joining cross-posted threads. Quite a few of them cross-posted from sci.physics seem very junky lately-- has sci.physics been going through a noisy phase lately? |Interesting learning problem: after the first time it happens and you |lose your careful reply, you get in the habit of copying your post |before hitting post. This behavior is reinforced so long as the |process fails regularly, but soon rejected as unnecessary overhead |after a winning streak ... until the next failure. | |See, Google is like life. ;-) With mondo storage devices with file systems optimized for big appends and deletes! I compose longer posts in notepad, then copy and paste, so keeping a copy is nearly automatic. A fixed-width font makes it easier to avoid extra long lines and to compose ASCII graphics (see below). :-) |I also learned though that even though your post seems to be lost, |hitting refresh and then resend from the dialogue box trys the |post again ... complete with your text: you text is in some memory |buffer somewhere, even if you can't see it. I had just had to refresh a couple of previous pages, so I figured Google might actually be too busy or something.... It also seems sometimes to claim it's failed, but then you find it succeeded after all (d'oh) and the retry posted a duplicate. |But enough administrivia -- well, besides the annoying point that we |seem to have incommensurate line lengths; which is odd, considering we |are both apparently posting from Google. Yeah, I find I can compose something in notepad going all the way out to 80 columns, cut and paste it into Google's window, and it looks fine, but gets posted with wrapped lines all over. Perhaps that's what the preview page is for. I think I'll reformat what you quoted from me, below, if you don't mind. [...] |> |This was the first and only intuitive example I came up with as an |> |example of an impossible putative correlation structure for 3 random |> |variables ... | |> it's a nice enough example of what it means to violate |> Bell's inequality that I've seen it used as an |>illustration, | |Oh. And I was _so_ proud of it. But are you sure seen it used is |not recall seeing Ed Green use it? ;-) My bag of tricks is so |small, and growing so slowly of late, I trot this one out every chance |I get. I thought I'd seen some other usenet poster use it, but I could be wrong. |> in spite of not being the kind of |> violation of Bell's inequality allowed by quantum mechanics. | |Well, I'm not sure what the difference in kind is. Just that quantum mechanics predicts the possibility of making certain violations and not others. Quantum mechanics doesn't permit three variables to perfectly anticorrelate in pairs like that. For one thing, the correlation matrix is not positive semidefinite, and I just tried to show that the correlation matrix you get according to quantum mechanics always is. [...] |> |In the famous so-called Bell inequalities we consider 4 random |> |variables (of which only two may be observed at a time), and in |> effect |> |test whether our observations are compatible with masked simultaneous |> |observations of 4 random variables. |> One complication is that not all pairs taken from these |> four are distant from each other, so we lose our reason |> for expecting that the hypothetical actual values of the |> pair of variables can be measured independently of each |> other. | |I don't follow you here. Probably there is some orthogonality to our |understanding -- since I normally find this -- but also I'm not even |sure if you really mean distant or if that's a typo for distinct. I'm saying that not only that, but some of the pairs of variables can't be observed at the same time. The original poster had given a matrix and asked whether it was the matrix of covariances of a set of random variables. We don't get all the entries of a matrix from our data; we just get some of the entries, and have the question of whether those entries could be the covariances of a set of random variables. The criterion for that is in principle just whether it's possible to fill in the remaining entries of the matrix in such a way that it satisfies the condition: positive semidefiniteness. It's a bit messy to determine that in general (apparently), but I found a way to fill in certain matrices, below, based on analogous features of quantum mechanics. Typically, the experiment involves creating a pair of then plenty of theories would say that the one measurement affects the result of the other measurement-- just by Generally, theories that say making one measurement affects the results of the other measurements can easily get away with predicting just the same kind of correlations as quantum mechanics does. That's the way some formulations of quantum mechanics work, after all. Usually the reason why a theory does not predict this kind of mutual influence is because it doesn't predict such influences between measurements being made at a distance from each other. Ideally, the measurements should be made at a space-like separation (i.e., far enough apart and close enough to the same time that a signal travelling at the speed of light does not have time to get from the one measurement to the other one). |Anyway, in lose our reason ..., etc: this may touch on my |idiosyncratic approach, but our reason for expecting I think need be |nothing more than an argument that, for a given class of theories, the |data would be compatible with a single joint distribution. IMO the |argument is often bollixed at this point. We don't have to assign |meaning, for example, to what we would have gotten if we measured |the remaining values: a common philosophical tongue twister. Sure. The derivation of the fact that the data is compatible with a joint distribution does typically use the fact that the simultaneous measurements we're making are being made on joint distribution-- but it could easily just give some arbitrary other results caused by the two measurements interfering with each other. |> In these experiments these pairs of variables are usually |> also incompatible according to quantum mechanics, so we |> can't even ask what distribution of joint values of both |> variables it says we should get. | |Right. But then my take is ... which of course I'm happy to repeat |;-) ... that ... scratch that. We don't even really care what quantum |mechanics says we should get. The test stands alone ... ah, you jar |loose old rants from the attic ... as a test on a class of classical |theories. We are _not_ testing quantum mechanics, we are testing a |class of theories pretty much defined by the test itself. In a certain sense, I would say we are testing it, because the experiment we're trying to (re)design is one which would either falsify a bunch of other theories, or falsify quantum mechanics. If the results come out as predicted by quantum mechanics, then whether they falsify another theory depends solely on what the results are, and what the other theory predicts they should be, of course. If all the measurements we use in the experiment are compatible with each other, according to quantum mechanics, then quantum mechanics will predict results that come from a joint distribution of all the corresponding variables, which is no good. If some of them are incompatible according to quantum mechanics, then since space-like separated ones are compatible, it must be that some of our measurements are not space-like separated. But, turning back to the classical theories, if two measurements are not space-like separated, it's easy to produce a theory in which the first measurement ruins the second one. So we are stuck with designing an experiment where on each run we only make measurements that are separate in space, but where the measurements made on different runs are neither space-like separated nor compatible in quantum mechanics' terms. [...] |I also liked to point out that this mysterious class of theories |would not seem very mysterious in some more prosaic situations: like a |smallish black box, e.g., which produced output at either end in the |from of flashing lights. A violation of a Bell-like inequality for |such a box would simply mean that our choice of which flashing lights |to observe at either end was somehow communicated to the opposite end, |and influenced the outcomes there. If the box were, say, .3m on a |side, we wouldn't find this possibility very spooky! Indeed, if a science museum, say, were to make a quantum mechanics exhibit with a box like that in it, one would tend to assume they'd just wired it up to work like that. Finding examples of theories lying squarely inside or outside is straightforward. The mystery comes in only when you try to categorize all possible explanations. It seems to be so easy to think one has found all the ways that a theory can predict such violations only to find out that there are other very different ones too. I think I've seen several times, for instance, a person react to a description of one experiment by saying that such experimental results would HAVE to be the result of some kind of signal being sent from one measuring station to the other. I'm not sure I know of anyone who's come up with a local explanation, one where no signal is sent, without knowing how quantum mechanics predicts it first. I find it especially hard to say how one should categorize such things as theories with quantum logic, or nonstandard probability theory. Over here on sci.math :-) we can talk about random variables safe in the knowledge that we mean functions on measure spaces (with the measure of the whole space being 1), for which the usual laws can be rigorously deduced, but over there in sci.physics it seems like some kind of caution is in order. |> For other readers, it might be worth pointing out that the correlation |> matrix is positive semidefinite just when the covariance matrix is. | |I wondered about that, and I must say the equivalence is plausible but |not obvious. It's not obvious to me that the weighting might somehow |shift a positive determinant to a negative one, or vice versa -- |though of course derivation is the key to this difficulty. :-) A symmetric matrix M is positive definite when for each nonzero vector v, we have v^T M v > 0, where v^T is the transpose of v (v written as a row vector). The function v^T M v is called the quadratic form associated with M. The 2 by 2 case is good enough to show why the equivalence holds. Suppose the variables have standard deviations of s1 and s2, and the correlation matrix is (a11 a12) (a21 a22). Then if v=(r s)^T, then v^T M v = a11r^2 + (a12+a21)rs + a22s^2. Now the covariance matrix M' is instead (s1^2*a11 s1s2*a12) (s1s2*a21 s2^2*a22) and v^T M' v = s1^2*a11*r^2 + s1s2(a12+a21)rs + s2^2*a22*s^2 = a11(r*s1)^2 + (a12+a21)(r*s1)(s*s2) + a22(s*s2)^2, which is just v'^T M v, where v' = (r*s1 s*s2)^T. So it's >0. To put it in words, the value of the quadratic form associated with the covariance matrix evaluated on v is the same as the value of the quadratic form associated with the correlation matrix evaluated at a different vector, namely v with its coordinates scaled by the standard deviations of the variables. The reverse also holds of course, when the standard deviations are nonzero, by scaling by the reciprocals of the standard deviations. [...] |> I tried to figure out whether or not this works. Apparently, |> it doesn't work because (and here's the punch line) quantum |> mechanics also predicts that the matrix is positive |> semi-definite! To put it another way, if the experiment |> works the way quantum mechanics says it will, we can partially |> fill in the entries of the correlation matrix with the values |> we can get directly from the data (and put 1's on the |> diagonal), and the resulting partial matrix is consistent |> with being filled out to being a positive semidefinite matrix. | |I'm not sure what you're talking about ... it's quite possible there |is more than one matrix that we might think is the matrix associated |with the system: but I'm fairly clear on this: if quantum mechanics |really does predict something which falls outside our nominal class |of theories, then there must be _some_ matrix of putative |correlations which _cannot_ be so filled out. Otherwise there would |be no discrimination between interesting outcomes, andno point. The experimental data includes more information than just the correlations, however. Consider the version of the experiment in which there are three axes 1, 2, and 3. One correlation matrix obtained is A1 A2 A3 B1 B2 B3 A1 ( 1 ? ? 1 -1/2 -1/2) A2 ( ? 1 ? -1/2 1 -1/2) A3 ( ? ? 1 -1/2 -1/2 1 ) B1 ( 1 -1/2 -1/2 1 ? ? ) B2 (-1/2 1 -1/2 ? 1 ? ) B3 (-1/2 -1/2 1 ? ? 1 ). We don't get correlations for the ? entries because it would which as described above is not so helpful. Now for any random variables A1, A2, A3, B1, B2, and B3 which have the correlations given above, the perfect correlations between A1 and B1, A2 and B2, and A3 and B3 imply that those pairs are linearly related. That makes them equivalent as far as taking correlations is concerned. So finding six such variables is equivalent to finding three with a correlation matrix ( 1 -1/2 -1/2) (-1/2 1 -1/2) (-1/2 -1/2 1 ). There's no problem in doing so; if we let A1=B1=1, A2=A3=B2=B3=0 with probability 1/3, A2=B2=1, A1=A3=B1=B3=0 with probability 1/3, A3=B3=1, A1=A2=B1=B2=0 with probability 1/3, then we get just such a correlation matrix. But the measurements tell us that each of these variables is 1 half of the time, not a third of the time! |It did occur to me right away there is some difficulty with positive |semi-definite vs. positive definite in discriminating between |cases. Because ... given some set of putative jointly distributed |random variables which illegal correlation structure, we could always |flesh them out with _more_ putative random variables, linear |combinations of the first set, such that the determinate of the |correlation matrix was zero. So zero just isn't good enough. If you extend a matrix which isn't positive semidefinite, however, the extension still is not positive semidefinite. If v^T M v < 0, and we extend M with more rows and columns, and extend v with 0's at the end, then v^T M v is still <0. [Much quotation deleted.] |You've lost me completely -- I need to go review/learn some linear |algebra. But just give me a precis of what you say you've just shown, |and how it relates to our discussion? First, I gave the example I describe a little differently above, of hypothetical experimental results as predicted by quantum mechanics which violate Bell's inequality, together with a non-Bell-violating way in which the same correlations could be produced. This shows that the violation of Bell's inequalities in this case is not detectable just by looking at measured correlations between variables. Second, I sketched a proof that this is not an isolated case: the correlations you would get from any other experiment whose results were consistent with quantum mechanics could also be reproduced by a joint distribution of random variables. It's crucial, of course, that the correlations are not all of the data we get from the experiment. The data as a whole is incompatible with having a joint distribution, but the correlations can be generated by one. |Anyway, as I said, you've revived an old rant in my head, one which |include the coinage Rutherford test. Simply, although the whole EPR |tradition culminating in Bell's work is or course intimate with |quantum mechanics, nothing in the final test need be or ought to be on |some irreducible level. As I said, we are _not_ testing quantum |mechanics, we are testing a class of theories best characterized as |... ta dum ... the class of theories tested for by investigation of |compatibility with a single joint distribution given a particular |experimental set-up -- which is a slightly less circular way of saying |the class of theories tested for by this test. | |What we want to convince ourselves of ... which I've made no argument |of here for brevity ... is the equivalence of this test to the vaguer |but more motivated idea of sending out some undefined information |from a central source to some antipodes of our apparatus where, |interacting with local chance influence but not anymore backreacting |on the information which travelled to the other end of our apparatus, |produces results. Any theory/statistical observations _not_ in this |class is ... most cautiously and tautologically, but correctly ... |_not_ a result which is compatible with some undefined information |propagating outward from a source to two antipodes, not further |interacting except with local influence ... etc. | |Anyway, I invoke Rutherford as somebody who would not have known about |qm, but could definitely understand the physical reasoning in ruling |out this class of theories based on some statical experimental |observations: and if we are really ruling this class of theories out |of nature, then we have better be able to build a black box ... a very |long baseline black box ... producing outcomes falsifying the |indicated class of models without any more need for special pleading |or interpretation. I'm not sure if any contemporary quantum optics |test meets this test or not. Every now and then I have a fantasy of bringing some well respected thinker from the past to the present day and taking them on a little tour. We now have artificial lighting not requiring fire, see? Let me warn you about the rather fast-moving oil-powered vehicles we now use, before we go out by the street. You are right, there are no musicians there, there's a device in there for repeating sounds. We have big machines that fly, and many of us have travelled in them many times...! When it comes to these experiments, I sometimes imagine describing them to Einstein, Podolsky and Rosen. I think they had a lot of the same motivation as Bell, and would have enjoyed seeing how attempts at investigating the issue had progressed experimentally. Keith Ramsay === Subject: Re: limitations on mutual anti-correlation? | |Interesting to see how many people do post from Google. I do it when I decide to cross-post something, which aol's own newsreader doesn't do. I sometimes think of getting a better ISP, but it hasn't been a priority-- and using Google for just when I cross-post is ok. Doing it this way has the useful feature of giving me a slight disincentive to joining cross-posted threads. Quite a few of them cross-posted from sci.physics seem very junky lately-- has sci.physics been going through a noisy phase lately? |Interesting learning problem: after the first time it happens and you |lose your careful reply, you get in the habit of copying your post |before hitting post. This behavior is reinforced so long as the |process fails regularly, but soon rejected as unnecessary overhead |after a winning streak ... until the next failure. | |See, Google is like life. ;-) With mondo storage devices with file systems optimized for big appends and deletes! I compose longer posts in notepad, then copy and paste, so keeping a copy is nearly automatic. A fixed-width font makes it easier to avoid extra long lines and to compose ASCII graphics (see below). :-) |I also learned though that even though your post seems to be lost, |hitting refresh and then resend from the dialogue box trys the |post again ... complete with your text: you text is in some memory |buffer somewhere, even if you can't see it. I had just had to refresh a couple of previous pages, so I figured Google might actually be too busy or something.... It also seems sometimes to claim it's failed, but then you find it succeeded after all (d'oh) and the retry posted a duplicate. |But enough administrivia -- well, besides the annoying point that we |seem to have incommensurate line lengths; which is odd, considering we |are both apparently posting from Google. Yeah, I find I can compose something in notepad going all the way out to 80 columns, cut and paste it into Google's window, and it looks fine, but gets posted with wrapped lines all over. Perhaps that's what the preview page is for. I think I'll reformat what you quoted from me, below, if you don't mind. [...] |> |This was the first and only intuitive example I came up with as an |> |example of an impossible putative correlation structure for 3 random |> |variables ... | |> it's a nice enough example of what it means to violate |> Bell's inequality that I've seen it used as an |>illustration, | |Oh. And I was _so_ proud of it. But are you sure seen it used is |not recall seeing Ed Green use it? ;-) My bag of tricks is so |small, and growing so slowly of late, I trot this one out every chance |I get. I thought I'd seen some other usenet poster use it, but I could be wrong. |> in spite of not being the kind of |> violation of Bell's inequality allowed by quantum mechanics. | |Well, I'm not sure what the difference in kind is. Just that quantum mechanics predicts the possibility of making certain violations and not others. Quantum mechanics doesn't permit three variables to perfectly anticorrelate in pairs like that. For one thing, the correlation matrix is not positive semidefinite, and I just tried to show that the correlation matrix you get according to quantum mechanics always is. [...] |> |In the famous so-called Bell inequalities we consider 4 random |> |variables (of which only two may be observed at a time), and in |> effect |> |test whether our observations are compatible with masked simultaneous |> |observations of 4 random variables. |> One complication is that not all pairs taken from these |> four are distant from each other, so we lose our reason |> for expecting that the hypothetical actual values of the |> pair of variables can be measured independently of each |> other. | |I don't follow you here. Probably there is some orthogonality to our |understanding -- since I normally find this -- but also I'm not even |sure if you really mean distant or if that's a typo for distinct. I'm saying that not only that, but some of the pairs of variables can't be observed at the same time. The original poster had given a matrix and asked whether it was the matrix of covariances of a set of random variables. We don't get all the entries of a matrix from our data; we just get some of the entries, and have the question of whether those entries could be the covariances of a set of random variables. The criterion for that is in principle just whether it's possible to fill in the remaining entries of the matrix in such a way that it satisfies the condition: positive semidefiniteness. It's a bit messy to determine that in general (apparently), but I found a way to fill in certain matrices, below, based on analogous features of quantum mechanics. Typically, the experiment involves creating a pair of then plenty of theories would say that the one measurement affects the result of the other measurement-- just by Generally, theories that say making one measurement affects the results of the other measurements can easily get away with predicting just the same kind of correlations as quantum mechanics does. That's the way some formulations of quantum mechanics work, after all. Usually the reason why a theory does not predict this kind of mutual influence is because it doesn't predict such influences between measurements being made at a distance from each other. Ideally, the measurements should be made at a space-like separation (i.e., far enough apart and close enough to the same time that a signal travelling at the speed of light does not have time to get from the one measurement to the other one). |Anyway, in lose our reason ..., etc: this may touch on my |idiosyncratic approach, but our reason for expecting I think need be |nothing more than an argument that, for a given class of theories, the |data would be compatible with a single joint distribution. IMO the |argument is often bollixed at this point. We don't have to assign |meaning, for example, to what we would have gotten if we measured |the remaining values: a common philosophical tongue twister. Sure. The derivation of the fact that the data is compatible with a joint distribution does typically use the fact that the simultaneous measurements we're making are being made on joint distribution-- but it could easily just give some arbitrary other results caused by the two measurements interfering with each other. |> In these experiments these pairs of variables are usually |> also incompatible according to quantum mechanics, so we |> can't even ask what distribution of joint values of both |> variables it says we should get. | |Right. But then my take is ... which of course I'm happy to repeat |;-) ... that ... scratch that. We don't even really care what quantum |mechanics says we should get. The test stands alone ... ah, you jar |loose old rants from the attic ... as a test on a class of classical |theories. We are _not_ testing quantum mechanics, we are testing a |class of theories pretty much defined by the test itself. In a certain sense, I would say we are testing it, because the experiment we're trying to (re)design is one which would either falsify a bunch of other theories, or falsify quantum mechanics. If the results come out as predicted by quantum mechanics, then whether they falsify another theory depends solely on what the results are, and what the other theory predicts they should be, of course. If all the measurements we use in the experiment are compatible with each other, according to quantum mechanics, then quantum mechanics will predict results that come from a joint distribution of all the corresponding variables, which is no good. If some of them are incompatible according to quantum mechanics, then since space-like separated ones are compatible, it must be that some of our measurements are not space-like separated. But, turning back to the classical theories, if two measurements are not space-like separated, it's easy to produce a theory in which the first measurement ruins the second one. So we are stuck with designing an experiment where on each run we only make measurements that are separate in space, but where the measurements made on different runs are neither space-like separated nor compatible in quantum mechanics' terms. [...] |I also liked to point out that this mysterious class of theories |would not seem very mysterious in some more prosaic situations: like a |smallish black box, e.g., which produced output at either end in the |from of flashing lights. A violation of a Bell-like inequality for |such a box would simply mean that our choice of which flashing lights |to observe at either end was somehow communicated to the opposite end, |and influenced the outcomes there. If the box were, say, .3m on a |side, we wouldn't find this possibility very spooky! Indeed, if a science museum, say, were to make a quantum mechanics exhibit with a box like that in it, one would tend to assume they'd just wired it up to work like that. Finding examples of theories lying squarely inside or outside is straightforward. The mystery comes in only when you try to categorize all possible explanations. It seems to be so easy to think one has found all the ways that a theory can predict such violations only to find out that there are other very different ones too. I think I've seen several times, for instance, a person react to a description of one experiment by saying that such experimental results would HAVE to be the result of some kind of signal being sent from one measuring station to the other. I'm not sure I know of anyone who's come up with a local explanation, one where no signal is sent, without knowing how quantum mechanics predicts it first. I find it especially hard to say how one should categorize such things as theories with quantum logic, or nonstandard probability theory. Over here on sci.math :-) we can talk about random variables safe in the knowledge that we mean functions on measure spaces (with the measure of the whole space being 1), for which the usual laws can be rigorously deduced, but over there in sci.physics it seems like some kind of caution is in order. |> For other readers, it might be worth pointing out that the correlation |> matrix is positive semidefinite just when the covariance matrix is. | |I wondered about that, and I must say the equivalence is plausible but |not obvious. It's not obvious to me that the weighting might somehow |shift a positive determinant to a negative one, or vice versa -- |though of course derivation is the key to this difficulty. :-) A symmetric matrix M is positive definite when for each nonzero vector v, we have v^T M v > 0, where v^T is the transpose of v (v written as a row vector). The function v^T M v is called the quadratic form associated with M. The 2 by 2 case is good enough to show why the equivalence holds. Suppose the variables have standard deviations of s1 and s2, and the correlation matrix is (a11 a12) (a21 a22). Then if v=(r s)^T, then v^T M v = a11r^2 + (a12+a21)rs + a22s^2. Now the covariance matrix M' is instead (s1^2*a11 s1s2*a12) (s1s2*a21 s2^2*a22) and v^T M' v = s1^2*a11*r^2 + s1s2(a12+a21)rs + s2^2*a22*s^2 = a11(r*s1)^2 + (a12+a21)(r*s1)(s*s2) + a22(s*s2)^2, which is just v'^T M v, where v' = (r*s1 s*s2)^T. So it's >0. To put it in words, the value of the quadratic form associated with the covariance matrix evaluated on v is the same as the value of the quadratic form associated with the correlation matrix evaluated at a different vector, namely v with its coordinates scaled by the standard deviations of the variables. The reverse also holds of course, when the standard deviations are nonzero, by scaling by the reciprocals of the standard deviations. [...] |> I tried to figure out whether or not this works. Apparently, |> it doesn't work because (and here's the punch line) quantum |> mechanics also predicts that the matrix is positive |> semi-definite! To put it another way, if the experiment |> works the way quantum mechanics says it will, we can partially |> fill in the entries of the correlation matrix with the values |> we can get directly from the data (and put 1's on the |> diagonal), and the resulting partial matrix is consistent |> with being filled out to being a positive semidefinite matrix. | |I'm not sure what you're talking about ... it's quite possible there |is more than one matrix that we might think is the matrix associated |with the system: but I'm fairly clear on this: if quantum mechanics |really does predict something which falls outside our nominal class |of theories, then there must be _some_ matrix of putative |correlations which _cannot_ be so filled out. Otherwise there would |be no discrimination between interesting outcomes, andno point. The experimental data includes more information than just the correlations, however. Consider the version of the experiment in which there are three axes 1, 2, and 3. One correlation matrix obtained is A1 A2 A3 B1 B2 B3 A1 ( 1 ? ? 1 -1/2 -1/2) A2 ( ? 1 ? -1/2 1 -1/2) A3 ( ? ? 1 -1/2 -1/2 1 ) B1 ( 1 -1/2 -1/2 1 ? ? ) B2 (-1/2 1 -1/2 ? 1 ? ) B3 (-1/2 -1/2 1 ? ? 1 ). We don't get correlations for the ? entries because it would which as described above is not so helpful. Now for any random variables A1, A2, A3, B1, B2, and B3 which have the correlations given above, the perfect correlations between A1 and B1, A2 and B2, and A3 and B3 imply that those pairs are linearly related. That makes them equivalent as far as taking correlations is concerned. So finding six such variables is equivalent to finding three with a correlation matrix ( 1 -1/2 -1/2) (-1/2 1 -1/2) (-1/2 -1/2 1 ). There's no problem in doing so; if we let A1=B1=1, A2=A3=B2=B3=0 with probability 1/3, A2=B2=1, A1=A3=B1=B3=0 with probability 1/3, A3=B3=1, A1=A2=B1=B2=0 with probability 1/3, then we get just such a correlation matrix. But the measurements tell us that each of these variables is 1 half of the time, not a third of the time! |It did occur to me right away there is some difficulty with positive |semi-definite vs. positive definite in discriminating between |cases. Because ... given some set of putative jointly distributed |random variables which illegal correlation structure, we could always |flesh them out with _more_ putative random variables, linear |combinations of the first set, such that the determinate of the |correlation matrix was zero. So zero just isn't good enough. If you extend a matrix which isn't positive semidefinite, however, the extension still is not positive semidefinite. If v^T M v < 0, and we extend M with more rows and columns, and extend v with 0's at the end, then v^T M v is still <0. [Much quotation deleted.] |You've lost me completely -- I need to go review/learn some linear |algebra. But just give me a precis of what you say you've just shown, |and how it relates to our discussion? First, I gave the example I describe a little differently above, of hypothetical experimental results as predicted by quantum mechanics which violate Bell's inequality, together with a non-Bell-violating way in which the same correlations could be produced. This shows that the violation of Bell's inequalities in this case is not detectable just by looking at measured correlations between variables. Second, I sketched a proof that this is not an isolated case: the correlations you would get from any other experiment whose results were consistent with quantum mechanics could also be reproduced by a joint distribution of random variables. It's crucial, of course, that the correlations are not all of the data we get from the experiment. The data as a whole is incompatible with having a joint distribution, but the correlations can be generated by one. |Anyway, as I said, you've revived an old rant in my head, one which |include the coinage Rutherford test. Simply, although the whole EPR |tradition culminating in Bell's work is or course intimate with |quantum mechanics, nothing in the final test need be or ought to be on |some irreducible level. As I said, we are _not_ testing quantum |mechanics, we are testing a class of theories best characterized as |... ta dum ... the class of theories tested for by investigation of |compatibility with a single joint distribution given a particular |experimental set-up -- which is a slightly less circular way of saying |the class of theories tested for by this test. | |What we want to convince ourselves of ... which I've made no argument |of here for brevity ... is the equivalence of this test to the vaguer |but more motivated idea of sending out some undefined information |from a central source to some antipodes of our apparatus where, |interacting with local chance influence but not anymore backreacting |on the information which travelled to the other end of our apparatus, |produces results. Any theory/statistical observations _not_ in this |class is ... most cautiously and tautologically, but correctly ... |_not_ a result which is compatible with some undefined information |propagating outward from a source to two antipodes, not further |interacting except with local influence ... etc. | |Anyway, I invoke Rutherford as somebody who would not have known about |qm, but could definitely understand the physical reasoning in ruling |out this class of theories based on some statical experimental |observations: and if we are really ruling this class of theories out |of nature, then we have better be able to build a black box ... a very |long baseline black box ... producing outcomes falsifying the |indicated class of models without any more need for special pleading |or interpretation. I'm not sure if any contemporary quantum optics |test meets this test or not. Every now and then I have a fantasy of bringing some well respected thinker from the past to the present day and taking them on a little tour. We now have artificial lighting not requiring fire, see? Let me warn you about the rather fast-moving oil-powered vehicles we now use, before we go out by the street. You are right, there are no musicians there, there's a device in there for repeating sounds. We have big machines that fly, and many of us have travelled in them many times...! When it comes to these experiments, I sometimes imagine describing them to Einstein, Podolsky and Rosen. I think they had a lot of the same motivation as Bell, and would have enjoyed seeing how attempts at investigating the issue had progressed experimentally. Keith Ramsay === Subject: Re: JSH: Distributive property, math argument >Consider that if I have f(x) = x + 3, where 3 has itself as a factor >and f(x) has 3 as a factor, then x *must* have 3 as a factor. Then I would conclude that f(x)/3 is in the ring you are in, by definition of factor. >Now let's say x=7. Then, 7/3 must be in the ring. >See? Is this supposed to prove that 7/3 is an algebraic integer? To do so, you must start from the assumption that 10/3 is an algebraic integer, which it isn't. - Randy === Subject: Re: JSH: Distributive property, math argument > ... > The start is simple enough, where the x shown is in the ring of > algebraic integers. > Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that you > use. At the start of the post I point out that if f(x) = x + 3, and f(x) > has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at > least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true, > without regard to the commutative ring. yes, you did. > You said Consider f(x) = x+3, where x is an algebraic integer. > Yup, you're right. I'd forgotten that quickly and didn't go back to > check before making the reply. > In any event, maybe I should say that it doesn't matter what the ring > is. > Consider that if I have f(x) = x + 3, where 3 has itself as a factor > and f(x) has 3 as a factor, then x *must* have 3 as a factor. > Now let's say x=7. Then, 7/3 must be in the ring. > See? > No, I don't understand . You said it was the ring of algebraic integers > and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of > concluding that this places a *restricition* on the values that x can take > (i.e. x must be an algebraic integer that is divisible by 3) you choose x to > be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of > algebraic integers.!!!! OOPS! In my previous post I have f(x) = 2((x+1)/2 + 1) when I should have had f(x) = 2(x(x+1)/2 + 1). James Harris === Subject: Re: JSH: Distributive property, math argument > ... > The start is simple enough, where the x shown is in the ring of > algebraic integers. > Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that you > use. At the start of the post I point out that if f(x) = x + 3, and f(x) > has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at > least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true, > without regard to the commutative ring. yes, you did. > You said Consider f(x) = x+3, where x is an algebraic integer. > Yup, you're right. I'd forgotten that quickly and didn't go back to > check before making the reply. > In any event, maybe I should say that it doesn't matter what the ring > is. > Consider that if I have f(x) = x + 3, where 3 has itself as a factor > and f(x) has 3 as a factor, then x *must* have 3 as a factor. > Now let's say x=7. Then, 7/3 must be in the ring. > See? > No, I don't understand . You said it was the ring of algebraic integers > and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of > concluding that this places a *restricition* on the values that x can take > (i.e. x must be an algebraic integer that is divisible by 3) you choose x to > be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of > algebraic integers.!!!! Nope. I'm saying that it doesn't matter what you might *say* the ring is, given f(x) = x+3, if you also have that f(x) has 3 as a factor and 3 has itself as a factor, as then necessarily x has 3 as a factor, which means that if x=7, which IS an algebraic integer, it *still* must have 3 as a factor for the other statements to be true. Look at it as a logical argument: 1. f(x) = x + 3 2. 3 has itself as a factor 3. f(x) has 3 as a factor 4. Therefore, x has 3 as a factor. You see, 1., 2., and 3., logically force 4. from the distributive property, understand? Here's another example to help you out. Now consider the factorization f(x) = 2((x+1)/2 + 1) where you'll notice that the factorization exists in the ring of integers, but does NOT always exist in the ring of algebraic integers. Think about what I said does NOT always exist and consider the purpose of the use of words. James Harris === Subject: Re: JSH: Distributive property, math argument > ... > The start is simple enough, where the x shown is in the ring of > algebraic integers. > Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that you > use. At the start of the post I point out that if f(x) = x + 3, and f(x) > has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at > least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true, > without regard to the commutative ring. yes, you did. > You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to > check before making the reply. In any event, maybe I should say that it doesn't matter what the ring > is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor > and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring. See? No, I don't understand . You said it was the ring of algebraic integers > and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of > concluding that this places a *restricition* on the values that x can take > (i.e. x must be an algebraic integer that is divisible by 3) you choose x to > be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of > algebraic integers.!!!! > Nope. I'm saying that it doesn't matter what you might *say* the ring > is, given f(x) = x+3, if you also have that f(x) has 3 as a factor and > 3 has itself as a factor, as then necessarily x has 3 as a factor, > which means that if x=7, which IS an algebraic integer, it *still* > must have 3 as a factor for the other statements to be true. > Look at it as a logical argument: > 1. f(x) = x + 3 > 2. 3 has itself as a factor > 3. f(x) has 3 as a factor > 4. Therefore, x has 3 as a factor. > You see, 1., 2., and 3., logically force 4. from the distributive > property, understand? > Here's another example to help you out. > Now consider the factorization > f(x) = 2((x+1)/2 + 1) > where you'll notice that the factorization exists in the ring of > integers, but does NOT always exist in the ring of algebraic integers. As you noted elsewhere, what you meant to say here was: f(x) = 2*(x*(x + 1)/2 + 1) This can be re-written as f(x) = x^2 + x + 3. Your statement is diametrically wrong. This polynomial does NOT factor with integer coefficients, but it DOES factor in the algebraic integers: f(x) = (x - r1)*(x - r2), where r1 = (-1 + sqrt(-11))/2 and r2 = (-1 - sqrt(-11))/2, and of course both r1 and r2 are algebraic integers. The factorization that you gave is not a factorization in EITHER the integers or the algebraic integers (dividing by 2 is the same as multiplying by 1/2). Are you totally losing it, or what ??? > Think about what I said does NOT always exist and consider the > purpose of the use of words. Uh ... right. Whatever you say, Einstein. It might also be worth noting that if x is an algebraic integer, f(x) = x^2 + x + 3 is also: the set of A.I.'s is closed under multiplication and addition. Nora B. > James Harris === Subject: Re: JSH: Distributive property, math argument Adjunct Assistant Professor at the University of Montana. [.snip.] >> Now consider the factorization >> f(x) = 2((x+1)/2 + 1) >> where you'll notice that the factorization exists in the ring of >> integers, but does NOT always exist in the ring of algebraic integers. > As you noted elsewhere, what you meant to say here was: > f(x) = 2*(x*(x + 1)/2 + 1) > This can be re-written as > f(x) = x^2 + x + 3. > Your statement is diametrically wrong. This polynomial does >NOT factor with integer coefficients, but it DOES factor in the >algebraic integers: > f(x) = (x - r1)*(x - r2), where > r1 = (-1 + sqrt(-11))/2 and > r2 = (-1 - sqrt(-11))/2, >and of course both r1 and r2 are algebraic integers. > The factorization that you gave is not a factorization in >EITHER the integers or the algebraic integers (dividing by >2 is the same as multiplying by 1/2). > Are you totally losing it, or what ??? James is once again confusing a polynomial factorization and a numeric factorization; that is, the difference between the values of a polynomial always dividing the values of another polynomial, and a polynomial divind the other polynomial. Here, we have: F(x) = x^2 + x + 3. P(x) = 2 Q(x) = (1/2)(x^2+x+3). Then F(x) = P(x)*Q(x) in the ring of all polynomials over Q. IN ADDITION, Q(x) is a function from Z to Z, and P(x) is a function from Z to Z, with the properties that, AS FUNCTIONS, F(x) = P(x)*Q(x). That is, for each value of x, Q(x) divides F(x) in Z, which is a different statement from the polynomial Q(x) divides the polynomial F(x) in Z[x]. Amazing, that after only about two years he finally realizes what he was told so long ago: that a factorization of the VALUES of a polynomial does not always correspond to the a factorization of the POLYNOMIAL. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Distributive property, math argument Now consider the factorization > f(x) = 2((x+1)/2 + 1) That should be f(x) = 2(x(x+1)/2 + 1). > where you'll notice that the factorization exists in the ring of > integers, but does NOT always exist in the ring of algebraic integers. > Think about what I said does NOT always exist and consider the > purpose of the use of words. James Harris === Subject: Re: JSH: Distributive property, math argument > ... > The start is simple enough, where the x shown is in the ring of > algebraic integers. > Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that you > use. At the start of the post I point out that if f(x) = x + 3, and f(x) > has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at > least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true, > without regard to the commutative ring. yes, you did. > You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to > check before making the reply. In any event, maybe I should say that it doesn't matter what the ring > is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor > and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring. See? > No, I don't understand . You said it was the ring of algebraic integers > and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of > concluding that this places a *restriction* on the values that x can take > (i.e. x must be an algebraic integer that is divisible by 3) you choose x to > be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of > algebraic integers.!!!! > Nope. I'm saying that it doesn't matter what you might *say* the ring > is, given f(x) = x+3, if you also have that f(x) has 3 as a factor and > 3 has itself as a factor, as then necessarily x has 3 as a factor, > which means that if x=7, which IS an algebraic integer, it *still* > must have 3 as a factor for the other statements to be true. Perhaps I misunderstood the example as in the original discussion of which this is an overly simplified example it was obvious that f(x) indeed had such a factor. What I don't understand is why you think (in either the ring of integers or algebraic integers), that given f(x) = x+3, you can simply *assert* as you did that f(x) has 3 as a factor without investigating whether for various values of x it indeed does so. In particular, for your example of x = 7, f(7) = 10, which does not have 3 as a factor, nor does 7 have 3 as a factor. > Look at it as a logical argument: > 1. f(x) = x + 3 > 2. 3 has itself as a factor > 3. f(x) has 3 as a factor > 4. Therefore, x has 3 as a factor. > You see, 1., 2., and 3., logically force 4. from the distributive > property, understand? No, given 1, 2 and 3, you can say: 4. Therefore x is constrained to only those values that have 3 as a factor OR you can say: 3. IF x has 3 as a factor THEN 4. f(x) has 3 as a factor but you can't turn it around the other way as you did The distributive property: a(y+z) = ay + bz, applied to your equation is: 3(x/3 +1) = 3(x/3) + 3(1) It says nothing at all about x. Pick an x, any x. Pick 7. It does *not* have 3 as a factor although the distributive property is upheld. KeithK > Here's another example to help you out. > Now consider the factorization > f(x) = 2((x+1)/2 + 1) > where you'll notice that the factorization exists in the ring of > integers, but does NOT always exist in the ring of algebraic integers. > Think about what I said does NOT always exist and consider the > purpose of the use of words. > James Harris === Subject: Re: JSH: Distributive property, math argument > ... > The start is simple enough, where the x shown is in the ring > of > algebraic integers. > Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that > you > use. At the start of the post I point out that if f(x) = x + 3, and > f(x) > has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at > least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true, > without regard to the commutative ring. yes, you did. > You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to > check before making the reply. In any event, maybe I should say that it doesn't matter what the ring > is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor > and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring. See? No, I don't understand . You said it was the ring of algebraic > integers > and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of > concluding that this places a *restriction* on the values that x can > take > (i.e. x must be an algebraic integer that is divisible by 3) you choose > x to > be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of > algebraic integers.!!!! > Nope. I'm saying that it doesn't matter what you might *say* the ring > is, given f(x) = x+3, if you also have that f(x) has 3 as a factor and > 3 has itself as a factor, as then necessarily x has 3 as a factor, > which means that if x=7, which IS an algebraic integer, it *still* > must have 3 as a factor for the other statements to be true. > Perhaps I misunderstood the example as in the original discussion of which > this is an overly simplified example it was obvious that f(x) indeed had > such a factor. > What I don't understand is why you think (in either the ring of integers or > algebraic integers), that given f(x) = x+3, you can simply *assert* as you > did that f(x) has 3 as a factor without investigating whether for various > values of x it indeed does so. It's a logical point, and perhaps a subtle one. If you have f(x) = x + 3, that f(x) has 3 as a factor, and you have that 3 has itself as a factor, then it *must* be true that x has 3 as a factor. There is no investigation as it's logically forced. Now you may wish to challenge that f(x) has 3 as a factor, or that 3 has itself as a factor, but the logical argument itself is not challengeable. It's like: 1. All dogs wear shoes. 2. Ralph is a dog. 3. Therefore Ralph wears shoes. The logical argument is ok, it's premise 1. that's faulty. So your investigation has to do with the premises, not with the conclusion. > In particular, for your example of x = 7, f(7) = 10, which does not have 3 > as a factor, nor does 7 have 3 as a factor. Yes it does, in a ring where 7 is a unit. For instance, in reals, 3(7/3) = 7, so 3 is a factor of 7. > Look at it as a logical argument: > 1. f(x) = x + 3 > 2. 3 has itself as a factor > 3. f(x) has 3 as a factor > 4. Therefore, x has 3 as a factor. > You see, 1., 2., and 3., logically force 4. from the distributive > property, understand? > No, given 1, 2 and 3, you can say: > 4. Therefore x is constrained to only those values that have 3 as a factor No. You can't go to the end, you have to go to the beginning. > OR you can say: > 3. IF x has 3 as a factor THEN > 4. f(x) has 3 as a factor > but you can't turn it around the other way as you did Yes I can. It's a *logical* argument. > The distributive property: a(y+z) = ay + bz, > applied to your equation is: 3(x/3 +1) = 3(x/3) + 3(1) > It says nothing at all about x. Pick an x, any x. Pick 7. It does *not* > have 3 as a factor although the distributive property is upheld. > KeithK It does in the field of reals. James Harris === Subject: Re: JSH: Distributive property, math argument Noting your correction: > Now consider the factorization > f(x) = 2((x+1)/2 + 1) Corrected to: > f(x) = 2(x(x+1)/2 + 1) > where you'll notice that the factorization exists in the ring of > integers, but does NOT always exist in the ring of algebraic integers. Lessee. f(x) = 2(x(x + 1)/2 + 1) = x(x + 1) + 2 = x^2 + x + 3. What is the factorisation in the integers? In the algebraic integers there is a ready factorisation for each and every algebraic integer x. -- 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: JSH: Distributive property, math argument >> It's amazing how easily people can question very basic concepts when >> the stakes are high, so here I am again to help you out by reminding >> you just how basic the math argument I've been giving recently is. >> Consider f(x) = x+3, where x is an algebraic integer. >> Now then, if f(x) has 3 as a factor, then x *must* have 3 as a factor >> as well. >> That follows easily enough from the distributive property: >> a(b+c) = ab + ac >> and for quite a few months, I've noticed people willing to debate it, >> as if that property can just go away at a whim. >> That's not mathematics, and many of you are showing a puzzling lack of >> rationality by fighting mathematics itself, rather than just accepting >> the truth. >> After all, it's not like the math is mad at you, or that there's some >> conspiracy to screw you over as mathematical truth is just true. >> If you believed something else in the past that was wrong, it was just >> wrong. >> Now here's the argument, where those people questioning it have >> basically been questioning the distributive property. >> The start is simple enough, where the x shown is in the ring of >> algebraic integers. >> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >> which you'll notice has a constant term that is 1078. >> Well moving things around with P(x) gives you >> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 >> which is a deliberate form to allow me to factor P(x), so that I have >> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >> where the a's are roots of >> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >> To use the distributive property as I wish I need to have constant >> terms. >> And it *appears* that the constant terms for the three factors are all >> 7, but that can't be right, as the constant term of P(x) is 1078. >> So I need to do another step, and analysis offers the simple technique >> of setting x=0 to pull out the constant terms, so setting x=0, I find >> that >> P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) >> as the cubic defining the a's at x=0 is >> a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and >> a_2(0) to equal 0, which leaves a_3(0) with a value of 3. >> So let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have >> P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) >> P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) >> and now my constant terms work out correctly. >> Now I can use the distributive property, as P(x) has 49 as a factor of >> each term in >> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >> so P(x) has 49 as a factor, so I can divide by 49, and dividing 1078 >> by 49 gives me 22, as the new constant term. >> Well that means that >> P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) >> is the only way that 49 can divide out and keep the constant terms >> matching, in what is a simple exercise dependent on the distributive >> property. >> For the following below, I am going to define g1, g2, and g3 as: >> g1 = 5*a_1 + 7 >> g2 = 5*a_2 + 7 and >> g3 = 5*b_3 + 22 = 5*a3 + 7. >> OK, you have that P(x)/49 has constant term 22, and indeed >> g3 = 5*b_3 + 22 has constant term 22 also. It is almost irresistible >The proof shows what *must* be the case, which is how math works. >Human emotion is irrelevant. >> to conclude that the only way you can factor P(x)/49 is in >> the form >> P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22). >> As you say, the constant terms multiply together perfectly >> to give 22. Plus 22 is coprime to 7. >> But it's possible, for any given value of x, that 5*b3 + 22 is >> NOT coprime to 7. For example, what if b3 = 4 ? >I focus on constant terms because they are constant, so it hardly >matters about changes from the value of x. This is such a blatantly ridiculous statement! I said that even though 22 is coprime to 7, it is not necessarily true that 5*b3 + 22 is coprime to 7. I gave you the example of b3 = 4, for which 5*b3 + 22 = 5*4 + 22 = 20 + 22 = 42, which is actually DIVISIBLE by 7. Then you come back and say: I focus on constant terms because they are constant, so it hardly matters about changes from the value of x. Have you completely lost your marbles ? You know that b3 is dependent on x. What happens if for a given value of x, b3 = b3(x) = 4 ???? How have you eliminated that possibility ? Not to mention the *many* other ways that 5*b3 + 22 could be non-coprime to 7 in the algebraic integers. In view of this example your statement, and especially the phrase hardly matters is simply senseless. Clearly you are just plain assuming that if the constant term g(0) of a function g(x) is coprime to 7, then g(x) is coprime to 7 for all x. This is so obviously false, yet you seem unable to grasp it. Why? >The proper approach is to look over the proof, step-by-step, and if >you see some step that bugs you, ask me about it. Which is exactly what I did. Your answer is not just inadequate; it is totally ridiculous! Plus you deleted completely my main objection! Your implied offer to respond when we ask about some step that bugs us is clearly insincere. You just go through the motions of flapping your lips and claim you have responded. It's dishonest. >I'm not interested in a lot of time wasted on tangents and wild goose >chases. It appears more like you are not interested in answering substantive objections in a substantive way. What I brought up is *central* to your argument, not a tangent or a wild goose chase. It is right down the middle. You are evading. Nora B. >James Harris === Subject: Lp norm exercise I'm trying to do one exercise of Rudin's Real and Complex Analysis chapter on Lp spaces: Show that if 0 < m(E) < infty, 0 < ||f||_infty < infty then a_{n+1}/a_n -> ||f||_infty where a_n = int_E |f|^n dm, with Esubset R^k, m lebesgue measure. One can easily show lim sup a_{n+1}/a_n <= ||f||_infty ... but the other side is giving me some trouble ... Any hints ? nojb. === Subject: Re: Lp norm exercise >I'm trying to do one exercise of Rudin's Real and Complex Analysis >chapter on Lp spaces: Show that if 0 < m(E) < infty, 0 < ||f||_infty >< infty then a_{n+1}/a_n -> ||f||_infty where a_n = int_E |f|^n dm, >with Esubset R^k, m lebesgue measure. >One can easily show lim sup a_{n+1}/a_n <= ||f||_infty ... but the >other side is giving me some trouble ... Any hints ? Hint: if n is large enough nearly all of a_n comes from the integral over those x for which |f(x)| is very near ||f||_infty. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Sum of series Trishia Rose > Chintan M. Shah > (SNIP) > I wanted to know how to get the formula for sum of series. > (SNIP) > The result can be proves by Mathematical Induction but how to actually get > the RHS needed to make use of the proof. (SNIP) > For the special case where the exponent is 1, theres a remarkably > elementary demonstration (though not really rigorous) which I'm > amazed was not included in the mathworld link. ... Here's another bit that MathWorld missed. The polynomials S_p defined by S_p(n)=sum{k=1 to n}k^p satisfy this recursion relation (when we extend them to polynomial functions on R, so that they can be integrated): S_{p+1}(n) = (p+1) int{t=0 to n} S_p(t)dt + kn where the constant k is such that S_{p+1}(1) = 1. Explicitly, k = 1 - (p+1) int{t=0 to 1} S_p{t}dt. I've never seen exactly this formula in print or on the web. LH === Subject: Re: Sum of series > Later on he was >correcting his dad's checking and banking accounts at age 12. :*) Any reasonably intelligent 12-year-old would be able to do that. But Gauss did it when he was 3. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Sum of series A Robert Israel Ucoaaa ooi iUiuia > Later on he was >correcting his dad's checking and banking accounts at age 12. :*) > Any reasonably intelligent 12-year-old would be able to do that. Maybe _you_ were able to do that at age 12, who we know were reasonable with math, but I still can't do it at age 40 :*) > But Gauss did it when he was 3. Yes. Didn't recall the source right. It is indeed 3. By the way, my source indicates sum(i,i=1..100), for the kid Gauss assignment, contrary to what Ignatio seems to quote. I assume his is a more recent source, so I will go with that, although it seems to me a little unreasonable for a teacher of elementary school to have given such a task to the kids. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: finding the extremity of a function of a matrix As we know, for a function y = f(x), the extremities can be found by letting dy/dx = 0. Now I have a scalar function, J = f(X), where J is a scalar and X is a NxN matrix, firstly, D = (partial J/partial X(n1,n2)) leads to an NxN matrix. Secondly, setting D to zero and to find the extremity of the X. Does it make sense? Lin. S. === Subject: Re: finding the extremity of a function of a matrix S. Lin grava .88 la saucisse et au marteau: > As we know, for a function y = f(x), the extremities can be found by > letting dy/dx = 0. > Now I have a scalar function, J = f(X), where J is a scalar and X is a > NxN matrix, firstly, D = (partial J/partial X(n1,n2)) leads to an NxN > matrix. Secondly, setting D to zero and to find the extremity of the X. > Does it make sense? Yes, as you simply consider f as a function from R^{N^2} to R and compute its gradient. -- Nicolas === Subject: Product of Reals Is it possible to determine what the uncountable product of all real numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1, but is this concept studied in general anywhere (e.g. the concept of uncountably infinite products or sums of real numbers)? === Subject: Re: Product of Reals > Is it possible to determine what the uncountable product of all real > numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1, > but is this concept studied in general anywhere (e.g. the concept of > uncountably infinite products or sums of real numbers)? Lookup the term summable family. This is a concept which allows definition of convergence for any numbers a_i with i in I where I is an arbitrary index set. By applying exp-log-formalism (transform your product into a sum) you get the analogue theory for products. A necessary condition for such a family to converge is that its support (the largest set J subset I such that a_j <> 0 forall j in J) is countable - this means that your product cannot converge. Because equally well than your reasoning I could argue that it converges to any other real number <> 1. Cf. Riemann's rearrangement theorem. -- reverse my forename for mail! === Subject: Re: Product of Reals >Is it possible to determine what the uncountable product of all real >numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1, >but is this concept studied in general anywhere (e.g. the concept of >uncountably infinite products or sums of real numbers)? I could argue that it's going to be zero, if it's defined (which it isn't). Write the above product as: (Product)_{0.5<=x<=1} (x * 1-x) Thus 0.5 times 1.5, 0.7 times 1.3, 0.9 times 1.1, etc. (This takes care of every product except for the middle term one, so multiply the answer by one when you're finished if it makes you feel better). Each term in this product is strictly less than one, and there are an infinite number of them. Hence the product is zero. Doug === Subject: Re: Product of Reals > Each term in this product is strictly less than one, and there are an > infinite number of them. Hence the product is zero. Actually (1/2)*(3/4)*(7/8)*(15/16)*... > 0. === Subject: Re: Product of Reals > Is it possible to determine what the uncountable product of all real > numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1, > but is this concept studied in general anywhere (e.g. the concept of > uncountably infinite products or sums of real numbers)? No, no, and no. It's possible to define something that works as the *average* of all the real numbers in the interval, and that's the integral of x over the interval, divided by the length of the interval. That works out, no surprise, to 1. That's average in the sense of arithmetic mean. There's also an average in the sense of geometric mean, which is something like what do want a product to do, and that's e^A, where A is the arithmetic mean of log x. So: exp ( integral from .5 to 1.5 of log x dx). I'll let you work it out. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Product of Reals > Is it possible to determine what the uncountable product of all real > numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1, > but is this concept studied in general anywhere (e.g. the concept of > uncountably infinite products or sums of real numbers)? > No, no, and no. > It's possible to define something that works as the *average* of > all the real numbers in the interval, and that's the integral of > x over the interval, divided by the length of the interval. That > works out, no surprise, to 1. > That's average in the sense of arithmetic mean. There's also an > average in the sense of geometric mean, which is something like > what do want a product to do, and that's e^A, where A is the > arithmetic mean of log x. So: > exp ( integral from .5 to 1.5 of log x dx). > I'll let you work it out. What does this number represent? Can you point me to a reference which discusses this number? I got that that the above expression equals 3^(1.5)/(2e) which is approximately equal to .955877. I guess a better question is what does the geometric mean of two numbers represent, and then I can perhaps apply this analogy to the above expression. === Subject: Re: Product of Reals > I guess a better question is what does the geometric mean of two > numbers represent, and then I can perhaps apply this analogy to the > above expression. The geometric mean is to multiplication what the arithmetic mean is to addition. Suppose shares of toolshed.com multiplied by 4 in value during 1999 and by 9 in 2000. The geometric mean of 4 and 9 is 6, and the net effect is the same as if the shares had multiplied in value by 6 each year. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Product of Reals Nobody grava .88 la saucisse et au marteau: > equals 3^(1.5)/(2e) which is approximately equal to .955877. I guess > a better question is what does the geometric mean of two numbers > represent, and then I can perhaps apply this analogy to the above > expression. The arithmetic mean c of a and b (a<= b) is such that a,c and b are 3 consecutive terms of an arithmetic sequence, that is to says c-a = b-c The geometric mean c of a and b is such that a,c and b are 3 consecutive terms of a geometric sequence, that is to say c/a = b/c (if a<>0). hth -- Nicolas === Subject: Re: Product of Reals >Is it possible to determine what the uncountable product of all real >numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1, >but is this concept studied in general anywhere (e.g. the concept of >uncountably infinite products or sums of real numbers)? No, such a product is not well-defined. Note that we can convert this to a question of sums by taking logarithms. First (in contradistinctiion to this example) consider a sum of uncountably many positive terms. Then, for some positive integer k, uncountably many of these terms which exceed 1/k. Hence, the sum is infinite. In the case at hand, after taking logarithms, there are uncountable many negative terms and uncountaby many infinite terms. Hence, you are trying to compute the difference infinity - infinity. And why would you think this product is 1 in the first place? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Product of Reals >Is it possible to determine what the uncountable product of all real >numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1, >but is this concept studied in general anywhere (e.g. the concept of >uncountably infinite products or sums of real numbers)? No, such a product is not well-defined. > Note that we can convert this to a question of sums by taking > logarithms. First (in contradistinctiion to this example) consider a > sum of uncountably many positive terms. Then, for some positive integer > k, uncountably many of these terms which exceed 1/k. Hence, the sum > is infinite. > In the case at hand, after taking logarithms, there are uncountable many > negative terms and uncountaby many infinite terms. Hence, you are > trying to compute the difference infinity - infinity. > And why would you think this product is 1 in the first place? I guess I was thinking of an average, which is of course a sum not a product. But how about this (which I'm sure is still wrong). What if you consider the infinite product of the numbers in the interval A=[2/3,3/2]. Then is this product 1? I would attempt to prove this as follows: Let x be in A. If x is 1 then it contributes nothing to the product, otherwise assume 2/3 <= x < 1. Then 1 < 1/x <= 3/2. So 1/x is in A. Likewise if 1 < x <= 3/2, then 1/x is in A. So for any number x in A, 1/x is in A. Hence the product of all of them is 1. Or is there some problem with this? === Subject: Re: Product of Reals >I guess I was thinking of an average, which is of course a sum not a >product. But how about this (which I'm sure is still wrong). What if >you consider the infinite product of the numbers in the interval >A=[2/3,3/2]. Then is this product 1? I would attempt to prove this >as follows: >Let x be in A. If x is 1 then it contributes nothing to the product, >otherwise assume 2/3 <= x < 1. Then 1 < 1/x <= 3/2. So 1/x is in A. >Likewise if 1 < x <= 3/2, then 1/x is in A. So for any number x in A, >1/x is in A. Hence the product of all of them is 1. Assume for simpleness that we are only concerned with rationals and that the infinite sum of rationals on a closed interval is well-defined. They are countable, so we can order them into a product where the product of two consecutive rationals equals 1. Then, taking the logarithm of this product gives an alternating series whose terms approach zero, so the series converges conditionally (to 0, of course). However, if we change the ordering of the series and go back to our original product we can obtain a different value for the product. But we know that multiplication of rationals is supposed to commutate, so we have run into a contradiction. I'm not sure either reasoning even works for reals. But you can see how it's easy to reach very confusing results by assuming such a product is well-defined. >Or is there some problem with this? No problem that doesn't already exist with infinite sums. === Subject: help me proof I'm a student in the mathematics department of College of Natural Sciences, Vietnam National University, Ho Chi Minh City I have a problem with my homework Can professor help me proof ?: Assume E ={ (a1 ,a2 ,....,an , ... ) ; a1 ,a2 ,....,an , ... are complex numbers} Define vector addition and scalar multiplication on E such that every x,y in E ,every p is complex number x+y=(x1 +y1 ,x2 +y2 ,....,xn +yn , ... ) p .x= (px1 ,px2 ,....,pxn , ... ) such that E is a vector space. Proof no exist a norm on E such that if every sequence x(n) of elements in E converge to 0 with this norm then limn->infinite x_m(n)=0 with m=1,2,3......... thank professor very much ,professor can send to me by email my email : nguyencaotrongduy2@yahoo.com === Subject: Re: Parabolic, Elliptical, and Hyperbolic Geometry. > How did these three geometries get their names? What does it have do to > with the conics by the same names? Not much. Elliptic means not enough, hyperbolic means too much. Elliptic geometry doesn't have enough parallel lines, hyperbolic has too many. Parabolic geometry is new to me. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Parabolic, Elliptical, and Hyperbolic Geometry. >> How did these three geometries get their names? What does it have do to >> with the conics by the same names? >Not much. Elliptic means not enough, hyperbolic means too much. >Elliptic geometry doesn't have enough parallel lines, hyperbolic >has too many. >Parabolic geometry is new to me. According to http://members.aol.com/jeff570/mathword.html : The terms HYPERBOLIC GEOMETRY, ELLIPTIC GEOMETRY, and PARABOLIC GEOMETRY were introduced by Felix Klein (1849-1925) in 1871 in .86ber die sogenannte Nicht-Euklidische Geometrie (On so-called non-Euclidean geometry), reprinted in his Gesammelte mathematische Abhandlungen I (1921) p. 246 (Ken Pledger and Smart, p. 301). I have not read that reference, but a likely reason for the terminology is that the 2nd-order local approximation of a surface in R^3 with positive, negative, or zero Gaussian curvature is an ellipsoid, hyperboloid, or paraboloid (or plane), respectively. John Mitchell === Subject: Re: Parabolic, Elliptical, and Hyperbolic Geometry. >Not much. Elliptic means not enough, hyperbolic means too much. >Parabolic geometry is new to me. That must be the one Goldilocks sat in. === Subject: Re: Homological algebra > Can someone suggest a good book on Homological algebra?? > I'm not sure where you are starting from, but I have been > reading Hilton and Wu's _A_Course_in_Modern_Algebra_. The > final chapter introduces homological techniques (Ext and Tor). > This book has much to recommend it, but for now let me just > say that it is a reasonably thin book that starts out nice > and slow--groups--and concentrates basically on > concepts which will lead to homological techniques, without > much else. For example, there is a chapter on category theory > including a discussion of abelian categories and adjoint functors. > Projective and injective modules are discussed at length. > The major problem with this book is that it is currently being > offered by in the Wiley Classic's Library at approximately $100, > which I suspect gives them a healthy profit margin. However, copies > are available on line. > Best wishes, > Mike Homological Algebra and that it was a very clear exposition. I didn't get very far into it, but that was not the fault of the book. Algebraic topology teachers are usually pretty clear that the homological machinery is slick, gets the results out, and is not at all intuitive. It takes time and searching (I have been told) to understand the history and intuition behind the modern developments. Achava === Subject: Re: No Perfect Cuboid Exists: - Proof > See also - http://www.bearnol.pwp.blueyonder.co.uk > **************************************** > a^2+b^2=d^2 > b^2+c^2=e^2 > c^2+a^2=f^2 > a^2+b^2+c^2=g^2 > Smallest solution hcf(a,b,c)=1 > a^2==b^2==c^2==0,1 [mod 4] > therefore a^2+b^2+c^2==0 [mod 4] > => a^2==b^2==c^2==0 [mod 4] > => a==b==c==0 [mod 2] > => hcf(a,b,c)>=2 > therefore g==1 [mod 2] (1) > but hcf(a,b,c)=1 > => only one (max) even > therefore all a,b,c odd [from (1)] > therefore all d,e,f even > => d^2==e^2==f^2==0 [mod 4] > => d^2+e^2+f^2==0 [mod 4] > but d^2+e^2+f^2=2.g^2 > and g^2==1 [mod 4] > => d^2+e^2+f^2==2 [mod 4] > => contradiction, from which result follows > ***************************************** People a lot brighter than you and I have been working on this problem for decades with no success. Do you really believe that there is any chance that there is a proof this short and using nothing beyond the 1st week of a number theory course and that all those bright people have missed it? You are wasting your time. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Making Star Trek Real > sarfatti@pacbell.net says... > Note some minor typographical corrections to the macro-quantum > geometrodynamical equations for the metric engineering of Star Gate Time > Travel Machines and Weightless FTL Warp Drives that underlie the physics > of UFO super-technology that I was put to work on in 1954 at the age of > 14 in a USG Project headed by Eugene Mc Dermott a co-founder of Texas > Instruments and a high level Defense Intelligence Honcho since WWII. > Can you say that in one breath? Sarfatti has yet to learn that merely writing an equation (we won't argue the correctness of the equation, we'll assume they are correct) gives absolutely no clue as to how one goes about building devices that engineer the metric to conform to those equations. Check it out. Section 19.3 in MTW's GRAVITATION describes Mass and Angular momentum of a fully relativistic source. Even when all the normal conservation laws are assumed to apply (and we don't know this for sure since matter doesn't occur at such densities and velocities simultaneously (fast moving things tend to fly apart)) there are some interesting frame dragging properties of fact moving masses - as vonStockum showed. But where the hell do you get an infinitely long mass that's several hundreds of kilometers in diameter (think of a really big and dense string of spaghettie) whose density approaches that of a neutron star and which is spinning at half light-speed? *That's* de minimus in providing a means to change the metric in interesting ways. Assuming of course there aren't changes that must be made when and if we actually go to engineer the metric. That is, without a priori knowledge its not certain things will go our way. Sure, fully relativistic masses seem to have some interesting properties, now. Enthusiasts like Sarfatti would have us believe that our ignorance means there may be even more interesting things possible than anyone believes. Unfortunately, in the absence of sure knowledge, the universe may end up being less interesting than anyone believes as well. Which is just another way of saying, that despite possibilities that may occur on the hairy edge of reality - we just don't know what's possible. This includes the possibility that very little in the way of Star Trek Engineering is possible, just as very little in the way of Witchcraft is possible. Making our future in space a reality involves making what is known genrally available to the business world in an environment where missile and nuclear proliferation are non-issues. This is a tall order. But one that may be brought about by a successful conclusion of our war on Terror. Given that, we also need to create a business and economic environment more favorable to business development. This involves allowing businesses and people to own assets and resources off-world, and to develop those assets without undue interference from terrestrial governments and without undue taxation. Given these preconditions we can clearly show that investment in space propulsion will decrease the cost of momentum - and as this cost decreases we will see the following occur; (1) suborbital ballistic high-value package and personell delivery (2) orbiting satellites (comsats, spysats, navsats, weathersats, etc) (3) large orbiting satellites (powersats) (4) very large interplanetary spacecraft (asteroid capture, factorysats) then, the same development arc at far lower prices (5) suborbital ballistic low-value package and personnel delivery (6) orbiting satellites (space homes) (7) large orbiting satellites (full integration of Earth/Earth orbit) (8) very large interplanetary spaceraft (mobile spacehomes) then, flight to the stars at about 1/3 light speed; (9) interstellar space homes then, once a number of colonies are set up surrounding sol, we can imagine research along the lines proposed by Sarfatti - basically, we'll collide shaped pieces of Iron-56 at 1/3 light speed - in an effort to find out more about the metric, and see how we may engineer it to do some interesting things. Things like time travel, tapping zero point energy, and gravity drives and so forth. Once you have these things, even given the limitations of existing time machine designs (you cannot visit a time before the first machine it built, or after the last machine ceases to function) its possible to create an effective superluminal travel (assuming all the physics we think we know remains unchanged under extreme conditions) - here's how that works; (a) Before you leave make a time violating region that lets you travel to any point in time the region exists - do this by establishing a ring of tiny black holes, preferrably by tapping the zero point energy of the vacuum in some way. (b) Then, use your gravity drive, and zero point energy generator to fly off at very high accelerations to a distant locale, say 10,000 light years away. (c) Very little shiptime later, but 10,000+ years later, arrive at your destination. Explore, and so forth. Then return. (d) Arrive back at your starting point 20,000+ years later, while only a few days pass on the ship due to very high speeds most of the way. (e) Use the time machine to travel back in time to the point just after your departure. For everyone's sanity exit the causality violating region so that the local time is synchronized with your ship time. (You may send radio telescope messages back and forth this way as well, to that everything is synchronized between the two frames. The reply messages from the moving ship are picked up by the time machine in the future and routed, like internet packets, to the right time exit, based on time stamps. Thus establishing instantaneous messaging) This is very much like Star Trek - but has nothing to do with the 'science' of Star Trek. Anymore than the flying in an airplane has to do with the 'science' of flying by Witchcraft. They're both talking about flight - but one achieves it while the other does not. === Subject: Re: Making Star Trek Real If someone was dieing and wanted to see somewhere we can show them by using the holodeck. If we can travel to other worlds will the war between human will it stop? === Subject: Re: WMD in Iraq , UFO & Doomsday Fact or Fiction? http://www.amazon.com/gp/reader/0471265179/ref=sib_dp_pt/102-9493486-1170531 #reader-link Check out the reality of our foreign policy before making such insane and foolish statements as those made here. The link above is to ALL THE SHAH'S MEN: An American Coup and the Roots of Middle East Terror. === Subject: EvoGraph: a package for graph synthesis by genetic programming I am glad to announce that a package for evolving generic graphs and networks using genetic programming is available for download. http://www.egr.msu.edu/~hujianju/evograph/evograph.htm EvoGraph is a package for graph synthesis by genetic programming. It is developed based on strongly-typed lilgp, originally developed at GARAGe, MSU and later patched by Sean Luke with the strongly-typed feature. Basically, EvoGraph provides a set of GP functions and terminals along with a C++ graph library for evolving arbitrary graphs. The same technique is widely used in developmental genetic programming for evolving electric circuits, bond graphs, neural networks. A benchmark problem: the wireless access point configuration problem is introduced in this package. This problem requires simultaneous search of both topology and parameters, the same as in circuit synthesis. And there could be a series of variations of this problem and the problem is also scalable in terms of problem size. This package is still in its starting stage and there is no complete or very detailed documents. But it is easy to use and understand after reading related reference papers. Jianjun Hu (George) Genetic Algorithm Research & Application Group (GARAGe) Department of Computer Science & Engineering Michigan State University hujianju@msu.edu Web: www.egr.msu.edu/~hujianju Phone: 517-355-3796(o) === Subject: Re: Cardinal Arithmetic > Let k be a cardinal number. Let 0 = null and 1 = {0}. > Show that k^0 = 1 and k^1 = k for all k. There is only one function with domain of empty set and that is the empty function, which is the same as the empty set. Thus k^0 = 1. Let K be a set with cardinality k. Thus k^1 = |{ {(0,x)} | x in K }| and { {(0,x)} | x in K } bijects with K by {(0,x)} -> x === Subject: Re: Leibniz & Newton > Also, the dy/dx notation is a little misleading in that it makes you > think Fraction! > Actually, it's the explications of it, that are. > It's not difficult at all to come up with an explication that matches > what Leibnitz intended [... followed by an explication...] >Hi Alfred! >classroom? I'm very impressed with your ability. Just to a web search on infinitesimals. It's all over the place. === Subject: Re: Leibniz & Newton >There is also Chandrasekhar's book Newton's _Principia_ for the >Common Reader, Oxford University Press, in which the most important >parts of Newton's great book are rewritten using modern notation and >terminology. I actually did that with Maxwell's treatise, rewriting a large chunk of it in the boldface, 3-D vector notation. Entire sections melted away into nothingness, with about a 50% size reduction and proportionate increase in transparency. === Subject: Re: Leibniz & Newton > It's not difficult at all to come up with an explication that matches > what Leibnitz intended and is perfectly consistent (and no, I'm NOT > talking about Non-Standard analysis here). > Consider the algebra given by: > A = (x + d y: x, y in R; d^2 = 0 }, If d^2 = 0, and the analysis is *not* non-standard, then how can d differ from 0? === Subject: Re: Leibniz & Newton >> and is perfectly consistent (and no, I'm NOT talking about Non-Standard >> analysis here). >If d^2 = 0, and the analysis is *not* non-standard, ... which has nothing to do with anything that was said that you're replying to... http://search.yahoo.com/search?p=How To Search The Web http://search.yahoo.com/search?p=Non-Standard Analysis http://search.yahoo.com/search?p=Wikipedia >then how can d differ from 0? *Sigh.* http://search.yahoo.com/search?p=Associative Linear Algebras http://search.yahoo.com/search?p=Vector Spaces http://search.yahoo.com/search?p=Nilpotent Algebras http://search.yahoo.com/search?p=Finitely Presented Algebras http://search.yahoo.com/search?p=Ideals+Infinitesimals http://search.yahoo.com/search?p=Hypercomplex Numbers+Infinitesimals === Subject: Re: Leibniz & Newton Kavon >> and is perfectly consistent (and no, I'm NOT talking about Non-Standard >> analysis here). >If d^2 = 0, and the analysis is *not* non-standard, > ... which has nothing to do with anything that was said that you're > replying to... > http://search.yahoo.com/search?p=How To Search The Web > http://search.yahoo.com/search?p=Non-Standard Analysis > http://search.yahoo.com/search?p=Wikipedia >then how can d differ from 0? > *Sigh.* > http://search.yahoo.com/search?p=Associative Linear Algebras > http://search.yahoo.com/search?p=Vector Spaces > http://search.yahoo.com/search?p=Nilpotent Algebras > http://search.yahoo.com/search?p=Finitely Presented Algebras > http://search.yahoo.com/search?p=Ideals+Infinitesimals > http://search.yahoo.com/search?p=Hypercomplex Numbers+Infinitesimals === Subject: Re: Leibniz & Newton I second that. r.e.s. > Kavon >> and is perfectly consistent (and no, I'm NOT talking about Non-Standard >> analysis here). >If d^2 = 0, and the analysis is *not* non-standard, > ... which has nothing to do with anything that was said that you're > replying to... > http://search.yahoo.com/search?p=How To Search The Web > http://search.yahoo.com/search?p=Non-Standard Analysis > http://search.yahoo.com/search?p=Wikipedia >then how can d differ from 0? > *Sigh.* > http://search.yahoo.com/search?p=Associative Linear Algebras > http://search.yahoo.com/search?p=Vector Spaces > http://search.yahoo.com/search?p=Nilpotent Algebras > http://search.yahoo.com/search?p=Finitely Presented Algebras > http://search.yahoo.com/search?p=Ideals+Infinitesimals > http://search.yahoo.com/search?p=Hypercomplex Numbers+Infinitesimals === Subject: How to prove that sqrt(2) exist? I.e., how to prove that there exists a real number x such that x^2 = 2? I'm getting started with the rigourous side of maths (as a hobby -- being an engineer, with several years of experience, you understand that I haven't really needed it), and really enjoyed reading a proof that sqrt(2) can not be a rational number. However, the justification they give is that from Pythagoras' theorem, looking at a triangle with sides 1 and 1, then the hypothenuse is such that its square is 2. I don't find that convincing at all (I mean, sure, intuitively yes -- I wouldn't even need that evidence to convince me -- I have always accepted that sqrt(2) exists, just because, well, it has to be there, somewhere between 1.4 and 1.5 ... You know... engineers!! :-)) But anyway, they don't really prove that sqrt(2) is a real number: they prove that there is no rational number p/q such that (p/q)^2 is 2. So, I'm curious: how can one prove that sqrt(2) exists and it is a real number? Carlos -- === Subject: Re: How to prove that sqrt(2) exist? > I.e., how to prove that there exists a real number x > such that x^2 = 2? > I'm getting started with the rigourous side of maths > (as a hobby -- being an engineer, with several years > of experience, you understand that I haven't really > needed it), and really enjoyed reading a proof that > sqrt(2) can not be a rational number. > However, the justification they give is that from > Pythagoras' theorem, looking at a triangle with > sides 1 and 1, then the hypothenuse is such that > its square is 2. I don't find that convincing at > all ... If we allow that real numbers are in one-to-one correspondence with points on the x-axis, then we can mark off sqrt(2) on this real number line, using compass and straightedge. Voila. There it is. Or do you think this line has holes or gaps? === Subject: Re: How to prove that sqrt(2) exist? >I.e., how to prove that there exists a real number x >such that x^2 = 2? >So, I'm curious: how can one prove that sqrt(2) >exists and it is a real number? [The first phrasing is better. Whatever existence means in a mathematical discussion, it isn't really productive to say sqrt(2) exists and then try to prove it's real; for example, in the set of integers modulo 7, sqrt(2) certainly exists -- it's equal to 3 -- but I wouldn't say it's a real number...] You've already gotten some fine answers. Just to add another one: there are several models you can use for what real numbers _are_, but the key property which all of them share, which distinguishes the real numbers from the rational numbers, is this: every nonempty set of real numbers which has any upper bound at all will have a smallest upper bound. (It's reasonable to be skeptical of this assertion but it can be proved once you do provide a model of what the real numbers are. Or, you can simply take it as an axiom and see what its consequences are.) Once you buy in to this axiom, you can easily prove that there exist real solutions to all kinds of problems. In the case of the square root of 2, you look at the set S of all real numbers x with x^2 <= 2. It's certainly not empty (e.g. x=0) and it's easily seen to have upper bounds (e.g. whenever x^2 <= 2, we certainly have x < 2; x=3/2 also serves as an upper bound for S, as does x= 17/12, etc.) So by the axiom, there is a _least_ upper bound L. The fact that it's an upper bound turns out to be enough to show that L^2 <=2 ; the fact that it's smaller than all the other upper bounds is enough to prove L^2 >=2 ; so in fact L^2 equals 2 on the nose. This same example shows that the set of rational numbers does NOT enjoy the least-upper-bound property. The set S has many _rational_ upper bounds, as noted above. But there is no smallest one (for every rational upper bound there's another smaller rational number which is also an upper bound for S). dave === Subject: Re: How to prove that sqrt(2) exist? In sci.math, Carlos Moreno : > I.e., how to prove that there exists a real number x > such that x^2 = 2? > I'm getting started with the rigourous side of maths > (as a hobby -- being an engineer, with several years > of experience, you understand that I haven't really > needed it), and really enjoyed reading a proof that > sqrt(2) can not be a rational number. > However, the justification they give is that from > Pythagoras' theorem, looking at a triangle with > sides 1 and 1, then the hypothenuse is such that > its square is 2. I don't find that convincing at > all (I mean, sure, intuitively yes -- I wouldn't > even need that evidence to convince me -- I have > always accepted that sqrt(2) exists, just because, > well, it has to be there, somewhere between 1.4 > and 1.5 ... You know... engineers!! :-)) > But anyway, they don't really prove that sqrt(2) is > a real number: they prove that there is no rational > number p/q such that (p/q)^2 is 2. > So, I'm curious: how can one prove that sqrt(2) > exists and it is a real number? > Carlos > -- I'm not sure exists works in this context, as no numbers exist physically, be they the natural number 1, the rational number 8/17, the radical sqrt(2), transcendental pi, or the complex number 3 + 4i. However, presumably what Dedekind did (I'd have to look) is assume that one defines an entity using two sets over Q (the lower set having no greatest member), and show that these setpairs could be manipulated in straightforward ways to define addition, subtraction, multiplication, and division, over the real numbers (which contain, of course, the rational numbers as a subfield). In the case of sqrt(2), one simply uses S1={x in Q:x^2<2} and S2={x in Q:x^2>=2}. It turns out S2 has no least member, either, in this case. Addition is easy. Subtraction may require addition of the negative. Multiplication requires some care, because of silly behavior around 0. Division may require multiplication of the reciprocal. I see a potential problem with Dedekind cuts, if only because one has to define the sets carefully lest one inadvertantly. In fact, you can probably see the error I've made already; a more proper definition might be S1={x in Q: x < 0} union {x in Q: x >= 0 and x^2 < 2} S2={x in Q: x >= 0 and x^2 >= 2} which makes S1 into the desired form, as opposed to an open interval centered around 0. A Cauchy sequence as suggested by Mathworld.wolfram.com can also be used. Basically, it's a sequence (in this case, over Q) such that lim (min(m,n)->+oo) d(a_m,a_n) = 0 where d(a,b) is a metric over Q (the typical one is abs(a-b)). Or one can go the slightly silly route: sqrt(2) exists because (sqrt(2)^2) = 2; this treats the problem as a rather abstract but somewhat useful formalism. For example, (sqrt(2) - 1)^2 = sqrt(2)^2 - 2 * sqrt(2) * 1 + 1 = 3 - 2 * sqrt(2). Or one goes the engineering route, which you've stated you don't like: sqrt(2) = 1.4142; pi = 3.1416; e=2.71828. http://mathworld.wolfram.com/DedekindCut.html http://mathworld.wolfram.com/CauchySequence.html http://mathworld.wolfram.com/Metric.html -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: How to prove that sqrt(2) exist? > In sci.math, Carlos Moreno > : I.e., how to prove that there exists a real number x > such that x^2 = 2? I'm getting started with the rigourous side of maths > (as a hobby -- being an engineer, with several years > of experience, you understand that I haven't really > needed it), and really enjoyed reading a proof that > sqrt(2) can not be a rational number. However, the justification they give is that from > Pythagoras' theorem, looking at a triangle with > sides 1 and 1, then the hypothenuse is such that > its square is 2. I don't find that convincing at > all (I mean, sure, intuitively yes -- I wouldn't > even need that evidence to convince me -- I have > always accepted that sqrt(2) exists, just because, > well, it has to be there, somewhere between 1.4 > and 1.5 ... You know... engineers!! :-)) But anyway, they don't really prove that sqrt(2) is > a real number: they prove that there is no rational > number p/q such that (p/q)^2 is 2. So, I'm curious: how can one prove that sqrt(2) > exists and it is a real number? > Carlos > -- I'm not sure exists works in this context, as no numbers > exist physically, be they the natural number 1, the > rational number 8/17, the radical sqrt(2), transcendental > pi, or the complex number 3 + 4i. All, surreal numbers exist physically, since it's well-known and well-documented that they were invented by idiots. You are right about 1 though, since nobody has yet proven than it exists at all. > However, presumably what Dedekind did (I'd have to look) > is assume that one defines an entity using two sets over Q > (the lower set having no greatest member), and show that > these setpairs could be manipulated in straightforward > ways to define addition, subtraction, multiplication, and > division, over the real numbers (which contain, of course, > the rational numbers as a subfield). > In the case of sqrt(2), one simply uses S1={x in Q:x^2<2} > and S2={x in Q:x^2>=2}. It turns out S2 has no least > member, either, in this case. > Addition is easy. Subtraction may require addition > of the negative. Multiplication requires some care, > because of silly behavior around 0. Division may require > multiplication of the reciprocal. > I see a potential problem with Dedekind cuts, if only > because one has to define the sets carefully lest > one inadvertantly. In fact, you can probably see the > error I've made already; a more proper definition might > be > S1={x in Q: x < 0} union {x in Q: x >= 0 and x^2 < 2} > S2={x in Q: x >= 0 and x^2 >= 2} > which makes S1 into the desired form, as opposed to an > open interval centered around 0. > A Cauchy sequence as suggested by Mathworld.wolfram.com > can also be used. Basically, it's a sequence (in this > case, over Q) such that > lim (min(m,n)->+oo) d(a_m,a_n) = 0 > where d(a,b) is a metric over Q (the typical one is abs(a-b)). > Or one can go the slightly silly route: sqrt(2) exists because > (sqrt(2)^2) = 2; this treats the problem as a rather abstract > but somewhat useful formalism. For example, (sqrt(2) - 1)^2 > = sqrt(2)^2 - 2 * sqrt(2) * 1 + 1 = 3 - 2 * sqrt(2). > Or one goes the engineering route, which you've stated you > don't like: sqrt(2) = 1.4142; pi = 3.1416; e=2.71828. You're right, only mathematicans go the engineering route. Engineers go the route: sqrt(2)=5 - $3.4858, for every time you ask another stupid question. > http://mathworld.wolfram.com/DedekindCut.html > http://mathworld.wolfram.com/CauchySequence.html > http://mathworld.wolfram.com/Metric.html === Subject: Re: How to prove that sqrt(2) exist? > Or one can go the slightly silly route: sqrt(2) exists because > (sqrt(2)^2) = 2 You can always define it that way (for instance, the way you define the imaginary unit). But that does not mean that that number you just defined is in the set of real numbers. In particular, I could arbitrarily come up with such definition for the rational numbers: I could simply say: sqrt(2) is the rational number X such that X^2 is 2. I could claim that the number exists because I defined it, and that means that it exists. You can, however, prove that if such number exists, it is not a rational number (and thus, not an integer or natural number either). The question remains -- when trying to be rigourous, at least the way I'm understanding it, as a beginner in the rigourous side of mathematics, you ask yourself the question: how do I really know that there is such number? How do I know that it will not be the case that whatever number I can come up with, its square will be either greater than 2, or less than 2? > Or one goes the engineering route, which you've stated you > don't like: sqrt(2) = 1.4142; pi = 3.1416; e=2.71828. Well, yes. There's the practical side to it, which is what I have all my life lived with -- it's not like now I want to become an actual mathematician; but I'm enjoying that side of maths, and am trying to adapt my mind to work in the skeptical mode. Carlos -- === Subject: Re: How to prove that sqrt(2) exist? >I.e., how to prove that there exists a real number x >such that x^2 = 2? >I'm getting started with the rigourous side of maths >(as a hobby -- being an engineer, with several years >of experience, you understand that I haven't really >needed it), and really enjoyed reading a proof that >sqrt(2) can not be a rational number. >However, the justification they give is that from >Pythagoras' theorem, looking at a triangle with >sides 1 and 1, then the hypothenuse is such that >its square is 2. I don't find that convincing at >all The fact that you don't find that convincing is a good thing, from a purely mathematical point of view. There are various ways to show that a real number exists - the simplest is first to give a rigorous definition of continuous function, prove the theorem that if a < b, f is continuous, f(a) < 0 and f(b) > 0 then f(x) = 0 for some x between a and b, and then apply this theorem with f(x) = x^2 - 2, a = 0, b = 2. There are a lot of details there - you can find proofs in many calculus books (especially old calculus books - they used to include a lot more proofs). Take the calculus book you used in school and look for the Intermediate Value Theorem... > (I mean, sure, intuitively yes -- I wouldn't >even need that evidence to convince me -- I have >always accepted that sqrt(2) exists, just because, >well, it has to be there, somewhere between 1.4 >and 1.5 ... You know... engineers!! :-)) >But anyway, they don't really prove that sqrt(2) is >a real number: they prove that there is no rational >number p/q such that (p/q)^2 is 2. >So, I'm curious: how can one prove that sqrt(2) >exists and it is a real number? >Carlos ************************ David C. Ullrich === Subject: Re: How to prove that sqrt(2) exist? > The fact that you don't find that convincing is a > good thing, from a purely mathematical point of view. :-) > There are various ways to show that a real number > exists - the simplest is first to give a rigorous > definition of continuous function, prove the > theorem that if a < b, f is continuous, f(a) < 0 > and f(b) > 0 then f(x) = 0 for some x between > a and b, and then apply this theorem with > f(x) = x^2 - 2, a = 0, b = 2. I actually thought about this (not in this much detail, but what I mean is that I thought of the continuity issue, but then I kind of dismissed it, since I thought the very notion of continuity relies on the existence of numbers... So, the function has a chance at being continuous because the values exist -- how could I use that to prove that a value exists such that f(x) = 0? What I find curious is that this continuity issue seems to rely on the fact that between any two distinct real numbers, there is at least one other real number -- but this fact is not a sufficient condition for the existence of such number for which f(x) = 0: it is also true for rational numbers that between any two distinct rational numbers, there is at least one other rational number. (at this point, I'm guessing my doubt becomes sort of philosophical? Something akin to how do we prove that 0 exists? or how do we prove that 1 is greater than 0?, or that sort of thing?) Carlos -- === Subject: Re: How to prove that sqrt(2) exist? >> The fact that you don't find that convincing is a >> good thing, from a purely mathematical point of view. >:-) >> There are various ways to show that a real number >> exists - the simplest is first to give a rigorous >> definition of continuous function, prove the >> theorem that if a < b, f is continuous, f(a) < 0 >> and f(b) > 0 then f(x) = 0 for some x between >> a and b, and then apply this theorem with >> f(x) = x^2 - 2, a = 0, b = 2. >I actually thought about this (not in this much >detail, but what I mean is that I thought of the >continuity issue, but then I kind of dismissed it, >since I thought the very notion of continuity >relies on the existence of numbers... So, the >function has a chance at being continuous because >the values exist -- how could I use that to prove >that a value exists such that f(x) = 0? >What I find curious is that this continuity issue >seems to rely on the fact that between any two >distinct real numbers, there is at least one other >real number -- but this fact is not a sufficient >condition for the existence of such number for >which f(x) = 0: it is also true for rational >numbers that between any two distinct rational >numbers, there is at least one other rational >number. It's a long story. Start with that calculus book, as I suggested. >(at this point, I'm guessing my doubt becomes >sort of philosophical? Something akin to how >do we prove that 0 exists? or how do we >prove that 1 is greater than 0?, or that sort >of thing?) >Carlos ************************ David C. Ullrich === Subject: Re: How to prove that sqrt(2) exist? > I.e., how to prove that there exists a real number x > such that x^2 = 2? > I'm getting started with the rigourous side of maths > (as a hobby -- being an engineer, with several years > of experience, you understand that I haven't really > needed it), and really enjoyed reading a proof that > sqrt(2) can not be a rational number. > However, the justification they give is that from > Pythagoras' theorem, looking at a triangle with > sides 1 and 1, then the hypothenuse is such that > its square is 2. I don't find that convincing at > all (I mean, sure, intuitively yes -- I wouldn't > even need that evidence to convince me -- I have > always accepted that sqrt(2) exists, just because, > well, it has to be there, somewhere between 1.4 > and 1.5 ... You know... engineers!! :-)) > But anyway, they don't really prove that sqrt(2) is > a real number: they prove that there is no rational > number p/q such that (p/q)^2 is 2. > So, I'm curious: how can one prove that sqrt(2) > exists and it is a real number? > Carlos > -- The set of rationals q, such that q^2 > 2, defines a Dedekind cut, x, whose square is neither less than nor greater than 2. === Subject: Re: How to prove that sqrt(2) exist? >I.e., how to prove that there exists a real number x >such that x^2 = 2? ... >So, I'm curious: how can one prove that sqrt(2) >exists and it is a real number? It depends on which flavour of definition of real number you're using. Using Dedekind cuts it's fairly simple: you just split the rationals into {x: x <= 0 or x^2 < 2} and {x: x >= 0 and x^2 > 2}. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Injective Mapping of Finite Set into Itself |Say I have a finite set S, and a mapping f: S -> S, which I know is |injective. How can I prove that f is surjective? By induction on the number of elements of f. If S is empty, then f is surjective because there are no elements that need to be mapped to. If S has n+1 elements, and the claim is true for all sets with n elements, then let x be an element of S and consider the function g(y) = f(x) if f(y)=x and g(y) = f(y) if f(y)<>x. The function g is from S-{x} to S-{x}: if f(y)=x then by the injectivity of f, one can't have f(x)=x also. Also g is injective, as g(y1) and g(y2) for y1<>y2 are either f(y1)<>f(y2) or f(x)<>f(y2) or f(y1)<>f(x); the case f(y1)=f(y2)=x does not occur because f is injective. So by induction g is surjective. This implies that the image of f on S contains S-{x}. If for some y in S-{x}, f(y)=x, then the image of f also contains x, and we are done. If for no y in S-{x} does f(y)=x, then g(y)=f(y) for all y in S-{x}, and f maps S-{x} surjectively to S-{x}. If so, then f(x) can't be an element of S-{x}, because each of them is an image of an element of S-{x} under f, and f is injective. So f(x)=x and we are done. Keith Ramsay === Subject: Campbell's theorem for filtered point processes I am looking for an english reference for the Campbell theorem for stationary filtered point processes (not only Poisson). I have only a reference in Koenig/Schmidt: Einfuehrung in Punktprozesse, have a reference which is easily accessible everywhere. -- Karl Breitung Remove nospam to get my reply address === Subject: Re: Replacing sum of gaussians by a single one? > Given a function f(x) constructed by a linear combination of two > gaussians: > f(x) = a*exp(-(x-m1)^2/(2*s1*s1)) + b*exp(-(x-m1)^2/(2*s1*s1)) I assume that should be f(x) = a*exp(-(x-m1)^2/(2*s1*s1)) + b*exp(-(x-m2)^2/(2*s2*s2)) > where m denotes mean and s denotes standard deviation. If the two > gaussians are close enough, such that replacing their superposition by > a single gaussian gives tolerable error, what are the optimal parameters > (c,m3,s3) for minimizing the error? > Formally: > f^(x) = c*exp(-(x-m3)^2/(2*s3*s3)) > Error = integral { (f^(x)-f(x))^2 dx} ,x from -inf to +inf. > After spending a few hours, I computed the error analytically and then > tried to make its partial derivatives (with respect to c,m3 and s3) > equal to zero. This resulted in a very complicated non-linear equation > that I couldn't solve. Do you know any solution for (c,m3,s3) ? If > the optimal solution is too complicated, even a good sub-optimal is > helpful. The simplest approach to approximating a mixture of variables by a single variable is to give the approximating variable the correct mean and variance. Rewrite the true density as f(x) = ((p1/s1)*exp(-(x-m1)^2/(2*s1*s1))+ (p2/s2)*exp(-(x-m2)^2/(2*s2*s2)))/sqrt(2*Pi), where p1 + p2 = 1. The mean of the mixture is m3 = p1*m1 + p2*m2. The variance is s3^2 = p1*(s1^2 + m1^2) + p2*(s2^2 + m2^2) - m3^2. The normal density with the same mean and variance is g(x) = exp(-(x-m3)^2/(2*s3*s3))/(s3*sqrt(2*Pi)). === Subject: Re: Replacing sum of gaussians by a single one? > Given a function f(x) constructed by a linear combination of two > gaussians: > f(x) = a*exp(-(x-m1)^2/(2*s1*s1)) + b*exp(-(x-m1)^2/(2*s1*s1)) > where m denotes mean and s denotes standard deviation. If the two > gaussians are close enough, such that replacing their superposition by > a single gaussian gives tolerable error, what are the optimal parameters > (c,m3,s3) for minimizing the error? Well, mean square error makes life difficult in this case. If f(x) is normalized (so that int_{-infty}^{infty} f(x) dx = 1) and the approximation (call it g(x)) is also assumed to be normalized, then you can consider using the cross entropy, -int f(x) log g(x) dx as the goodness of fit instead of mean square error. If g(x) is a single Gaussian bump, setting its mean equal to the mean of f(x) and its variance equal to the variance of f(x) minimizes the cross entropy. (You can verify this by computing derivatives of the cross entropy and setting derivatives to zero.) If f(x) is not normalized, you can of course compute a scaling factor to make it normalized, and divide g(x) by the same factor. Offhand, I don't know if that yields the same result as minimizing the cross entropy with respect to the mean, variance, and scaling factor; I don't even know if cross entropy is really meaningfully applied to nonnormalized functions. For what it's worth, Robert Dodier -- Far better an approximate answer to the right question, which is often vague, than an exact answer to the wrong question, which can always be made precise. -- John W. Tukey === Subject: Re: Replacing sum of gaussians by a single one? Could try making it linear. write m2=m1+delta m and m3 = m1 + delta M ditto s1 this may work for the two initial gaussians being very close because when they are the same you know the answer. > [Hossein could not access to this group and asked me to send it. So > this is Hossein's message.] > Hello > Given a function f(x) constructed by a linear combination of two > gaussians: > f(x) = a*exp(-(x-m1)^2/(2*s1*s1)) + b*exp(-(x-m1)^2/(2*s1*s1)) > where m denotes mean and s denotes standard deviation. If the two > gaussians are close enough, such that replacing their superposition by > single gaussian gives tolerable error, what are the optimal parameters > (c,m3,s3) for minimizing the error? > Formally: > f^(x) = c*exp(-(x-m3)^2/(2*s3*s3)) > Error = integral { (f^(x)-f(x))^2 dx} ,x from -inf to +inf. > After spending a few hours, I computed the error analytically and then > tried to make its partial derivatives (with respect to c,m3 and s3) > equal to zero. This resulted in a very complicated non-linear equation > that I couldn't solve. Do you know any solution for (c,m3,s3) ? If > the > optimal solution is too complicated, even a good sub-optimal is > helpful. > --Hossein Mobahi === Subject: Re: Real Roots of Polynomial Maybe i was a bit unclear as to what i was asking. I know that we can apply the Descartes' Sign Rule to determine the max number (upper limit) of positive and negative real roots of a polynomial. But from this we cannot (always) determine if a polynomial of degree n has n real roots. Now we can apply Sturms Theorem to determine the exact number of distinct real roots of a polynomial on an interval(a,b). However we must limit ourselves to an interval here. So is there any specific condition whereby given a polynomial of degree n we can always state that, given this specific condition holds, all its roots are real? Maybe what i am asking is rather silly/uneducated and i am missing some basic points here, if so please tell :) >hi, >could anyone tell me what conditions are neccessary in order for a >polynomial to have real roots only? > Look for Sturm sequences in an old Theory of equations > book, or elsewhere. They tell you exactly how many > real roots a real polynomial has. === Subject: Re: Real Roots of Polynomial > Maybe i was a bit unclear as to what i was asking. > I know that we can apply the Descartes' Sign Rule to determine the > max number (upper limit) of positive and negative real roots of a > polynomial. But from this we cannot (always) determine if a polynomial > of degree n has n real roots. > Now we can apply Sturms Theorem to determine the exact number of > distinct real roots of a polynomial on an interval(a,b). However we > must limit ourselves to an interval here. It is also possible, using synthetic division, to determine that no real roots are larger than a given value or no real roots smaller than a given value. There are other ways, too, of finding bounds on real roots. Thexe, together with Sturm's Theorem, do it all. === Subject: Re: Real Roots of Polynomial > Now we can apply Sturms Theorem to determine the exact number of > distinct real roots of a polynomial on an interval(a,b). However we > must limit ourselves to an interval here. Yes, but one can effctively find a number M such that all zeroes of the polynomial satisfy |z| < M. Apply Sturm to (-M,M). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Real Roots of Polynomial Now we can apply Sturms Theorem to determine the exact number of > distinct real roots of a polynomial on an interval(a,b). However we > must limit ourselves to an interval here. > Yes, but one can effctively find a number M such that all zeroes > of the polynomial satisfy |z| < M. Apply Sturm to (-M,M). Yes obviously we can set our range to (-infinity, +infinity). But that is just a method for finding out the number of real zeros. What are the conditions that guarantee, given a polynomial of degree n, that the polynomial has n real zeros? For example given polynomial of degree 2, ax^2 + bx + c, The condition that, b^2 - 4ac > 0 guarantees that this polynomial shall have 2 real zeros. Does there exist such a condition that holds for polynomials of arbitary degree? === Subject: Re: Real Roots of Polynomial > For example given polynomial of degree 2, > ax^2 + bx + c, > The condition that, > b^2 - 4ac > 0 > guarantees that this polynomial shall have 2 real zeros. > Does there exist such a condition that holds for polynomials of > arbitary degree? no === Subject: Re: A 'basic' topology question about interiors Of interest perhaps are these formulas: When U subset A subspace S cl_A (U) = A / cl_S (U) int_A (U) = A / int_S (SA / U) / / , intersect union ---- === Subject: Re: A 'basic' topology question about interiors > Of interest perhaps are these formulas: > When U subset A subspace S > cl_A (U) = A / cl_S (U) > int_A (U) = A / int_S (SA / U) Ben Scott === Subject: Re: Graduate algebra book > What's a good book for a first year graduate algebra course? > Something with alot of emphasis on factorization, polynomial rings, > fields, PIDs, Galois Theory, and of course all the more basic topics > of algebra as well like groups, ideals, integral domains, etc. That's graduate algebra? There's a recent book by Joe Rotman Advanced Modern Algebra which does all that, and more. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Graduate algebra book > What's a good book for a first year graduate algebra course? > Something with alot of emphasis on factorization, polynomial rings, > fields, PIDs, Galois Theory, and of course all the more basic topics > of algebra as well like groups, ideals, integral domains, etc. > That's graduate algebra? > There's a recent book by Joe Rotman Advanced Modern Algebra > which does all that, and more. Uhh... Yeah. I mean do you feel like the topics I mentioned should be covered at the undergraduate level and not at the graduate level? If so, then why do both Lang and Hungerford treat all of the above topics in detail in their GTM books? Granted, of course groups, ideals, and integral domains should be covered at the undergraduate level, but should free groups, finitely generated abelian groups, and galois theory? Maybe if you go to Harvard or MIT. === Subject: Re: Graduate algebra book > What's a good book for a first year graduate algebra course? > Something with alot of emphasis on factorization, polynomial rings, > fields, PIDs, Galois Theory, and of course all the more basic topics > of algebra as well like groups, ideals, integral domains, etc. >> That's graduate algebra? >> There's a recent book by Joe Rotman Advanced Modern Algebra >> which does all that, and more. > Uhh... Yeah. I mean do you feel like the topics I mentioned should > be covered at the undergraduate level and not at the graduate level? > If so, then why do both Lang and Hungerford treat all of the above > topics in detail in their GTM books? Granted, of course groups, > ideals, and integral domains should be covered at the undergraduate > level, but should free groups, finitely generated abelian groups, and > galois theory? Maybe if you go to Harvard or MIT. All the listed topics can (and should) be taught in any good first course in abstract algebra. Anyone lacking such fundamental algebraic knowledge would be poorly prepared for graduate-level studies. -Bill Dubuque === Subject: Re: Graduate algebra book > What's a good book for a first year graduate algebra course? > Something with alot of emphasis on factorization, polynomial rings, > fields, PIDs, Galois Theory, and of course all the more basic topics > of algebra as well like groups, ideals, integral domains, etc. >> That's graduate algebra? >> There's a recent book by Joe Rotman Advanced Modern Algebra >> which does all that, and more. Uhh... Yeah. I mean do you feel like the topics I mentioned should > be covered at the undergraduate level and not at the graduate level? > If so, then why do both Lang and Hungerford treat all of the above > topics in detail in their GTM books? Granted, of course groups, > ideals, and integral domains should be covered at the undergraduate > level, but should free groups, finitely generated abelian groups, and > galois theory? Maybe if you go to Harvard or MIT. > All the listed topics can (and should) be taught in any good first course > in abstract algebra. Anyone lacking such fundamental algebraic knowledge > would be poorly prepared for graduate-level studies. > -Bill Dubuque Are you serious? If so, maybe I'm just a complete idiot. But are you telling me that Galois theory can and should be taught in a first course in abstract algebra? Granted, all the topics I mentioned (groups, rings, ideals, integral domains, fields, PIDs, factorization) should be mentioned in a first course on abstract algebra, but how much detail can you possibly go into in such a short amount of time? Should free abelian groups, sylow theory, and solvable groups be included in this introductory course? How about Splitting Fields and Galois groups? If you say yes then you've lost your noodle, however these still fall under the category of group theory or field theory. I said in my original post that I'll have finished hungerford by the time I start reading the new book, so when I refer to group theory obviously I'm not talking about the definition of a group, or a homomorphism, although many books will probably mention that anyway. === Subject: Re: Graduate algebra book >> What's a good book for a first year graduate algebra course? >> Something with alot of emphasis on factorization, polynomial rings, >> fields, PIDs, Galois Theory, and of course all the more basic topics >> of algebra as well like groups, ideals, integral domains, etc. >> That's graduate algebra? >> There's a recent book by Joe Rotman Advanced Modern Algebra >> which does all that, and more. > Uhh... Yeah. I mean do you feel like the topics I mentioned should > be covered at the undergraduate level and not at the graduate level? I don't feel anything, but all of these were covered in the first two years of my undergraduate degree. > If so, then why do both Lang and Hungerford treat all of the above > topics in detail in their GTM books? Why did the publisher put them in the graduate texts series? > Granted, of course groups, > ideals, and integral domains should be covered at the undergraduate > level, but should free groups, finitely generated abelian groups, and > galois theory? Maybe if you go to Harvard or MIT. I went to neither. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Graduate algebra book >> What's a good book for a first year graduate algebra course? >> Something with alot of emphasis on factorization, polynomial rings, >> fields, PIDs, Galois Theory, and of course all the more basic topics >> of algebra as well like groups, ideals, integral domains, etc. >> That's graduate algebra? >> There's a recent book by Joe Rotman Advanced Modern Algebra >> which does all that, and more. Uhh... Yeah. I mean do you feel like the topics I mentioned should > be covered at the undergraduate level and not at the graduate level? > I don't feel anything, but all of these were covered in the first > two years of my undergraduate degree. Well it certainly isn't a fact that they should be covered at the undergraduate level, hence it's a matter of opinion. Either you feel 1) that they should be covered at the graduate level, 2) that they should be covered at the undergraduate level, or 3) you have no opinion or can't decide If 3) were the case, you would not respond with such a snotty remark as That's graduate algebra? because you would have no opinion. Thus one can only conclude that you either feel 1) or you feel 2). In your opinion, what topics SHOULD be covered in a first year graduate algebra course? And what textbook is suitable for such a course? === Subject: Re: Graduate algebra book >> I don't feel anything, but all of these were covered in the first >> two years of my undergraduate degree. > Well it certainly isn't a fact that they should be covered at the > undergraduate level, hence it's a matter of opinion. Either you feel > 1) that they should be covered at the graduate level, 2) that they > should be covered at the undergraduate level, or 3) you have no > opinion or can't decide > If 3) were the case, you would not respond with such a snotty remark > as snotty? that's abuse. > That's graduate algebra? > because you would have no opinion. > Thus one can only conclude that you either feel 1) or you feel 2). > In your opinion, what topics SHOULD be covered in a first year > graduate algebra course? And what textbook is suitable for such a > course? Don't be asinine. I was merely expressing my surprise that such elementary topics should be considered as graduate. And I did suggest a book. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Graduate algebra book > I don't feel anything, but all of these were covered in the first > two years of my undergraduate degree. That's Europe. In America, undergraduates study liberal arts, and squeeze in only a little of their specialty in the first two years. === Subject: Re: Graduate algebra book I would also recommend Dummit & Foote's book Abstract Algebra. It covers a lot of material. And I have often heard this book praised by graduate students. Check it out! Lurch > What's a good book for a first year graduate algebra course? > Something with alot of emphasis on factorization, polynomial rings, > fields, PIDs, Galois Theory, and of course all the more basic topics > of algebra as well like groups, ideals, integral domains, etc. > Something that has lots and lots of _good_ exercises? Hungerford is > good, but I'll have finished most of the exercises by the time I want > to start doin the exercises in this new book. Lang is probably good, > but there don't seem to be very many exercises. === Subject: How to solve this differential equation? I've encountered such a D.E., a(x,y)(dx/dt)^2+b(x,y)(dx/dt)(dy/dt)+c(x,y)(dy/dt)^2=f(t) where x=x[t],y=y[t] and a(x,y),b(x,y),c(x,y),f(t) are all known functions. How to derive the equation in x,y,t? e.g. dx/dt+dy/dt=t => x+y=(1/2)t^2+const. (dx/dt)^2+(dy/dt)^2=t => plane curve with arc length s(t)=(1/2)t^2+const, t is the parameter. Any comments or suggestions? === Subject: Re: Time complexity of algorithms >> helped me understand the reasoning behind algorithms. >> I realize this might not be the best place to post such a question but >> since it basically involved math (what doesn't :)) and was independant >> of computer languages, I thought someone might be able to help me out. >> There still remains one problem.. how do I reason with the algorithm >> that I presented? >// part of code example by Rick > i := 1 > while i <= n and A[i] = 0 do > i := i + 1 > endwhile > return 1 + HelpMe(A, n-i) >First, the code is wrong (that could be part of the confusion). If A[i]<>0 >for some i, then your while-loop is endless. This while loop is obviously NOT endless for any combination of values in A, as i is always increased, and the while loop stops if i>n. -- Wim Benthem === Subject: Re: Time complexity of algorithms >> helped me understand the reasoning behind algorithms. >> I realize this might not be the best place to post such a question but >> since it basically involved math (what doesn't :)) and was independant >> of computer languages, I thought someone might be able to help me out. >> There still remains one problem.. how do I reason with the algorithm >> that I presented? >// part of code example by Rick > i := 1 > while i <= n and A[i] = 0 do > i := i + 1 > endwhile > return 1 + HelpMe(A, n-i) >First, the code is wrong (that could be part of the confusion). If A[i]<>0 >for some i, then your while-loop is endless. > This while loop is obviously NOT endless for any combination of values in > A, as i is always increased, and the while loop stops if i>n. How stupid. Of course it ends. I misunderstood the meaning of it. It stops at the first nonzero entry of A (instead of just not doing the i:=i+1). The complexity is lowered by this. I think it'll be O(n) now, but take a closer look (I don't have much time now). - Arthur > -- > Wim Benthem === Subject: Re: Time complexity of algorithms good links/websites that explain how to form recurrence relationships Rick === Subject: Re: normal form of matrices over PID > Let R = Z[x], where x is 1/2 times > 1 + sqrt{-19}. I'm looking for a > 1 by 2 matrix over R that is not > equivalent by any sequence of > elementary column operations to > a matrix of the form (z 0). Is > (x+1 x-1) such a matrix? (Any other > would be fine!) I'm sure there is one: whether your example is one, I don't know. You should consult this paper Swan, Richard G. Generators and relations for certain special linear groups. Advances in Math. 6 1971 1--77 (1971). I don't have this to hand, so the following is based on vague memories. You are really asking if SL(2,R) = E(2,R) where E(2,R) is the subgroup of SL(2,R) generated by elementary matrices (1 0 /a 1) and (1 a/0 1). If R is Euclidean the answer is yes. If R is not Euclidean, even if it's a Dedekind domain of class number 1 then then answer may be no. Swan analyses the situation by looking at an action of SL(2,R) (R imaginary quadratic) on hyperbolic 3-space. By finding a fundamental domain you can read off generators and relations, and see that the elementary matrices are inadequate as generators. Ther connection with your problem is that if (a b) is the top row of a matrix in SL(2,R) but outside E(2,R), this is not equivalent under elementary column operations to (1 0) (despite a and b being coprime). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? How do you integrw do you integrate the function f(x) = x/(tanx) ? -- Georg Goerg === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? > How do you integrw do you integrate the function f(x) = x/(tanx) ? Please forget my previous reply. It is wrong, as A. N. Niel pointed out. Your function is bounded, of course. Jose Carlos Santos === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? > How do you integrw do you integrate the function f(x) = x/(tanx) ? The behaviour around 0 is like 1/x; therefore, the integral of f(x) in ]0,Pi/2] is equal to +oo. Jose Carlos Santos === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? > How do you integrw do you integrate the function f(x) = x/(tanx) ? > The behaviour around 0 is like 1/x; therefore, the integral of f(x) > in ]0,Pi/2] is equal to +oo. > Jose Carlos Santos I think you are in error. === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? > How do you integrw do you integrate the function f(x) = x/(tanx) ? Maple does it in terms of a polylogarithm. Suggesting that the indefinite integral is not an elementary function. === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? > How do you integrw do you integrate the function f(x) = x/(tanx) ? > Maple does it in terms of a polylogarithm. Suggesting that the > indefinite integral is not an elementary function. Mathematica does similarly, giving x*Log[1 - E^((2*I)*x)] - (I/2)*(x^2 + PolyLog[2, E^((2*I)*x)]) for the indefinite integral. However, based on the title of the thread, perhaps Georg really wants the definite integral from x = 0 to Pi/2. The value of that is Pi/2*Log[2]. David Cantrell === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? > How do you integrw do you integrate the function f(x) = x/(tanx) ? > Maple does it in terms of a polylogarithm. Suggesting that the > indefinite integral is not an elementary function. > Mathematica does similarly, giving > x*Log[1 - E^((2*I)*x)] - (I/2)*(x^2 + PolyLog[2, E^((2*I)*x)]) > for the indefinite integral. However, based on the title of the thread, > perhaps Georg really wants the definite integral from x = 0 to Pi/2. The > value of that is Pi/2*Log[2]. Try the substitution u= tan x to reduce it to standard nonelementary form. Ray Steiner > David Cantrell === Subject: Re: simple question about subspaces | at 08:46 PM, kramsay@aol.com (KRamsay) said: | | >Deriving it from the group properties does not indicate a way to | >eliminate references to 0 from the usual axioms for vector spaces. | | It's precisely because of the axioms that it is derived. A vector | space is defined as an ordered set with certin properties. Not necessarily ordered. | There are | various ways to do it, but some of them involve redundant | constituents. Okay, but what about how to eliminate the references to 0 in the definition without using an additional axiom? small trick one could use: I considered mentioning that, plus the fact that one could eliminate the references to 0 in the definition by having the axiom (x+(-x))+y=y and so on. I had explained previously why it was I wanted to get rid of references to 0 in the axioms. One answer to the original poster's question was that the empty subspace is ruled out by the definition which accords with the definition of vector space by conventions of universal algebra. I wanted to elaborate on this explanation a bit. The reason why the universal algebra definition rules out the empty set as a substructure is that in the universal algebra treatment, the 0 is considered a 0-ary function, and closure under this function means including 0 in the structure. Since it's possible to tweak the axioms slightly, so that they still fit into the framework of universal algebra, but lack these references to zero (and are equivalent aside from allowing the empty set as a structure), just following the conventions of universal algebra isn't *quite* a complete explanation of what the original poster asked about. We are in fact going the usual route and requiring the presence of 0 in the structure for other considerations like elegance. It's not an attempt to improve on the axioms at all; it's an exploration of why we don't go down this other road. Now, for such a purpose, just lumping together the axioms referring to 0 under the rubric of is an abelian group under + does absolutely nothing. It appears as though you consider doing such a lumping-together eliminates the reference to 0 in a manner worth mentioning, however. You replied: You don't need it; you can derive it from the group properties. | >The usual group axioms contain references to the identity | >(0 for the additive group), | | A group is nomally defined as an ordered pair (E,*) satisfyning | certain properties. One of those properties is that there exists an | identity element. Defining a group as (E,e,*) would be redundant. This helps me get rid of the references to the identity in the axioms how? Imagine, if you will, that someone has been giving a brief explanation of how it's possible to build a car without using spark plugs to ignite the fuel. Someone pipes up, But you don't need to use this sort of trick. Just use an engine. | >and if you figure out a set of axioms for groups which don't, [speaking hopefully] So you've also figured out some way to build the engine without needing spark plugs? | And if your grandmother had wheels she'd be a wagon. Please don't | bother rebutting things that I never claimed. I said nothing about building an engine without using spark plugs! | >because the references to 0 in the usual axioms for vector spaces | >are in the axioms for the space as an additive group. But the spark plugs normally are part of the engine. | Nu, so what else is new? Who gives a ! |There is no need to include 0 as an explicit |element of the vector space precisely because the group axioms require |that it exist. Just use the engine! The engine will ignite and burn the fuel for you! |Similarly, there is no need to explicitly require |uniqueness, because the group axioms already suffice to prove it. | | >After some thought, I suspect you mean to say that we can derive | >(x+(-x))+y=y from the other axioms for VECTOR SPACES, not GROUPS. Perhaps you mean something other than the engine? |No. What I mean to say is that the group axioms are sufficient to |prove the existence and uniqueness of 0 and the existence and |uniqueness of -x for any x. | |>but I don't see how to get from there |>to 0*v being an additive identity without using the existence of |>additive inverses (in some form). | |Then it's a good thing that I didn't suggest that you don't need |is a group and on top of that make the inverse one of the constituent |operations, since the group axioms already ensure that it exists and |is unique. No, just use the engine, like I keep telling you! | >So unless I'm missing something here, | | Quite a bit. Former professors these days, boy, they just don't listen, do they? | To start out, you use the constants 0 and the operator | - without defining them. This is commonplace. Some authors phrase the axiom as saying that for each x there exists a y such that..., either assume or prove that it's unique and then introduce the notation -x for this y. Others introduce a function unary - and give identities for it. (The latter is what fits in with universal algebra, which we had been discussing.) |More important, the vector space axioms |include being a group under +, which your system is not; there is an |identity element but there are elements for which no inverse exists. Perhaps you think I've given at some point what I think are all the axioms of a vector space. I was addressing those in the audience who would not need to have a list provided to them. The details of the remaining axioms are irrelevant to the question of how to rephrase the ones involving 0 so as not to refer to 0. | >So please be less terse. I might have said instead: kindly explain how what you are saying makes anything at all unnecessary for any meaningful purpose whatsoever. | A vector space is defined as an ordered You don't mean ordered. | set containing some | combination of sets, some of them being functions, satisfying certain | conditions. There are various ways in which this can be done; the | definitions are formally equivalent. Some of the ways in which it can | be done include elements that are not present in orther ways. In | particular, any definition involving either the 0 vector or the unary | negation operator as elements of the ordered list is equivalent to | another definition that does not include them. | | Thus, If I defined[1] a vector space V to be an ordered set | (K,V_E,V_0,V_+,V_-,V_*) and gave the obvious axioms, that would be | equivalent to defining it as V'=(K,V_E,V_+,V_*) with the obvious | axioms. Similarly, it would be equivalent to V''=(K,V_G,V_*). | | [1] I'm using the notation V_E for the elements of V, V_0 for the null | vector, V_+ for vector addition, V_G for the group (V_E,V_+), etc. Great. Do me a favor, and the next time I'm trying to explain how it is possible to do something, and you have some piece of nearly content-free trivia, please phrase it more like this: This does absolutely nothing to help you accomplish what you're trying to do here, but.... Keith Ramsay === Subject: Re: grasshopper consciousness? > NOVA | The Elegant Universe | PBS > http://www.pbs.org/wgbh/nova/elegant/ A Theory of Everything? Some physicists believe string theory may unify the forces of nature by Brian Greene physicists have identified -- electrons, neutrinos, quarks, and so on -- are the letters of all matter. Just like their linguistic counterparts, they appear to have no further internal substructure. String theory proclaims otherwise. According to with even greater precision -- a precision many orders of magnitude beyond our present technological capacity -- we would find that each is not pointlike but instead consists of a tiny, one-dimensional loop. contains a vibrating, oscillating, dancing filament that physicists have named a string. [...] http://www.pbs.org/wgbh/nova/elegant/everything.html _________ Viewpoints on String Theory Steven Weinberg http://www.pbs.org/wgbh/nova/elegant/view-weinberg.html The Elegant Universe homepage [For more on a final theory, see A Theory of Everything? .] > AL-JILWAH (The Revelation) > Yezidi Scripture > Before all creation, this revelation was with Melek Taus, > who sent Abd Taus to this world that he might separate > truth known to his particular people. This was done, first > of all, by means of oral tradition, and afterward by means > of this book, Al-Jilwah, which the outsiders may neither > read nor behold. [...]-{{snipped}}-[...] > [CHAPTER IV.] > I will not give my rights to other gods. I have allowed the > creation of four substances, four times, and four corners, > because they are necessary things for creatures. The books > of Jews, Christians, and Moslems, as those who are without, > accept in a sense, so far as they agree with and conform to, > my statutes. Whatsoever is contrary to these they have > altered; do not accept it. Three things are against me, and > I hate three things. But those who keep my secret shall > receive the fulfillment of my promises. It is my desire that > all my followers shall unite in a bond of unity, lest those > who are without prevail against them. Now, then, all ye who > have followed my commandments and my teachings, reject all > the teachings and sayings of such as are without. I have not > taught these teachings, nor do they proceed from me. O ye > that have believed in me, honor my symbol and my image, for > they remind you of me. Observe my laws and my statutes. Obey > my servants and listen to whatever they may dictate to you > of the hidden things. Do not mention my name nor my > attributes, lest ye regret it; for ye do not know what those > who are without may do. > [CHAPTER V.] > O ye that have believed in me, honor my symbol and my image, > for they remind you of me. Observe my laws and my statutes. > Obey my servants and listen to whatever they may dictate to > you of the hidden things. Receive that that is dictated, and > do not carry it before those who are without, Jews, Christians, > Moslems, and others; for they know not the nature of my > teaching. Do not give them your books, lest they alter them > without your knowledge. Learn by heart the greater part of > them, lest they be altered. > Thus endeth the book of Al-Jilwah. MEMRI: The Middle East Media Research Institute * http://memri.org/ [{}] Special Dispatch Series - No. 601 2nd Issue of 'Voice of Jihad' Al-Qa'ida Online Magazine: Strategy to Avoid Clashes with Saudi Security Forces, Convert the World's Countries to Islam The second issue of The Voice of Jihad, the new biweekly on-line magazine identified with Al-Qa'ida has been posted. The following are excerpts from the latest issue, which includes a sequel to an interview with Abd Al-'Aziz bin 'Issa bin Abd Al-Mohsen, also known as Abu Hajjer, who is one of the high-ranking Al-Qa'ida members on Saudi Arabia's most-wanted list. Lead Editorial: Combat Jews and Americans, Not Saudi Security Forces The second issue of The Voice of Jihad opened with an editorial by Suleiman Al-Dosari: ...Our war with the enemies of Allah continues everywhere... We will not let the Americans occupy the land of the two holy places [i.e. the Arabian Peninsula] [and feel] secure and safe, and we will not cease our Jihad until we liberate every inch of Muslim land. We draw the attention of the Mujahideen to the strategy adopted by the Sheikh of the Mujahideen, Abu Abdallah Osama bin Laden, and Sheikh Dr. Ayman Al-Zawahiri, and agreed to by many of the great Mujahideen, regarding combat against the enemy: Our number one enemy is the Jews and the Christians, and we must free ourselves to invest all our efforts until we annihilate them ... and we are able do this if Allah allows us to do it ... because they are the main obstacle to establishing the Islamic state. ...We must take note of the ploy used by the tyrants [i.e. Arab rulers] in many countries. They attempted to stop the Jihad project in these countries by shifting the confrontation with the occupying enemy (the masters) to confrontation with his guards (slaves) [meaning Muslims], because the tyrants see the killing of one American or Westerner as more serious than the killing of a hundred of their country's soldiers; the blood of an American, in their view, is worth the blood of all Muslims. They are ready to cast hundreds to their deaths so that Americans, in exchange, will enjoy security and comfort. ...We must guard ourselves against this ploy and avoid, as much as possible, confrontations with the armies and forces of the state, so that we can strike lethal blows to the occupiers, Allah willing. This does not mean we will surrender to those who defend the Crusaders if they attack us; on the contrary, in this case we must resist with all the strength we have, and we must punish them so that they turn their swords toward the Americans and fight among our ranks, or refrain from entering [into] a confrontation with us ... or they will stand against us and wait for what lies in store for them [at our hands], Sheikh Nasser Al-Najdi: Jihad Must Continue Until All Infidels Convert to Islam or Pay a Poll Tax for people and for regimes... At a time when people are given the choice [of believing] in Islam or paying Jizya [a poll tax paid by non-Muslims living under Muslim rule], Islam is the only alternative for the countries [of the world.]... Therefore, the crime of the tyrants in infidel [i.e. non-Muslim] countries, who do not rule according to Allah's law, is an enormous sin... and we are obliged to fight them and initiate until they convert to Islam, or until Muslims rule the country and he who does not convert to Islam pays Jizya. That is the religious ruling with regard to infidel countries and all the more so with regard to those who rule Muslim countries by way of the cursed law [i.e. a man-made law]... Al-Qa'ida Leaders Praise Jihad The latest Voice of Jihad also features selections from Al-Qa'ida leaders and supporters, including an excerpt of a letter by Sheikh Abu Muhammad Al-Maqdisi (currently imprisoned in Jordan) justifying the suicide bombing perpetrated by Anas Al-Kandari and Jasem Al-Hajeri in the early days of the war against Iraq. The excerpt contains a link to the full letter. (the title of Sheikh Ayman Al-Zawahiri'sbook ), Al-Qa'ida's military leader, the Egyptian Seif Al-'Adl Hamasri, told of a C-130 air attack in Afghanistan in which the wives of Al-Qa'ida members were killed. Also included in the second issue is news coverage of the U.S. military presence in the Persian Gulf, taken from the website www.islammemo.cc/news. One page of the issue is devoted to quotes in praise of Jihad by the Imam of the Mujahideen, Osama bin Laden, Sheikh Dr. Ayman Al-Zawahiri, and the Mujahid Sheikh Mohammad bin Abdallah Al-Seif. The final name most likely refers to Sheikh Abu Omar Muhammad Al-Seif, who appeared recently in a film distributed by Al-Qa'ida that also included the final statements of the Riyadh suicide bombers. Al-Seif has been identified on Islamic websites as the number two leader of the Chechnyan Jihad. The Murder of Ahmed Shah Massoud: Ordered by bin Laden The magazine continued its biography of Sheikh Yousef Al-'Ayyiri, who served as personal bodyguard to Osama bin Laden and manager of the Al-Qa'ida website, until he was killed by Saudi security forces. The biography addresses Osama bin Laden's order for the murder of Ahmed Shah Massoud. According to the biographer, after Sheikh Al-'Ayyiri's release from a Saudi prison, he recruited youths and encouraged them to wage Jihad in Afghanistan and take part in the training camps there: Afterwards, the greatest event in Afghan history occurred ... the assassination of the despicable commander, Ahmed Shah Massoud, and there was no describing Sheikh Al-'Ayyiri's joy. I remember asking him, 'What happened?' And he replied by saying that Sheikh Osama asked the brothers: 'Who will take it upon himself to deal with Ahmed [Shah] Massoud for me, because he harmed Allah and His sons?' A few brothers volunteered to assassinate Massoud and be rewarded by Allah, and you heard the good news. Afterwards, the happy events took place in America [the September 11 attacks], the bastion of disbelief, and the Sheikh was so joyous he nearly floated on air. I called the Sheikh, and he told me he was in a meeting with the religious scholars of Al-Quseim, because a few of them had been a bit critical of the events that occurred in America. He told me about the discussions and about the meetings conducted with them, which persuaded them to support the Jihad and the Mujahideen... The biography ends with a description of the final period of Al-'Ayyiri's life, when he was being sought by Saudi support of Jihad. Two Options for Allah's Enemies - the Jews and Christians: Conversion to Islam or Death This issue also includes a sequel to an interview with Abd Al-Aziz bin 'Issa bin Abd Al-Mohsen, also known as Abu Hajjer, a name high on Saudi Arabia's most wanted list. Abu Hajjer stated that his goal was to wave the banner of monotheism... and expel the enemies of Allah ... the Jews and the Crusaders ... from the land of the two holy places, to conquer the Muslim nations and restore them to their previous state. And may Allah lengthen our days to allow us to infuriate the enemies of Allah, kill them, and strike them by the sword until they either join the religion of Allah or we kill every last one of them. Our model is [the Prophet] Mohammad, who said to the infidels of Qureish: 'I have brought the slaughter upon you.' After characterizing his relationship with Osama bin Laden as similar to that of father and son, Abu Hajjer described his reaction upon hearing that his own name was among the top 19 on the Saudi's most wanted list: I was in a course in which we trained the brothers in one of the valleys. This was a special course, a practical urban warfare course, and after we got back [to the city]... I met with one of the brothers, who told me... they published a list and my picture was on it... I said, 'Allah be praised'... When I entered the land of the two holy places, I felt ... and I also told the brothers ... that the day would come when we will get exposed. We had done everything in our power to delay that day. But now that Allah had put us to this test, we needed to act patiently. Controversy Within Al-Qa'ida Over the Bombings in Saudi Arabia and their Effect on Fundraising Abu Hajjer described a debate within the Jihad movement over the need to attack Americans in Saudi Arabia, which could lead to a more serious entanglement with the Saudi security forces and could hamper fundraising efforts: Jihad members and lovers of Mujahideen were split: There were those who said we must attack the invading forces that defile the land of the two holy places, and that we must turn the Americans' concerns to themselves and their bases, so they would not take off from there to crush Muslim lands and countries, one by one. There were others who said we had to preserve the security of this base and this country [i.e. Saudi Arabia], from which we recruit the armies, from which we take the youth, from which we get the [financial] backing. It must therefore remain safe. My opinion is midway between the two. It is true that we need to keep the enemy occupied with himself and not give him a sense of security, because as soon as he secures his bases and his supply lines, he will have an opportunity to use them to attack our brothers in different parts of countries in the Islamic world. We must therefore make the proper arrangements and prepare ourselves for this momentous event in the best possible way we can. We told them: Wait, we are preparing ourselves, and then we attacked the Americans. It is also true that we must use this country [Saudi Arabia] because it is the primary source of funds for most Jihad movements, and it has some degree of security and freedom of movement. However, we must strike a balance between this and the American invasion of the Islamic world and its [strangling of] the Jihad movement and even other Islamic movements... We must understand that the situation in the land of the two holy places is deteriorating day by day for the Mujahideen and their financial sources, as well as regarding the secularization of the country and the attempt by the treacherous rulers to lead it to moral collapse, in accordance with the dictates of the White House. We tell them [those opposed to the bombings in Saudi Arabia]: If only you were to see the prisons filled with Mujahideen youth and pro-Jihad preachers. We have not carried out even one attack. All operations that took place were defensive operations. On the contrary, the brothers try as much as possible to avoid confrontations with the army and security forces. Nevertheless, the government is escalating the war, and it is trying to uproot me, uproot you, and uproot all Islamists. Abu Hajjer concluded the interview with this message to the youth he trained: You must do something, you must fight the enemies of Allah, the Crusaders and the Jews, and become a bone in their throats and hearts... Following the interview was a communiqu.8e published by the Mujahideen in the Arabian Peninsula reacting to recent statements made by the Saudi interior minister: The communiqu.8e is a response to the Saudi Interior Ministry announcement about the confiscation of a large quantity of weapons. The communiqu.8e claims that many Saudis carry personal weapons, automatic rifles or hand guns, and that the confiscations were actually confiscations of goods possessed by known weapon dealers, who the Interior Ministry falsely connected to the Jihad movement. The Voice of Jihad's next issue will feature an interview with the wanted Sheikh Abdallah bin Muhammad Al-Rashuod. --------------------------------------------- [{}] * MEMRI: The Middle East Media Research Institute http://memri.org/ The Middle East Media Research Institute (MEMRI) explores the Middle East through the region's media. MEMRI bridges the language gap which exists between the West and the Middle East, providing timely translations of Arabic, Farsi, and Hebrew media, as well as original analysis of political, ideological, intellectual, social, cultural, and religious trends in the Middle East. Founded in February 1998 to inform the debate over U.S. policy in the Middle East, MEMRI is an independent, nonpartisan, nonprofit, 501 (c)3 organization. MEMRI's headquarters is located in Washington, DC with branch offices in Berlin, London, and Jerusalem, where MEMRI also maintains its Media Center. MEMRI research is translated to English, German, Hebrew, Italian, French, Spanish, Turkish, and Russian. http://memri.org/aboutus.html memri@memri.org MEMRI P.O. Box 27837 Phone: (202) 955-9070 Fax: (202) 955-9077 Email to: donations@memri.org > Timothy Ferris > URL: http://www.timothyferris.com/ > The Hunger Site: > http://www.thehungersite.com/ > ~o~ > Jeff Mishlove > Thinking Allowed Productions > URL: http://www.thinking-allowed.com/ > ~o~ > What we may Learn from Future Physics Experiments > Andris Skuja, University of Maryland > Professor Andris Skuja > Web Page: http://www.physics.umd.edu/hep/ > http://www.physics.umd.edu/people/faculty/skuja.html > C. SCOTT LITTLETON > http://www.oxy.edu/~yokatta/home.htm Any sufficiently advanced technology is > indistinguishable from magic > --Sir Arthur C. Clarke I think we're property. . . > --Charles Fort Einstein's gravitons and Tony Smith's conformal gravitons. > Tony Smith's conformal gravitons > http://www.innerx.net/personal/tsmith/SegalConf.html > Conformal Groups are related to Moebius Transformations > SPECIAL RELATIVITY PROHIBITS SPACELIKE CAUSATION > http://www.mindspring.com/~cerebroscopic/Angelidis.html > Sloan Digital Sky Survey (SDSS) The First Detailed Full Sky Picture > of the Oldest Light in the Universe. > http://map.gsfc.nasa.gov/m_mm.html The Wilkinson Microwave Anisotropy Probe (WMAP) > }))) > MSU University News > [...] > Cornish, Spergel and Starkman have posted > their findings on the Internet > ()* > where other scientists can evaluate them. > Ultimately, Cornish said, it's the scientific > community that will decide. The most amazing thing about this whole story is the > fact that we can go out there and try and figure out > the shape of the universe, Cornish said. ... Regardless > of what the results are, I think it's great for people to > know this is even a possibility. > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% * Constraining the Topology of the Universe > http://arxiv.org/abs/astro-ph/0310233 1 parsec = 3.2 light years distance > 1 Megaparsec (Mpc) = 3.2 light years distance times 1,000,000 > 1 Gigaparsec (Gpc) = 3.2 light years distance times 1,000,000,000 > 24 Gpc = 24 times 3.2 light years times one billion (size) > 76,800,000,000 Light Years!! > [Uhhhmmm... is that radius or diameter?] > Our universe, so they say, is 13.7 billion years old since the Big > Bang... > [[Cosmology Tutorial: > ]] > Wilkinson Microwave Anisotropy Probe > WMAP: http://map.gsfc.nasa.gov/ > Data Analysis http://lambda.gsfc.nasa.gov/ > So what the hell are WE doing here??!! > 0101010101010101010101010101010101010101 > A high resolution foreground cleaned CMB map from WMAP > http://arxiv.org/abs/astro-ph/0302496 Prof. Max Tegmark > http://www.hep.upenn.edu/~max/wmap.html Brane Worlds, the Subanthropic Principle and the > Undetectability Conjecture by Beatriz Gato-Rivera > http://www.unknowncountry.com/mindframe/opinion/?id=96 > Generally ... http://xxx.arxiv.cornell.edu/find/physics/1/au:+Gato_Rivera_B/0/1/0/all/0/1 Dark Matter, Extra Dimensions Related And Possibly Detectable > http://www.spacedaily.com/news/cosmology-03o.html Neuroblast Wormhole Behaviorism, a.k.a. > The Portal, to Dodecahedron Head > http://pw1.netcom.com/~mthorn/mt90.htm Meanwhile... n a n o t u b e s . . . > A New Alternative To Depleted Uranium?! http://popularmechanics.com/science/space/2002/7/going_up/print.phtml The promise of inexpensive access to space is so important > to the human race that we are ready to meet these challenges > head on. Viewed in one way, the space elevator will be the > largest civil engineering project ever attempted, > Laubscher said. [...] Crawling to space: ... [N]ew work in China that suggests carbon nanotubes > can be fused together, without need of a matrix material. > [...] http://www.space.com/businesstechnology/technology/space_elevator_030917.htm l Space Elevator, Nanotubes Aumakua _ Uhane _ Unihipili Electromagnetic Mind Field > http://pw1.netcom.com/~mthorn/magnmind.htm Committee for Surrealist Investigation of Claims of the Normal > [ CSICON ] http://www.rawilson.com/csicon.shtml < C O N T A C T Uppaluri Gopala Krishnamurti (Born 9 July 1918) > http://www.well.com/user/jct/mystiq1.htm KA: > http://fivebodies.removetheveil.net/ka/ Brainwave Frequency Listing > http://www.lunarsight.com/freq.htm a. Electrons, e^1- > b. Protons, p^1+ > c. Neutrons, n^o > http://falcon.sbuniv.edu/~ggray.wh.bol/CHE1104/pcp3outl.html > The three flavors of quarks--up, down and strange--would > also come in three different colors--red, white, and blue. > [...] Murray [Gell-Mann] picked the arbitrary labels in honor > of the French flag. [...] The metaphor seemed even more elegant > after Murray's colors were replaced with the labels that > came to predominate, the primary colors of light: red, green, > and blue. On a TV screen, the three mixed to form colorless > 'white' light. STRANGE BEAUTY, Murray Gell-Mann and the > Revolution in Twentieth-Century Physics by George Johnson > 1999 ISBN0-679-75688-4 THE MYSTIQUE OF ENLIGHTENMENT > Part One [Excerpt] > U.G. Krishnamurti > [CLICK--> ] > The Archetype and the Beast '98 > http://pw1.netcom.com/~mthorn/arcbeast.htm > [~][^][~] > Disingenuous Demagogues Deteriorate Daily > All Politicians are Demagogues, yet not all > Demagogues are Politicians... > <> ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ > <> Sagittarius assimilated by Milky Way: > <> Resistence Is Futile! > <> http://www.astro.virginia.edu/~mfs4n/sgr/ > <> ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ > <> FREE THE YEZIDI!!! > http://www.songsouponsea.com/Promenade/GnosisE.html > The Purification will begin > shortly after humans build > a great house in the sky. > By then there will be fires > everywhere and greedy, selfish, > power-mad leaders, internal wars. > http://www.deoxy.org/omega.htm > Yes, and you can't escape it by closing your eyes either, > as your whole 'being' is thrust into a powerful spotlight > where every photon is a gently tugging, beaming consciousness; > searing traumatically a heightened essence at work on your > subjective optical/emotional/cognitive space. It's an > unwavering, unflinching wake-up call to the brain. > Reality is infinite, and sometimes untenable, during & > afterwards, upon return to the so-called mundane plane > of existence, looping, inextricably bound by gravitational > forces. We should, could, maybe, love each other a bit > more as we're surrounded by strangeness imbedded in > a whirling biosphere of fellow travelers. > Heading either Somewhere new, or Nowhere... > _________ > STARS and STRIPES: http://www.estripes.com/ > Report pending on rising rate of GI suicide in Iraq > _________ > Why? The Neuroscience of Suicide > Despite the inconsistencies, the bulk of evidence points strongly to a > problem in the brains of suicides involving the serotonin system. > [Cont... ] > _________ > Meme: (pron. 'meem') A contagious idea that > replicates like a virus, passed on from mind > to mind. Memes function the same way genes > and viruses do, propagating through > communication networks and face-to-face > contact between people > David S. Bennahum. > _________ > What a tragic world this is, he reflected. > Those down here are prisoners, and the > ultimate tragedy is that they don't know it; > they think they are free because they never > have been free, and do not understand what > it means. This is a prison, and few men have > guessed. But I know, he said to himself. > Because that is why I am here. To burn the > walls, to tear down the metal gates, to > break each chain. Thou shalt not muzzle > the Ox as he treadeth out the corn, he > thought, remembering the Torah. > The Divine Invasion, Phillip K Dick. > HarperCollins 1981. > http://www.philipkdick.com/weirdo.htm > _________ > <> ... The Inuit of northern Canada have one of the highest suicide > <> rates in the world, Yale University research affiliate Michael > <> Kral found in a study based on a 1998 visit to the Arctic Circle. > <> http://www.suicideinfo.ca/csp/go.aspx?tabid=59 > _________ > In work known to Western culture. The reason is that the critical > time has arrived. I told you that the unhealed traumatic > experiences which gain status of their own and become spirits > of trauma continue their existence throughout generations. If > they are not healed, they build up, connect, accelerate, enforce > and support each other, and become collective entities... > At some point, they accumulate on the collective level and become > very dangerous. This dangerous time has come. It can become many > times more dangerous now than in previous times, when world wars > were induced by accumulated spirits of trauma. The purpose of my > visit is to tell you that there is a great danger for all the > people on the planet, but there are effective means to overcome > it. That's why I invite you to our place in Central Asia, to > Samarkand, to learn about our practice... > The Master of Lucid Dreams by Olga Kharitidi, M.D. > (... accumulated spirits of trauma...) > &$&$&$&$&$& $&$&$&$&$&$ &$&$&$&$&$&$[Sterling ]&$&$ > -------- Original Message -------- === > Subject: The men we are being attacked by, he said, are > Syrian-trained terrorists and local freedom fighters. > *Robert Fisk: One, two, three, what are they fighting for? [{snip}] T h e Y e z i d i s o f K u r d i s t a n http://www.songsouponsea.com/Promenade/GnosisE.html Sufi Martyrs ...the stormiest and most controversial early movements within Sufism were led by Husayn ibn Mansur Haflaj (crucified AD 922),'Ain al-QudAt Hamaddni (crucified AD 1131), and Shah.89b al-Din Suhrawardi (crucified AD 1191). Hamaddni's idea of successive reincarnation, and the redemption of Lucifer, added to his other non-Islamic preachings, qualified him for burning on a cross by the Muslim authorities when he was 33. Al Hall.89j claimed himself to be an avatar of the divinity... and that all creatures are ultimately the manifestations of the same original Universal Spirit. He thus also declared Lucifer to have been redeemed and elevated to the highest universal station, as in Yezidism. He was subjected to exquisite tortures before being crucified in Baghdad. At present there is a shrine dedicated to Hall.89j in the sacred Yezidi religious center and shrine complex at L.89lish, next to the tomb of Shaykh Adi. - Sufism The Yezidis of Kurdistan have been called many things, most notoriously 'devil-worshippers,' a term used both by unsympathetic neighbours and fascinated Westerners. This sensational epithet is not only deeply offensive to the Yezidis themselves, but quite simply wrong. Yezidism is not devil-worship, but something far more elusive, and interesting. This religion, which has aroused so much antagonism, is difficult to summarize succinctly. To generalize, seven Holy Beings are venerated, chiefly Melek Tawus, the Peacock Angel. - The Evolution of Yezidi Religion Melek Tawus is an angel who has fallen from God's grace, and for his repentance, is restored to God's favor. While in hell, Melek Tawus, in seven thousand years filled seven jars with his tears of repentance which extinguished the fires of hell, or at the last judgment will be used to do so. Melek Tawus was banished from the sight of God because he refused to bow down, as asked, before God's creation, Adam. But Yezidis do not identify the fallen angel with a spirit of evil. The act of disobedience has been forgiven by god as a father forgives a wayward son. For the Yezidis, the internal aspect of Tawus Melek remains alive. They hold that Tawus Melek is as fire with two elementary qualities: fire as light and fire to burn-and the apparently dual aspects of the good and the evil exist within the same Person. Every human being is a mixture of good and evil, and every Yezidi has Tawus Melek in himself. This is one of the meanings evoked in a 19th century report by Haji Zain al-Abidin al-Shirwani, who, in a conversation with a Yezidi leader concerning Tawus Melek, asked, Who is Iblis (Satan)? 'What is the reason behind your worship of him and fear of him? The Yezidi replied, The learned are perplexed by the comprehension of his reality and scholars are in ignorance from detecting an atom of the valley of his realization but the men of knowledge said in his description, if he shows mankind his light they would have worshipped him as god. He is hidden but near everyone. All-knowing in regard to men and their deeds. Satan sometimes comes to you from the walls and other times from the roofs. Occasionally he resides in the deepest folds of the heart and sometimes he joins your body and goes in it like blood. And it is related in the tradition that Satan is from the fire of glory, meaning that his glory was created from the fire of the glory of God.' - Gurdjieff and Yezidism There is a myth that Prometheus created humans by mixing Earth and Water to create the gross body; Athena breathed Air into it, imbuing it with a Spirit-Soul. Prometheus added the Higher Soul, which is the Fire that He took from the Wheel of the Sun and brought to humanity in a Narth.90x (giant fennel) stalk. (The Narth.90x corresponds to Shushumna, the esoteric spinal column of yoga philosophy, which contains the Fiery Kundalini power.) Recall also that the Thursos, the sacred Bacchic wand, is made from the Narth.90x and holds Promethean Fire. As implied in the myth of our Promethean origins, the Fire in our souls is akin to the Celestial Fire (As above, so below). - The Elements: Fire Gnostic Christians maintained that the light Lucifer brought was true enlightenment, which He gave humanity against God's will, as Prometheus stole the fire of heaven to bring civilization to mankind against the will of Zeus. God denied Adam and Eve the fruit of the tree of knowledge, desiring to keep them ignorant; but Lucifer, in the form of the serpent, gave them the light of wisdom. He became the Prince of the Power of the Air (Ephesians2:2) who threw his lightning bolts at church towers. He wielded the trident which, in Eastern symbolism is a triple lightning-phallus destined to fertilize the Triple Goddess. In Tantric Hindu symbology, He is the bindu, the spark that ignites the sacred serpent and awakens the kundalini. - Lucifer Rising Gurdjieff's choice of Beelzebub as hero of the First Series and the principal expositor of the Teaching, surely has put those who would take it up in a position similar to the one the Yezidis have had to endure for many centuries. Externally, when contacted by outsiders or as Gurdjieff sometimes referred to them, natives, the Teaching might seem, like that of the Yezidis to be at bottom devil worship. They would entirely miss, for instance, the prayers scattered through the book, let alone recognize their sacred quality. In a world that is becoming more and more divided between a narrow sectarian fundamentalism and a liberal scientism, the very hint of Beelzebub as hero would bring anathema from one camp and a condescending dismissal from the other. That the Teaching may not be casually spoken of with either sector, helps to protect it, making it difficult for the student to take for granted, or worse, profane. Gurdjieff's Beelzebub, like the Yezidis Melek Tawus is a being fallen from nearness to God, who redeems himself and is restored to grace. Each has fallen because in self-will, following the dictates of his reason, he acted in disobedience to what was higher. For the Yezidis, the Devil is everywhere, a practitioner, in the sense of the third of the types of school about which Gurdjieff told Ouspensky, fit for the business of the world. In his nearness and familiarity, he is an archetype of the returning prodigal, a model for the rest of us who have fallen, who are much closer to devils than angels. The Devil is also what none of us wish to look at in ourselves, but must nevertheless see, for that looking actualizes the possibility of redemption. - Gurdjieff and Yezidism The distressing struggle of integrating the shadow side of ourselves, of wrestling with the dark angel within, is ongoing and never ending for those who are willing to take it on. Jung considers it a moral task of the first magnitude, and given the godlike power conferred upon us most notably by the ability to unleash atomic energy, the struggle assumes an urgency and necessity for the future survival of our humankind. Let us hope enough of us are able and willing to enter the ring. - Thoughts on Evil by James M. Shultz If you bring forth what is within you, what you bring forth will save you. If you do not bring forth what is within you, what you do not bring forth will destroy you. - Jesus, Gospel of Thomas Come now, and let us reason together, saith the Lord: though your sins be as scarlet, they shall be as white as snow; though they be red like crimson, they shall be as wool. - Isaiah 1:18 === Subject: can not show that.....(see text) Hi All, I dont know where is mistake: a and r are vectors, a=a1*i+a2*j+a3*k, a1,a2,a3-const, r=x*i+y*j+k*z fi=fi(r) rot(fi*a x r))=(grad(fi).r).a-(grad(fi).a).r+2*fi*a I use rot(fi*v)=grad(fi)xv+fi*rot(v) and can't get 2*fi*a, always have 3*fi*a. What am I missing??? (the problem from electrodynamics,(field of magnetic dipole)) Tnx Ihor === Subject: Re: can not show that.....(see text) > Hi All, > I dont know where is mistake: > a and r are vectors, a=a1*i+a2*j+a3*k, a1,a2,a3-const, > r=x*i+y*j+k*z > fi=fi(r) > rot(fi*a x r))=(grad(fi).r).a-(grad(fi).a).r+2*fi*a > I use rot(fi*v)=grad(fi)xv+fi*rot(v) and can't get 2*fi*a, always have > 3*fi*a. > What am I missing??? > (the problem from electrodynamics,(field of magnetic dipole)) Use suffix notation, D{i} = partial d/dx{i}, e{ijk} is alternating tensor, d{ij} is Kroenecker delta[1]. Summation over all values of a repeated index is implied (e.g. a{i}a{i} = (a1)^2 + (a2)^2 + (a3)^2 = |a|^2): e{ijk} D{j} e{kpq} f a{p} x{q} = (d{ip}d{jq} - d{iq}d{jp})a{p}D{j}(fx{q}) = a{i} ( fD{j}x{j} + x{j}D{j}f) - a{j} (fd{ij} + x{i}D{j}f) ^^^^^^ = 3a{i}f + a{i}(x.grad)f - a{i}f - x{i} (a.grad)f = 2a{i}f + a{i}(x.grad)f - x{i}(a.grad)f It looks like you're missing the under-^-ed term. [1]: e{ijk} = 1 if (i,j,k) is an even permutation of (1,2,3), -1 if (i,j,k) is an odd permutation of (1,2,3), and 0 if i=j, j=k or i=k. d{ij} = 1 if i = j and 0 otherwise. It can be shown that e{ijk}e{kpq} = d{ip}d{jq} - d{iq}d{jp}. It can also be shown that D{i}x{j} = d{ij} = D{j}x{i}. Vector operations: (a cross b){i} = e{ijk}a{j}b{k} and a.b = a{i}b{i}. (rot F){i} = e{ijk}D{j}F{k}, div(F) = D{i}f{i}, (grad f){i} = D{i}f. This is by far the easiest method for deriving vector derivative identities. -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will find these people and we will bring them to justice. === Subject: No Perfect Cuboid Exists: - Proof (take 2!) No Perfect Cuboid Exists See also - http://www.bearnol.pwp.blueyonder.co.uk **************************************** b^2+c^2=d^2 c^2+a^2=e^2 a^2+b^2=f^2 a^2+b^2+c^2=g^2 Smallest solution hcf[a,b,c]=1 g^2==a^2==b^2==c^2==0,1 [mod 4] => g^2==1 [mod 4] => g==1 [mod 2] g^2=a^2+d^2=b^2+e^2=c^2+f^2 one of (a,d) even, one of (b,e) even, one of (c,f) even therefore WLOG all a,b,c even => hcf[a,b,c]>=2 => contradiction, from which result follows ***************************************** === Subject: Re: No Perfect Cuboid Exists: - Proof (take 2!) This is funny! : No Perfect Cuboid Exists : See also - http://www.bearnol.pwp.blueyonder.co.uk : **************************************** : b^2+c^2=d^2 : c^2+a^2=e^2 : a^2+b^2=f^2 : a^2+b^2+c^2=g^2 : Smallest solution hcf[a,b,c]=1 Okay, but you've destroyed lots of symmetry by this. : g^2==a^2==b^2==c^2==0,1 [mod 4] : => g^2==1 [mod 4] : => g==1 [mod 2] : g^2=a^2+d^2=b^2+e^2=c^2+f^2 : one of (a,d) even, one of (b,e) even, one of (c,f) even : therefore WLOG all a,b,c even No, you've got LOG because of destroyed symmetry. At this point you *can* say: *If* a,b,c all even, *then* hcf[a,b,c]>1, which contradicts smallest solution stuff, *hence* a,b,c *are not all even*. So a,b,c are not all even, what now? Justin === Subject: No Perfect Cuboid Exists(?) - non-Proof (was Re: No Perfect Cuboid Exists: - Proof (take 2!)) > No Perfect Cuboid Exists Heh-heh! I copy it below! > b^2+c^2=d^2 > c^2+a^2=e^2 > a^2+b^2=f^2 > a^2+b^2+c^2=g^2 > Smallest solution hcf[a,b,c]=1 > g^2==a^2==b^2==c^2==0,1 [mod 4] > => g^2==1 [mod 4] > => g==1 [mod 2] OK. > g^2=a^2+d^2=b^2+e^2=c^2+f^2 > one of (a,d) even, one of (b,e) even, one of (c,f) even > therefore WLOG all a,b,c even Non-sequitur! Why not a even, b even, c odd, d odd, e odd, f even? > => hcf[a,b,c]>=2 > => contradiction, from which result follows No. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists(?) - non-Proof (was Re: No Perfect Cuboid Exists: - Proof (take 2!)) Pls see below for response(s)... J > No Perfect Cuboid Exists > Heh-heh! I copy it below! > b^2+c^2=d^2 > c^2+a^2=e^2 > a^2+b^2=f^2 > a^2+b^2+c^2=g^2 > Smallest solution hcf[a,b,c]=1 > g^2==a^2==b^2==c^2==0,1 [mod 4] > => g^2==1 [mod 4] > => g==1 [mod 2] > OK. > g^2=a^2+d^2=b^2+e^2=c^2+f^2 > one of (a,d) even, one of (b,e) even, one of (c,f) even > therefore WLOG all a,b,c even > Non-sequitur! Why not a even, b even, c odd, d odd, e odd, > f even? ****************************************** Because the equation above (g^2=a^2+d^2=b^2+e^2=c^2+f^2) is _totally_ symmetric in a/d, b/e, c/f, so we do have a free choice - I _choose_ a,b,c even to complete the proof ****************************************** > => hcf[a,b,c]>=2 > => contradiction, from which result follows > No. > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists(?) - non-Proof (was Re: No Perfect Cuboid Exists: - Proof (take 2!)) > Pls see below for response(s)... >> No Perfect Cuboid Exists >> Heh-heh! I copy it below! >> b^2+c^2=d^2 >> c^2+a^2=e^2 >> a^2+b^2=f^2 >> a^2+b^2+c^2=g^2 >> Smallest solution hcf[a,b,c]=1 >> g^2==a^2==b^2==c^2==0,1 [mod 4] >> => g^2==1 [mod 4] >> => g==1 [mod 2] >> OK. >> g^2=a^2+d^2=b^2+e^2=c^2+f^2 >> one of (a,d) even, one of (b,e) even, one of (c,f) even >> therefore WLOG all a,b,c even >> Non-sequitur! Why not a even, b even, c odd, d odd, e odd, >> f even? > ****************************************** > Because the equation above (g^2=a^2+d^2=b^2+e^2=c^2+f^2) is _totally_ > symmetric in a/d, b/e, c/f, so we do have a free choice - I _choose_ a,b,c > even to complete the proof > ****************************************** Nice to have that that sort of choice. Do you choose to have gravity work in reverse when travelling upstairs too? More seriously: your equations also have d^2 = b^2 + c^2 etc. which destroys your total symmetry. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists(?) - non-Proof (was Re: No Perfect Cuboid Exists: - Proof (take 2!)) Pls see below for response(s)... J > Pls see below for response(s)... > J >> No Perfect Cuboid Exists >> Heh-heh! I copy it below! >> b^2+c^2=d^2 >> c^2+a^2=e^2 >> a^2+b^2=f^2 >> a^2+b^2+c^2=g^2 >> Smallest solution hcf[a,b,c]=1 >> g^2==a^2==b^2==c^2==0,1 [mod 4] >> => g^2==1 [mod 4] >> => g==1 [mod 2] >> OK. >> g^2=a^2+d^2=b^2+e^2=c^2+f^2 >> one of (a,d) even, one of (b,e) even, one of (c,f) even >> therefore WLOG all a,b,c even >> Non-sequitur! Why not a even, b even, c odd, d odd, e odd, >> f even? > ****************************************** > Because the equation above (g^2=a^2+d^2=b^2+e^2=c^2+f^2) is _totally_ > symmetric in a/d, b/e, c/f, so we do have a free choice - I _choose_ a,b,c > even to complete the proof > ****************************************** > Nice to have that that sort of choice. Do you choose to have gravity > work in reverse when travelling upstairs too? > More seriously: your equations also have d^2 = b^2 + c^2 etc. > which destroys your total symmetry. ***************************************** The symmetry stands - let me explain why: [B: g^2=a^+d^2=b^2+e^2=c^2+f^2] [A: d^2=b^2+c^2 etc.] Since the proof is by contradiction, once I arrive at the later point (B), we can ignore the previous logic (A) It is sufficient to disprove B to disprove A, if A=>B (by whatever route) ****************************************** > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists(?) - non-Proof (was Re: No Perfect Cuboid Exists: - Proof (take 2!)) > ***************************************** > The symmetry stands - let me explain why: > [B: g^2=a^+d^2=b^2+e^2=c^2+f^2] > [A: d^2=b^2+c^2 etc.] > Since the proof is by contradiction, once I arrive at the later point > (B), we can ignore the previous logic (A) > It is sufficient to disprove B to disprove A, if A=>B (by whatever route) Except that you can't disprove B: try a = 576, b = 1104, c = 744, d = 943, e = 47, f = 817, g = 1105. These satisfy B (but not of course A). You really shoudl follow Gerry Myerson's advice. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists(?) - non-Proof (was Re: No Perfect Cuboid Exists: - Proof (take 2!)) Pls see below for response(s)... [as usual!] J > ***************************************** > The symmetry stands - let me explain why: > [B: g^2=a^+d^2=b^2+e^2=c^2+f^2] > [A: d^2=b^2+c^2 etc.] > Since the proof is by contradiction, once I arrive at the later point > (B), we can ignore the previous logic (A) > It is sufficient to disprove B to disprove A, if A=>B (by whatever route) > Except that you can't disprove B: try > a = 576, > b = 1104, > c = 744, > d = 943, > e = 47, > f = 817, > g = 1105. > These satisfy B (but not of course A). **************************************** Notice that the example given here has hcf[a,b,c]>=2 - in support of my thesis, and therefore of my disproof of the existence of a minimal set - I challenge you to give me a set of seven similar where hcf[a,b,c]=1, as you wish to claim **************************************** > You really shoudl follow Gerry Myerson's advice. > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists(?) - non-Proof (was Re: No Perfect Cuboid Exists: - Proof (take 2!)) > Pls see below for response(s)... [as usual!] >> ***************************************** >> The symmetry stands - let me explain why: >> [B: g^2=a^+d^2=b^2+e^2=c^2+f^2] >> [A: d^2=b^2+c^2 etc.] >> Since the proof is by contradiction, once I arrive at the later point >> (B), we can ignore the previous logic (A) >> It is sufficient to disprove B to disprove A, if A=>B (by whatever > route) >> Except that you can't disprove B: try >> a = 576, >> b = 1104, >> c = 744, >> d = 943, >> e = 47, >> f = 817, >> g = 1105. >> These satisfy B (but not of course A). > **************************************** > Notice that the example given here has hcf[a,b,c]>=2 - in support of my > thesis, and therefore of my disproof of the existence of a minimal set - I > challenge you to give me a set of seven similar where hcf[a,b,c]=1, as you > wish to claim Didn't see that in claim B.! How about d = 576, e = 1104, f = 744, a = 943, b = 47, c = 817, g = 1105? Did you miss Gerry's advice? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: No Perfect Cuboid Exists: - Proof Statement B: is g^2=a^2+d^2=b^2+e^2=c^2+f^2 which is symmetric in a/d, b/e, c/f, therefore a=576 b=1104 c=744,... [as you have already (kindly :-) provided] These have a hcf[a,b,c]>=2 hence B: is not a minimal set, and, (because A=>B, as already stated) ~A: is confirmed > Pls see below for response(s)... [as usual!] > J >> ***************************************** >> The symmetry stands - let me explain why: >> [B: g^2=a^+d^2=b^2+e^2=c^2+f^2] >> [A: d^2=b^2+c^2 etc.] >> Since the proof is by contradiction, once I arrive at the later point >> (B), we can ignore the previous logic (A) >> It is sufficient to disprove B to disprove A, if A=>B (by whatever > route) >> Except that you can't disprove B: try >> a = 576, >> b = 1104, >> c = 744, >> d = 943, >> e = 47, >> f = 817, >> g = 1105. >> These satisfy B (but not of course A). > **************************************** > Notice that the example given here has hcf[a,b,c]>=2 - in support of my > thesis, and therefore of my disproof of the existence of a minimal set - I > challenge you to give me a set of seven similar where hcf[a,b,c]=1, as you > wish to claim > Didn't see that in claim B.! How about > d = 576, > e = 1104, > f = 744, > a = 943, > b = 47, > c = 817, > g = 1105? > Did you miss Gerry's advice? > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists: - Proof > Statement B: is > g^2=a^2+d^2=b^2+e^2=c^2+f^2 > which is symmetric in a/d, b/e, c/f, therefore and soluble by a = 576, b = 1104, c = 744, d = 943, e = 47, f = 817, g = 1105, and by a = 943, b = 47, c = 817, d = 576, e = 1104, f = 744, g = 1105, and by lots of other 7-tuples (a,...,g) of distinct positive integers. > [as you have already (kindly :-) provided] > These have a hcf[a,b,c]>=2 Not mentioned in (B) above :-( > hence B: is not a minimal set, Minimal set? Private language time :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists: - Proof ... > Statement B: is > g^2=a^2+d^2=b^2+e^2=c^2+f^2 > which is symmetric in a/d, b/e, c/f, therefore > and soluble by > a = 576, > b = 1104, > c = 744, > d = 943, > e = 47, > f = 817, > g = 1105, > and by > a = 943, > b = 47, > c = 817, > d = 576, > e = 1104, > f = 744, > g = 1105, **************************************** For the purposes of this proof these are not distinct sets... **************************************** > and by lots of other 7-tuples (a,...,g) of distinct positive integers. **************************************** ...all of which have hcf[a,b,c]>=2 **************************************** > [as you have already (kindly :-) provided] > These have a hcf[a,b,c]>=2 > Not mentioned in (B) above :-( > hence B: is not a minimal set, > Minimal set? Private language time :-( > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists: - Proof > ... >> Statement B: is >> g^2=a^2+d^2=b^2+e^2=c^2+f^2 >> which is symmetric in a/d, b/e, c/f, therefore >> and soluble by >> a = 576, >> b = 1104, >> c = 744, >> d = 943, >> e = 47, >> f = 817, >> g = 1105, >> and by >> a = 943, >> b = 47, >> c = 817, >> d = 576, >> e = 1104, >> f = 744, >> g = 1105, > **************************************** > For the purposes of this proof these are not distinct sets... proof? what proof? > **************************************** >> and by lots of other 7-tuples (a,...,g) of distinct positive integers. > **************************************** > ...all of which have hcf[a,b,c]>=2 apart from a = 943, b = 47, c = 817, d = 576, e = 1104, f = 744, g = 1105, -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists: - Proof Let it go Robin - you know you've lost it... James Wanless [bouncing back 15 times :-)] > ... >> Statement B: is >> g^2=a^2+d^2=b^2+e^2=c^2+f^2 >> which is symmetric in a/d, b/e, c/f, therefore >> and soluble by >> a = 576, >> b = 1104, >> c = 744, >> d = 943, >> e = 47, >> f = 817, >> g = 1105, >> and by >> a = 943, >> b = 47, >> c = 817, >> d = 576, >> e = 1104, >> f = 744, >> g = 1105, > **************************************** > For the purposes of this proof these are not distinct sets... > proof? what proof? > **************************************** >> and by lots of other 7-tuples (a,...,g) of distinct positive integers. > **************************************** > ...all of which have hcf[a,b,c]>=2 > apart from > a = 943, > b = 47, > c = 817, > d = 576, > e = 1104, > f = 744, > g = 1105, > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists: - Proof > Let it go Robin - you know you've lost it... Lost what? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists: - Proof ... > ... >> Statement B: is >> g^2=a^2+d^2=b^2+e^2=c^2+f^2 >> which is symmetric in a/d, b/e, c/f, therefore >> and soluble by >> a = 576, >> b = 1104, >> c = 744, >> d = 943, >> e = 47, >> f = 817, >> g = 1105, >> and by >> a = 943, >> b = 47, >> c = 817, >> d = 576, >> e = 1104, >> f = 744, >> g = 1105, > **************************************** > For the purposes of this proof these are not distinct sets... > proof? what proof? > **************************************** >> and by lots of other 7-tuples (a,...,g) of distinct positive integers. > **************************************** > ...all of which have hcf[a,b,c]>=2 > apart from > a = 943, > b = 47, > c = 817, > d = 576, > e = 1104, > f = 744, > g = 1105, ***************************************** Which is equivalent to: a=576, b=1104, c=744, ... (as you very well know! - I suggest) ***************************************** > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists: - Proof > ... >> ... >> Statement B: is > g^2=a^2+d^2=b^2+e^2=c^2+f^2 > which is symmetric in a/d, b/e, c/f, therefore >> and soluble by >> a = 576, > b = 1104, > c = 744, > d = 943, > e = 47, > f = 817, > g = 1105, >> and by >> a = 943, > b = 47, > c = 817, > d = 576, > e = 1104, > f = 744, > g = 1105, >> **************************************** >> For the purposes of this proof these are not distinct sets... >> proof? what proof? >> **************************************** >> and by lots of other 7-tuples (a,...,g) of distinct positive integers. >> **************************************** >> ...all of which have hcf[a,b,c]>=2 >> apart from >> a = 943, >> b = 47, >> c = 817, >> d = 576, >> e = 1104, >> f = 744, >> g = 1105, > ***************************************** > Which is equivalent to: equivalent? > a=576, > b=1104, > c=744, Your statement B is g^2=a^2+d^2=b^2+e^2=c^2+f^2 (your words not mine). I presumed that the a, ..., g are intended to be positive integers (you didn't say that). Under this stipulation there are (infinitely) many solutions: here's another (a,...,g) = (3,4,4,3,3,4,5). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: No Perfect Cuboid Exists: - Proof > Statement B: is > g^2=a^2+d^2=b^2+e^2=c^2+f^2 > which is symmetric in a/d, b/e, c/f, therefore > a=576 > b=1104 > c=744,... > [as you have already (kindly :-) provided] > These have a hcf[a,b,c]>=2 > hence B: is not a minimal set, You can't disprove B. You may attempt to disprove B together with hcf(a,b,c) = 1, but in that case there is *no* symmetry, so you can not use it. -- 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: No Perfect Cuboid Exists: - Proof I'm _not_ trying to disprove B (in general), I'm just showing there's no _minimal_ B This is then sufficient to complete the proof - we've already used (successfully) all the symmetry we need to > Statement B: is > g^2=a^2+d^2=b^2+e^2=c^2+f^2 > which is symmetric in a/d, b/e, c/f, therefore > a=576 > b=1104 > c=744,... > [as you have already (kindly :-) provided] > These have a hcf[a,b,c]>=2 > hence B: is not a minimal set, > You can't disprove B. You may attempt to disprove B together with > hcf(a,b,c) = 1, but in that case there is *no* symmetry, so you can > not use it. > -- > 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: No Perfect Cuboid Exists: - Proof > Statement B: is > g^2=a^2+d^2=b^2+e^2=c^2+f^2 > which is symmetric in a/d, b/e, c/f, therefore > a=576 > b=1104 > c=744,... > [as you have already (kindly :-) provided] > These have a hcf[a,b,c]>=2 > hence B: is not a minimal set, You can't disprove B. You may attempt to disprove B together with > hcf(a,b,c) = 1, but in that case there is *no* symmetry, so you can > not use it. > I'm _not_ trying to disprove B (in general), I'm just showing there's no > _minimal_ B > This is then sufficient to complete the proof - we've already used > (successfully) all the symmetry we need to But B is minimal if hcf(a,b,c,f,e,f) = 1. Which was the case in Robin's example. -- 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: What is codomain of cardinal function...? > Hi folks, > A question. > We can define card:{A | A is a subset of Q}-->{0, 1, ..., aleph_0} as the > cardinal function restricted to the subsets of Q. This operator is a > whole function in sense that the domain and codomain are classicals sets > (we have not afraid with categories and etc...; we would do it with the > codomain the universal set). > There is no problem there. But I point one problem: there are a natural > bijection between the natural numbers and it's cardinal: > 0 as number has 0 as a cardinal, > 1 as a number has 1 as a cardinal,... > but why not we have not the cardinal -1, -2, ..., or 3/4?. This is no > stupid idea: essentially the division operation is the intention to, given > one cardinal a, divide a in cardinals of the same cardinal. It's under 6 > is 2a3: the cardinal 6 can divide in cardinals of the same cardinal: 2 > times cardinal 3. > So, if we divide 7 by 2 and we get 3.5, why we can't think that there are > sets of cardinal 3.5?. We can interpret these sets as fuzzy sets in any > way?. > No, cardinals (less than aleph_0) are defined as follows. 0 is the empty > set. 1 is s(0) or the {{0},0}. Two is s(1) or {{1},1}. Likewise s(a) is > defined as {{a},a}. You cannot have a cardinal between a and s(a), so 1/2 > has no natural cardinal. What does exist is the surreal numbers, invented > by John Conway. I have had the good fortune of learning this from him in > person. Imagiine that in the beginning, there was nothing, well there was > the empty set. On the zeroeth (he counts like that) day, he creates the set > containing the empty set. This is denoted { | } or 0. On the first day, he > creates +1 and -1. +1 is denoted {0| }, and -1 is { |0}. Now on the second > day, he creates -2, -1/2, 1/2, and +2. For example 1/2 is denoted {0|1}, so > {a|b} means the simplest number between a and b. A few days later he is > able to create your number 3.5: {3|4}. However, Conway was very patient > because he had to wait to day aleph_0 to create, say, 1/3, but his patience > payed off, for on that day he created all of the surreal numbers. For > example he had created the number sqrt(3)+pi*w, and it is differents from > pi*w. > Hope I answered your question. > Steven. > I know the construction of the rational numbers, but I want to view the > rational number from another point of view. > Xan. I know the construction of cardinals less than aleph_0, but this construction suppose the existence of the concept of set and implicitly the theory supposses that there are only integer (and positive) cardinals. The construction could be the most elaborated theory as you want, but the implicit ideas of the theory, the ideas that have all the mathematicions when (we or) they think about it, are simpler: the concept of the (finite) set have to be integer cardinal. Respecting very much John Conway, and from my knowless, I think that it's only a way to construct numbers, but not the way to contruct sets, because if we want to extend the idea of cardinal, I think that we have to extend the idea of set. And the problem is doing this without perversion-circle problem. Xan. === Subject: Re: What is codomain of cardinal function...? > Hi folks, A question. > We can define card:{A | A is a subset of Q}-->{0, 1, ..., aleph_0} as the > cardinal function restricted to the subsets of Q. This operator is a > whole function in sense that the domain and codomain are classicals sets > (we have not afraid with categories and etc...; we would do it with the > codomain the universal set). There is no problem there. But I point one problem: there are a natural > bijection between the natural numbers and it's cardinal: > 0 as number has 0 as a cardinal, > 1 as a number has 1 as a cardinal,... but why not we have not the cardinal -1, -2, ..., or 3/4?. This is no > stupid idea: essentially the division operation is the intention to, given > one cardinal a, divide a in cardinals of the same cardinal. It's under 6 > is 2a3: the cardinal 6 can divide in cardinals of the same cardinal: 2 > times cardinal 3. So, if we divide 7 by 2 and we get 3.5, why we can't think that there are > sets of cardinal 3.5?. We can interpret these sets as fuzzy sets in any > way?. > No, cardinals (less than aleph_0) are defined as follows. 0 is the empty > set. 1 is s(0) or the {{0},0}. Two is s(1) or {{1},1}. Likewise s(a) is > defined as {{a},a}. You cannot have a cardinal between a and s(a), so 1/2 > has no natural cardinal. What does exist is the surreal numbers, invented > by John Conway. I have had the good fortune of learning this from him in > person. Imagiine that in the beginning, there was nothing, well there was > the empty set. On the zeroeth (he counts like that) day, he creates the set > containing the empty set. This is denoted { | } or 0. On the first day, he > creates +1 and -1. +1 is denoted {0| }, and -1 is { |0}. Now on the second > day, he creates -2, -1/2, 1/2, and +2. For example 1/2 is denoted {0|1}, so > {a|b} means the simplest number between a and b. A few days later he is > able to create your number 3.5: {3|4}. However, Conway was very patient > because he had to wait to day aleph_0 to create, say, 1/3, but his patience > payed off, for on that day he created all of the surreal numbers. For > example he had created the number sqrt(3)+pi*w, and it is differents from > pi*w. > Hope I answered your question. > Steven. I know the construction of the rational numbers, but I want to view the > rational number from another point of view. Xan. > I know the construction of cardinals less than aleph_0, but this > construction suppose the existence of the concept of set and > implicitly the theory supposses that there are only integer (and > positive) cardinals. > The construction could be the most elaborated theory as you want, but > the implicit ideas of the theory, the ideas that have all the > mathematicions when (we or) they think about it, are simpler: the > concept of the (finite) set have to be integer cardinal. > Respecting very much John Conway, and from my knowless, I think that > it's only a way to construct numbers, but not the way to contruct > sets, because if we want to extend the idea of cardinal, I think > that we have to extend the idea of set. And the problem is doing this > without perversion-circle problem. > Xan. These are actually games, not sets. For example if you have 3 games of -1/3 and one game of 1, it is equivalent to the game of zero (a second player win). Likewise, if you a large finite amount of negative infitismals and one game of one, it is going to be a game where right wins. I don't think it is possible to have non whole numbered cardinals like you want. === Subject: is hausdorff dimension a topological invariant?? given two homeomorphic metric spaces, are their hausdorff dimensions identical? I have no idea what the answer is, so as always: proof or -- reverse my forename for mail! === Subject: Re: is hausdorff dimension a topological invariant?? > given two homeomorphic metric spaces, are their hausdorff dimensions > identical? No, not necessarily. For example, you can construct Cantor sets on the real line with each Hausdorff dimension between 0 and 1; just delete different-sized pieces instead of the middle third of each interval. === Subject: Re: is hausdorff dimension a topological invariant?? > given two homeomorphic metric spaces, are their hausdorff dimensions > identical? I have no idea what the answer is, so as always: proof or No. Two Cantor sets are always homeomorphic. But you can find two Cantor sets who have distinct Haussdorff dimensions. === Subject: Re: is hausdorff dimension a topological invariant?? -- > given two homeomorphic metric spaces, are their hausdorff dimensions > identical? I have no idea what the answer is, so as always: proof or > -- > reverse my forename for mail! It is not, as should be clear from the definition: the metric is used heavily. There are Cantor sets (so homeomorphic to the standard middle third set) that have about any Haudorff dimension tat you like... Henno Brandsma === Subject: Re: homoeomorphims between different dimensions? > Proof the following proposition or give a counterexample: > For all homoeomorphisms (not sure about the spelling, I mean an open > and closed bijective mapping) between two open subsets of (of perhaps > two different) euclidean vector spaces, the dimensions of the two > euclidean spaces are identical. Maybe some of you could be interested in a proof of this without algebraic topology: I read something about the concept of topological dimension and without working out the details I am pretty sure that this will work: homeomorphic sets must have identical top. dim., which is not the case for open subsets of euclidean spaces with different algebraic dimensions because the top. dim. coincides with the dim. as a vector space over R. === Subject: Re: homoeomorphims between different dimensions? > Proof the following proposition or give a counterexample: For all homoeomorphisms (not sure about the spelling, I mean an open > and closed bijective mapping) between two open subsets of (of perhaps > two different) euclidean vector spaces, the dimensions of the two > euclidean spaces are identical. Maybe some of you could be interested in a proof of this without > algebraic topology: I read something about the concept of topological > dimension and without working out the details I am pretty sure that > this will work: homeomorphic sets must have identical top. dim., which > is not the case for open subsets of euclidean spaces with different > algebraic dimensions because the top. dim. coincides with the dim. as > a vector space over R. As far as I know, proofs that R^m and R^n are not homeomorphic rely on algebraic topology. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: homoeomorphims between different dimensions? > As far as I know, proofs that R^m and R^n are not homeomorphic rely on > algebraic topology. That follows from say the Borsuk-Ulam theorem: if m < n, there is a map from S^m to R^n identifying no antipodal points; but not from S^m to R^m. You may say that the B-U theorem is algenaric topology, but there are tricks to prove it that don't make use of the standard apparatus of algebraic topology (homology/ homotopy groups etc.), for instance the Nullstellensatz based proof in Pfister's book on quadratic forms. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Jihad Neurosis (Was: Re: UFOs and Dark Energy) ----- Original Message ----- Astronaut Ed Lu returned last Monday from a six-month tour as > science officer on the International Space Station with many > memories and at least one nagging puzzle: what caused the > mysterious flashes of light he saw while studying Earth's aurora > from orbit? [...] > http://www.spaceref.com/news/viewsr.html?pid=10308 > .... So that > leaves us with a mystery... > http://www.spaceref.com/news/viewsr.html?pid=10308 > _________ http://www.crystalinks.com/antimatter.html > ... Matter and antimatter [...] > http://www.crystalinks.com/antimatter.html > _________ NOVA | The Elegant Universe | PBS > http://www.pbs.org/wgbh/nova/elegant/ > A Theory of Everything? > Some physicists believe string theory > may unify the forces of nature > http://www.pbs.org/wgbh/nova/elegant/everything.html > Viewpoints on String Theory > Steven Weinberg > http://www.pbs.org/wgbh/nova/elegant/view-weinberg.html > The Elegant Universe homepage > http://www.pbs.org/wgbh/nova/elegant/ > _________ The Religious Experience of Philip K. Dick by R. Crumb > http://www.philipkdick.com/weirdo.htm THE POWER of NOW > by Eckhart Tolle > http://www.eckharttolle.com/ > ... It is finding your true nature beyond > name and form. The inability to feel this connectedness > gives rise to the illusion of separation, from yourself > and from the world around you. You then perceive yourself, > consciously or unconsciously, as an isolated fragment. > Fear arises, and conflict within and without becomes > the norm. ... > -- Eckhart Tolle http://www.eckharttolle.com/ http://adidam.org/ > Yes! There is no religion, no Way of God, no Way of Divine > Realization, no Way of Enlightenment, and no Way of > Liberation that is Higher or Greater than Truth Itself. > .... Therefore, Reality (Itself) Is Truth, > and Reality (Itself) Is the Only Truth. > -- Adi Da Samraj, a.k.a. Bubba Free John, > a.k.a. Da Free John, a.k.a. Da Kalki, > a.k.a. the Ruchira Avatar, Adi Da Love-Ananda Samraj, > a.k.a. Franklin Jones > http://adidam.org/ > http://www.daplastique.com/home.html Let me share with you this little model I've worked out > about who we are as human beings. I call it the > Three-Plane Consciousness Model. [...] > That's the mystic I, because in Number Three there's > actually only one of us. Your Number Three isn't merely > like my Number Three---They're the same thing. > -- Baba Ram Das, a.k.a. Richard Alpert > http://ramdasstapes.org/index.htm > _________ Committee for Surrealist Investigation of Claims of the Normal > [ CSICON ] http://www.rawilson.com/csicon.shtml < C O N T A C T Uppaluri Gopala Krishnamurti (Born 9 July 1918) > http://www.well.com/user/jct/mystiq1.htm > THE MYSTIQUE OF ENLIGHTENMENT > [1968] SCIENTIFIC STUDY OF UNIDENTIFIED FLYING OBJECTS > Conducted by the University of Colorado > Under contract No. 44620-67-C-0035 With > the United States Air Force > Dr. Edward U. Condon, Scientific Director > [1968] > C O N T E N T S > http://ncas.sawco.com/condon/text/contents.htm Calls to Jihad Are Said to Lure Hundreds of Militants Into Iraq Don Van Natta Jr. and Desmond Butler http://209.157.64.200/focus/f-news/1012912/posts LONDON, Oct. 31 -- Across Europe and the Middle East, young militant Muslim men are answering a call issued by Osama bin Laden and other extremists, and leaving home to join the fight against the American-led occupation in Iraq, according to senior counterterrorism officials based in six countries. The intelligence officials say that since late summer they have detected a growing stream of itinerant Muslim militants headed for Iraq. They estimate that hundreds of young men from an array of countries have now arrived in Iraq by crossing the Syrian or Iranian borders. But the officials say this influx is not necessarily evidence of coordination by Al Qaeda or other terrorist groups, since it remains unclear if the men are under the control of any one leader or what, if any, role they have had in the kind of deadly attacks that shook Baghdad on Monday. A European intelligence official called the foreign recruits foot soldiers with limited or no training. A senior British official, who was in Iraq in September, said most of the foreign men captured there were from the Middle East -- Syria, Lebanon and Yemen -- or North Africa. He described them as young, angry men motivated by the anti-British, anti-American rhetoric that fills their ears every day. Signs of a movement to Iraq have also been detected in Europe. Jean-Louis Brugui.8fre, France's top investigative judge on terrorism, said dozens of poor and middle-class Muslim men had left France for Iraq since the summer. He said some of them appeared to have been inspired by exhortations of Qaeda leaders, even if they were not trained by Al Qaeda. Mr. Brugui.8fre, who earlier this year opened an investigation of young men leaving France to fight on the side of Muslims in Chechnya, said the traffic to Iraq was now a similar problem. He called the changing pattern a new threat. The rising agitation in parts of the Muslim world over the American-led occupation in Iraq was clear at Friday Prayers at Al Nur Mosque in a working-class section of Berlin. Dr. Izzeldin Hamad, the director of the Saudi-financed mosque, said political discussion was banned there. But outside, a 21-year-old man who identified himself as Akmed said that while Saddam Hussein was unpopular, now there are people who are angry about the American occupation. He and others said that inside the mosque, collections usually requested for Muslims in Palestine and Chechnya were now being offered for Iraq as well. An initial hint that Iraq would become a magnet for foreign recruits came just before the war began in March, with the arrest in Syria of four Algerian men, who had been living in Hamburg and attending a mosque frequented by three of the Sept. 11 hijackers. The authorities believed that the men intended to fight in Iraq. One of them, Abderazak Mahdjoub, whom German investigators have linked to a Spanish-based terror network, is under investigation for alleged involvement in a planned terror strike on a tourist location on the Costa del Bravo in Spain. Syria deported the men to Germany, but none of the four men is in custody, since there is no German law against going to Iraq. A senior German intelligence official, who spoke on condition of anonymity, said the authorities had detected other cases of immigrants in Germany trying to go to Iraq. We know that in Germany there are people in the militant Muslim scene who are willing to go other places to participate in jihad, including Iraq, the official said. There are scattered reports from other places, including Saudi Arabia, where a senior Saudi official said two Saudi militants, believed to have ties to Al Qaeda, were missing from the kingdom and believed by the authorities to have gone to Iraq. Intelligence officials, who base their assessment of the traffic into Iraq on surveillance of mosques and Islamic centers and on interrogations of terrorist suspects captured inside Iraq, say they have found no connections between the recruits. Nobody is organizing this move from Europe to Iraq, a senior European counterterrorism official said. At least it is difficult to analyze and know who is organizing this. This may be just the beginning of a new phenomenon. http://209.157.64.200/focus/f-news/1012912/posts ********** ********** http://216.239.41.104/search?q=cache:w6anT1ve2JYJ:209.157.64.200/focus/f-new s/978372/posts+&hl=en&ie=UTF-8 Al-Qa'ida Website, Back On-Line Publishes Book About its War on the U.S. and Bombing in Saudi Arabia Al-Qa'ida's website, which was originally called Al-Nida and then changed its name and URL every few days to avoid being hacked, had disappeared from the Internet following the killing in Saudi Arabia. The website has now been reactivated, at http://www.faroq.org/news/. Among the postings on the site is a book called The Raid of the 11th of Rabi' Al-Awwal - The Eastern Riyadh Operation and Our War on America and Its Agents. The book's foreword states, among other things, that the raid of attack on Western residential housing complexes in Riyadh]... was but the opening shot, Allah willing, and the Mujahiddeen had a need for this detailed communiqu.8e to present the reasons for the Jihad activity in the Arabian Peninsula and to remove some of the religious and military problems regarding it. The following are excerpts from each of the book's four chapters:(1) Chapter I: The Reality of the Islamic World ...The Muslim countries are today subject to overt or covert colonialism. The last century was the century of direct colonialism in the Muslim countries... By the end of the colonialist era, the colonialist states were no longer capable of withstanding the painful blows they were taking from the hands of the Mujahiddeen... At this point, Zionism intervened, [suggesting] that they would protect the colonialists' interests and rescue them from the complicated situation in which they found themselves. It concealed colonialism in the most naive way... It did not change the schemes of colonialism, rather the figures leading it: Why should we not comply with the desire of the resistance and replace those with blue eyes and blond hair with our own people, who speak our language, and wear sheepskins over a wolf's heart? Why should we not replace the name John or Napoleon with the name Muhammad, Anwar, or Abd Al-Aziz?... A 'Karzai' regime exists officially in all the Muslim countries. All rulers are crowned in the Karzai way... The legitimacy of any of these Karzais is no different than that of his brothers... The rulers of the land of the two holy places [Saudi Arabia] are no different than the others... The real ruler is Crusader America; the subjugation of these rulers [to America] is no different than the subjugation of district rulers to the king or president of their country... Anyone who fights them is in effect fighting the one who has given them authority and made them rulers over the Muslims... The breasts of the Jews - Allah's curses upon them - are filled with an arrogance that is not present in others, and therefore they have not settled for the kind of covert colonialism that satisfies the Crusader countries. Likewise, their occupation of Muslim Palestine stems from the belief that they cannot give it up, or else they would be apostates from their Judaism, exactly like the Arab rulers were apostates from Islam to which they belonged... The issue of Palestine is [actually] the issue of the Islamic world... but the Zionist media and the collaborating media neutralized the non-Arab Muslims by calling it 'the Arab issue.' Repeating [this term] has a powerful effect on the distortion of consciousness and eradication of the facts, and thus every non-Arab Muslim has been excluded of those interested in the Palestine issue... The establishment of the Palestinian state wrested the issue even from [the hands of] the countries of confrontation, and made it an issue of the Palestinian state and its treacherous government, headed by the vilest of agents ever in history - Yasser Arafat, who will get what he deserves from Allah... The situation of the armies in the collaborating countries proves clearly that resistance to aggression, primarily to the American [aggression], by these feeble armies is nearly impossible... The sensitivity of [the land of the two holy places, namely Saudi Arabia] requires it to have a powerful army - let alone that this is what all the Muslims are ordered to carry out, as Allah said: 'Make ready for them all armed strength and of mounted pickets that you can, whereby you may scare the enemy of Allah and your enemy'... Regarding the army of the government of the land of the two holy places, after consecutive years of profligate squandering of money that, they claimed, was going for arming it, when the country felt endangered by Saddam the rulers simply announced that the army was incapable of defending the country and brought in, in an act that was a precedent of its kind, the American Crusader armies in fortified bases, on the pretext of defending the country... While we see that the Israeli army includes over one million soldiers in a midget country whose inhabitants number six million, while its reserve army includes all men capable of bearing arms - the army in the land of the two holy places is the smallest and weakest [in the region] with regard to armaments and equipment and with regard to training and willingness to defend the religion and the honor... Chapter II: What is the Solution? ...We have presented some of the American crimes against the Muslims. Even if the religion [of Islam] were not stirring the Muslims to fight America, their [i.e. the Americans'] crimes would be enough to arouse in them [i.e. Muslims] the courage and gallantry to retaliate... [For] the Muslim countries in this situation - Jihad is what they need more than anything else, except food and water. Jihad is... a commandment that applies personally to every Muslim. Even if people disagree today regarding which of these two groups - the apostate traitor agents [i.e. the Arab governments] or the colonialist enemies [i.e. the Americans] - is more worthy to wage Jihad against, there need be no disagreement that Jihad is the solution for dealing with both of them... Although the Al-Qa'ida organization fights to defend the [Islamic] nation, it does not fight on the nation's behalf; therefore, anyone who cannot join Al-Qa'ida is not exempt from the obligation of Jihad [until] he has done everything possible to search for Jihad and did not succeed in joining any of its fronts, and after he has invested his best efforts, as he would in search for a medical specialist for a dread disease for someone dear to him... Chapter III: Why Riyadh? Many of those who esteem Jihad and the Mujahiddeen were amazed at and even condemned the Riyadh bombing, although they support bombings against America and American interests in other countries. Therefore, we would like to answer the question that [was] on the lips of many: Why Riyadh? Anyone who asks this question must recall the religious texts that command fighting all polytheists. Allah said: ... 'When the four months in which fighting is forbidden have passed, fight the polytheists everywhere you find them...' We found the Americans in Riyadh, and killed them in Riyadh. Similarly, it must also be noted that in its war with America, the Al-Qa'ida organization adopted the strategy of expanding the battle arena... This strategy has priceless advantages; the enemy who had only his country to defend realized that he now must defend his enormous interests in every country... While this strategy might cause some damages to Muslims in the process of defending the nation, this happens all the time, and in every Jihad. The Afghans suffered long years of war because they stood up to the Communist invasion that sought to conquer, among other things, the land of the two holy places, and that even managed to set foot in the southern Arabian Peninsula by means of the Communist country in Southern Yemen that is now extinct. Even the collaborating countries operate according to this rule... The government of the land of the two holy places charges [its] public for the expenses of its attack on the Mujahiddeen in the so-called war on terror... on the pretext that there is no other way to implement the interest of the people. It is better that some damage be caused to the people in the interests of religion than in the interests of establishing the thrones of the despots and agents... Chapter IV: The Eastern Riyadh Operation On the 11th of Rabi' Al-Awwal, 1424, a group of the youths of Islam set out and attacked Crusader complexes in eastern Riyadh, in one of the highest quality operations that even a few American officials were forced to admit was a well-planned commando operation... This operation reminded the Americans that they cannot dream of security before the Muslims in Palestine experience it, and before all Crusader countries leave the peninsula of the Prophet Muhammad... In the past, it was agreed between the collaborating government in the land of the two holy places and America that these complexes would be part of the U.S. The Americans have religious freedom in them; nothing is forbidden to them. There are churches there, bars, dance clubs, mixed swimming pools, and all sorts of heresy. They are not subject to [Islamic] religious law; furthermore, they are not subject to the sovereignty of the [Saudi] government itself. The police and security forces do not enter there, and neither does the Commission for the Promotion of Virtue and Prevention of Vice [i.e. the Saudi religious police]... The number [of those carrying out the bombing] reported by the press is inaccurate. Some of them who were not predestined for martyrdom continue to cause losses to the enemies of Allah, the Americans, in Iraq, and carried out acts of heroism there... The Americans and the agent governments that support them, like the Karzai government in Afghanistan, the [Pervez] Musharraf government in Pakistan, the Fahd government in the land of the two holy places, and the Ali Abdallah Saleh government in Yemen, are permissible targets for Mujahiddeen, according to [Islamic] religious law... They [i.e. the Arab governments] and the Americans are identical, both in their war on [Islam] and in the extent to which they are a target for the Mujahiddeen... Endnote: (1) http://www.faroq.org/news/news.php?id http://216.239.41.104/search?q=cache:w6anT1ve2JYJ:209.157.64.200/focus/f-new s/978372/posts+&hl=en&ie=UTF-8 > The Condon Report - 1968 CE > http://ncas.sawco.com/condon/index.html > http://www.csicop.org/klassfiles/posner_klass.html > .... You see, I don't know how to put it. Never does > the thought that I am different from anybody come into my > consciousness. [...] > ((({}))) > Continued at: The Archetype and the Beast '98 > http://pw1.netcom.com/~mthorn/arcbeast.htm > [~][^][~] > Disingenuous Demagogues Deteriorate Daily > All Politicians are Demagogues, yet not all > Demagogues are Politicians... > ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ > Sagittarius assimilated by Milky Way: > Resistence Is Futile! > http://www.astro.virginia.edu/~mfs4n/sgr/ > ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ > FREE THE YEZIDI!!! > http://www.songsouponsea.com/Promenade/GnosisE.html The Purification will begin > shortly after humans build > a great house in the sky. > By then there will be fires > everywhere and greedy, selfish, > power-mad leaders, internal wars. > http://www.deoxy.org/omega.htm Yes, and you can't escape it by closing your eyes either, > as your whole 'being' is thrust into a powerful spotlight > where every photon is a gently tugging, beaming consciousness; > searing traumatically a heightened essence at work on your > subjective optical/emotional/cognitive space. It's an > unwavering, unflinching wake-up call to the brain. > Reality is infinite, and sometimes untenable, during & > afterwards, upon return to the so-called mundane plane > of existence, looping, inextricably bound by gravitational > forces. We should, could, maybe, love each other a bit > more as we're surrounded by strangeness imbedded in > a whirling biosphere of fellow travelers. > Heading either Somewhere new, or Nowhere... > _________ > STARS and STRIPES: http://www.estripes.com/ > Report pending on rising rate of GI suicide in Iraq > _________ > Why? The Neuroscience of Suicide > Despite the inconsistencies, the bulk of evidence points strongly to a > problem in the brains of suicides involving the serotonin system. > [Cont... ] > _________ > Meme: (pron. 'meem') A contagious idea that > replicates like a virus, passed on from mind > to mind. Memes function the same way genes > and viruses do, propagating through > communication networks and face-to-face > contact between people > David S. Bennahum. > _________ > What a tragic world this is, he reflected. > Those down here are prisoners, and the > ultimate tragedy is that they don't know it; > they think they are free because they never > have been free, and do not understand what > it means. This is a prison, and few men have > guessed. But I know, he said to himself. > Because that is why I am here. To burn the > walls, to tear down the metal gates, to > break each chain. Thou shalt not muzzle > the Ox as he treadeth out the corn, he > thought, remembering the Torah. > The Divine Invasion, Phillip K Dick. > HarperCollins 1981. > http://www.philipkdick.com/weirdo.htm > _________ > ... The Inuit of northern Canada have one of the highest suicide > rates in the world, Yale University research affiliate Michael > Kral found in a study based on a 1998 visit to the Arctic Circle. > http://www.suicideinfo.ca/csp/go.aspx?tabid=59 > _________ > In (... accumulated spirits of trauma...) > &$&$&$&$&$& $&$&$&$&$&$ &$&$&$&$&$&$[Sterling ]&$&$ > [{snip}] An Atlas of The Universe > http://www.anzwers.org/free/universe/ SDSS: Sloan Digital Sky Survey > http://www.sdss.org/ WMAP: Wilkinson Microwave Anisotropy Probe > http://map.gsfc.nasa.gov/ http://lambda.gsfc.nasa.gov/ The Origins of the Fear of Death and Dying > http://primal-page.com/death.htm ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ > Hilarious, terrifying & brilliant. > One of TCB's absolute best! I laughed... I cried... > I excommunicated my DNA! http://pw1.netcom.com/~mthorn/tcbkrsna.htm ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ > [{snip}] The Religious Experience of Philip K. Dick by R. Crumb > http://www.philipkdick.com/weirdo.htm Parallel Universes [...] http://www.sciam.com/issue.cfm > Not just a staple of science fiction, other universes > are a direct implication of cosmological observations > By Max Tegmark w e a r e a l l a n o m a l o u s > http://www.sparkchamber.co.uk/quantum.html *********************** > International Space Station > NASA Human Space Flight > http://spaceflight.nasa.gov/home/index.html *********************** > Dreaming: A Neurocognitive Approach > http://munshi.4t.com/wbt/stickgold.html > http://www.daplastique.com/suite_images/SIM_Entry_254.jpg Dr Paul Bennewitz > http://exopolitics.org/Dulce-Report.htm Bennewitz's report, Project Beta > http://www.paraarchives.com/documents/p/beta01.htm > Project Beta - Alien Command and Control Alien Command and Control > Because of the aliens apparent logic system, a key decision > cannot be made without higher clearance. All are under, yet > it would appear that even this is not the final authority. > Delays as long. How long the time delay is during battle > conditions is uncertain. Because of this apparent control, > individual decision making by the 'greys' are limited. > If the 'plan' goes even slightly out of balance or context, > they become confused. Faced with this, possibly, the humanoids > they have created would be the first to break and run. Psychological Aspects > Psychologically, at present, their morale is down. There is > pronounced discension in the ranks- even with the humanoids. > Communication can encourage this problem due to their own > internal vulnerability mind-wise, and therein lies a prime > weakness.... > CSICOP > Box 703 > Amherst, NY, 14226 > 716-636-1425 > subscriptions@csicop.org > Problems with your magazine subscription > press@csicop.org > Media inquiries > donors@csicop.org > Donations to CSICOP > europe@csicop.org > European representative of Skeptical Inquirer > letters@csicop.org > Letters to the editor of Skeptical Inquirer > Copyright requests > info@csicop.org > General inquiries > webmaster@csicop.org > Problems with the website > Secure On-line Store, or 800-634-1610 > Subscribe to Skeptical Inquirer magazine, > or become a CSICOP associate and subscribe > to the Skeptical Briefs newsletter. === Subject: turbulence and drag in motion math want to add turbulence and drag to the motion. I don't know if those are the right words to use, but what I'm looking for is a randomized variation in motion similar to turbulence and drag in air or water. Following are the equations I'm using x = velocity*cos(angle)*time + (wind*(time)^2) y = velocity*sin(angle)*time + (gravity*(time)^2) ...where wind is an arbitrary force from the side and gravity is an arbitrary downward force. This is for a simulation, so appearing as the viewer expects it to appear is more important than accuracy. I'm also limited in the number of iterations of time I can use (usually under 100), so it doesn't need to be all that precise either (to 1 decimal place). Where do I begin to understand how turbulence might be added to this system? Do I think of it only as a random variation in position, and if so, how do I calculate the amount of variation on x and y, since it's not uniform? Is it some kind of field, causing randomized acceleration and deceleration, and if so, how would I map that field? air, and how they roil as they rise, or how a mushroom cloud, well... mushrooms. How do I go about incorporating that type of motion into the above equations? If this relates to spin I also have a value for spin in degrees per unit of time, which is currently being used to rotate the I just don't know how to approach these problems, and lack the references to even begin looking into it. Any links to online explanations or math tutorials, examples, or just relatively simple explanations would be greatly helpful. I haven't been in a classroom in more than 10 years, but I have always been good at math (particularly algebra and trig), so while it doesn't need to be a 3rd grade level explanation, it does need to be simple enough that someone who isn't studying math can grasp it. TIA, ryanm === Subject: Re: turbulence and drag in motion math To a first approximation, you could model drag as a force which increases as the square of the airspeed. I > want to add turbulence and drag to the motion. I don't know if those are > the right words to use, but what I'm looking for is a randomized variation > in motion similar to turbulence and drag in air or water. Following are the > equations I'm using > x = velocity*cos(angle)*time + (wind*(time)^2) > y = velocity*sin(angle)*time + (gravity*(time)^2) > ...where wind is an arbitrary force from the side and gravity is an > arbitrary downward force. This is for a simulation, so appearing as the > viewer expects it to appear is more important than accuracy. I'm also > limited in the number of iterations of time I can use (usually under 100), > so it doesn't need to be all that precise either (to 1 decimal place). > Where do I begin to understand how turbulence might be added to this > system? Do I think of it only as a random variation in position, and if so, > how do I calculate the amount of variation on x and y, since it's not > uniform? Is it some kind of field, causing randomized acceleration and > deceleration, and if so, how would I map that field? the > air, and how they roil as they rise, or how a mushroom cloud, well... > mushrooms. How do I go about incorporating that type of motion into the > above equations? If this relates to spin I also have a value for spin in > degrees per unit of time, which is currently being used to rotate the > I just don't know how to approach these problems, and lack the > references to even begin looking into it. Any links to online explanations > or math tutorials, examples, or just relatively simple explanations would be > greatly helpful. I haven't been in a classroom in more than 10 years, but I > have always been good at math (particularly algebra and trig), so while it > doesn't need to be a 3rd grade level explanation, it does need to be simple > enough that someone who isn't studying math can grasp it. > TIA, > ryanm === Subject: Re: turbulence and drag in motion math > want to add turbulence and drag to the motion. I don't know if those are > the right words to use, but what I'm looking for is a randomized variation > in motion similar to turbulence and drag in air or water. Following are the > equations I'm using > x = velocity*cos(angle)*time + (wind*(time)^2) > y = velocity*sin(angle)*time + (gravity*(time)^2) > ...where wind is an arbitrary force from the side and gravity is an > arbitrary downward force. This is for a simulation, so appearing as the > viewer expects it to appear is more important than accuracy. I'm also > limited in the number of iterations of time I can use (usually under 100), > so it doesn't need to be all that precise either (to 1 decimal place). Accuracy isn't a huge deal, but I still suspect you'll get better results if you use a more physically-correct model. That would be modeling forces as things that cause accelerations, which means that at every time t you do this: ax = sum of forces in x direction ay = sum of forces in y direction vx = vx + ax*dt vy = vy + ay*dt x = x + vx*dt y = y + vy*dt Now this kind of iteration is notoriously unstable (i.e. the errors get progressively worse over time) but over 100 iterations you should be fine. > Where do I begin to understand how turbulence might be added to this > system? Do I think of it only as a random variation in position, and if so, > how do I calculate the amount of variation on x and y, since it's not > uniform? Is it some kind of field, causing randomized acceleration and > deceleration, and if so, how would I map that field? You could model both wind and turbulence as random forces, but perhaps wind is more horizontal and has more correlation from time to time, where turbulence is independent from time to time. > Where do I begin to understand drag? Usually as a force which is proportional to velocity, or to the square of velocity. > I just don't know how to approach these problems, and lack the > references to even begin looking into it. Any links to online explanations > or math tutorials, examples, or just relatively simple explanations would be > greatly helpful. I haven't been in a classroom in more than 10 years, but I > have always been good at math (particularly algebra and trig), so while it > doesn't need to be a 3rd grade level explanation, it does need to be simple > enough that someone who isn't studying math can grasp it. Probably you should look at resources designed for game designers, who are often people looking to model realistic physics but with very little physics or math background. For instance, googling for game physics gave me this: as well as a bunch of references to books. - Randy === Subject: Re: Another optimization question Peter or whomever is knowledgeable, > no numerical procedure can reasonably deal with infinity. hence you must > use a barrier-approach (adding terms like -log(g_i(x)) for each function > g_i which describes a boundary part of C as g_i(x)=0 , with g_i(x)>0 in > interior(C), provided such interior exists. otherwise you first must reduce > the problem to a subspace where this is the case) . next you could use > unconstrained optimization using only function values (e.g. uobyqa from > powell or its newer successor, described in a recently published paper in > but you must modify this code in order to maintain feasiblitiy, that means you > must restrict the possible moves such that feasibility is maintained > (via checking the g_i individually). > also a grid-search a la torczon comes into mind, again with the moves > restricted to the feasible part of the grid. do the Powell, Torczon etc derivative free direct search methods also work if the actual minimizer lies on the boundary of the feasible instead of being an interior point of the feasible set and if the minimizer is not unique? above, but i will soon). cheers, J. === Subject: Re: Another optimization question >Peter or whomever is knowledgeable, >> no numerical procedure can reasonably deal with infinity. hence you must >> use a barrier-approach (adding terms like -log(g_i(x)) for each function >> g_i which describes a boundary part of C as g_i(x)=0 , with g_i(x)>0 in >> interior(C), provided such interior exists. otherwise you first must reduce >> the problem to a subspace where this is the case) . next you could use >> unconstrained optimization using only function values (e.g. uobyqa from >> powell or its newer successor, described in a recently published paper in >> but you must modify this code in order to maintain feasiblitiy, that means you >> must restrict the possible moves such that feasibility is maintained >> (via checking the g_i individually). >> also a grid-search a la torczon comes into mind, again with the moves >> restricted to the feasible part of the grid. > do the Powell, Torczon etc derivative free direct search methods >also work if the actual minimizer lies on the boundary of the feasible >instead of being an interior point of the feasible set and if the > minimizer is not unique? > above, but i will soon). >cheers, >J. if I understood your problem correctly, you cannot evaluate the objective function outside the feasible domain. therefore I proposed using the interior point approach. if your constraints are linear then you can use the Torczon grid search too, with proper consideration of feasibility. (for the nonlinear constraints case there exists an analysis of dennis & audet, whose results are not completely satisfactory however. indeed things are complicated here because the directions of the grid must be related to the orientation of the boundary. powells approach uses interpolation and you cannot restrict the interpolation to lower dimensional subsets. It should however be possible to restrict interpolation to boundary and interior points. (not considered in the original paper) if the minimizer is nonunique _locally_ (that means you have a continuum of minimizers, this will not impeed torczon but powell, since the powell model assumes a locally unique minimizer . hth peter === Subject: Equation Object converting? I got a big MS Word document (Word 2.0) with a lot of mathematical equations. The equations does not look properly, so I have to ->->->. To do that one equation per converting takes a time. My question: How I may convert all equations object at once? === Subject: Re: Equation Object converting? Probably can do so with a quick macro. To what version of Word are you converting? -- http://www.standards.com/; See Howard Kaikow's web site. > I got a big MS Word document (Word 2.0) with a lot of mathematical > equations. The equations does not look properly, so I have to Object>->->->. To do that > one equation per converting takes a time. > My question: How I may convert all equations object at once? === Subject: Re: Two coin flip/ clarification for C Bond > [...] Out of curiosity, do you think that Dr. Ullrich's statement was > complimentary? >Oh no! His favorite expression for me is 'dumb '. Not >>complimentary. >> Classic misrepresentation. Locate the first instance you can of me >> referring to you as a dumb and I'll show you the previous posts >> in which you called me the same and/or worse. >> Dishonest as well as dumb - for _you_ to complain about >> people being less than complimentary is very funny. >>He underestimates my intelligence and refuses to >>consider my argument. His mistake. He's on the naive side and doesn't >>wish to admit it. >> ************************ >> David C. Ullrich >I told him you were not complimentary to me. I rest my case. > Tee-hee. I suspect that the many people who've been complaining > that you don't seem to read more than a few key words in messages > before replying to them are resting _their_ cases at this point as > well... The question asked was whether or not I thought you were complimenting me? I honestly didn't think you were. You, as usual, paid no attention to the question which was asked. Eldon Moritz >Eldon Moritz > ************************ > David C. Ullrich === Subject: Re: Two coin flip/ clarification for C Bond >[...] can't fill in >> that blank, prior to some kind of inspection, or 'look'. >> The short answer, which I'm sure you've been given many times, is >> simply Sorry, Chum, no, we don't mean the statement with a blank was >> made, and then the blank was unprejudicially filled. If we _did_ mean >> that, of course your value of 50% is correct. But we hereby clarify >> that that is _not_ what we mean. End of discussion. >> We need to clarify who _we_ is. >> I hope it's not too presumptuous of me to inform you that by we I >> mean, er, everyone involved in the discussion except yourself. >I'll accept that, except that we don't have a concensus. If we >did, it wouold be even more embarrassing, as, either Moritz is >absolutely correct, or, he's wrong. [...] >> [...] >No, it's a little probability question, and the majority have it >wrong. We are answering the question which we think the questioner >meant. Moritz is by no means alone in his quest. [...] >First, it's a matter of right and wrong. Then, Moritz has spent a lot >of time on this question. He understands it well. Maybe, if we get >off our high horse, and pick his brain a little, we can get more out >of it than just this one question. [...] >We know that there had to be a look. Was court card decided prior to >the look? If so, how were we supposed to know, we weren't told about >it. This one gets a little ambiguous. Moritz doesn't say that >ambiguity doesn't exist, he just says that our question isn't >ambiguous. >At first gasp Moritz thinks the complement would be, Two decks were >cut and* at least one is not a court card. [...] >Here, we make false statement. Listen to Moritz, he'll show you what >can be said prior to the look. [...] > An off-topic comment, but one you should be interested in. As has > been explained to you before, when you refer to yourself in the > third person this way it comes off sounding _extremely_ bizarre. Doctor Ullrich, I'm writing this real slow. (I know that you think slow). I'm bizarre, I'm extremely bizarre, that's okay with me. The truth of my argument is my only concern. You have dodged my argument, all you do is kibitz about me. The following is my argument: Assumptions: Heads and Tails are equally likely. The problem statement is true. Question: Two coins were flipped and at least one is a head. What are the chances for two heads? Solution: I. Heads and tails were equally likely. II. The problem statement is true. III. The problem statement is complete. IV. There must have been an inspection of at least one coin. V. The statement gives no evidence of inspecting more than one coin. To flip two coins three ways, both coins must be inspected. VI. The flip was a first time event. VII. The statements at least one is a head and at least one is a tail were equally likely. VIII. To get an answer other than one half, heads must have been chosen, prior to inspection.. IX. To get an answer other than one half, there must be more information. The answer is 1/2. Either my argument is correct, or, one of those nine statements is wrong. Prove one wrong and I'll send the $1000 to the American Heart Association in your honor. We'll let Mitchell Jones be the referee. Eldon > ************************ > David C. Ullrich === Subject: Re: Two coin flip/ clarification for C Bond >>[...] can't fill in > that blank, prior to some kind of inspection, or 'look'. The short answer, which I'm sure you've been given many times, is > simply Sorry, Chum, no, we don't mean the statement with a blank was > made, and then the blank was unprejudicially filled. If we _did_ mean > that, of course your value of 50% is correct. But we hereby clarify > that that is _not_ what we mean. End of discussion. >> We need to clarify who _we_ is. I hope it's not too presumptuous of me to inform you that by we I > mean, er, everyone involved in the discussion except yourself. >I'll accept that, except that we don't have a concensus. If we >>did, it wouold be even more embarrassing, as, either Moritz is >>absolutely correct, or, he's wrong. [...] > [...] >>No, it's a little probability question, and the majority have it >>wrong. We are answering the question which we think the questioner >>meant. Moritz is by no means alone in his quest. [...] >First, it's a matter of right and wrong. Then, Moritz has spent a lot >>of time on this question. He understands it well. Maybe, if we get >>off our high horse, and pick his brain a little, we can get more out >>of it than just this one question. [...] >We know that there had to be a look. Was court card decided prior to >>the look? If so, how were we supposed to know, we weren't told about >>it. This one gets a little ambiguous. Moritz doesn't say that >>ambiguity doesn't exist, he just says that our question isn't >>ambiguous. >>At first gasp Moritz thinks the complement would be, Two decks were >>cut and* at least one is not a court card. [...] >Here, we make false statement. Listen to Moritz, he'll show you what >>can be said prior to the look. [...] >> An off-topic comment, but one you should be interested in. As has >> been explained to you before, when you refer to yourself in the >> third person this way it comes off sounding _extremely_ bizarre. >Doctor Ullrich, >I'm writing this real slow. (I know that you think slow). >I'm bizarre, I'm extremely bizarre, that's okay with me. The truth of >my argument is my only concern. You have dodged my argument, all you >do is kibitz about me. Uh, no. I've explained very clearly what I think of your argument, many times. The fact that you're willing to repeat the argument ad infinitum, regardless of what anyone says about it, but I'm not interested in infinitely many repetitions of what I have to say about it, does not imply that I've dodged it. Regarding the kibitzing about you, it's just something I thought you might want to know. Not that you're going to convince people you're Right regardless, but I thought you might want to know that talking about yourself in the third person like that doesn't help - it makes >The following is my argument: >Assumptions: >Heads and Tails are equally likely. >The problem statement is true. >Question: > Two coins were flipped and at least one is a head. What are the >chances for two heads? >Solution: >I. Heads and tails were equally likely. >II. The problem statement is true. >III. The problem statement is complete. >IV. There must have been an inspection of at least one coin. >V. The statement gives no evidence of inspecting more than one coin. >To flip two coins three ways, both coins must be inspected. >VI. The flip was a first time event. >VII. The statements at least one is a head and at least one is a >tail were equally likely. >VIII. To get an answer other than one half, heads must have been >chosen, prior to inspection.. >IX. To get an answer other than one half, there must be more >information. The answer is 1/2. >Either my argument is correct, or, one of those nine statements is >wrong. Prove one wrong and I'll send the $1000 to the American Heart >Association in your honor. >We'll let Mitchell Jones be the referee. >Eldon >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Two coin flip/ clarification for C Bond >>To flip two coins three ways, both coins must be inspected. I still can't make head nor tail of this, but as there are three ways mentioned, perhaps I should say I can't make head nor tail nor edge of this. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Two coin flip/ clarification for C Bond Eldon Moritz grava .88 la saucisse et au marteau: > Question: > Two coins were flipped and at least one is a head. What are the > chances for two heads? If you understand French, I suggest you go see http://www.eleves.ens.fr:8080/home/madore/math/proba.html which deals about all the traps in probability. A problem similar to yours is treated. I'll try to translate it into English: I visit the Martin's Family. I know that the parents have two children. I ring at the door and a girl answers. What is the probability for the other child to be a kid? The natural answer is 1/2. The sophisticated one is No, because there are four likewise solutions for the composition of the family (G/G, G/B, B/B, B/G) and the fact that a girl answered excludes the third possibility, which makes the probability for the other child to be a boy be 2/3 and 1/3 for a girl. Explication: The problem is too vague to determine which of the solutions is correct. If we say the older child answers to the door and it's a girl or any of the child (with the same probability) opens the door and it's a girl, then the first proposed solution is correct. If, on the other hand, one says boys never answer to the door and I've been answered to the door or boys answer to the door only if there is no girl in the house, and a girl has answered, then the second solution is correct. I tend to think the first solution is correct, but without any further information, we can't answer to this question. To sum up: Among families with two children, one of which is a girl, 2/3 have a boy for other child. But among families with two children, the oldest being a girl, 1/2 have a boy for second child. hth -- Nicolas === Subject: Re: Two coin flip/ clarification for C Bond > Eldon Moritz grava .88 la saucisse et au marteau: > Question: > Two coins were flipped and at least one is a head. What are the > chances for two heads? > If you understand French, I suggest you go see > http://www.eleves.ens.fr:8080/home/madore/math/proba.html which deals > about all the traps in probability. > A problem similar to yours is treated. I'll try to translate it into > English: > I visit the Martin's Family. I know that the parents have two children. > I ring at the door and a girl answers. What is the probability for the > other child to be a kid? You meant for the other child to be a girl, I think. We know this problem in English, and Eldon is similarly obsessed with it and has his own ideosyncratic interpretation. > The natural answer is 1/2. > The sophisticated one is No, because there are four likewise solutions > for the composition of the family (G/G, G/B, B/B, B/G) and the fact that > a girl answered excludes the third possibility, which makes the > probability for the other child to be a boy be 2/3 and 1/3 for a girl. My psychic powers tell me that Eldon will tell you there is only one possible interpretation of the problem, it is one not shared by anyone on earth, and it results in a probability of 1/2. - Randy === Subject: Re: Having trouble with sine > This sounds like a standard sound-synthesis exercise. If you need > more help, try looking for a textbook on computer music. Yes, you guessed it right away. I just didn't want to write it as if it was a computer homework problem-- I'm doing all that myself. > If you want table[0]=127 and table[1]=128, you get 1=127(sin(2*pi/N)), > which you can solve (approximately) for N. This has made the most sense. I tried that with, L(x) = f(a) + f'(a)(x - a) and I got L(x) = x when a = 0. Then, and T(1) = 127 (L(x)) = 1 (+ 127 since T(0) = 127) So 127x = 1, x = 1/127, thus T(1) = sine(1/127) * 127. If I continued approximating along this tangent I imagine every increment in x by x = n/127 and n is an integer, will make T change by 1. But I want it to behave like sine, the slope getting smaller, so I just sample sine instead, and this will lead to the points that produce a change in T. This method works if I want a table whose values can assume integers clarifications, I really appreciate it, this problem was bugging me for so long... Now I feel very relieved! -HG -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Probability of a Run >If you mean sci.math, it is not and has never been moderated; >sci.math.research is moderated, and AFAIK its moderators are very >much alive. >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 question? You are one of the web' s best resources on probability and I know that you have a firm grasp on The Probability of Runs. === Subject: Re: Probability of a Run >question? You are one of the web' s best resources on probability and >I know that you have a firm grasp on The Probability of Runs. The question being: |I am trying to calculate the probability that a gambler with capital C |and who uses a Martingale betting strategy will be wiped out in m |turns at the game. This would happen with a run of n consecutive |losses and I want to calculate the probability of this. |The textbooks treat this problem at an advanced level, invoking |generating functions or difference equations, but I wonder if a |solution satisfactory for the purpose could be arrived at in a simpler |way. All I need is the probability that there would be AT LEAST one |run of length AT LEAST n. |If in n independent trials the probability of a loss at a single trial |is q, the probability of all losses in n trials would be q^n. A string |of n losses could begin at any trial from the first up to the m - n + |1 th, so would the probability be (m - n + 1) * q^n? Brief answer: no, because these are not mutually exclusive events. The correct probability is rather complicated: so there's a reason for the advanced level treatment! There have been a number of threads on this topic in sci.math, e.g. Consecutive runs in Bernoulli trials in June 1997. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: minimum foam Hi Stephen >Does anyone disagree that space fills with irregular tetrahedra >such that each vertex is shared by 20 tetradrahedra? NO? Okay. >Then.... Place a vertex at the center of each tetrahedron in that maximum >foam and connect the dots. Each vertex has 4 edges connected. Voila! >Space fills with irregular pentagonal dodecahedra. Minimum foam. Yes? No? >Does anyone know if space fills with irregular pentagonal dodecahedra? >Why or why not? > If you go to 4 dimensions, then the tetrahedra can be regular. If I 3 go with dimensions, can I fill space with irregular tetrahedra such that all edges share 5 tetrahedra and all vertexes share 20 tetrahedra? > If you try to do this with distortion in Euclidean 3-space, you will > get a distorted version of the same result. It is analogous to > trying to tile the plane with triangles, while allowing only five > around each vertex, or tiling with three pentagons around each vertex. So your saying, yes, irregular tetrahedra and irregular dodecahedra are both space fillers? Dick === Subject: Is there a name for this series? I'm studying a series that is generated by the product of a number's arithmetic subparts. Given N = X + Y The series is generated by the product of X and Y, for all posible non-negative values for X and Y bounded by N. As an exmple: N = 10 X Y XY 1 9 9 2 8 16 3 7 21 4 6 24 5 5 25 6 4 24 7 3 21 8 2 16 9 1 9 10 0 0 Is there a formal name for this series of XY? My ultimate quest is to understand the relationships between series for different values of N, and to find efficient algorithms (if any) for performing Union, Intersection, Subtraction, etc... between them I'm not looking for quick answers, but rather just the guidance to steer me in the right direction. Any help will be much appreciated. --MathNeophite-- === Subject: Re: Is there a name for this series? > I'm studying a series that is generated by the product of a number's > arithmetic subparts. > Given N = X + Y > The series is generated by the product of X and Y, for all posible > non-negative values for X and Y bounded by N. > As an exmple: > N = 10 > X Y XY > 1 9 9 > 2 8 16 > 3 7 21 > 4 6 24 > 5 5 25 > 6 4 24 > 7 3 21 > 8 2 16 > 9 1 9 > 10 0 0 > Is there a formal name for this series of XY? > My ultimate quest is to understand the relationships between series for > different values of N, and to find efficient algorithms (if any) for > performing Union, Intersection, Subtraction, etc... between them > I'm not looking for quick answers, but rather just the guidance to steer > me in the right direction. > Any help will be much appreciated. I will change your notation slightly. For some fixed N, the two numbers that you multiply together are x and N-x. If we call this product y, we have the equation y=x(N-x). For example, if N=10 and x=2 we have y=2*(10-2)=2*8=16. So far you have, perhaps, only let x be an integer such that 1<=x<=N. But we might choose other values of x, for example (for N=10) if x=-1.2, y=-1.2(10-(-1.2))=-1.2*11.2=-13.44 x=0.5, y=0.5*9.5=4.75 x=20, y=-200 For some N, you might plot the graph of y against x. This would be a parabola. Hope this helps. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Is there a name for this series? >>I'm studying a series that is generated by the product of a number's >>arithmetic subparts. >>Given N = X + Y >>The series is generated by the product of X and Y, for all posible >>non-negative values for X and Y bounded by N. >>As an exmple: >>N = 10 >>X Y XY >>1 9 9 >>2 8 16 >>3 7 21 >>4 6 24 >>5 5 25 >>6 4 24 >>7 3 21 >>8 2 16 >>9 1 9 >>10 0 0 >>Is there a formal name for this series of XY? >>My ultimate quest is to understand the relationships between series for >>different values of N, and to find efficient algorithms (if any) for >>performing Union, Intersection, Subtraction, etc... between them >>I'm not looking for quick answers, but rather just the guidance to steer >>me in the right direction. >>Any help will be much appreciated. > I will change your notation slightly. > For some fixed N, the two numbers that you multiply together are x and N-x. > If we call this product y, we have the equation y=x(N-x). For example, if > N=10 and x=2 we have y=2*(10-2)=2*8=16. > So far you have, perhaps, only let x be an integer such that 1<=x<=N. But we > might choose other values of x, for example (for N=10) > if x=-1.2, y=-1.2(10-(-1.2))=-1.2*11.2=-13.44 > x=0.5, y=0.5*9.5=4.75 > x=20, y=-200 > For some N, you might plot the graph of y against x. This would be a > parabola. > Hope this helps. It's clear to me now that I'm looking at parabolas. To get to my ultimate quest mentioned above, I will move on to studying the laws governing parabolas. --MathNeophite-- === Subject: Re: Is there a name for this series? > I'm studying a series that is generated by the product of a number's > arithmetic subparts. > Given N = X + Y > The series is generated by the product of X and Y, for all posible > non-negative values for X and Y bounded by N. > As an exmple: > N = 10 > X Y XY > 1 9 9 > 2 8 16 > 3 7 21 > 4 6 24 > 5 5 25 > 6 4 24 > 7 3 21 > 8 2 16 > 9 1 9 > 10 0 0 So it looks like you are picking N, then generating X*(N-X) for X = 0, 1, ..., N The values of X*(N-X) are points on a parabola which crosses the X axis at 0 and N, and has a maximum of (N/2)^2 at X = N/2. - Randy === Subject: Re: JSH: Expanding out proof >>Ok, now for yet another test. Some people are still trying to >>challenge my proof, and I've seen posts that indicate that some of the >>usual suspects think that they are still winning, >> some posts with this property? How many posts have you >> seen that agree you're right? >What the does agreement matter? If agreement doesn't matter then why does it seem to bother you that _nobody_ agrees that you're right? (Note qualifications on nobody in the original post). >You are a low-life David Ullrich, who clearly couldn't care less about >mathematics, as you seem to live in some political world, where all >that matter is what people think, and not what's mathematically >correct. >You're a goddamn anti-mathematician David Ullrich. Some weird monster >dressed up with tenure at Oklahoma State University, free to infect >young minds. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Expanding out proof ... > Well, let's see if you can be reasonable Dik Winter. Oh, I always am. > 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > coprime to 22. Therefore, *none* of the constant terms can have 7 as > a factor. 6. However, the constant term of Q(x)/49 is 22, which is verified by again setting x=0, which gives Q(0)/49 = 22. But for two of the factors of Q(x), the constant terms is 7, which is coprime to 22. Therefore, *none* of the constant terms can have 7 as a factor. I see nothing wrong in that part. > Did you see that word coprime in there Dik Winter? Yes, I did. > If in fact 22 is not coprime to 7 then the rest of the argument does > in fact, not follow. > I guess I could change the wording to reflect that fact, but it's kind > of obvious, you know. Yes. Also with my polynomial 7 is coprime to 22. So what are you talking about? -- 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: JSH: Expanding out proof > [cut] > 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant terms is 7, which is > coprime to 22. Therefore, *none* of the constant terms can have 7 as > a factor. > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. > I don't see why it must be the case that 7 divides (5 a_1(x)+ 7) and > 7 divides (5 a_2(x)+ 7). > Can you give more detail? > -- Bill Hale OOPS! In a previous post I had F(x) = 2((x+1)/2 + 1) is always in the ring of integers, when I should have had F(x) = 2(x(x+1)/2 + 1) is always in the ring of integers, but not in algebraic integers. Sorry for the mistake. James Harris === Subject: Re: JSH: Expanding out proof > OOPS! In a previous post I had F(x) = 2((x+1)/2 + 1) is always in the > ring of integers, when I should have had F(x) = 2(x(x+1)/2 + 1) is > always in the ring of integers, but not in algebraic integers. Can you provide a specific example where F(x) = 2(x(x+1)/2+1) = x^2+x+1 is not in the ring of algebraic integers when x is an algebraic integer? -- 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: JSH: Expanding out proof > OOPS! In a previous post I had F(x) = 2((x+1)/2 + 1) is always in the > ring of integers, when I should have had F(x) = 2(x(x+1)/2 + 1) is > always in the ring of integers, but not in algebraic integers. > Can you provide a specific example where F(x) = 2(x(x+1)/2+1) = x^2+x+1 > is not in the ring of algebraic integers when x is an algebraic integer? No, it's not F(x) that's not in the ring, it's the *factorization* and your confusion here, granted my sentence was confusing, shows that as I've said Dik Winter, you're out of your depth. The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in general valid in the ring of algebraic integers, as while with x an integer, x(x+1)/2 is always an integer, but with x an algebraic integer, it's not always an algebraic integer. Now then Dik Winter, do you think you can find a *different* factorization to invalidate the factorization 2(x(x+1)/2 + 1) in the ring of integers? James Harris === Subject: Re: JSH: Expanding out proof > The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in > general valid in the ring of algebraic integers, as while with x an > integer, x(x+1)/2 is always an integer, but with x an algebraic > integer, it's not always an algebraic integer. > Now then Dik Winter, do you think you can find a *different* > factorization to invalidate the factorization 2(x(x+1)/2 + 1) in the > ring of integers? > James Harris I think your conclusion is false. If 'x' is an algebraic integer, then 'x + 1' is also an algebraic integer. So is x(x + 1) an algebraic integer. Expanding your expression: 2(x(x + 1)/2 + 1) = x(x+1) + 2 you have: x(x+1) + 2, which is also an algebraic integer. Hence, for all 'x', 2(x(x + 1)/2 + 1) *is* an algebraic integer. QED -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Expanding out proof > The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in > general valid in the ring of algebraic integers, as while with x an > integer, x(x+1)/2 is always an integer, but with x an algebraic > integer, it's not always an algebraic integer. > Now then Dik Winter, do you think you can find a *different* > factorization to invalidate the factorization 2(x(x+1)/2 + 1) in the > ring of integers? > James Harris > I think your conclusion is false. If 'x' is an algebraic integer, then 'x + 1' is also an algebraic > integer. So is x(x + 1) an algebraic integer. Expanding your expression: > 2(x(x + 1)/2 + 1) = x(x+1) + 2 > you have: x(x+1) + 2, which is also an algebraic integer. > Hence, for all 'x', 2(x(x + 1)/2 + 1) *is* an algebraic integer. As near as I can tell, James is saying that x integer => x(x+1)/2 integer, but x algebraic integer does not imply x(x+1)/2 is an algebraic integer. This is true. What bearing this is supposed to have on why James' factorization method is OK for his polynomial but not OK for Dik's is a mystery of course. I think it has something to do with James claiming to be assured (by divine inspiration, presumably) that his terms a_1/7, b_1/7 are algebraic integers, while Dik's (nyaah, nyaah) are not. Or something. - Randy === Subject: Re: JSH: Expanding out proof > The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in > general valid in the ring of algebraic integers, as while with x an > integer, x(x+1)/2 is always an integer, but with x an algebraic > integer, it's not always an algebraic integer. > Now then Dik Winter, do you think you can find a *different* > factorization to invalidate the factorization 2(x(x+1)/2 + 1) in the > ring of integers? > James Harris > I think your conclusion is false. If 'x' is an algebraic integer, then 'x + >1' is also an algebraic > integer. So is x(x + 1) an algebraic integer. Expanding your expression: It looks like that example has hit on what has turned out to be a subtle or confusing point for many of you, which is the fact that a factorization necessarily separates out an expression into factors. So when you see F(x) = 2(x(x+1)/2 + 1) what you're seeing is a factorization into the factors 2 and x(x+1)/2 + 1 which is distinct from the product, as the product has an INFINITY of factorizations, and here the product, of course, is F(x) = x^2 + x + 2 but it's distinct from its factorization in what apparently, for some of you, is a subtle way. Clearly C. Bond is already lost on the distinction, choosing to ignore the factorization and instead focus exclusively on the product, which is like refusing to see 4(3) and only 12. > 2(x(x + 1)/2 + 1) = x(x+1) + 2 > you have: x(x+1) + 2, which is also an algebraic integer. Yes, the poster C. Bond is correct about the *product* but seems to want to force the factorization to be one particular thing. Pay close attention to how the confusion of C. Bond leads the poster to a false conclusion from a correct one. > Hence, for all 'x', 2(x(x + 1)/2 + 1) *is* an algebraic integer. > QED Yes, for all x, where x is an algebraic integers--which I'll add for the poster--it IS true that F(x) is an algebraic integer, which doesn't change the fact about the factorization given, which is that it is NOT in general valid in the ring of algebraic integers. The poster tried to prove his previous assertion above challenging my position that is based on x(x+1)/2 not being valid in general in the ring of algebraic integers, but the poster just kept giving the product. Now the poster finishes up with QED and apparently feels confident. Consider the posters tagline, which follows: > -- > There are two things you must never attempt to prove: the unprovable -- and the obvious. > -- > Democracy: The triumph of popularity over principle. Now to me that's very fascinating as it shows the very human ability to not only be wrong, but spectacularly wrong *and* spectacularly confident in being right, despite the truth. Here the poster C. Bond shows you how that can all play out with somewhat humorous results. James Harris === Subject: Re: JSH: Expanding out proof > The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in > general valid in the ring of algebraic integers, as while with x an > integer, x(x+1)/2 is always an integer, but with x an algebraic > integer, it's not always an algebraic integer. Now then Dik Winter, do you think you can find a *different* > factorization to invalidate the factorization 2(x(x+1)/2 + 1) in the > ring of integers? James Harris > I think your conclusion is false. If 'x' is an algebraic integer, then 'x + >1' is also an algebraic > integer. So is x(x + 1) an algebraic integer. Expanding your expression: > It looks like that example has hit on what has turned out to be a > subtle or confusing point for many of you, which is the fact that a > factorization necessarily separates out an expression into factors. > So when you see > F(x) = 2(x(x+1)/2 + 1) > what you're seeing is a factorization into the factors > 2 and x(x+1)/2 + 1 > which is distinct from the product, as the product has an INFINITY of > factorizations, and here the product, of course, is > F(x) = x^2 + x + 2 > but it's distinct from its factorization in what apparently, for some > of you, is a subtle way. > Clearly C. Bond is already lost on the distinction, choosing to > ignore the factorization and instead focus exclusively on the product, > which is like refusing to see 4(3) and only 12. > 2(x(x + 1)/2 + 1) = x(x+1) + 2 > you have: x(x+1) + 2, which is also an algebraic integer. > Yes, the poster C. Bond is correct about the *product* but seems to > want to force the factorization to be one particular thing. > Pay close attention to how the confusion of C. Bond leads the poster > to a false conclusion from a correct one. > Hence, for all 'x', 2(x(x + 1)/2 + 1) *is* an algebraic integer. > QED > Yes, for all x, where x is an algebraic integers--which I'll add for > the poster--it IS true that F(x) is an algebraic integer, which > doesn't change the fact about the factorization given, which is that > it is NOT in general valid in the ring of algebraic integers. > The poster tried to prove his previous assertion above challenging my > position that is based on x(x+1)/2 not being valid in general in the > ring of algebraic integers, but the poster just kept giving the > product. > Now the poster finishes up with QED and apparently feels confident. > Consider the posters tagline, which follows: > -- > There are two things you must never attempt to prove: the unprovable -- and the obvious. > -- > Democracy: The triumph of popularity over principle. > Now to me that's very fascinating as it shows the very human ability > to not only be wrong, but spectacularly wrong *and* spectacularly > confident in being right, despite the truth. > Here the poster C. Bond shows you how that can all play out with > somewhat humorous results. > James Harris Simmer down, Wacky. That post was a TEST! Since you rarely answer my posts unless I am able to clobber with the '2x4-of-Truth' hard enough to leave a dent, I wanted to see if you would answer a post which gave you an opportunity to launch one of your arrogant diatribes. Bingo! I *know* you were talking about factoring, not about whether P(x) was an algebraic integer whenever 'x' was an integer, so I posted the previous off-topic and irrelevant text to see if you would respond. Viola! You did. At the same time, I posted a pair of legitimate questions in another thread and, guess what? You did *not* respond to that post! Your posting patterns and the contents of your posts reveals much more about you than it does about me. QED -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Expanding out proof > C. Bond but it's distinct from its factorization in what apparently, for some > of you, is a subtle way. One thing I'll give JSH is that he uses its and it's correctly, something I wish others would learn how to do. === Subject: Re: JSH: Expanding out proof Adjunct Assistant Professor at the University of Montana. >> The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in >> general valid in the ring of algebraic integers, as while with x an >> integer, x(x+1)/2 is always an integer, but with x an algebraic >> integer, it's not always an algebraic integer. >> Now then Dik Winter, do you think you can find a *different* >> factorization to invalidate the factorization 2(x(x+1)/2 + 1) in the >> ring of integers? >> James Harris >> I think your conclusion is false. If 'x' is an algebraic integer, then 'x + >1' is also an algebraic >> integer. So is x(x + 1) an algebraic integer. Expanding your expression: >It looks like that example has hit on what has turned out to be a >subtle or confusing point for many of you, which is the fact that a >factorization necessarily separates out an expression into factors. Yeah, very subtle. That's probably why it was mentioned several months ago; although what was used was the fact that 2 divides x(x+1) for every integer value of x; that is, that one can factor x^2 + x as 2* (x(x+1)/2) for any integer value of x; that was brought up as the difference between saying divides in the ring of polynomials and saying divides in the ring of functions. That is, the difference between saying: The polynomial 2 divides the polynomial x^2+x in the ring of polynomials with integer coefficients (a false statement); and saying: The function f:Z->Z with f(x) = 2 for all x divides the function p:Z->Z with value p(x) = x^2+x, in the ring of functions from Z to Z; that is, the value of f is always a divisor of the value of p, for each x in the integers. You know who brought this up? Here's one hint: Setting that aside, you are still confusing polynomials, functions, and their values. Polynomials, qua polynomials, are not functions, though they can be used to define functions. However, their properties as functions do not always translate to properties as polynomials: the constant function 2 from the integers to the integers divides the function x|->(x^2+x) in the ring of all functions from the integers to the integers, but the polynomial 2 does not divide the polynomial x^2+x in the ring of all polynomials with integer coefficients. Apparently, you are finally close to getting the point. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Expanding out proof that is just the sort of parity investigation that I thought I could use to solve the Perfect Box problem, just by comparing three right trigona in it; doesn't work! >> The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in >> general valid in the ring of algebraic integers, as while with x an >> integer, x(x+1)/2 is always an integer, but with x an algebraic >> integer, it's not always an algebraic integer. > was used was the fact that 2 divides x(x+1) for every integer value of > x; that is, that one can factor > The polynomial 2 divides the polynomial x^2+x in the ring of > polynomials with integer coefficients (a false statement); > and saying: > The function f:Z->Z with f(x) = 2 for all x divides the function > p:Z->Z with value p(x) = x^2+x, in the ring of functions from Z to Z; > that is, the value of f is always a divisor of the value of p, for > each x in the integers. --ils duces d'Enron! http://www.movisol.org/ http://members.tripod.com/~american_almanac/ === Subject: Re: JSH: Expanding out proof >>The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in >>general valid in the ring of algebraic integers, as while with x an >>integer, x(x+1)/2 is always an integer, but with x an algebraic >>integer, it's not always an algebraic integer. >>Now then Dik Winter, do you think you can find a *different* >>factorization to invalidate the factorization 2(x(x+1)/2 + 1) in the >>ring of integers? >>James Harris > I think your conclusion is false. If 'x' is an algebraic integer, then 'x + 1' is also an algebraic > integer. So is x(x + 1) an algebraic integer. Expanding your expression: > 2(x(x + 1)/2 + 1) = x(x+1) + 2 > you have: x(x+1) + 2, which is also an algebraic integer. > Hence, for all 'x', 2(x(x + 1)/2 + 1) *is* an algebraic integer. Actually, what he was pointing out is that (braces added for clarity) x^2 + x + 1 = 2({x(x+1)/2} + 1) Is a factorization in integers for any integral value of x, i.e., {x(x+1)/2} is an integer for every integer x. However, the equation above is not a factorization in algebraic integers for all algebraic integers x, i.e., there are some algebraic integers for which {x(x+1)/2} isn't an algebraic integer. In a similar way, one could say that x^2 + x + 1 = (x - (-1 + sqrt(-3))/2)(x - (-1 - sqrt(-3))/2)) is a factorization in algebraic integers for any algebraic integer x, but is not a factorization in rational integers for some rational integers x. Why this observation is important is something one would have to query James about. Rick === Subject: Re: JSH: Expanding out proof >>The factorization 2(x(x+1)/2 + 1) is valid in integers, but not in >>general valid in the ring of algebraic integers, as while with x an >>integer, x(x+1)/2 is always an integer, but with x an algebraic >>integer, it's not always an algebraic integer. >>Now then Dik Winter, do you think you can find a *different* >>factorization to invalidate the factorization 2(x(x+1)/2 + 1) in the >>ring of integers? >>James Harris > I think your conclusion is false. If 'x' is an algebraic integer, then 'x + 1' is also an algebraic > integer. So is x(x + 1) an algebraic integer. Expanding your expression: > 2(x(x + 1)/2 + 1) = x(x+1) + 2 > you have: x(x+1) + 2, which is also an algebraic integer. > Hence, for all 'x', 2(x(x + 1)/2 + 1) *is* an algebraic integer. > Actually, what he was pointing out is that (braces added for clarity) > x^2 + x + 1 = 2({x(x+1)/2} + 1) > Is a factorization in integers for any integral value of x, i.e., > {x(x+1)/2} is an integer for every integer x. However, the equation > above is not a factorization in algebraic integers for all algebraic > integers x, i.e., there are some algebraic integers for which > {x(x+1)/2} isn't an algebraic integer. > In a similar way, one could say that > x^2 + x + 1 = (x - (-1 + sqrt(-3))/2)(x - (-1 - sqrt(-3))/2)) > is a factorization in algebraic integers for any algebraic integer > x, but is not a factorization in rational integers for some rational > integers x. > Why this observation is important is something one would have > to query James about. > Rick Sh-h-h-h. The post was a TEST. (I know he was discussing factors, not whether P(x) was an algebraic integer if 'x' was an algebraic integer.) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Expanding out proof > [cut] > 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant terms is 7, which is > coprime to 22. Therefore, *none* of the constant terms can have 7 as > a factor. > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. > I don't see why it must be the case that 7 divides (5 a_1(x)+ 7) and > 7 divides (5 a_2(x)+ 7). > Can you give more detail? > -- Bill Hale Certainly. The key word in 6. is coprime as I have the following statement: But for two of the factors of P(x), the constant terms is 7, which is coprime to 22. Therefore, *none* of the constant terms can have 7 as a factor. Previously I'd isolated constant terms in my factors by setting x=0, so I already know that 5a_1(0) = 0, so 7 is a constant term, *and* a factor of the constant term of P(x), which is 1078. Now I find that dividing P(x) by 49 gives me 22 as the constant term. Well, the constant terms are *still* factors of the constant term of P(x), but now it's coprime to 7, as it's 22. So those constant terms have to be coprime to 7, which means that 7 is divided out, which means that 5a_1(x) + 7 has to have a factor of 7. I know that some of you are confused on situations where the factorization means that 7 becomes a unit in the ring, but it's the *factorization* that decides and here's a quick example. Consider F(x) = x+3, with the factorization F(x) = 2((x+1)/2 + 1) and notice that in *integers* the factorization exists, but it doesn't always in algebraic integers for algebraic integer x. What I've found is a way to factor polynomials, and not all of those factorizations necessarily will lie in the ring of algebraic integers, or any particular ring, just like with my example of F(x). You see, you can't just hack together a factorization. It's mathematics, so there's detail and subtlety. Look carefully at that factorization F(x) = 2((x+1)/2 + 1), and think what's happening. Ultimately, *trust* logic and mathematics. If you can step through a proof, starting from a truth, it MUST be true. Some of you may not have realized until now that you didn't really believe in mathematics, but in people. But you see, none of your opinions actually matter at all to mathematical truth. I hope I've at least made that point. James Harris === Subject: Re: JSH: Expanding out proof F(x) = 2((x+1)/2 + 1) My error as that should be F(x) = 2(x(x+1)/2 + 1). > and notice that in *integers* the factorization exists, but it doesn't > always in algebraic integers for algebraic integer x. > What I've found is a way to factor polynomials, and not all of those > factorizations necessarily will lie in the ring of algebraic integers, > or any particular ring, just like with my example of F(x). > You see, you can't just hack together a factorization. > It's mathematics, so there's detail and subtlety. > Look carefully at that factorization F(x) = 2((x+1)/2 + 1), and think Again, should be F(x) = 2(x(x+1)/2 + 1). > what's happening. > Ultimately, *trust* logic and mathematics. If you can step through a > proof, starting from a truth, it MUST be true. > Some of you may not have realized until now that you didn't really > believe in mathematics, but in people. But you see, none of your > opinions actually matter at all to mathematical truth. > I hope I've at least made that point. James Harris === Subject: Re: JSH: Expanding out proof Adjunct Assistant Professor at the University of Montana. > Consider F(x) = x+3, with the factorization >> F(x) = 2((x+1)/2 + 1) >My error as that should be F(x) = 2(x(x+1)/2 + 1). This is still wrong. If x = 5, then x+3 = 8, but 2(x(x+1)/2 + 1) = 2(5*6/2 + 1) = 2(15+1) = 2(16) = 32. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Expanding out proof > [cut] > 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant terms is 7, which is > coprime to 22. Therefore, *none* of the constant terms can have 7 as > a factor. Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. I don't see why it must be the case that 7 divides (5 a_1(x)+ 7) and > 7 divides (5 a_2(x)+ 7). Can you give more detail? -- Bill Hale > Certainly. The key word in 6. is coprime as I have the following > statement: > But for two of the factors of P(x), the constant terms is 7, which is > coprime to 22. Therefore, *none* of the constant terms can have 7 as > a factor. > Previously I'd isolated constant terms in my factors by setting x=0, > so I already know that > 5a_1(0) = 0, > so 7 is a constant term, *and* a factor of the constant term of P(x), > which is 1078. Now I find that dividing P(x) by 49 gives me 22 as the > constant term. > Well, the constant terms are *still* factors of the constant term of > P(x), I think you mean P(x)/49 here. P(x)/49 = (5 a_1(x) / 7 + 1)(5 a_2(x) / 7 + 1)(5 b_3(x) + 22) Ok, you have: 1, 1, and 22 are factors of 22. I agree. > but now it's coprime to 7, as it's 22. Ok, 22 is coprime to 7. I agree. > So those constant terms > have to be coprime to 7, You lost me here. The constant terms are 1, 1, 22. They don't have to be coprime to 7: they are coprime to 7. Can you rephrase the above to So those constant terms are coprime to 7? Are is the phrase have to be significant? > which means that 7 is divided out, I don't see how this follows from the above. The sylogism is the following: 1, 1, and 22 are factors of 22. 22 is coprime to 7. Thus, 7 is divided out of (5 a_1(x) / 7 + 1). Actually, I am not sure what that last statement even means. Can you give me the definition of divided out? For example, can I say 7 is divided out of (sqrt(5)/7 + 1)? > which means that 5a_1(x) + 7 has to have a factor of 7. I still don't understand why 5a_1(x) + 7 has to have a factor of 7. -- Bill Hale === Subject: Re: JSH: Expanding out proof > [cut] > 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant terms is 7, which is > coprime to 22. Therefore, *none* of the constant terms can have 7 as > a factor. Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. I don't see why it must be the case that 7 divides (5 a_1(x)+ 7) and > 7 divides (5 a_2(x)+ 7). Can you give more detail? -- Bill Hale Certainly. The key word in 6. is coprime as I have the following > statement: But for two of the factors of P(x), the constant terms is 7, which is > coprime to 22. Therefore, *none* of the constant terms can have 7 as > a factor. Previously I'd isolated constant terms in my factors by setting x=0, > so I already know that 5a_1(0) = 0, so 7 is a constant term, *and* a factor of the constant term of P(x), > which is 1078. Now I find that dividing P(x) by 49 gives me 22 as the > constant term. Well, the constant terms are *still* factors of the constant term of > P(x), > I think you mean P(x)/49 here. OOPS! Good catch. Yes, they are factors of the constant term of P(x)/49. > P(x)/49 = (5 a_1(x) / 7 + 1)(5 a_2(x) / 7 + 1)(5 b_3(x) + 22) > Ok, you have: > 1, 1, and 22 are factors of 22. > I agree. > but now it's coprime to 7, as it's 22. > Ok, 22 is coprime to 7. I agree. > So those constant terms > have to be coprime to 7, > You lost me here. The constant terms are 1, 1, 22. > They don't have to be coprime to 7: they are coprime to 7. Hmmm...looks like I lost you on a subtle point. You can divide 49 through the three factors of P(x) an INFINITE number of ways. Understand William Hale? However, there's only ONE WAY to divide it out leaving constant terms coprime to 7. That's the subtle but simple point which is the linchpin to understanding the proof. If you can't grasp it, you'll wander and wonder never quite getting it. > Can you rephrase the above to So those constant terms are coprime > to 7? Are is the phrase have to be significant? If the factorization is valid in a ring where 7 is not a unit, then the only way that the constant terms can be coprime to 7 is to divide 7 from the two constant terms which are *both* 7 from the factors (5 a_1(x)+ 7), (5 a_2(x) + 7), and (5 b_3(x) + 22) where you'll notice that I've isolated out constant terms as at x=0, you have 5a_1(0) = 5a_2(0) = 5b_3(0) = 0. > which means that 7 is divided out, > I don't see how this follows from the above. It follows that for that *ONE* way to divide out 49 from the constant terms, as remember 7(7)(22) is the constant term of P(x), such that the resultant is coprime to 7, is to divide 7 from TWO of the constant terms, which gives you P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22). And remember, the subtle point is that 7 is NOT a unit, as if it is, then you can't have those constant terms coprime to 7, as then, 22 is NOT coprime to 7, and neither is 1. > The sylogism is the following: > 1, 1, and 22 are factors of 22. > 22 is coprime to 7. > Thus, 7 is divided out of (5 a_1(x) / 7 + 1). > Actually, I am not sure what that last statement even means. > Can you give me the definition of divided out? If you have 7 and divide it by 7, then you have divided out a factor of 7. If you have 12 and divide it by 3, then you have divided out a factor of 3. Understand William Hale? > For example, can I say 7 is divided out of (sqrt(5)/7 + 1)? > which means that 5a_1(x) + 7 has to have a factor of 7. > I still don't understand why 5a_1(x) + 7 has to have a factor of 7. > -- Bill Hale Well William Hale, whether or not you can understand will depend on your ability to grasp that 49 can divide out an infinite number ways, dependent on your ability to understand divide out, as well as what it means for the constant terms to be coprime to 7, as well as what it means if 7 is a unit in the ring in which the factorization is valid. I'll elaborate on validity of a factorization. Consider F(x) = x^2 + x + 1, with the *factorization* F(x) = 2(x(x+1)/2 + 1) which is valid in the ring of integers, but not in general in the ring of algebraic integers. But note F(x) is ok in BOTH rings. It's the FACTORIZATION. That may be the most subtle point to grasp, but if you can't, or won't grasp it William Hale, you won't understand. James Harris === Subject: Re: JSH: Expanding out proof [cut] > So those constant terms > have to be coprime to 7, [cut added comments] > which means that 7 is divided out, [cut] > Actually, I am not sure what that last statement even means. > Can you give me the definition of divided out? > If you have 7 and divide it by 7, then you have divided out a factor > of 7. > If you have 12 and divide it by 3, then you have divided out a factor > of 3. And if you have 10 and divide it by 7, then you have divided out a factor of 7. Is that ok? I would suspect not. Or, should the first number be a multiple of 7 in order to say that if you have w and divide it by 7, then you have divided out a factor of 7. Is is necessary that 7 divide w? I will assume that this is the correct meaning for my further comments below. So, let me rephrase your original statement into the phrase that you did explain. In your rephrase proof, then you would write: You have 5a_1(x) + 7 and divide it by 7, then you have divided out a factor of 7. According to the above, you require that 7 divides 5a_1(x) + 7. Is this correct? If so, then it seems to me that you have already arrived at the conclusion that you are trying to prove before you make the statement which means that 7 is divided out. Thus, you could end the proof prior to making such a statement. I am sure that this is not correct and that I am misunderstanding what you are saying. All this seems very confusing. I think it would be easier if you just make declarative statements like 7 divides 5a_1(x) + 7, rather than make operational(?) statements like If you have 5a_1(x) + 7 and divide it out by 7, then you have divided out a factor of 7. I find declarative statements easier to check whether they are true or false. I find operational statements more difficult or even impossible to check whether they are true or false. -- Bill Hale === Subject: Re: JSH: Expanding out proof A8EwTYfhf*u~,Eu,tf6$HN*MY&)u0G =N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@J6m5.EN?>Zh Xh;Y V|',x(js'Jfq02joVpj|#x linux) > Understand William Hale? [...] > Understand William Hale? [...] > Well William Hale, whether or not [...] > That may be the most subtle point to grasp, but if you can't, or won't > grasp it William Hale, you won't understand. AAGH! James, Mr. Big Shot Writer Who's Even Been Published In Times Magazines Letters Pages, please, please, please learn to use commas when addressing someone. You want to continually use full names, no matter how odd that seems? Fine. You want to use quotations around some names[1]? Whatever floats your boat. But, if you intend, James S. Harris,[2] on repeatedly referring to your correspondent by name, learn that one ought to use commas to bracket the reference. Otherwise, you sully my quotation trove with bad punctuation. We can't have that. Footnotes: [1] Presumably to show that you're so bright that you noticed Nora Baron is a palindrome and hence likely a pseudonym *right after* being told so. [2] I know that many sources would say that the comma always goes within the quotations, but I tend to follow or neglect that rule dependent on context. Here, I can't imagine why the comma should be inside the quotations. Others may disagree, but that is *not* my obligatory typo in this punctuation flame. Must be somewhere, though. -- Jesse Hughes Such behaviour is exclusively confined to functions invented by mathematicians for the sake of causing trouble. -Albert Eagle's _A Practical Treatise on Fourier's Theorem_ === Subject: Re: SymbMath.com: web-based computer algebra system > SymbMath For Java is web-based symbolic math and computer > algebra > system, > which runs in any computer with Java. You can play it online. www.SymbMath.com >> I don't see any point to this. > Note the source address: nospam@nospam.com. > I think it's an art project, some sort of post-modernist celebration of > irony. It's not. The guy has been peddling his system for over ten years now, periodically posting irritating messages like those. His attitude throughout the years has made me, for one, extremely biased against his toy. === Subject: Re: SymbMath.com: web-based computer algebra system > I think it's an art project, some sort of post-modernist celebration of > irony. > It's not. The guy has been peddling his system for over ten years now, > periodically posting irritating messages like those. His attitude > throughout the years has made me, for one, extremely biased against his > toy. I was bored so I had a look - it was either that or configure a dos utility to print over the network. And I was overwhelmed by the impression of how pointless it was. In fact both are pointless. Everything's pointless dammit. === Subject: Re: Racist Nonsense <3F82F182.992FEC19@hate.spam.net> <3F832520.71D6778@hate.spam.net> <3f89f614$40$fuzhry+tra$mr2ice@news.patriot.net> <1482GpA5WNj$EwIC@aglow.demon.co.uk> <3f8fd502$6$fuzhry+tra$mr2ice@news.patriot.net> <3f9865ad$12$fuzhry+tra$mr2ice@news.patriot.net> <3f9ef4ef$5$fuzhry+tra$mr2ice@news.patriot.net> at 05:05 PM, Jan said: >Well, if we stick strictly to rape or murder and not worry about any >other possibilities. No. There are other variables that are known. >Extensive rape over many generations obviously can, and craniofacial >evidence shows obviously has, changed Jewish DNA. For every question there is an answer that is simple, obvious and wrong. >Are you denying this? I am denying that your claim is consistent with DNA studies. >Simple mass murder of Jews would not have changed their DNA by >itself, with no host gene influx. Do you have any idea of how genetics operates? >If you have any evidence relating the phases of the moon to this, do >post it. Do the phases of the moon account for your lunacy? >Are you suggesting phases of the moon as a third variable? Are you illiterate? >You seem to be basically hoping the craniofacial evidence will >simply disappear, No, but I was hoping that you woould look at some of the published studies instead of pulling crap out of your ass. >How does the DNA evidence conflict? By showing genetic correlations among geographically dispersed populations. >I think the difference is in the DNA. What do you think? I think that study of the actual DNA takes precedence over your preconceived notions. I think that you are unwilling to look at those data. I think that you are an ignorant fool with delusions of adequacy. >That it`s the food they eat? ESAD. I have better uses for my time than to argue with the willfully ignorant. Mr. Shakran, meet Mr. Filter. *PLONK* -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Geometry of 2nd derivative of a Vector function <6GFob.20$au6.21407@monger.newsread.comAren't manifolds locally parameterized by orthogonal coordinate >vectors? No. >Don't the partials with respect to the coordinate curves form an >orthogonal basis for the tangent space at each point? No. >I think that if s and t are >locally orthogonal, then d2x/dsdt would then be normal to the >surface being parameterized. Is this right? It may be right that you think so. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be 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: Re: Geometry of 2nd derivative of a Vector function >Because the derivative u'=du/ds of a unit vector u, uu=1, is always >orthogonal to u: d(uu)/ds = 0, udu/ds=0. Why do you assume that either df/ds or df/dt is a unit vector? The OP certainly didn't stipulatre that. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be 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: Re: Geometry of 2nd derivative of a Vector function >Because the derivative u'=du/ds of a unit vector u, uu=1, is always >orthogonal to u: d(uu)/ds = 0, udu/ds=0. > Why do you assume that either df/ds or df/dt is a unit vector? The OP > certainly didn't stipulatre that. I do not have to and neither does he but I may and I can :) === Subject: Re: The need of geodesics in physics <3F9C2E05.AC871CC9@hate.spam.net> <3F9C5A9D.ECC0A6F0@yahoo.com> <3F9C6EDF.CD8B5D@hate.spam.net> <09iupvkca3n0ur4e7bcvroeviu9isfbbit@4ax.combut QT requires m to change discretely, No. QT allows both discrete and continuous spectra. Further, even in systems with discrete spectra, QT is based on differential equations involving continuous variables. The Devil is in the details; you can't understand what is going on in QT by reading a popularization. >changing orbits That would be the old QT, obsolete even before QED came along. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be 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: Re: The need of geodesics in physics >>changing orbits >That would be the old QT, obsolete even before QED came along. The equations of motion for r(t), v(t) are still the same r'(t) = v(t), v'(t) = -kr(t)/|r(t)|^3 only now with the boundary condition [r(0), v(0)] = i h-bar/m The orbital parameters h = r x v, e = v x h/k - r/|r| are still there. They're just q-numbers now. In fact, the commutation relations are those [h_1, e_1] = [h_2, e_2] = [h_3, e_3] = 0 [h_1, h_2] = i h-bar h_3 & cyc. [h_1, e_2] = i h-bar e_3 & cyc. [e_1, e_2] = i h-bar K h_3 & cyc. (K depends on orbit type) are just those of SO(4), Galilei(2) or SL(2, C) depending whether the orbit is elliptical (|e| < 1), parabolic (|e| = 1) or hyperbolic (|e| > 1). The bound states are just the (k,k) representatives of the Lie algebra so(4) (the restriction on the range of representations arising from the constraint e.h = 0). Hence, the term Hydrogen SO(4). === Subject: Re: Fundamental Reason for High Achievements of Jews <43knb.5455$P%1.4496100@newssvr28.news.prodigy.com> due to their religion, and will only feel 'one' when they die in a >nuclear bonfire. This is because they consider both the book of >Revealation, and the book of Nostradamus, to have predicted it, so >they are on a 'holy mission', to die. To the majority of the people in the Middle East, neither of those books is particularly relevant. The first is relevant only to the small number of Christians, and the second is not relevant even to the Christians. The majority of the inhabitants neither believe, care or know what is predicted therein. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be 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: Re: Fundamental Reason for High Achievements of Jews >>All of the people of the middle east, instinctively hate each other >>due to their religion, and will only feel 'one' when they die in a >>nuclear bonfire. This is because they consider both the book of >>Revealation, and the book of Nostradamus, to have predicted it, so >>they are on a 'holy mission', to die. >To the majority of the people in the Middle East, neither of those >books is particularly relevant. The first is relevant only to the >small number of Christians, and the second is not relevant even to the >Christians. The majority of the inhabitants neither believe, care or >know what is predicted therein. Note : Most people do not hate each other. Most people want to get on with their lives in peace and harmony. Nutters hate each other ... it's a pointless activity. Bruce ----------------------------------------------------------------------- It was so much easier to blame it on Them. It was bleakly depressing to think that They were Us. If it was Them, then nothing was anyones fault. If it was Us, what did that make Me ? After all, Im one of Us. I must be. Ive certainly never thought of myself as one of Them. No-one ever thinks of themselves as one of Them. Were always one of Us. Its Them that do the bad things. <=> Terry Pratchett. Jingo. === Subject: Re: Fundamental Reason for High Achievements of Jews > Scratch's philosophy reminded me of the story > about a farmer who lived next to another farm > that was up for sale. A potential buyer stopped by and asked him how the neighbors were > in the community, and the farmer asked him how they were > in his community. The man replied, > The people where I live now, are no good, > bastards, etc. The farmer replied: > Yep, that's the way they are around here. Another potential buyer stopped by and asked him how the neighbors were > in the community, and the farmer asked him how they were > in his community. The man replied, > The people where I live now, > are really great. They help out when folks are sick > or have financial problems, they are friendly, etc. The farmer replied: > Yep, that's the way they are around here. Who would you like to have for a neighbor? > People as two-faced as that, don't deserve to be neighbors. > The question therefore arises, as to wheather my minions even deserve to worship me. > I will contemplate this. > -Satan The farmer wasn't two-faced. He was letting the devils filter themselves out, as he did not want them for neighbors. I agree with you that you should direct your worshippers to someone, or something, else, or destroy them. -- Tom Potter === Subject: Re: Hard tensor question (kst). at 01:04 PM, dynamics@vianet.on.ca (Ken S. Tucker) said: >1) Suppose K= |g^uv|, where g^uv is the >contravariant metric tensor, and K is >it's determinant and is a relative tensor. >And then form a covariant tensor K_uv, >by K_uv = K*g_uv where g_uv is a the >covariant metric, That doesn't give you a covariant tensor. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be 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: Re: Question: Obtaining a Basis for a subspace Subspace >If I know that the basis for a fininite-dimensional vector space V is >B , There's no such thing as *the* basis, only a basis. The choice of a basis is arbitrary. >then is there always a method to find the basis of a subspace of V >using the basis B? Not in general. However, if you are given additional constraints or additional structure than there may be. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be 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: Key core error argument, stepped out Mathematicians are *still* apparently hell-bent on running from the result which is key in showing an over hundred year old problem, but at least arguing with them gives me a sense of places where people get confused. So here's the argument again with slight changes based on what I've gathered. I also number out the main steps, so if anyone thinks there's a problem and wishes to reply back, they need to give at least *some* numbers. 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is in the ring of algebraic integers, notice that P(x) has a constant term that is 1078. 2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 where the *same* polynomial has been put in a form which allows a factorization into non-polynomial factors so that I have P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). 3. Now let x=0, so P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's at x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and a_2(0) to equal 0, which leaves a_3(0) with a value of 3. 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. Most posters seem ok with the steps up to the final two. In what follows it's important to understand the word coprime. In integers, 2 and 3 are coprime as are 12 and 13, as it simply means they don't share non-unit factors. However, in reals, NO numbers are coprime, as for instance, 2(3/2) = 3, so 2 is a factor of 3. That's an important point to consider going forward. 6. However, the constant term of P(x)/49 is 22, which is verified by again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant terms is 7, which is coprime to 22. Therefore, *none* of the constant terms can have 7 as a factor. (By saying that 7 is coprime to 22, I'm making a choice as to where the proof is going. Since I've been talking about algebraic integers, where 7 is coprime to 22, it's natural to go with a choice where 7 is coprime to 22.) Given that the constant terms are independent of x's value, it must be the case that dividing P(x) by 49 divides the two constant terms equal to 7, by 7. 7. But to divide 7 from those constant terms requires dividing through two of the factors, so (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 from reverse use of the distributive property, which gives a constant term coprime to 7, as required. For some odd reason I've STILL had mathematicians refusing to acknowledge the truth. But, surprisingly, the proof here, from what I read in a link posted further down, is tighter than what mathematicians usually claim is a proof, and represents perfection at a level most mathematicians, even professional mathematicians, don't even attempt. To understand what I mean, you might want to see When is a proof? http://www.maa.org/devlin/devlin_06_03.html Here's an excerpt: What is a proof? The question has two answers. The right wing (right-or-wrong, rule-of-law) definition is that a proof is a logically correct argument that establishes the truth of a given statement. The left wing answer (fuzzy, democratic, and human centered) is that a proof is an argument that convinces a typical mathematician of the truth of a given statement. While valid in an idealistic sense, the right wing definition of a proof has the problem that, except for trivial examples, it is not clear that anyone has ever seen such a thing. What I've shown you in my post is irrefutable proof, yet mathematicians seem to think they can just ignore it, while they themselves have far vaguer works, which they expect the world to celebrate. They're cheating. James Harris http://mathforprofit.blogspot.com === Subject: Re: Key core error argument, stepped out Adjunct Assistant Professor at the University of Montana. Newsgroups trimmed. Again. >1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is >in the ring of algebraic integers, notice that P(x) has a constant >term that is 1078. >2. It can be shown that >P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 >where the *same* polynomial has been put in a form which allows a >factorization into non-polynomial factors so that I have >P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >where the a's are roots of >a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). For the following you are actually REQUIRING that the a_i(x) be given as roots of that polynomial. This is not something that you allow, it is something you require. It's a nitpick, but it is important for what follows. >3. Now let x=0, so >P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) >as the cubic defining the a's at x=0 is >a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and >a_2(0) to equal 0, which leaves a_3(0) with a value of 3. >4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I >have >P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) >P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). Note that all you have done is add and subtract 3 to define b_3(x); that is, you are writing b_3(x) = (a_3(x) - 3) so 5a_3(x) + 7 = 5(a_3(x)-3+3) + 7 = 5(a_3(x)-3) + 15 + 22 = 5b_3(x) + 22. The exact same process that you decried when I used it. You claimed that doing this ma[de] no sense mathematically. Do you still make that claim? Just curious. >5. Now P(x) has a factor of 49 as >P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 >which means that >(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) >has a factor of 49. >Most posters seem ok with the steps up to the final two. In what >follows it's important to understand the word coprime. You need to understand that coprime is CONTEXTUAL, just as divides. Saying a and b are coprime per se has no meaning; one must say a and b are coprime IN SUCH AND SUCH A RING, unless the ring is understood from context. >In integers, 2 and 3 are coprime as are 12 and 13, as it simply means >they don't share non-unit factors. >However, in reals, NO numbers are coprime, as for instance, 2(3/2) = >3, so 2 is a factor of 3. But 2 is a unit, so your claim here is false. Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE COPRIME. That's because, given any two nonzero real numbers x and y, any common divisor of x and y is a unit. In a field, any two nonzero elements are coprime. >That's an important point to consider going forward. If it is such an important point, should you not get it RIGHT? >6. However, the constant term of P(x)/49 is 22, which is verified by >again setting x=0, which gives P(0)/49 = 22. >But for two of the factors of P(x), the constant terms is 7, which is >coprime to 22. Therefore, *none* of the constant terms can have 7 as >a factor. >(By saying that 7 is coprime to 22, I'm making a choice as to where >the proof is going. Really? Do you mean, you are making a choice as to what ring you are working on? > Since I've been talking about algebraic integers, >where 7 is coprime to 22, it's natural to go with a choice where 7 is >coprime to 22.) 7 and 22 are coprime in ANY ring that contains the integers. That's because there exist integers a and b such that 7a+22b = 1 (namely, a=-3, b=1); so any common divisor of 7 and 22 in ANY ring that contains the integers must be a divisor of 1, hence a unit. But whatever. Just more evidence that you don't really know what you are talking about. >Given that the constant terms are independent of x's value, it must be >the case that dividing P(x) by 49 divides the two constant terms equal >to 7, by 7. No. It is not the case. This claim does not follow, and is not the case. In general, 49 will factor out as 49 = w_1(x)*w_2(x)*w_3(x), where w_1(x), w_2(x), w_3(x) are functions with algebraic integer values, such that w_1(0)=w_2(0)=7, and w_3(0)=1. Then the factorization of P(x)/49 will look as follows: P(x)/49 = [(5a_1(x)+7)/w_1(x)]*[(5a_2(x)+7)/w_2(x)]*[(5b_3(x)+22)/w_3(x)]. To see how this relates to your constant terms, it is enough to realize one thing: ANY function f(x), defined at x=0, can be written as g(x) + f(0), where g(0)=0; namely, we MUST have g(x)=f(x)-f(0). That is, it amounts to writing f(x) = [f(x)-f(0)] + f(0). So, to write the factors as something plus a constant term, we MUST add and subtract their value at x=0. So we have: P(x)/49 = (h_1(x) + 1)*(h_2(x) + 1)*(h_3(x) + 22) where h_1(x) = (5a_1(x)+7)/w_1(x) - (5a_1(0)+7)/w_1(0) = (5a_1(x)+7)/w_1(x) - (7/7) = (5a_1(x)+7)/w_1(x) - 1. h_2(x) = (5a_2(x)+7)/w_2(x) - (5a_2(0)+7)/w_2(0) = (5a_2(x)+7)/w_2(x) - (7/7) = (5a_2(x)+7)/w_3(x) - 1. h_3(x) = (5b_3(x)+22)/w_3(x) - (5b_3(0)+22)/w_3(0) = (5b_3(x)+22)/w_3(x) - (22/1) = (5b_3(x)+22)/w_3(x) - 22. However, this DOES NOT imply that w_1(x)=7, w_2(x)=7, and w_3(x)=1 for all values of x. The equations simply DO NOT REQUIRE THAT. No matter how many times and in how many ways you claim they do. In general, you will have that 7 = r(x)*s(x)*t(x), where r(x), s(x), and t(x) take values in the algebraic integers, for each x they are pairwise coprime (in the ring of algebraic integers), and w_1(x) = s(x)*t(x); w_2(x)=r(x)*t(x); w_3(x)=r(x)*s(x). When x=0, you have r(x)=1, s(x)=1, t(x)=7. When, for a specific value of x, P(x)/49 is irreducible over Q, you will have that r(x), s(x), and t(x) are all non-units (in the ring of algebraic integers) In order for you to conclude that we must have h_1(x) = 5a_1(x)/w_1(x) + 1, as you claim, you must ASSUME that w_1(x)=7 for all x. You are trying to PROVE that w_1(x)=7 for all x, so you cannot assume this is the case. You are engaging in a circular argument, which is a logical fallacy. What follows is based on this flawed assumption, and therefore it is invalid. >7. But to divide 7 from those constant terms requires dividing >through two of the factors, so >(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 >from reverse use of the distributive property, which gives a constant >term coprime to 7, as required. You are assuming w_1(x)=7 w_2(x)=7, w_3(x)=1 for all x; this is what you are trying to PROVE, so you cannot assume it. Your argument here about constant terms is a red herring, and does not yield the conclusion you claim it does. >For some odd reason I've STILL had mathematicians refusing to >acknowledge the truth. No, we acknowledge the truth all right: that your argument is wrong. What you mean is that we refuse to acknowledge your proof is correct; and the reason there is also trivial: we refuse to acknowledge it as correct for the simple reason that it is NOT correct. [.rest deleted.] Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Key core error argument, stepped out that is so cool, adding three to an equation, AND THEN SUBTRACTING IT to transform the equation, although I thought taht you'd done it, before. when you say nonpolynomial factorization, do you just mean an all-monomial one? >3. Now let x=0, so >P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) >as the cubic defining the a's at x=0 is >a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and >a_2(0) to equal 0, which leaves a_3(0) with a value of 3. >4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I >have >P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) >P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). > Note that all you have done is add and subtract 3 to define b_3(x); > that is, you are writing > b_3(x) = (a_3(x) - 3) > so > 5a_3(x) + 7 = 5(a_3(x)-3+3) + 7 > = 5(a_3(x)-3) + 15 + 22 > = 5b_3(x) + 22. > Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE > COPRIME. That's because, given any two nonzero real numbers x and y, > any common divisor of x and y is a unit. > In a field, any two nonzero elements are coprime. --ils duces d'Enron! http://www.movisol.org/ http://members.tripod.com/~american_almanac/ === Subject: Re: Key core error argument, stepped out > Newsgroups trimmed. Again. >1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is >in the ring of algebraic integers, notice that P(x) has a constant >term that is 1078. >2. It can be shown that >P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 >where the *same* polynomial has been put in a form which allows a >factorization into non-polynomial factors so that I have >P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >where the a's are roots of >a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > For the following you are actually REQUIRING that the a_i(x) be given > as roots of that polynomial. This is not something that you allow, > it is something you require. It's a nitpick, but it is important for > what follows. The poster Arturo Magidin *should* be apologizing to this newsgroup and others for leading so many of you astray for YEARS, but instead, he's quibbling as he appears hell-bent on continuing to obfuscate. It takes a rather extreme hatred of common decency to behave as this poster has over such a long period of time, and STILL, even at the end, keep trying to play the same old games. >3. Now let x=0, so >P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) >as the cubic defining the a's at x=0 is >a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and >a_2(0) to equal 0, which leaves a_3(0) with a value of 3. >4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I >have >P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) >P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). > Note that all you have done is add and subtract 3 to define b_3(x); > that is, you are writing > b_3(x) = (a_3(x) - 3) > so > 5a_3(x) + 7 = 5(a_3(x)-3+3) + 7 > = 5(a_3(x)-3) + 15 + 22 > = 5b_3(x) + 22. > The exact same process that you decried when I used it. You claimed > that doing this ma[de] no sense mathematically. Do you still make > that claim? Just curious. That is a lie from Arturo Magidin as in fact he just subtracted and added 3 on the same line. I'm focusing on constant terms, not trying to hide a correct argument with meaningless operations like subtracting and adding 3. Here by switching to b_3(x) I have 7,7, and 22, the three constant term factors of the main constant term 1078 of P(x), shown so that there's less room for confusion. I'm adding additional steps to handle people like Arturo Magidin who have made it their business to lie for so long, and not surprisingly, he's trying to find fault with the process. Posters like Arturo Magidin waste a lot of other people's time. >5. Now P(x) has a factor of 49 as >P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 >which means that >(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) >has a factor of 49. >Most posters seem ok with the steps up to the final two. In what >follows it's important to understand the word coprime. > You need to understand that coprime is CONTEXTUAL, just as > divides. Saying a and b are coprime per se has no meaning; one > must say a and b are coprime IN SUCH AND SUCH A RING, unless the > ring is understood from context. Like I said, posters like Arturo Magidin waste a LOT of people's time. >In integers, 2 and 3 are coprime as are 12 and 13, as it simply means >they don't share non-unit factors. >However, in reals, NO numbers are coprime, as for instance, 2(3/2) = >3, so 2 is a factor of 3. > But 2 is a unit, so your claim here is false. Hmmm...now that's interesting, every real but 0 is a unit, eh? Fascinating perspective. > Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE > COPRIME. That's because, given any two nonzero real numbers x and y, > any common divisor of x and y is a unit. Well, then every real number but 0 is a unit follows from that position. I guess you could say that and it doesn't change things in any meaningful way. But it's *fascinating* that Arturo Magidin pushed that position! Well consider then, he's saying that 3 is not a factor of 6 in reals because they're coprime! And you know what? I think the way mathematicians usually go, he's right!!! Fun stuff, eh? James Harris === Subject: Re: Key core error argument, stepped out ... >In integers, 2 and 3 are coprime as are 12 and 13, as it simply means >they don't share non-unit factors. However, in reals, NO numbers are coprime, as for instance, 2(3/2) = >3, so 2 is a factor of 3. But 2 is a unit, so your claim here is false. > Hmmm...now that's interesting, every real but 0 is a unit, eh? Yup. Every element of a field, except 0, is a unit. > Fascinating perspective. Isn't it? > Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE > COPRIME. That's because, given any two nonzero real numbers x and y, > any common divisor of x and y is a unit. > Well, then every real number but 0 is a unit follows from that > position. As that is the definition of a field, it looks about right. > I guess you could say that and it doesn't change things in any > meaningful way. > But it's *fascinating* that Arturo Magidin pushed that position! Are you contradicting it? > Well consider then, he's saying that 3 is not a factor of 6 in reals > because they're coprime! > And you know what? I think the way mathematicians usually go, he's > right!!! And indeed. But apparently you have no idea about the mathematical meaning of coprime. > Fun stuff, eh? Yeah. -- 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: Key core error argument, stepped out Adjunct Assistant Professor at the University of Montana. >> Newsgroups trimmed. Again. >>1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is >>in the ring of algebraic integers, notice that P(x) has a constant >>term that is 1078. >>2. It can be shown that >>P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 >>where the *same* polynomial has been put in a form which allows a >>factorization into non-polynomial factors so that I have >>P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >>where the a's are roots of >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >> For the following you are actually REQUIRING that the a_i(x) be given >> as roots of that polynomial. This is not something that you allow, >> it is something you require. It's a nitpick, but it is important for >> what follows. >The poster Arturo Magidin *should* be apologizing to this newsgroup >and others for leading so many of you astray for YEARS, but instead, >he's quibbling as he appears hell-bent on continuing to obfuscate. This is your claim. It is based on your solemn and firm belief that you are correct. That belief is based on the lies you tell yourself, and has little or no relation to reality. >It takes a rather extreme hatred of common decency to behave as this >poster has over such a long period of time, and STILL, even at the >end, keep trying to play the same old games. Alas, James, the only one who has exhibited extreme hatred, lack of common decency, and pretty much all the things you accuse others of, is you. >>3. Now let x=0, so >>P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) >>as the cubic defining the a's at x=0 is >>a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and >>a_2(0) to equal 0, which leaves a_3(0) with a value of 3. >>4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I >>have >>P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) >>P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). >> Note that all you have done is add and subtract 3 to define b_3(x); >> that is, you are writing >> b_3(x) = (a_3(x) - 3) >> so >> 5a_3(x) + 7 = 5(a_3(x)-3+3) + 7 >> = 5(a_3(x)-3) + 15 + 22 >> = 5b_3(x) + 22. >> The exact same process that you decried when I used it. You claimed >> that doing this ma[de] no sense mathematically. Do you still make >> that claim? Just curious. >That is a lie from Arturo Magidin as in fact he just subtracted and >added 3 on the same line. ->WHAT<- is a lie? Is this not HOW you find b_3? You find b_3 by figuring out the value of a_3 at 0, and then subtracting it. Since you need the value 5a_3(X) + 7 to remain the same, you must write 5b_3(x) + 5(3) + 7; but all you did was take a_3(x)-3, and then add 3 more to get the right result. So what I said is NOT a lie; it is also not a lie that you claimed that adding and subtracting something made no mathematical sense. Therefore, your assertion that this is a lie is in fact false. And knowingly so. Hence, the assertion that I lied is in fact a lie from you. As I said, you seem to be the only one guilty of committing the offenses you try to impugne on others. > I'm focusing on constant terms, not trying >to hide a correct argument with meaningless operations like >subtracting and adding 3. No, YOU are hiding what you are doing by doing the switch to b_3; what you took a_3-3 and renamed it b_3, and then you took the +3, multiplied by 5, and added it to the 7. >Here by switching to b_3(x) I have 7,7, and 22, the three constant >term factors of the main constant term 1078 of P(x), shown so that >there's less room for confusion. Which is exactly what I did; but you are so hell bent on attacking me that you attack your own method, failing to recongize it for what it is, because hate is blinding you. >I'm adding additional steps to handle people like Arturo Magidin who >have made it their business to lie for so long, and not surprisingly, >he's trying to find fault with the process. >Posters like Arturo Magidin waste a lot of other people's time. You waste a lot of people's time by repeating the same mistake over, and over, and over, and over, and over. >>5. Now P(x) has a factor of 49 as >>P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 >>which means that >>(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) >>has a factor of 49. >>Most posters seem ok with the steps up to the final two. In what >>follows it's important to understand the word coprime. >> You need to understand that coprime is CONTEXTUAL, just as >> divides. Saying a and b are coprime per se has no meaning; one >> must say a and b are coprime IN SUCH AND SUCH A RING, unless the >> ring is understood from context. >Like I said, posters like Arturo Magidin waste a LOT of people's time. of time and completely incorrect. >>In integers, 2 and 3 are coprime as are 12 and 13, as it simply means >>they don't share non-unit factors. >>However, in reals, NO numbers are coprime, as for instance, 2(3/2) = >>3, so 2 is a factor of 3. >> But 2 is a unit, so your claim here is false. >Hmmm...now that's interesting, every real but 0 is a unit, eh? Duh. A unit is an element with a multiplicative inverse in the ring. Every real number except for 0 has a multiplicative inverse. Therefore, every real number except 0 is a unit. >Fascinating perspective. It's not a perspective, it's the DEFINITION. >> Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE >> COPRIME. That's because, given any two nonzero real numbers x and y, >> any common divisor of x and y is a unit. >Well, then every real number but 0 is a unit follows from that >position. No, it follows from the DEFINITION. >I guess you could say that and it doesn't change things in any >meaningful way. >But it's *fascinating* that Arturo Magidin pushed that position! >Well consider then, he's saying that 3 is not a factor of 6 in reals >because they're coprime! NO, I am NOT saying that. 3 is a factor of 6 in the reals; it is ALSO true that 3 and 6 are coprime in the reals. Because 3 (and 6) are units in the reals, and a unit is coprime to every number in the ring. Your assertion shows that you STILL don't know what you are talking about. I also notice that you REMOVED the rest, in which I point out (yet again) exactly where and why your argument is unfounded and incorrect. Scared? Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Key core error argument, stepped out Nothing. http://www.crank.net/harris.html It's not every jackass who rates an entire crank.net page. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Key core error argument, stepped out > Nothing. > http://www.crank.net/harris.html I suggest readers check the link. > It's not every jackass who rates an entire crank.net page. For those who don't know Uncle Al is a known critic troll who thinks that it's ok to post a LOT mainly insulting people. Yup, he's one of the stupider variety of posters. If you have any doubt about that you can go to and put in UncleAl0@hate.spam.net, which you see above into the author field on that page and hit search. I just did it and Google pulled up 28,600 posts, so you can get some ideas about this poster from that number, along with his demonstrated hatred. The fact is that the poster attaches himself like a leach to posts made by people who get attention, being someone who clearly lacks the ability to stand on his own two feet, and generate content that will draw readers on his own. He's a sad example of the dark side of Usenet. James Harris === Subject: Re: Key core error argument, stepped out > Nothing. > http://www.crank.net/harris.html > I suggest readers check the link. > It's not every jackass who rates an entire crank.net page. > For those who don't know Uncle Al is a known critic troll who thinks > that it's ok to post a LOT mainly insulting people. Yup, he's one of > the stupider variety of posters. If you have any doubt about that you > can go to....... > James Harris Well, I'll be damned. James has finally said something I agree with. === Subject: Re: Key core error argument, stepped out >Nothing. >http://www.crank.net/harris.html >>I suggest readers check the link. >It's not every jackass who rates an entire crank.net page. > For those who don't know Uncle Al is a known critic troll who thinks >>that it's ok to post a LOT mainly insulting people. Yup, he's one of >>the stupider variety of posters. If you have any doubt about that you >>can go to....... >>James Harris > Well, I'll be damned. James has finally said something I agree with. The boy must be growing up! === Subject: Re: Key core error argument, stepped out > Mathematicians are *still* apparently hell-bent on running from the > result which is key in showing an over hundred year old problem, but > at least arguing with them gives me a sense of places where people get > confused. So here's the argument again with slight changes based on > what I've gathered. > I also number out the main steps, so if anyone thinks there's a > problem and wishes to reply back, they need to give at least *some* > numbers. > 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. > 2. It can be shown that > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Can you clarify the above statement? Is it your position that the 'a's determined in this manner are the *only* values for a_1, a_2 and a_3 which are correct? Or do you mean that the 'a's determined in this manner are the only ones which preserve the constants? Or what??? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Key core error argument, stepped out > Mathematicians are *still* apparently hell-bent on running from the > result which is key in showing an over hundred year old problem, but > at least arguing with them gives me a sense of places where people get > confused. So here's the argument again with slight changes based on > what I've gathered. > I also number out the main steps, so if anyone thinks there's a > problem and wishes to reply back, they need to give at least *some* > numbers. > 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. > 2. It can be shown that > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Can you clarify the above statement? Is it your position that the 'a's > determined in this manner are the *only* values for a_1, a_2 and a_3 which > are correct? Or do you mean that the 'a's determined in this manner are > the only ones which preserve the constants? Or what??? The functions a_1(x), a_2(x), and a_3(x) are the roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). For instance, a_1(x)^3 + 3(-1 + 49x)a_1(x)^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0. James Harris === Subject: Re: Key core error argument, stepped out > Mathematicians are *still* apparently hell-bent on running from the > result which is key in showing an over hundred year old problem, but > at least arguing with them gives me a sense of places where people get > confused. So here's the argument again with slight changes based on > what I've gathered. I also number out the main steps, so if anyone thinks there's a > problem and wishes to reply back, they need to give at least *some* > numbers. 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. 2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Can you clarify the above statement? Is it your position that the 'a's > determined in this manner are the *only* values for a_1, a_2 and a_3 which > are correct? Or do you mean that the 'a's determined in this manner are > the only ones which preserve the constants? Or what??? > The functions a_1(x), a_2(x), and a_3(x) are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > For instance, > a_1(x)^3 + 3(-1 + 49x)a_1(x)^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0. > James Harris As is your usual failing, you did not answer either of the questions posed, but simply restated the ambiguous assertion you previously posted. However, pushing forward anyway, note that the functions a_1(x), a_2(x) and a_3(x) can also be found as the roots of: (5a + 7)^3 - P(x) = 0 where P(x) = 14706125x^3 - 900375x^2 - 17640x + 1078 In this case, a_1 = -7/5 + (1/5)7^(2/3)(22 - 360x - 18375x^2 + 300125x^3)^(1/3) a_2 = -7/5 - (1/10)7^(2/3)(1 - 3^(1/2)I)(22 - 360x - 18375x^2 + 300125x^3)^(1/3) a_3 = -7/5 - (1/10)7^(2/3)(1 + 3^(1/2)I)(22 - 360x - 18375x^2 + 300125x^3)^(1/3) Letting Q(x) = (5a_1 + 7)(5a_2 + 7)(5a_3 + 7) the following results are obtained: x P(x) Q(x) a_n (CB) (JSH) 0 1078 1078 a1 0.650704 3 a2 -2.42535 + 1.77596I 0 a3 -2.42535 - 1.77596I 0 1 13789188 13789188 a1 46.5597 25.5404 a2 -25.3798 + 41.5343I -138.21 a3 -25.3798 - 41.5343I -31.3299 2 114013298 114013298 a1 95.5799 51.6126 a2 -49.89 + 83.9871I -279.3 a3 -49.89 - 83.9871I -63.3123 3 388910158 388910158 a1 144.587 77.685 a2 -74.3933 + 126.429I -420.39 a3 -74.3933 - 126.429I -95.2947 So, clearly, there is more than one way to find a valid set of 'a's from the given equation: P(x) = 14706125x^3 - 900375x^2 - 17640x + 1078 = (5a_1 + 7)(5a_2 +7)(5a_3 +7) Would you now be so kind as to answer the questions originally asked, which are: 1) Do you claim that your solution for the 'a's is the only correct one? 2) Do you claim that your solution for the 'a's is the only one which preserves the constants? 3) Or what??? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Key core error argument, stepped out http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: JSH: Advanced Polynomial Factorization I've called the factoring of polynomial into non-polynomial factors advanced for a reason, understanding requires that you have the basics down, are VERY well-founded in mathematical logic, and realize that you will have to make some serious effort. Unfortunately, I have people who seem to think that replying to me is an easy exercise, or just some kind of a lark, which has no meaning or consequence. However, if you THINK you're going to be a mathematician, or think you already are a mathematician, what you say in reply to me, is a part of your work as a mathematician. And thankfully, as long as Google is around, it looks like it'll be part of the historical record till you die, and beyond. Now then, I've come across a nice example to give some perspective on advanced concepts in factoring polynomials, which has the nice feature of, I hope, being easy to understand. The polynomial is F(x) = x^2 + x + 2, and the factorization relies on something you probably know about, but didn't pursue, which is that x^2 + x is always even if x is an integer, so I can factor as F(x) = 2(x(x+1)/2 + 1) for a factorization valid in integers, but not in general valid in algebraic integers. To move beyond factoring polynomials into polynomial factors, you need to have a *thorough* understanding of coprimeness, ring operations, and logical argument. What I've found is a puzzling lack of a thorough foundation in these areas from LOTS of people in the math world, as I'll include Barry Mazur, Andrew Granville, and Ralph McKenzie, in with the posters who have continually posted in reply to me on this forum. Basically, quite a few of you have demonstrated a failure to understand mathematics at a level I'd call competent, as the basics necessary are what I'd *think* any mathematician would have. But here, in the 21st century, mathematicians are displaying a woeful lack of the basics, besides often displaying childish petulance, arrogance, and a stubborn refusal to acknowledge the math. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Advanced Polynomial Factorization >I've called the factoring of polynomial into non-polynomial factors >advanced for a reason, understanding requires that you have the basics >down, are VERY well-founded in mathematical logic, Tee-hee. As people used to try to explain, you have no idea what the term mathematical logic even means - it means something very specific, not just mathematical reasoning or the way logic is used in mathematics or some such. >and realize that >you will have to make some serious effort. >Unfortunately, I have people who seem to think that replying to me is >an easy exercise, or just some kind of a lark, which has no meaning or >consequence. >However, if you THINK you're going to be a mathematician, or think you >already are a mathematician, what you say in reply to me, is a part of >your work as a mathematician. And thankfully, as long as Google is >around, it looks like it'll be part of the historical record till you >die, and beyond. >Now then, I've come across a nice example to give some perspective on >advanced concepts in factoring polynomials, which has the nice feature >of, I hope, being easy to understand. >The polynomial is F(x) = x^2 + x + 2, and the factorization relies on >something you probably know about, but didn't pursue, which is that >x^2 + x is always even if x is an integer, so I can factor as >F(x) = 2(x(x+1)/2 + 1) >for a factorization valid in integers, but not in general valid in >algebraic integers. >To move beyond factoring polynomials Bzzt! You haven't been talking about factoring polynomials! The fact that F(x) = 2*(x(x+1)/2+1) for every integer x, and that the two factors are integers for every integer x, does _not_ show that you have a factorization in the ring of polynomials with integer coefficients. You don't. (Of course it is a factorization in the ring of polynomials with real coefficients but that's irrelevant.) You haven't been talking about factoring polynomials, you've been talking about factoring the _values_ of polynomial functions. Some day you may understand the difference. Or maybe not, the propects seem kinda dim, considering the history. >into polynomial factors, you need >to have a *thorough* understanding of coprimeness, ring operations, >and logical argument. >What I've found is a puzzling lack of a thorough foundation in these >areas from LOTS of people in the math world, as I'll include Barry >Mazur, Andrew Granville, and Ralph McKenzie, in with the posters who >have continually posted in reply to me on this forum. Yeah, it _is_ puzzling how the morons here share the very same lack of _basic_ mathematical understanding with those esteemed names you mention. You know, a lot of people would decide at this point that maybe there was another explanation, that maybe even if they didn't understand the explanations people were giving for why they were wrong it just _might_ be that _they_ were wrong for reasons they didn't understand, not that _all_ the mathematicians in the friggin _world_ were totally incompetent regarding _basic_ issues. >Basically, quite a few of you have demonstrated a failure to >understand mathematics at a level I'd call competent, as the basics >necessary are what I'd *think* any mathematician would have. >But here, in the 21st century, mathematicians are displaying a >woeful lack of the basics, besides often displaying childish >petulance, arrogance, and a stubborn refusal to acknowledge the math. >James Harris >http://mathforprofit.blogspot.com/ ************************ David C. Ullrich === Subject: Re: JSH: Advanced Polynomial Factorization > I've called the factoring of polynomial into non-polynomial factors > advanced for a reason, understanding requires that you have the basics > down, are VERY well-founded in mathematical logic, and realize that > you will have to make some serious effort. > Unfortunately, I have people who seem to think that replying to me is > an easy exercise, or just some kind of a lark, which has no meaning or > consequence. > However, if you THINK you're going to be a mathematician, or think you > already are a mathematician, what you say in reply to me, is a part of > your work as a mathematician. And thankfully, as long as Google is > around, it looks like it'll be part of the historical record till you > die, and beyond. > Now then, I've come across a nice example to give some perspective on > advanced concepts in factoring polynomials, which has the nice feature > of, I hope, being easy to understand. > The polynomial is F(x) = x^2 + x + 2, and the factorization relies on > something you probably know about, but didn't pursue, which is that > x^2 + x is always even if x is an integer, so I can factor as > F(x) = 2(x(x+1)/2 + 1) > for a factorization valid in integers, but not in general valid in > algebraic integers. > To move beyond factoring polynomials into polynomial factors, you need > to have a *thorough* understanding of coprimeness, ring operations, > and logical argument. > What I've found is a puzzling lack of a thorough foundation in these > areas from LOTS of people in the math world, as I'll include Barry > Mazur, Andrew Granville, and Ralph McKenzie, in with the posters who > have continually posted in reply to me on this forum. > Basically, quite a few of you have demonstrated a failure to > understand mathematics at a level I'd call competent, as the basics > necessary are what I'd *think* any mathematician would have. > But here, in the 21st century, mathematicians are displaying a > woeful lack of the basics, besides often displaying childish > petulance, arrogance, and a stubborn refusal to acknowledge the math. Since most of them can't understand group theory, it's surprises no one. And the rest of them believe that geometry is something Gauss invented it also surprises noone. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Do finite strings over a countable alphabet form a countable set? Given a countable set of letters A={L_1,L_2,L_3,...}, consider the set of finite strings S(A) which can be formed using letters in A. Is S(A) a countable set? === Subject: Re: Do finite strings over a countable alphabet form a countable set? > Given a countable set of letters A={L_1,L_2,L_3,...}, consider the set > of finite strings S(A) which can be formed using letters in A. Is S(A) a > countable set? yes -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Do finite strings over a countable alphabet form a countable set? > Given a countable set of letters A={L_1,L_2,L_3,...}, consider the set > of finite strings S(A) which can be formed using letters in A. Is S(A) a > countable set? > yes Yes, but it's not well-orderable is the correct answer. === Subject: Re: Do finite strings over a countable alphabet form a countable set? Given a countable set of letters A={L_1,L_2,L_3,...}, consider the set > of finite strings S(A) which can be formed using letters in A. Is S(A) a > countable set? yes > Yes, but it's not well-orderable is the correct answer. How could you have a countable set that's not well-orderable? If a set is countable there's a bijection to N, the natural numbers. That bijection induces a well-order on your original set. === Subject: Re: Do finite strings over a countable alphabet form a countable set? >> Given a countable set of letters A={L_1,L_2,L_3,...}, consider the set >> of finite strings S(A) which can be formed using letters in A. Is S(A) a >> countable set? >> yes > Yes, but it's not well-orderable is the correct answer. Uh, no. This set _is_ well-orderable - being countable you don't even need to assume the Axiom of Choice, you can write down an explicit well-ordering. Good point except for that. ************************ David C. Ullrich === Subject: Re: Delay differential equation Hi again, > Fred Brauer pointed out to me that it's easy to show that 0 is the > only root with nonnegative real part if a >= |bc|. Fortunately, this is always the case for me! Using this and the value for U(x) 00 as x->Inf. Numerical simulation indicates this at least. > Also, he pointed out R.E. Bellman and K.L. Cooke, > Differential-Difference Equations, as the standard reference in this > area. I will try to get hold of this book. Stefan Nilsson Marine ecology G.9ateborg university === Subject: Solving definite integral problem - newbie question I am studying for an exam and don't have access to my instructor today. Can someone help me with the following? I have a graph of the right half of a bell curve, a portion of which I must solve. The portion is from 0 to 1 (that is x=0 to x=1) and the equation is y=e^(-x^2/2). That is, e raised to (negative x squared, divided by two). I just need some help in the proper approach. I have been given the answer, but so far nothing that I have tried has helped me to get the answer. Pam === Subject: Re: Solving definite integral problem - newbie question > I am studying for an exam and don't have access to my instructor > today. > Can someone help me with the following? I have a graph of the right > half of a bell curve, a portion of which I must solve. > The portion is from 0 to 1 (that is x=0 to x=1) and the equation is > y=e^(-x^2/2). This can't be integrated in closed form. You have two main choices: 1. Note that the standard normal distribution is given by P(x) = e^(-x^2/2)/sqrt(2*pi). The integral of this from 0 to 1 is the probability that a standard normal random variable lies between 0 and 1. Since your y is sqrt(2*pi) P(x), then your answer is 2*pi*P(standard normal rv lies in 0<=x<=1). 2. Relate it to the error function erf(x) = (2/sqrt(pi))*integral(from 0 to x) e^(-t^2) dt How do you do that? integral(from 0 to 1) e^(-x^2/2) dx = ? Define t such that t^2 = x^2/2, i.e. t = x/sqrt(2) and dx = dt*sqrt(2) Substituting: integral(from x=0 to x=1) e^(-x^2/2) dx = integral(from t=0 to t=1/sqrt(2)) e^(-t^2) * sqrt(2) dt = sqrt(2) * integral(from 0 to 1/sqrt(2)) e^(-t^2) dt Which you can write in terms of erf(1/sqrt(2)). Thus, you can get the answer by looking up erf(1/sqrt(2)) and multiplying by an appropriate constant. - Randy === Subject: Matrix problem I got redirected to this newgroup for this problem: 5 customers (A thru E) requesting a load of supply: 25, 10, 3, 6 and 7 tons respectively. Delivery-cars (1 thru 6), capacity 8, 16, 24, 32, 32 and 32 tons resp. go each to 1 customer, go back and stay. In matrix are the haulage-costs: Cust: A B C D E Car: 1 6 7 8 9 10 2 2 6 9 11 14 3 6 7 10 11 12 4 6 8 10 12 13 5 8 10 11 13 13 6 7 8 10 12 13 Request:25 10 3 6 7 tons How do I solve this using Linear Programming (with objective function and constraints) so that total haulagecosts are minimal? I'm required to use Maple, tho in that newsgroup they say it can hardly be done that way. I have difficulties with this since it's an un-equal matrix. Huub === Subject: Derivatives f(p) = q where f(140) = 15000 and f'(140) = -100 R = pq What is dR/dp (p=140) I get (f'*p) + (f+f')*dp = (-100*140) + (14900*1) = -14000 + 14900 = +900 This solution is hardly eloquent and might even be wrong as I am unable to fit the rules for differentiation (sum, product or quotient) in a satisfactory manner. Anyone? Phil Holman === Subject: Re: Derivatives Phil Holman grava .88 la saucisse et au marteau: > f(p) = q > where f(140) = 15000 > and f'(140) = -100 > R = pq > What is dR/dp (p=140) > I get (f'*p) + (f+f')*dp > = (-100*140) + (14900*1) > = -14000 + 14900 > = +900 R = pf(p) dR/dp = f(p) + pf'(p) So dR/dp |p =140 = f(140) + 140*f'(140) = 15000 - 14000 = 1000 -- Nicolas === Subject: Re: Coordinates of World-Sheets in String theory > I'm trying to understand the manifold properties of world-sheets in string > theory. I'm told that manifolds are locally Euclidean. So I would like to > know the characteristics between the space-time coordinates of the > world-sheet given as x^i verses the 2D surface parameterized by (s,t). Are x > superscript i (written x^i) locally Euclidean? Are the coordinates (s,t) > locally Euclidean? Remember x^i are function of the parameters (s,t), or > x^i=x^i(s,t). There are manifolds that are not locally isomorphic to R^n, but rather to M^n, in other words to a Minkowski space of n dimension, one being timelike. Furthermore, you can have a manifold that is locally isomorphic to M^{n+m}, a Minkowski space with n timelike dimensions and m spacelike dimensions. The x^i are locally Minkowskian, they are the coordinates of the spacetime where the string lives. x^0 is taken timelike, the rest spacelike. The same applies to (s,t). s is taken spacelike and runs along the length of the string. t is taken timelike, short of internal time for the string. Kostas. === Subject: Re: Coordinates of World-Sheets in String theory at 11:42 PM, Mike said: >I'm trying to understand the manifold properties of world-sheets in >string theory. I'm told that manifolds are locally Euclidean. Their topological properties are locally Euclidean. They need not have a metric, and if they do it need not be locally Euclidean. In the case of String Theory, locally Minkowskian metrics are relevant. You don't ask about global properties, but various string theories impose global constraints on the world sheet. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be 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: Re: Factoring Integers near a power > I seem to recall that to factor an integer near a power was easier > than factoring arbitrary integers. Consider n = a^b +/- c with small c. I seem to recall there is a > (relatively) fast algorithm to factor numbers of this form (that takes > advantage of that property). Can anyone point me towards this--if it does indeed exist? > The Special Number Field Sieve (SNFS) can take advantage of this > special form. That's probably what you're thinking of. > I believe that there's a Special Quadratic Sieve too, but > this doesn't seem to have found its niche. > One thing that the n=a^b+/-c form can help is any calculation modulo n, > as the modular reduction can be performed very easily if an appropriate > representation is used. Therefore P-1/P+1/ECM/Rho and friends can all > be performed more swiftly for these than for arbitrary numbers. > Phil I am having trouble locating a reference to how this special form can === Subject: Re: Factoring Integers near a power > One thing that the n=a^b+/-c form can help is any calculation modulo n, > as the modular reduction can be performed very easily if an appropriate > representation is used. Therefore P-1/P+1/ECM/Rho and friends can all > be performed more swiftly for these than for arbitrary numbers. > I am having trouble locating a reference to how this special form can What's 12345678987654321 modulo 10^12+3? Grab Tom St. Denis' LibTomMath, and the accompanying book. I believe that Tom's code can take advantage of this quick modular reduction technique therein. Phil -- Unpatched IE vulnerability: Click hijacking Description: Pointing IE mouse events at non-IE/system windows Reference: http://safecenter.net/liudieyu/HijackClick/HijackClick-Content.HTM Exploit: http://safecenter.net/liudieyu/HijackClick/HijackClick2-MyPage.HTM === Subject: Re: Why is math so difficult for some people? > I dont get it.Im a perfect 0 at math. > Some people have no problems at all with it. I guess for the same reason that some people are tone-deaf, and the same reason that some people are not good for art or graphical design, and the same reason that some people are not able to make it through medicine school, and the same reason why some people simply can not learn other languages while some others can speak 5 languages with absolute mastery and with practically no noticeable accent... What is that universal reason? Each person has a completely different, unique brain/mind that works in a particular way -- this particular way is affected by the genetics AND by the exposure that the brain has had over the years of life (I guess the early years would have more impact on some areas, since maybe the brain is molded in irreversible or quasi-irreversible ways?) > Am I too dumb for math? You're asking the wrong question -- you're not dumb for math; you *might* have a brain/mind that works in a way that is not compatible with the way of thinking that you need for understanding maths (emphasis on the *might* -- you claim to be dumb for maths, but who knows, maybe you're much better newsgroup and just don't know it, or maybe you're a high-expectation kind of person, and then anything below Newton, or Gauss, or Fourier's brains means too dumb for math in your mind? :-)) Anyway, this, plus many of the things that have been already said (mainly about math being genuinely hard -- the more sophisticated level of math, the harder, of course) HTH, Carlos -- === Subject: Re: Why is math so difficult for some people? at 12:14 PM, Gib Bogle said: >Just as the >answer to a mathematical question is either right or right, and a >proof is either valid or invalid, Not where I went to school. A perfectly valid prrof would be marked down if it was not elegant; quite properly, IMHO. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be 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: Re: Why is math so difficult for some people? >> Well, that's already been proved wrong. >> Since if you find a chemist that knows anything >> about science, you'll be a first. >Are you saying that chemists are not scientists, therefore brain >chemistry does not influence intelligence? ZZBunker is a Markov text generator and can not be viewed as saying anything. It would not pass the classic Turing test for humanity. - Randy === Subject: Inverse of Matrix The Inverse of matrix A is normal defined as a matrix X with the property AX = XA = I(identity matrix) can the definition be reduced as an Inverse von A is a matrix X with AX = I ??? Can man conclude XA = I from AX = I ? when not, please give me an opposite example. pan === Subject: Re: Inverse of Matrix > The Inverse of matrix A is normal defined as a matrix X with the property > AX = XA = I(identity matrix) > can the definition be reduced as > an Inverse von A is a matrix X with AX = I ??? > Can man conclude XA = I from AX = I ? when not, please give me an opposite > example. A = (1 0), X = 2 x 1 matrix with entries 1, 0. AX = 1, the 1 x 1 identity matrix. XA = 2 x 2 matrix with 1 in upper left corner, zero everywhere else, not the 2 x 2 identity matrix. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Inverse of Matrix >The Inverse of matrix A is normal defined as a matrix X with the property > AX = XA = I(identity matrix) Assuming the identity matrix commutes, we can always write: AA^(-1)A = AI = IA = A^(-1)AA => AA^(-1) = A^(-1)A Now all you have to show is that the identity matrix in fact commutes. === Subject: Re: Inverse of Matrix >>The Inverse of matrix A is normal defined as a matrix X with the property >> AX = XA = I(identity matrix) > Assuming the identity matrix commutes, we can always write: > AA^(-1)A = AI = IA = A^(-1)AA => AA^(-1) = A^(-1)A > Now all you have to show is that the identity matrix in fact commutes. Unfortunately this does not show anything. In arbitrary rings, one cannot conclude from a*b = 1 that b*a = 1. Example: Let V be the vector space of sequences over some field K. Then we may define the linear maps right shift R: (a1, a2, a3, ...) |--> ( 0, a1, a2, ...) left shift L: (a1, a2, a3, ...) |--> (a2, a3, a4, ...) Obviously L*R = id, although R*L <> id because R is not surjective (nor is L injective). But in our case (finite dimensions), a linear mapping is injective if and only if it is surjective. So if we have A*B = id, we conclude that B is injective, thus surjective and so B^(-1) exists. So we get by multiplying with it from the right A = B^(-1). If you wish, mulitply again with B from the left so that you get B*A = id. -- reverse my forename for mail! === Subject: Re: Inverse of Matrix > But in our case (finite dimensions), a linear mapping is injective if and > only if it is surjective. So if we have A*B = id, we conclude that B is > injective, thus surjective and so B^(-1) exists. So we get by multiplying > with it from the right A = B^(-1). If you wish, mulitply again with B from > the left so that you get B*A = id. Oops, I forgot to mention that I actually meant a linear mapping from a vector space into itself, that is an endomorphism. See also Gerry's post; so your procedure is valid for (n x m) matrices if and only if n=m. -- reverse my forename for mail! === Subject: Riemann's Zeta Function by H. M. Edwards Have read through the Schaum's Outline on Complex Analysis. I am stuck on a step of Riemann's Zeta Function by H. M. Edwards. On p. 13 he derives the functional equation of the RZF. Zeta(s)=Gamma(1-s)(2*pi)^s-1(2sin(s*pi/2)Zeta(1-s). How does Sigma_1^infinity n^(s-1) translate to Zeta(1-s)? Isn't 1-s the negative of s-1? There are other things in this section I am struggling to understand, Paul === Subject: Re: Riemann's Zeta Function by H. M. Edwards PaulHjelmstad says... >Have read through the Schaum's Outline on Complex Analysis. I am stuck >a step of Riemann's Zeta Function by H. M. Edwards. On p. 13 he >derives the functional equation of the RZF. >Zeta(s)=Gamma(1-s)(2*pi)^s-1(2sin(s*pi/2)Zeta(1-s). >How does Sigma_1^infinity n^(s-1) translate to Zeta(1-s)? Isn't 1-s >the negative of s-1? I was puzzling over exactly this point about a year ago, and the best explanation that I saw was given in notes by D. Bump http://www.maths.ex.ac.uk/~mwatkins/zeta/bump-fnleqn.ps It's in postscript, if you have a reader for that. -- Daryl McCullough Ithaca, NY === Subject: Re: Riemann's Zeta Function by H. M. Edwards > PaulHjelmstad says... >Have read through the Schaum's Outline on Complex Analysis. I am stuck >on >a step of Riemann's Zeta Function by H. M. Edwards. On p. 13 he >derives the functional equation of the RZF. >Zeta(s)=Gamma(1-s)(2*pi)^s-1(2sin(s*pi/2)Zeta(1-s). >How does Sigma_1^infinity n^(s-1) translate to Zeta(1-s)? Isn't 1-s >the negative of s-1? > I was puzzling over exactly this point about a year ago, and the best > explanation that I saw was given in notes by D. Bump > http://www.maths.ex.ac.uk/~mwatkins/zeta/bump-fnleqn.ps > It's in postscript, if you have a reader for that. Duh! (I can't believe I got stuck on something so simple.) OK, so backtracking a few lines: Take Gamma(1-s)(-2*pi*n)^s-1. Summing over all integers n other than n=0 and using 3) -Zeta(s)-Sigma(Gamma 1-s/2*pi*i)Int_|x+-2*pi*n|=Epsilon ((-x)^s)/((e^x)-1) * dx/x=0 then gives Zeta(s)=Sigma_n=1^infinity(Gamma(1-s)[(-2*pi*n)^s-1 + (2*pi*n)^s-1]. I don't see how 3) is implemented to give this result. Sorry ahead of time for any Ughs. === Subject: Re: Riemann's Zeta Function by H. M. Edwards > Have read through the Schaum's Outline on Complex Analysis. I am stuck > on > a step of Riemann's Zeta Function by H. M. Edwards. On p. 13 he > derives the functional equation of the RZF. > Zeta(s)=Gamma(1-s)(2*pi)^s-1(2sin(s*pi/2)Zeta(1-s). > How does Sigma_1^infinity n^(s-1) translate to Zeta(1-s)? Isn't 1-s > the negative of s-1? Yes. It could also be written as Sum_1^infinity 1/n^(1-s). -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Riemann's Zeta Function by H. M. Edwards > Have read through the Schaum's Outline on Complex Analysis. I am stuck > on > a step of Riemann's Zeta Function by H. M. Edwards. On p. 13 he > derives the functional equation of the RZF. > Zeta(s)=Gamma(1-s)(2*pi)^s-1(2sin(s*pi/2)Zeta(1-s). > How does Sigma_1^infinity n^(s-1) translate to Zeta(1-s)? Cos it equals sum 1/n^{1-s}. > Isn't 1-s > the negative of s-1? Yes. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: someone told me ... === >Subject: someone told me ... That, somewhere in the decimal expansion of pi, >there is ASCII code for the date of my birth, the >date of my death, a thumb-scale sketch of my >life on this earth, and even a few JPEG's of me >at various activies during my life. This is sci.math, and that statement must be >true, or false, or Godel unknowable. Which is it? If you don't know the answer and >must speculate, please don't. If you do know >the answer, please post. Ben As others have said, the answer is maybe. But I wouldn't waste > any time looking for that JPEG of you with Pamela Anderson. > On my cheap digital camera, a good JPEG has over a million > bits. How big is a million bits? Try imagining something a little > smaller like 2000 bits: http://members.aol.com/mensanator666/fun/2000_bit.htm You'll never find a simple icon, let alone a JPEG. > Does this mean that if Pi is normal, then you could find _any_ JPEG in > it? If so, it seems like the number Pi would contain all information > known to mankind, and even that which has not yet been discovered, > encoded in its decimal expansion. Yes. So somewhere in it would be written in Russian and encoded in ASCII, the message the poster known as Nobody will die on Dec. 14, 2107. And the same message in English, in Spanish, in Chinese (using 16-bit coding), in Old French, and in Klingon. Is that the kind of thing that's bothering you? Would it help to know that the same message would appear, in all of those languages, using every other date that can be encoded including Christmas day, 1470938 BCE? The message George Bush will win the US Presidential Election in 2000 would appear, but so does Al Franken will win the US Presidential Election in 1996, and so does George Bush was eaten by ferrets in 1984. As prophecy goes, digits of normal numbers aren't extremely useful. - Randy === Subject: Re: someone told me ... >Does this mean that if Pi is normal, then you could find _any_ JPEG in >it? The first number to be proven normal is Champerowne's number, .123456789101112131415161718192021.... It contains all finite sequences of digits in it, with the advantage that given the pattern, it's much easier to compute where in the sequence it first appears. It doesn't help you find interesting ones. Jorge Luis Borges has a story about a library containing all possible books. It's not a useful library, because there's no guide to where the worthwhile ones are. Keith Ramsay === Subject: Re: someone told me ... > Should be easy. What do you look like? I look exactly like Santa Claus. What do you look like? Ben === Subject: Re: someone told me ... > ... do you think this would help us decypher the > laws of the universe? No, I just wanted to see that JPEG with Brittany Spears that someone mentioned. Well, actually I'm not that particular. Your sister? Is she down there in pi, or not? Ben === Subject: Re: someone told me ... > But I wouldn't waste any time looking for that JPEG > of you with Pamela Anderson. Never mind Pamela Anderson. How about that cute little lady I knew in high school, with all her equipment hanging out? Is THAT picture down there somewhere in pi? I think this is really a meaningful question. If it isn't, no one has yet offered a proof. Ben === Subject: infinite summation and complex integration When is it ok to say sum(complexcontourintegral(fj(z)dz),j=0..oo) = complexcontourintegral(sum(fj(z),j=0..oo)) === Subject: Re: infinite summation and complex integration > When is it ok to say > sum(complexcontourintegral(fj(z)dz),j=0..oo) = > complexcontourintegral(sum(fj(z),j=0..oo)) It's not always ok. So why don't you tell us what assumptions you're working with? === Subject: Re: infinite summation and complex integration > When is it ok to say sum(complexcontourintegral(fj(z)dz),j=0..oo) = > complexcontourintegral(sum(fj(z),j=0..oo)) > It's not always ok. So why don't you tell us what assumptions you're > working with? Sorry, I misread your when as why. Let E denote the range of the contour. I'll assume each fj is continuous. The desired conclusion follows if you assume sum(fj(z),j=0..oo) converges uniformly on E. And that is often shown by using the Weierstrass M-test: If |fj(z)| <= Mj for all z in E, and M1 + M2 + ... < oo, then sum(fj(z),j=0..oo) convergess uniformly on E. There are other conditions that will work, but the above is simple and quite common. === Subject: Bible 1, Darwin 0! And we are talking pure archeology? Forbidden archeology by Cremo, Thompson! Like the name says: Forbidden I urge you not to read it! === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? > Forbidden archeology by Cremo, Thompson! > Like the name says: Forbidden > I urge you not to read it! Hmm. Bible 1, Darwin 0! Since 0! = 1, I guess it's a tie so far. === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? > Forbidden archeology by Cremo, Thompson! > Like the name says: Forbidden > I urge you not to read it! If I wanted to read Bible propoganda I'd go to church. === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? > Forbidden archeology by Cremo, Thompson! > Like the name says: Forbidden > I urge you not to read it! > If I wanted to read Bible propoganda I'd go to church. Cremo's work has nothing to do with religion.. not that I agree with his conclusions... he does produce some interesting evidence for a divergent human time line... Paul R. Mays ---------------------------------------------------------------------------- - Some where within the Quantum State Http://Paul.Mays.Com/story.html http://paul.mays.com/mayday.html http://paul.mays.com/rainy.html Furthermore it becomes folly to seek a boundary between synthetic statements, which hold contingently on experience, and analytic statements, which hold come what may. Any statement can be held true come what may, if we make drastic enough adjustments elsewhere in the system. Even a statement very close to the periphery can be held true in the face of recalcitrant experience by pleading hallucination or by amending certain statements of the kind called logical laws. Conversely, by the same token, no statement is immune to revision. Revision even of the logical law of the excluded middle has been proposed as a means of simplifying quantum mechanics. - Willard Van Orman Quine === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? >Forbidden archeology by Cremo, Thompson! >Like the name says: Forbidden >I urge you not to read it! -- Let us learn to dream, gentlemen, then perhaps we shall find the truth... But let us beware of publishing our dreams before they have been put to the proof by the waking understanding. -- Friedrich August Kekul.8e === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? John Tapper says... >Forbidden archeology by Cremo, Thompson! >Like the name says: Forbidden >I urge you not to read it! Okay, I won't. -- Daryl McCullough Ithaca, NY === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? > Forbidden archeology by Cremo, Thompson! Physical reality, by nobody in particular. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? Forbidden archeology by Cremo, Thompson! > Physical reality, by nobody in particular. U again. Same attitude, often wrong. How much Physics did u memorize today? All encyclopedia, negligible independent thought. === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? did some sarious thank'n and scribbled: >> Forbidden archeology by Cremo, Thompson! >> Physical reality, by nobody in particular. >U again. Same attitude, often wrong. How much Physics did u memorize >today? All encyclopedia, negligible independent thought. If you guys are about to have a flame war might you unspam the NG line before you start. === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? John Tapper replied: > Forbidden archeology by Cremo, Thompson! Ain't this Rick Sobie's gig? > Like the name says: Forbidden > I urge you not to read it! I've looked into it (found it even though the book was misnamed). What I don't see is any connection between it and the Bible, despite the fact that bible thumpers think it disproves evolution. It neither disproves evolution nor proves creation. If it is all true, then clearly reality is more complex than we know (and I should think most here would admit that right off). Can you point out some specific item you think expecially telling, then tell me what you think it says? Rich === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? > John Tapper replied: > Forbidden archeology by Cremo, Thompson! > Ain't this Rick Sobie's gig? > Like the name says: Forbidden I urge you not to read it! > I've looked into it (found it even though the book was > misnamed). What I don't see is any connection between it > and the Bible, despite the fact that bible thumpers think > it disproves evolution. It neither disproves evolution > nor proves creation. If it is all true, then clearly > reality is more complex than we know (and I should think > most here would admit that right off). > Can you point out some specific item you think expecially > telling, then tell me what you think it says? > Rich Archaeology has never been able to give definitive proof, any more than any other science really can when you are looking at events that took place in a distant past or will take place in distant future. What science can do is establish probabilities, or provide evidence supporting a theory. For instance, were archaeologists to discover the ruins of Sodom and Gamorrah, or the remnants of Noah's Ark, this would support the fact that these PARTICULAR events did occur, and, by association, the bolster the case that for the rest of, in this instance, the book of Genesis; but by proving a single event in the book probably occurred, we cannot prove that the entire book is true on that count. Quite frankly though, I don't understand why Christians are wasting our time trying to disprove a theory like evolution; the Bible isn't about evolution, or even science, and spending time trying to convince people that it is seems rather fruitless. === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? > John Tapper replied: > Forbidden archeology by Cremo, Thompson! > Ain't this Rick Sobie's gig? > Like the name says: Forbidden I urge you not to read it! > I've looked into it (found it even though the book was > misnamed). What I don't see is any connection between it > and the Bible, despite the fact that bible thumpers think > it disproves evolution. It neither disproves evolution > nor proves creation. If it is all true, then clearly > reality is more complex than we know (and I should think > most here would admit that right off). > Can you point out some specific item you think expecially > telling, then tell me what you think it says? > Rich The aim of the book is not to prove things that the Bible said, I just put a catchy title. The aim is to question Darwin's theory. Nice theory, but is it true? The skeleton time periods and a lot of other things do not seem to be consistent with Darwin's claims. And I am putting it mildly. === Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology? > The aim of the book is not to prove things that the Bible said, I just put > a catchy title. The aim is to question Darwin's theory. Nice theory, but > is it true? Wrong question. One can only prove the falsehood of a scientific theory. > The skeleton time periods and a lot of other things do not > seem to be consistent with Darwin's claims. And I am putting it mildly. Well, there is a whole scientific community who disagrees with you. === Subject: Re: Naive Q: Set theory, logic - which comes first? >>A natural number is a sequence of symbols in which only the symbol '|' >>occurs. >> That is ONE representation of a natural number. >Indeed. But identifying mathematical objects with what are really >representations of them is very common, and I was just doing what >everyone else does, though maybe I should not have. > This is an extremely poor way of attempting to understand > mathematical objects. I agree. The first thing I do is always to abstract away from inessentials, which includes representations (or are there cases where it is necessary to use representations?). I do not see how people manage to do mathematics with so much garbage around. > Take, for example, the following: > A positive integer is a non-empty finite string of symbols > from a collection of 15 symbols, of which one of the > symbols cannot be the first element of the string. > Now what led me to produce this statement? You might > consider that it came from the base-15 representation of > integers, but what led me to this was the Sumerian > representation in base 60, in which a digit is either > the zero symbol, or (to use our terminology) one of five > symbols for multiples of ten, or one of nine symbols for > multiples of one, or a symbol for a multiple of ten > followed by one for a multiple of one. But a string > without attention to the spacing can be interpreted in > only one manner. > So it is not possible to tell from a description of a > representation how the representation is to be interpreted. This is a good example. However, I think that with the representation of natural numbers I gave, unlike this one, there is no risk of misunderstandings. Two questions whose answers I am seeking are: 1) What exactly might we mean by representation? 2) How do we define a mathematical object if we do not want to identify it with some representation of it? Natural numbers pose a special problem since they may be viewed in two different ways. We may view them either as something defined by induction, or as the smallest algebraic structure which is closed under the operations 0, 1 , +, and * (theese being subject to the usual laws). Mattias === Subject: Re: Naive Q: Set theory, logic - which comes first? >>A natural number is a sequence of symbols in which only the symbol '|' >>occurs. >> That is ONE representation of a natural number. >Indeed. But identifying mathematical objects with what are really >representations of them is very common, and I was just doing what >everyone else does, though maybe I should not have. This is an extremely poor way of attempting to understand > mathematical objects. > I agree. The first thing I do is always to abstract away from > inessentials, which includes representations (or are there cases where > it is necessary to use representations?). I do not see how people > manage to do mathematics with so much garbage around. > Take, for example, the following: A positive integer is a non-empty finite string of symbols > from a collection of 15 symbols, of which one of the > symbols cannot be the first element of the string. Now what led me to produce this statement? You might > consider that it came from the base-15 representation of > integers, but what led me to this was the Sumerian > representation in base 60, in which a digit is either > the zero symbol, or (to use our terminology) one of five > symbols for multiples of ten, or one of nine symbols for > multiples of one, or a symbol for a multiple of ten > followed by one for a multiple of one. But a string > without attention to the spacing can be interpreted in > only one manner. So it is not possible to tell from a description of a > representation how the representation is to be interpreted. > This is a good example. However, I think that with the representation > of natural numbers I gave, unlike this one, there is no risk of > misunderstandings. > Two questions whose answers I am seeking are: > 1) What exactly might we mean by representation? > 2) How do we define a mathematical object if we do not want to > identify it with some representation of it? > Natural numbers pose a special problem since they may be viewed in two > different ways. We may view them either as something defined by > induction, or as the smallest algebraic structure which is closed > under the operations 0, 1 , +, and * (theese being subject to the > usual laws). Well, that's not such a big problem. Since that's basically what Goedel numbering is all about. The problem is exclusively related to the phrase defined by induction. just as it has been for the last several thoushand years. Since you have to know something about geometry rather than natural numbers to define inductive logic. > Mattias === Subject: Re: Naive Q: Set theory, logic - which comes first? | Close; one reasons from (exceeds v ~exceeds). Even with an axiom | system in which exceeds is undecidable, the reasoning is valid. Why do you think so? Are the undecidable statements sometimes true or false? Keith Ramsay === Subject: Re: Math dependency logic >> I'd like to state for the record, so James doesn't accuse me, that by doing >> the above transformation, I do not change the equation. It is just in a more >> useful form for determining characteristics and integrating when needed. >> David Moran > You're still trying to **** with me David Moran because NOTHING in > what Arturo Magidin is doing makes any sense MATHEMATICALLY!!! > I don't know what's ******* wrong with you people, or why you think > that you can come here and ******** for eternity without consequence, > but I'm ******* tired of the stupidity. I'm ******* tired of the > nonsense. > I'm ******* tired of people coming in here and saying ******** as if > mathematics is just some useless game and the truth doesn't matter. > You need to quit ******* around David Moran. You, Randy Poe, and > Arturo Magidin among others needs to get a ******* CLUE!!! It's very easy to tell when James realizes that he's wrong and that he can't bluff his way past it. He loses the thin veneer of civilized behavior that he usually tries to assume while posting here, and reverts to the guttertrash language of his everyday life. It's a clear sign that his megalomania is taking a (temporary) back seat to his insecurities. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Math dependency logic > Readers should note the difference, and consider that Nora Baron is > deliberately trying to obfuscate. Not at all. I gave a proof. You call it obfuscation. Let > the audience decide. It's math Nora Baron and not a poll. My position all along has been that for certain values of x, it's > provable that two of the a's are not algebraic integers, which is the > error in core. In spite of this contradiction, which Harris has seen dozens of > times, and in spite of our identifying the exact line in his proof > where he makes a false claim, he continues to insist that he has found > a flaw in the *definition* of algebraic integers, that they >are incomplete. There is a claim that a line in my proof has been identified as having > an error. Then give it Nora Baron. > Certainly: > Well that means that P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 +1)(5 b_3/7 + 22) is the only way that the constant terms keep matching. That's the conclusion, which follows from previous steps. Are you saying that you agree with the previous steps Nora Baron? Just trying to make sure that you're claiming that only the *last* > statement is an error. Remember, a proof begins with a truth and > proceeds by logical steps to a conclusion which then must be true. Are you stating that all previous steps up to the conclusion are fine? > You may be surprised by this answer. Your argument is mostly correct. > No, I'm not surprised as you're playing psychological games Nora > Baron and it's a well-known tactic to try and seem reasonable by > making such a claim. > But there are a couple of places where problems occur with your argument. > One of these can be fixed. Below is a direct quote from one of your many recent posts. I have > inserted a few comments to address your question. > Notice how I'll be strongly emphasizing constant terms all the way >down. P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 which has a constant term that is 1078. > No problem so far. >Well P(x) can also be written out as [*] P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 so I can factor to get P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the a's are roots of [**] a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > This is mistake number 1. It is true that you *can* factor the polynomial > [*] above so that the a's are roots of [**]. However, in what you do later > you need something stronger. You need that the a's MUST be roots of [*]. > In general they are not. If you substitute a1 = 1, a2 = 1562, and a3 = 28, > you will find that P(1) = (5*a1 + 7)*(5*a2 + 7)*(5*a3 + 7). You can check the arithmetic. That's all it is. > Actually, there is no mistake there as I say that the a's from the > given factorization are roots of the given expression. > And where the can applies is to the factorization as there are an > *infinity* of factorizations for P(x), but I'm pointing out that one > of them is the one that follows. > So, there's a direct statement telling *exactly* what the a's are for > the factorization, which readers can see for themselves. If you had bothered to read further, you would see that I agree this is not really a problem, or at least it is a problem that is easily fixed. But as I mentioned at the top there are two problems. The second one is the real one. You will not be able to fix it. But you for some reason deleted off all reference to it. So here it is again. Evade it again if you want, but at least quit pretending that you are willing to respond to all objections. Deleted part of my post: This problem is easy to fix. Just replace 5 by y or some other symbol. You have then restored P(x) to being a polynomial in y. You would then be factoring it in the form P(x) = (a1*y + 7)*(a2*y + 7)*(a3*y + 7), and indeed you can then conclude CORRECTLY that the a's *must be* roots of [**]. Thus this is one place where you have a mistake, but it is easily fixed. I will pretend that you have made the right fix and proceed. >Notice it *appears* that the constant terms for the three factors are >all 7, which can't be right, as the constant term of P(x) is 1078, so >setting x=0, reveals >P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) >as the cubic defining the a's with x=0 is >a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2 >for 0, so that leaves a_3 with a value of 3 when x=0. >So let a_3 = b_3 + 3, where I keep indices matched. Then I have >P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7) >P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) >and now my constant terms work out correctly. All this is basically OK, conditional on the fix described above. >But P(x) has 49 as a factor as every term in >P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 >gives me 22, as the new constant term. Yes. Also true. No doubt about that. >Well that means that >P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) >is the only way that the constant terms keep matching. No. This last step is where your most serious error is. You are going to say But you accepted all the steps leading up to this!!! This last step is just summing up all that went before! You are trying to cheat me out of my proof!!! You are lying yet again, 'Nora Baron' !!! No again. I am not lying. Here is an analogy. Theorem: 2 + 2 = 5. Proof. Step 1. 2 + 0 = 2. Step 2. cos(0) = 1. Step 3. sin(pi) = 0. Step 4. A triangle with two equal sides has two equal angles. Step 5. Therefore 2 + 2 = 5, QED. All the steps preceding the last one are correct. The last one is a leap into space. The preceding steps do not justify the conclusion. That is EXACTLY what you have in your argument. There are conceivably many ways that the factorization of P(x) could occur. Note that, in particular, that 5*b3 + 22 is *not necessarily* coprime to 7 in the algebraic integers. Once you recognize this possibility, the rest of your factorization, which you think is forced, falls apart. It is NOT the only way P(x) can be factored. You appear to think that because 22 is constant and 22 is coprime to 7, and 5*b3 is nonconstant (i.e. is dependent on x), then 5*b3 + 22 must be coprime to 7. Just plain not true. Suppose, for example, for some specific x, b3 = 4. 5*4 + 22 = 42, which is assuredly not coprime to 7, right? Have you proved somewhere that b3(x) cannot be 4, for all possible values of x ? Not that I have seen. Saying 5*b3 + 22 is not coprime to 7, as you well know, is not the same as saying that 5*b3 + 22 is DIVISIBLE by 7. What it means is that there is an algebraic integer w3 which divides 5*b3 + 22 and also divides 7. Similarly, there are algebraic integers w1 and w2 which divide 7 and also divide a1 and a2. Finally, w1*w2*w3 = 49. The correct form of the factorization is P(x)/49 = (5*a1/w1 + 7/w1)*(5*a2/w2 + 7/w2)*(5*(b3+3)/w3 + 7/w3). This is in fact what actually happens. All of a1/w1, 7/w1, a2/w2, 7/w2, a3/w3 = (b3 + 3)/w3, and 7/w3 are algebraic integers. There is no reason to assume that the factorization needs to go outside the algebraic integers. You do not arrive at a contradiction which requires you to invoke objects. You have not shown an error in 'core' mathematics. Again: your main error is indeed in your last line. It is not at all justified by what preceded it. In particular, it is wrong to assume that 5*b3 + 22 is necessarily coprime to 7. Nora B. > James Harris === Subject: Re: Math dependency logic Adjunct Assistant Professor at the University of Montana. >> [.snip.] >> Since you have repeated this argument so many times this >>week, I guess it is worthwhile pointing out another >>place where it is incorrect. > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 which has a constant term that is 1078. Well P(x) can also be written out as P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 so I can factor to get P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the a's are roots of [**] a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > No. The a's do NOT have to be roots of this polynomial. >> In all fairness, he did not say they have to be. He said he CAN factor >> it with the a's being roots of that polynomial. That statement is >> accurate. >> [.rest deleted.] > He later concludes that the factorization MUST be of >the form >[*] (5*a_1/7 + 1)(5*a_2/7 + 1)(5*b3 + 22). Well, first, we both agree that this statement is incorrect, so it seems to me that it hardly matters. Rather, here is how I see his latest attempt, once he has hidden the variable that used to be called x... You MAY produce factors of his polynomial by choosing a1, a2, a3 to be functions of x, defined as the three roots of the polynomial [**]. IF we choose such a's, then certain things WILL happen at the value x=0. THEN he concludes (falsely), that certain things will also happen at arbitrary values of x, IF we choose a1(x), a2(x), and a3(x) to be defined as the three roots of [**]. That is, rather than state that the a's CAN be chosen, he should state that he ASSUMES the a's ARE chosen that way, and proceed. So, you are correct that a plain reading leads to the wrong conclusion; but then, so does a corrected reading. I did not mean to imply your example was wrong, but perhaps that he placing of your comment obscured it; the point at which you complained was not the point at which he was using that the a's ->were<- given by the polynomial. [.snip.] Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Math dependency logic In sci.math, James Harris Sorry to but in: >> ... >> My position all along has been that for certain values of x, it's >> provable that two of the a's are not algebraic integers, which is the >> error in core. >> The error is in your core. The a's are all algebraic integers. And in >> general all three are not coprime to 7 (in the algebraic integers). > Your first sentence is incorrect. The second is correct. The third > is correct as well, and is an important point. > Ultimately my argument relies on numbers like 7 being NUMBERS, not > variables dependent on x, and on the distributive property. What kind of numbers? Natural numbers (positive integers)? Integers? Rational numbers? Numbers constructible from Q given square roots and the standard arithmetic operations? (I don't know if there's an official name for this, but it's a proper subset of the algebraic numbers.) Algebraic integers? Algebraic numbers? Complex numbers? Please clarify. Also, please note that the polynomial x^3 - 49 can be factored (x - r1)(x - r2)(x - r3) but none of r1, r2, and r3 are divisible by 7, even if one liberalizes the concept of a divisible by b to include the existence of an equality a / b = c where all of a, b, and c are algebraic integers, (In this case b = 7.) > Such an argument can't have a hidden error, and in fact, as the > argument is now well worked out, it's perfect as required. > It is a proof. Well, I'll have to find it, then, in its fully well worked out form, to critique it properly. Or perhaps you could simply point me thereto? (Your P(x) was originally from an interesting source, for example.) I don't really like arguing disconnected subpoints. > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: PDE-related problem During my research, I ran into the following PDE: frac{partial I(x,y,t)}{partial t}=left{begin{array}{cc}Phi_{0}(I)&(x,y)in Omega_{0}Phi_{1}(I)&(x,y)in Omega_{1}end{array}right. where Omega=[0,M] times [0,N]=Omega_{0} cup Omega_{1}. Basically, it is the combination of two PDEs with a shared boundary (between Omega_{0} and Omega_{1}). I want to study if it is a well-posed problem. I understand maximum principle is the fundamental tool for studying the uniqueness of PDE-related problems. But I don't know if there are already known results on the above kind of problem. === Subject: Re: PDE-related problem > frac{partial I(x,y,t)}{partial > t}=left{begin{array}{cc}Phi_{0}(I)&(x,y)in > Omega_{0}Phi_{1}(I)&(x,y)in Omega_{1}end{array}right. > where Omega=[0,M] times [0,N]=Omega_{0} cup Omega_{1}. === Subject: Re: T1 space.......problem... > hello........ > ---------------------------- > let A : T1 space > show that derived set of A is closed set > ------------------------------- > if A is finite, it is trivial. > but in the infinite, i don't know. > advice ...please...thank you. Since A is T1, it follows that, if u is a limit point of A, then every neighborhood of u contains infinitely many elements of A. Let A' be the derived set of A and CA' its complement. If x is in CA', then x has a neighborhood V that contains only finitely many elements of A. Since V is a neighborhood of all of its elements, this same condition satisfied by x holds equally well for every y in V. Therefore, no element of V is a limit point of A, which shows V is contained in CA'. It follows every x in CA' has a neighborhood contained in CA', which implies CA' is open and A' is closed. Actually, this conclusion is more general than stated. If A is a T1 topological space, then the derived set of every subset of A is closed. The proof is exactly as above. Artur === Subject: Re: BEWARE: Springer books fall apart in 2 days, still NNTP-Posting-User: [rskvOLAmZ1kNcXOULRSDjeLXlbg6BTRb] > On July 19, 1996 I reported here [1] that a couple of Springer > UTM textbooks literally fell apart after being gently opened > only a handful of times. In that thread others reported similar > experiences. Since then I've had the same problem with other > Springer textbooks, and a recent thread here [2] indicates that > Springer has yet to fix the problems. I have had the same experience with the GTM series. It doesn't seem to be universal, nor was it so drastic as you indicate. In my case the book (Introduction to Smooth Manifolds, Lee) fell absolutely to pieces over the course of a 20-week class. Certainly I used the book heavily. But these books are, I can only suppose, meant either for extensive study or years of reference. I do not think it is so much to ask that a $50 paperback book last long enough to learn the material. On the other hand I own other Springer books that have fared well. None of them was used quite as hard as this one. We'll see how Hartshorne holds up. (snicker) It seems worth mentioning that most of the other students in the class had a similar experience with the Lee book, even those who splurged and got the hardcover edition. At least one person tired of trying to keep the book together and simply carried around multiple pieces. This feels like an even more egregious robbery when I think about the fact that this book was used in manuscript form for the same course here, prior to the Springer printing. Of course then it was bound in a copy shop and cost something like $10, and no pages fell out. What a rip. (Springer also insisted that the author of the book remove his online version at the time of publication.) And I need not even mention Dover... > Does Springer have no quality controls? > Does Springer have no pride? I will treat these as rhetorical. > What can we do about this? Well, the suggestion that comes to mind that is the least illegal is for instructors to try to use texts freely available as much as possible, Hatcher's algebraic topology book, for instance. I don't know what he had to go through with his publisher to be able to make it available for free online, but it's there. As far as the less-than-strictly-legal side goes, I imagine that students constitute a large part of Springer's market for these books. Especially graduate students. Poor graduate students. Poor graduate students who already don't really have a lot of dough to throw at books, even those that are well-constructed and will last for years. Let's photocopy them instead. I can't imagine the royalties are significant for any kind of math book other than those huge/ridiculous calculus tomes used for 100-level courses, and that's a racket in and of itself. Of course I have never published a book, mathematical or otherwise, and would be interested to hear if my assumptions are incorrect in this instance. I would even be willing to write checks to the authors directly, for some fraction of the difference in cost. So, samizdat mathematics texts, who's in? -david === Subject: Re: Limsup, Liminf > If a sequence of reals {s_n} converges, then the corresponding Limsup > equals its Liminf. These 2, however, appear to be independent of the > particular ordering of {s_n}. > This would be true, by looking at their definitions and considering that > for any epsilon, the s_i that cause a problem can be excluded finitely > far into the sequence, regardless of the order. > Is it accurate to state, then, that > Limsup=Liminf does not necessarily imply the convergence of {s_n}? > Any help/explanation appreciated. > {s_n} converges iff Limsup=Liminf. Think about epsilon/N proofs. for a given relationship (n,s_n). For an arbitrary e, there exists an N such that n>N ----> |s_n - A|A good Sunday :) >Sorry, that i'm firtsly posted in german, here the english post: >1. Do anyone of you know, when the dicrete log was invented ? >2. Know's anyone of you a prove of the diffe - hellman algorithm ? > (or can anyone of you explain me, why it does function ?) Have a look onto Andrew Odllyzko's page and publication list http://www.dtc.umn.edu/~odlyzko/index.html http://www.dtc.umn.edu/~odlyzko/doc/complete.html Hermann -- === Subject: FW: SAB problem > Hi all, > I have a problem finding a solution to an equation: > Some info first: S, A, and B are known numbers (and if this helps, S=Range[0,691], A=0.0003, B=1.975). > Solve for X: > S = 0.1*[(0.001+X)^(A*X^B) - 1] > I have come to a step where: > [(S/0.1)+1]^(1/A) = (0.001+X)^(X^B) > and cannot continue. I tried with raising e to the both sides and > taking the natural log after in order to remove that X^B somehow. I'm > stuck. > Martin === Subject: Re: UFOs and Dark Energy > Prof. Littleton agrees that the available UFO evidence is soft, but > thinks that the intelligence agencies have squirrelled away some hard > evidence from supposed crashes at Roswell and Kecksburg. The Roswell > crash yarns have been quite effectively debunked by Pflock, Klass, and > others, I have nothing to add. As for Kecksburg, is he truly unaware > that this crashing UFO was simply a fable spun from confusion over > the great fireball meteor of December 9, 1965? See my web page debunking > this at > http://www.debunker.com/kecksburg.html . > Truly a house of cards. You UFO > believers are *so easily* fooled. > And on such trivia are new scientific theories to be built?????? > I replied to Sheaffer's Straw Man tactic: > You seem to continually misunderstand that my physics theories in > http://qedcorp.com/APS/StarGate1.mov > are not motivated by the above incidents you mention. Indeed, this is > the first I even heard of Kecksburg. > Again I say for the Nth time, the motivations of my theory are: > I.1 Stability of the electron (Abraham, Becker, Lorentz ~ 1900) > Electron as extended object at e^2/mc^ ~ 10^-13 cm does not explode from > Coulomb self-replusion. > I.2 Universal slope of Regge trajectories of hadronic resonances as > Kerr-Newmann micro-geons in strong short range gravity (my work of 1973 > that caused Abdus Salam to invite me to Trieste). > 1.3 Electron and quarks look more an more pointlike ~ 10^-18 cm for > high-energy deep probes. > II Astrophysics - The Blackett Effect (see papers by Saul-Paul Sirag > including one in NATURE) > i.e. Effective charge of many rotating astrophysical objects that seem > to obey > Effective Electrical Circulating Charge Making Magnetic Moment = > G(Newton)^1/2 (Mass of Object) > III. Hermann Bondi's negative matter propulsion talk at Cornell ~ 1960 > when he was Chief Scientist of the British Ministry of Defense. > This timelike geodesic pre-Alcubierre warp drive uses both positive > and negative matter configurations. The negative matter (i.e. strong > negative pressure with w less than -1/3 note w = -1 for zero point > fluctuations) is what Kip Thorne called exotic matter in his 1986 > Contact (Sagan) traversable wormhole Star Gate paper with his Cal > Tech students. > Made more interesting by fact that Uncle Joe Stalin's Los Alamos Physics > Spy Master Y. Terletskii, Bondi's Soviet Doppelganger, worked on same > problem as noted by defense physicist Robert Forward in early 1990's. > IV. The New Precision Cosmology > The new anti-gravity dark energy + gravitating dark matter form 96% > of the stuff of the universe that is not in the form of real > off-mass-shell macro-quantum vacuum coherence effect IMHO. > My new equations for this are: > (Exotic Vacuum Zero Point Energy Density) = (String Tension)(Effective > Planck Area)^-1[1 - (Effective Planck Volume)|Vacuum Coherence Field|^2] > A positive LHS is anti-gravity dark energy with dominating negative > pressure. > A negative LHS is gravitating dark matter with dominating positive > pressure. > The Einstein source of gravity is ~ (G/c^2)(Energy Density + 3Pressure_ > Where Energy Density = - Pressure > for zero point fluctuations from > Einstein's Equivalence Principle + Heisenberg's Uncertainty Principle > Vacuum Coherence Field = (Higgs Amplitude Field)e^i(Goldstone Phase Field) > from dynamical instability of globally flat spacetime vacuum of quantum > electrodynamics. > Where Einstein's 1915 geometrodynamic field guv is simply Andrei > Sakharov's metric elasticity or Hagen Kleinert's world crystal > lattice strain tensor that obeys the Bohm pilot wave IT FROM BIT constraint > (Geometrodynamics of smooth curved space-time) = guv = Flat Space-Time + > (Effective Planck Area)(Goldstone Phase Field)(,u,v) > Where (,u,v) is the symmetrized mixed second order ordinary partial > derivatives. > Torsion field = (Effective Planck Area)(Goldstone Phase Field)[,u,v)] > where [ ] is the anti-symmetrizer > The non-gravitating classical vacuum has > (Exotic Vacuum Zero Point Energy Density) = - (Zero Point Pressure) = 0 > corresponding at large scale of the adaptive windowed Wavelet > Transform to a > Vanishing Einstein Cosmological Constant / = 0 in the FRW metric > post-inflationary metric where > Vacuum Coherence Field limits to Chaotic Inflation Field in sense of Max > Tegmark's Parallel Universes > Equilibrium Vacuum Coherence Field = (Effective Planck Volume)^-1 = transformations of the 15 parameter SU(2,2) conformal group for > Minkowski space-time using the 10 parameter Poincare sub-group. > That is maybe unified exotic vacuum dark energy/matter field is the > compensating gauge field of these special conformal transformations what > would be macro-quantum coherent squeezed states of Tony Smith's > conformal gravitons from modulating the Higgs amplitude of the Vacuum > Coherence. Einstein's gravitons cohere in the modulation of the > Goldstone phase of the Vacuum Coherence? For the physicists here: I have looked at Sarfatti's equations several times, but they look to me like the musings of someone deranged. Is there anything remotely === Subject: Re: Strings & Things > Higgs Field = Amplitude of the Vacuum Coherence Field from BCS > negentropic collapse of phase space of the unstable Dirac electron > vacuum to a much smaller volume of phase space that is macroscopically > occupied by bound lepto-quark pairs. Does this make any sense as a definition of the Higgs field? === Subject: Re: Integral of e^x^2 dx >For the indefinite integral, there is no elementary expression. It can be >expressed as an integral or an infinite power series but that's about it. >I've never seen a proof of this but that's what I'm told anyway. >For the improper, definite integral over the entire real line (i.e. >int(e^x^2,x=-infinity..infinity) ), the answer is of course infinite. >However, if you add in a little negative sign to make >int(e^(-x^2),x=-infinity..infinity) the answer is, incredibly, sqrt(pi). >The proof of this is one of my favorite proofs. You'll probably see it if >you take vector calculus. >Have a tolerable existence. Eli Kris Hermansen === Subject: Re: Integral of e^x^2 dx The indefinite integral of e^(-x^2) is no more elementary than that of e^x^2. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Integral of e^x^2 dx about picking up a higher level book so I can read some more about it. It really intrigues me that this one is not elementary, because it seems it should be. Integral of e^x is elementary, e^-(x^2) is elementary, e^x^2 is not. I need to read up some more about this. I did find an approximation > The indefinite integral of e^(-x^2) is no more elementary than that of e^x^2. > -- > Dave Seaman > Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > I was working this integral out last night (a definite integral) and power series didn't seem to work. What am I overlooking? David Moran === Subject: Re: Repeat: White Noise Dilemma > I find it odd (2n+1) that no one has responded to this. I'm guessing > that either a) everyone thinks it is a homework problem (it is not), > or b) no one knows the answer. Come on, all you big strong mathematicians - > certainly you can answer a little ol' question like this, can't you? ... > Here is a question that has had me in a quandary for several > years now. > Let X(t) be a zero-mean IID process with variance of > sigma^2. Now we know that since this function is > IID, it has a white PSD and therefore its autocorrelation > function at lag 0, R_XX(0), should be b*delta(tau), where > delta(tau) is the usual Dirac delta function and b is some > constant. What are you using as a definition of PSD (power spectrum density)? This would seem to be a problem for a process such as X(t) in which t--> X(t) is not measurable, or continuous in L^2. > However, we also know that the autocorrelation function R_XX(tau) > is defined to be E[X(t)*X(t+tau)], and therefore that R_XX(0) > is E[X^2] (where we have dropped the dependence on t since the > process is IID). Therefore in this sense R_XX(0) = E[X^2] = sigma^2, > which is not infinite. > How do you resolve this discrepancy? -- A. === Subject: Re: Repeat: White Noise Dilemma [...] > The way you've phrased your question, it is abundantly clear that > you're coming at things from an applied perspective. Unfortunately, > from a mathematical perspective, your question doesn't even make > sense. You state, let X(t) be a zero-mean IID process with variance > of sigma^2. By this, I assume you mean that X(t) is a mean zero > stationary process such that > (i) X(t) and X(s) are independent whenever t < s, and > (ii) E|X(t)|^2 = sigma^2 < infty. > Well, there is no reasonable process that satisfies these conditions. > (In this case, such a process would not even be measurable. If you > drop (ii), such a process would not be continuous.) Sure, there is; it is called Johnson or Nyquist noise and it is the noise generated in any resistor its power spectral density being kT from DC well into far infrared . That is a very well established physically correct mathematical model. Whether X(t) is really independent of X(s) for all t < s is a question of how close t is to s. If the density is kT from 0Hz all the way to 10^13 Hz then we are talking about possible violations of your criterion (i) for time separations |s-t| not exceeding 10^-13 sec = 100fsec. That is a small time interval not likely to be of electrical engineering importance for the foreseeable future (note electrical and not optical). > On the applied side, there is a lot of intuition about white noise. > Translating that into rigorous mathematical definitions is no small > task. The white noise mathematicians work with is a completely > different kind of object than you're probably used to. (In fact, it's > not even a process in the traditional sense.) Even in a mathematical not just engineering sense the process becomes continuous when an arbitrarily large but finite bandwidth filter is emplaced after the noise generator. As long as the bandwidth of that filter is larger then the reciprocal of the samllest meaningful time resolution in the system under investigation, it will be OK. It is a nuisance to put these words in front of every statement involving white noise, so after a while you just internalize it and remember it when needed. The nonexisting derivative of the Wiener process is a very fine practical physical model for thermal noise up to the infrared. I think Randy Yates, who is an electrical engineer, is quite used to white noise be it generated in the lab or in a management communications meeting... === Subject: Re: Repeat: White Noise Dilemma > I find it odd (2n+1) that no one has responded to this. I'm guessing > that either a) everyone thinks it is a homework problem (it is not), > or b) no one knows the answer. Come on, all you big strong mathematicians > - certainly you can answer a little ol' question like this, can't you? ... > Here is a question that has had me in a quandary for several > years now. > Let X(t) be a zero-mean IID process with variance of > sigma^2. Now we know that since this function is > IID, it has a white PSD and therefore its autocorrelation > function at lag 0, R_XX(0), should be b*delta(tau), where > delta(tau) is the usual Dirac delta function and b is some > constant. > However, we also know that the autocorrelation function R_XX(tau) > is defined to be E[X(t)*X(t+tau)], and therefore that R_XX(0) > is E[X^2] (where we have dropped the dependence on t since the > process is IID). Therefore in this sense R_XX(0) = E[X^2] = sigma^2, > which is not infinite. > How do you resolve this discrepancy? I read the answers of Jason and Robert. This looks like a classical engineer-mathematician dispute. On the danger of annoying everybody I will try to attempt a solution. Where white noise processes came from? In studying linear systems, you have S_YY (spectrum of output)=S_XX(spectrum of input)*H(system function). Setting the input spectrum equal to a constant proved to be a nice tool for getting nice results. But what input process has such a spectrum? Engineers ask the mathematicians and they said: Such a process does not exist, but since it was practical, engineers continued to use it and gave it the name white noise. But what they study is the output of systems with it as input and by integration and other smoothing one gets processes with a well defined PSD which exist in the usual sense. In the standard textbooks for engineers white noise is simply defined as process with constant PSD, without giving a damn about the involved mathematical difficulties. In the standard textbooks for mathematicians it is ignored or declared inexistent without giving a damn about its practical use. The problem with the properties of white noise is, that from the constant PSD one constructed by retroeingineering a process with this property and this gives problems. Since white noise is not a stochastic process in the classical sense, you can not expect that its R_XX follows the rules for autocorrelation functions of those, i.e. that R_XX(0)=sigma^2. Now my solution attempts: in Gradshteyn, Ryzhik: Table of integrals, Series and Products, Academic Press, 5th ed., 1993, p.1184 there are the Fourier transforms listed, for a constant we get a Dirac delta function, so we have the R_XX, since it is the transform of the PSD. 2) MAYBE BETTER: Unknown to many mathematicians there is a mathematically sound foundation for the white noise process, called theory of generalized functions/processes. In this theory one studies not the distributions of the process at time points, but the distributions of integrals over the process multiplied by a differentiable or step function, i.e. int_a^b g(t)x(t) dt. Now for what is white noise used? As input in a linear system it will be integrated and we are interested in such integrals.Therefore this theory should give all we need. If we now have the integral: I(T)=int_{-T}^T x(t) dt we have for its variance: var[I(T)]=int_{-2T}^{2T} (2T-|u|)R_XX(u) du (for the derivation: A. Papoulis: Prob., rand. var. and stoch. processes, McGraw-Hill, first edition 1965, p.325) If now we want to have white noise, the variance var[I(T)] should be 2T*sigma^2, since we add up independent rv's., i.e. R_XX=delta(0)*sigma^2. So we can define the generalized process white noise using this R_XX. -- Karl Breitung Remove nospam to get my reply address === Subject: Re: Repeat: White Noise Dilemma > [..] > from a mathematical perspective, your question doesn't even make > sense. You state, let X(t) be a zero-mean IID process with variance > of sigma^2. By this, I assume you mean that X(t) is a mean zero > stationary process such that > (i) X(t) and X(s) are independent whenever t < s, and Er, last I checked IID includes dependent, so you have simply repeated my conditions. > (ii) E|X(t)|^2 = sigma^2 < infty. Do you really have to be that anal? Actually, you're not being anal enough - if the covariance was not finite, then I would've stated that it does not exist. > Well, there is no reasonable process that satisfies these conditions. > (In this case, such a process would not even be measurable. If you > drop (ii), such a process would not be continuous.) And how did you get to these conclusions? > On the applied side, there is a lot of intuition about white noise. > Translating that into rigorous mathematical definitions is no small > task. The white noise mathematicians work with is a completely > different kind of object than you're probably used to. (In fact, it's > not even a process in the traditional sense.) Please, enlighten me, or point me to the subjects and/or books that will enlighten me. > I know this doesn't answer your question, but I hope it at least > partially explains why you haven't received more responses. In some > sense, you're posting to the wrong group. I am absolutely not posting to the wrong group. If the answer to this this question involves delving into pure math, so be it. > If you post your question in > an engineering group, they may be able to give you an answer based on > their own intuition and jargon, which may be what you're looking for. No, that's not what I'm looking for. And I HAVE posted it to an engineering group (comp.dsp) SEVERAL times now over the last two or three years. It is not resolvable by the folks there. > If you want a mathematical answer, then the question must be posed in > a more formal way. Hey Jason, bend a little - give me a clue as to what is lacking in formality here. -- % Randy Yates % ...the answer lies within your soul %% Fuquay-Varina, NC % 'cause no one knows which side %%% 919-577-9882 % the coin will fall. %%%% % 'Big Wheels', *Out of the Blue*, ELO http://home.earthlink.net/~yatescr === Subject: Re: Repeat: White Noise Dilemma Let me start by saying that I didn't mean to offend or annoy you in case you're offended or annoyed. Next, let me just say, for the record, that as a probabilist, in particular, one who works with stochastic ordinary and partial differential equations, I'm very familiar with white noise and the techniques for working with it in a mathematically rigorous fashion. I am, however, a student of pure mathematics, so anything I say will be from that perspective. I'll try to answer your questions below. > [..] > from a mathematical perspective, your question doesn't even make > sense. You state, let X(t) be a zero-mean IID process with variance > of sigma^2. By this, I assume you mean that X(t) is a mean zero > stationary process such that (i) X(t) and X(s) are independent whenever t < s, and > Er, last I checked IID includes dependent, so you have simply > repeated my conditions. > (ii) E|X(t)|^2 = sigma^2 < infty. > Do you really have to be that anal? Actually, you're not being anal > enough - if the covariance was not finite, then I would've stated that > it does not exist. It's true that, here, I am simply repeating your conditions. I'm trying to use the terminology you will be more likely to see in the pure math literature. For example, when it comes to a process of a real parameter, t, the term IID is less common. The reason is twofold: (1) postulating independence can lead to difficulties regarding the existence of the process, as I mention below; and (2) processes of a real parameter which are identically distributed are more commonly referred to as stationary. The < infty is there to emphasize that this is the important part of the assumption. If the variance is allowed to be infinite, then such a process will have a measurable version, though it won't be continuous. > Well, there is no reasonable process that satisfies these conditions. > (In this case, such a process would not even be measurable. If you > drop (ii), such a process would not be continuous.) > And how did you get to these conclusions? See the first page of Chapter III of Stochastic Differential Equations by Bernt Oksendal and the references therein. (Incidentally, white noise is *not* the non-continuous process constructed by dropping assumption (ii). See below.) > On the applied side, there is a lot of intuition about white noise. > Translating that into rigorous mathematical definitions is no small > task. The white noise mathematicians work with is a completely > different kind of object than you're probably used to. (In fact, it's > not even a process in the traditional sense.) > Please, enlighten me, or point me to the subjects and/or books > that will enlighten me. An ordinary process is a function X(t,omega) of two variables. For each omega, we have a function of t, and for each t, we have a random variable. It is possible to regard white noise as a process in the sense that for each t we have an object called a generalized random variable or a generalized Wiener functional. See Stochastic Partial Differential Equations: A Modeling White Noise Approach by Oksendal, et al, and the references therein for an introduction to this approach. More commonly, though, white noise is regarded as a random variable taking values in the space of distributions: for each omega, we get a distribution rather than a function of t. To be precise, white noise would be the distributional derivative of Brownian motion. In the former approach, white noise does not have pointwise values for each omega, in the latter it does not have pointwise values for each t. > I know this doesn't answer your question, but I hope it at least > partially explains why you haven't received more responses. In some > sense, you're posting to the wrong group. > I am absolutely not posting to the wrong group. If the answer to this > this question involves delving into pure math, so be it. > If you post your question in > an engineering group, they may be able to give you an answer based on > their own intuition and jargon, which may be what you're looking for. > No, that's not what I'm looking for. And I HAVE posted it to an engineering > group (comp.dsp) SEVERAL times now over the last two or three years. It > is not resolvable by the folks there. > If you want a mathematical answer, then the question must be posed in > a more formal way. > Hey Jason, bend a little - give me a clue as to what is lacking in formality > here. Okay, I'll give it a try. If someone said to me, consider an IID process whose covariance function is EX(s)X(t) = delta(t-s), I would understand that to mean the white noise I mentioned earlier, which I would think of as the distributional derivative of Brownian motion. The connection between the mathematics and the heuristics is this: if X_r(t) = [B(t+r)-B(t)]/r, where B(t) is Brownian motion, then EX_r(s)X_r(t) = f_r(t-s) for some functions f_r which coverge in the sense of distributions to the delta distribution. Now, I think you want to consider an IID process whose covariance function is EX(s)X(t) = sigma^2 if s=t, 0 otherwise. How should I interpret this in a rigorous sense? Is it the distributional derivative of some ordinary process? If so, which one? Okay, once we establish what the process is, we come to next part, which is that it has a white PSD. As I understand it, the PSD is simply the Fourier transform of the covariance function. But two functions which differ only a set of measure zero have the same Fourier transform. This process has a covariance function which is 0 almost everywhere. Therefore, its Fourier transform is the zero function. Notice that we don't have this problem when taking the Fourier transform of the delta function, even though the delta function is zero everywhere except the origin. This is because the delta function is not a function at all, but a distribution, so we must take the Fourier transform in the sense of distributions. So, in summary, the process you've introduced does not exist as an ordinary process. Perhaps it exists as a random variable taking values in the space of Schwartz distributions. If so, it's not clear exactly what that random variable should be, so you need to explicitly define it in the language of distribution theory. Once done, you need to be more explicit about the white PSD part, since, as it stands, the Fourier transform of your covariance function is the zero function, which is not white. A good place to start reading about distribution theory (if you're not already familiar with it) is Introduction to the Theory of Distributions by F. G. Friedlander. Anyway, I hope this helps. I enjoy discussing this stuff because I enjoy stochastic analysis. But I don't want to annoy or offend you and I don't want to engage in any kind of emotionally charged discussion over this stuff. Good luck in your investigations. Please let me know if you come upon the formal way to describe your process. I'd be very interested in seeing it. === Subject: Skeptickal Inquirer UFO My point in all of this as summarized in my semi-popular SLS talk in http://qedcorp.com/APS/StarGate1.mov does not depend on any particular alleged UFO contact incident like the interesting one you describe below. The probabilities have changed drastically since 1999 Type 1a supernovae data that ~ 70% of all the large scale stuff of the Universe's past light cone of our detectors is in the form of a dark energy anti-gravity field of powerful negative pressure that began to accelerate the expansion rate of the universe about half way though its history to now, i.e. about 7 billion years ago. This dark energy is the same as Kip Thorne's exotic matter in his 1986 Star Gate paper and it is the same stuff needed for Alcubierre's weightless warp drive metric of the early 1990's and it is the same negative matter needed for exotic propulsion that British Ministry of Defense Chief Scientist Herman Bondi told us Cornell students about in ~ 1960 that Stalin's top physics Spy Master Y. Terletskii was also very keen on. One need not be a rocket scientist to connect these dots! Any Gum Shoe and see the pattern here. :-) If what happened near Kecksburg, PA, on December 9, 1965, was the result of a fireball, it was certainly a most peculiar one. I've never heard of a fireball that maneuvered, changed course, slowed down, and came in for a landing, nor one that immediately attracted the military, a tight security cordon, and mean-faced guys wearing long black overcoats who intimidated witnesses. But then it seems that anything is possible when it comes to the wacky-worldview of knee-jerk debunking. Or does your highly public take on this and other UFO-related phenomena perhaps reflect a commitment to the cynical, pseudo-patriotic, and thoroughly black world of Government-sponsored disinformation? Scott C. SCOTT LITTLETON President, Phi Beta Kappa Alumni in Southern California Professor of Anthropology, Emeritus Occidental College Los Angeles, CA 90041 TEL (323) 255-5477 FAX (323) 982-0264 http://www.oxy.edu/~yokatta/home.htm Any sufficiently advanced technology is indistinguishable from magic --Sir Arthur C. Clarke I think we're property. . . --Charles Fort -----Original Message----- Cc: Rosanne C Losee; Richard Dolan; Don Ecker === Subject: Re: ppt & UFO debunking Prof. Littleton agrees that the available UFO evidence is soft, but thinks that the intelligence agencies have squirrelled away some hard evidence from supposed crashes at Roswell and Kecksburg. The Roswell crash yarns have been quite effectively debunked by Pflock, Klass, and others, I have nothing to add. As for Kecksburg, is he truly unaware that this crashing UFO was simply a fable spun from confusion over the great fireball meteor of December 9, 1965? See my web page debunking this at http://www.debunker.com/kecksburg.html . Truly a house of cards. You UFO believers are *so easily* fooled. And on such trivia are new scientific theories to be built?????? Robert Sheaffer ----- Original Message ----- Cc: Rosanne C Losee ; Richard Dolan ; Don Ecker === Subject: RE: ppt & UFO debunking Hi Jack, Let me take a moment away from the TV, where we've been watching the progress of the horrendous local fires and hoping they don't come any closer to Pasadena, to comment on Robert Sheaffer's recent comment that It is exceedingly unlikely that UFOs... are real entities, they are mostly due to 'random errors' occurring in the human perceptual and social-belief formation system. There is noise in every information channel, and our society has agreed to describe a certain kind of that noise as UFOs. While we may argue about their origin and purpose, it has long since become abundantly clear that what we call UFOs are not simply noise. Unless the thousands of internally consistent reports by credible witnesses, including abductees, are some form of mass hallucination, these phenomena clearly have an objective existence. Moreover, as a student of comparative mythology and folklore, I can firmly attest that the patterns reported by contemporary witnesses and experiencers jibe remarkably well with pre-modern accounts of flying shields, abductions to fairy-land by short creatures with big eyes and pointed, elfin chins, missing-time episodes, and rides in flying wheels a la Ezekiel's famous trip. To ignore all this evidence, soft though it most of it may be (although I strongly suspect that despite a half-century of stonewalling, some members of the U.S. and other intelligence communities-the CIA, MI6, the old KGB, etc.-have in their possession an abundance of hard evidence garnered from Roswell, Kecksburg, and other UFO crash sites), is to wallow in a state of denial so all-embracing that it beggars the imagination. Or do debunkers like Sheaffer, Oberg, Klass, and the rest have another, more devious agenda, that is, to further the Government's long-standing policy of keeping the lid on by systematically ridiculing those of us who suspect the truth about this phenomenon, all the while being privy to above-top-secret knowledge that would prove us right? One wonders.... All best wishes & Scott C. SCOTT LITTLETON President, Phi Beta Kappa Alumni in Southern California Professor of Anthropology, Emeritus Occidental College Los Angeles, CA 90041 TEL (323) 255-5477 FAX (323) 982-0264 http://www.oxy.edu/~yokatta/home.htm Any sufficiently advanced technology is indistinguishable from magic --Sir Arthur C. Clarke I think we're property. . . --Charles Fort === Subject: Re: Skeptickal Inquirer UFO > My point in all of this as summarized in my semi-popular SLS talk in > http://qedcorp.com/APS/StarGate1.mov > does not depend on any particular alleged UFO contact incident like the > interesting one you describe below. > The probabilities have changed drastically since 1999 Type 1a supernovae > data that ~ 70% of all the large scale stuff of the Universe's past > light cone of our detectors is in the form of a dark energy > anti-gravity field of powerful negative pressure that began to > accelerate the expansion rate of the universe about half way though its > history to now, i.e. about 7 billion years ago. This dark energy is the > same as Kip Thorne's exotic matter in his 1986 Star Gate paper and it > is the same stuff needed for Alcubierre's weightless warp drive metric > of the early 1990's and it is the same negative matter needed for > exotic propulsion that British Ministry of Defense Chief Scientist > Herman Bondi told us Cornell students about in ~ 1960 that Stalin's top > physics Spy Master Y. Terletskii was also very keen on. > One need not be a rocket scientist to connect these dots! Any Gum Shoe > and see the pattern here. :-) Seeing patterns where none exist is not evidence of clear and rational thinking. === Subject: Re: How I know, linchpin of my FLT proof > In fact one can easily show that integral extensions may be > characterized in terms of the lying-over (LO) property and > the incomparable (INC) property, e.g. consider the following > THEOREM For commutative rings R < T the following are equivalent > (1) R < T is an integral extension > (2) A < B has INC and LO for all A,B with R < A < B < T > (3) A < A[u] has INC and LO for all A,B with R < A < T, all u in T Below is an excerpt (intro and refs) of an online paper which has much more of interest on this. For further details see the full paper (link below) and those papers referenced (esp. [1]). SURVIVAL-PAIRS OF COMMUTATIVE RINGS HAVE THE LYING-OVER PROPERTY http://math.ndsu.nodak.edu/faculty/coykenda/paper26b.pdf ABSTRACT Let R < T be a unital extension of commutative rings. It is proved that (R,T) is a lying-over pair (in the sense that A < B satisfies LO for all rings R < A < B < T) if (and only if) (R,T) is a survival-pair (in the sense that PT != T for all rings R < A < T and all prime ideals P of A). As a consequence, T is integral over R if (and only if) (R[X],T[X]) is a survival-pair, where X is an indeterminate over T. It is also proved that a unital homomorphism of commutative rings satisfies LO if and only if it is universally a survival homomorphism. 1 INTRODUCTION All rings considered are commutative with identity; all inclusions of rings and all ring homomorphisms are unital. As in [6], if F is a property of some ring extensions, we say that (R,T) is a F-pair in case R < T are rings such that A < B satisfies F for all rings R < A < B < T. The pair approach figures in the result that helped to motivate both [5] and [6], namely, [6, Folklore Theorem, p. 454]: a ring extension R < T is integral if and only if (R,T) is both an INC-pair and an LO-pair. (Following [12, p.28], we let INC, LO, and GU denote the incomparable, lying-over, and going-up properties, respectively. In particular, a ring extension A < B satisfies LO if and only if each prime ideal P of A is of the form Q / A for at least one prime ideal Q of B.) Over the years, INC-pairs have become well-understood. They were introduced, without the terminology, in [5, Theorem, p.38], which implies that (R,T) is an INC-pair if and only if R < T are rings such that each element of T is a root of a polynomial over R having a unit coefficient. In other words, we have [5, Corollary 4]: (R,T) is an INC-pair if and only if R < T is a P-extension (in the sense of Gilmer-Hoffmann [10]). In addition, although the implication INC => MINC is non-reversible, the concepts of INC-pair and MINC-pair are equivalent [6, Example 2.2, Corollary 2.4(bis)]. More recently, Ayache-Jaballah [1] showed that INC-pairs are the same as the residually algebraic pairs. For a survey of current knowledge about INC-pairs, see [9, Section 6.5]. The present paper contributes to deepening our understanding of LO-pairs. LO-pairs were introduced in [6]. Subsequent work on this topic has included a study of LO-pairs of affine algebras [13] and a recent non-integral example [2] that answered a question raised in [6, Remark 3.12(b)]. Much of [6] was motivated by the following non-reversible implications connecting properties of ring extensions: GU => LO => Survival. (Following [12, p. 35], we say that if R < T are rings and I is a proper ideal of R, then I survives in T if IT != T, that is, if not 1 in IT; and that R < T is a survival extension if each proper (resp., prime) ideal of R survives in T.) In the spirit of [6, Corollary 2.4(bis)], it was shown in [6, Corollary 3.2] that LO-pairs are the same as GU-pairs. It seems irresistible to ask if they are also the same as the survival-pairs. Our main result, Theorem 2.2, settles the matter. Although each LO-pair must be a survival-pair, the converse does not seem obvious to us. In fact, although the property of being an LO-pair is easily seen to be a local property [6, Lemma 2.11(c)], we do not consider it to be obvious whether survival-pair is a local property. Nevertheless, there has long been compelling evidence suggesting that survival-pairs are the same as LO-pairs. For instance, a number of contexts have been identified in which the survival-pair and LO-pair concepts coincide: cf. [6, Theorem 2.7, Remark 2.8(a),(b),(c),(e)]. Another similarity of behavior arises from the fact that survival-pair plays the same role as LO-pair in sharpenings of the Folklore Theorem characterizing integrality [6, Theorem 2.1]. Similar phenomena occur in a nearly integral context [7, Theorem 2.14, Proposition 2.16] and in a result concerning complete integral closure [8, Proposition 2.1]. In our deepest result, Theorem 2.2, we give the underlying reason: survival-pairs _are_ the same as LO-pairs. In a number of contexts mentioned above (such as in [6, Theorem 2.7, Remark 2.8(a),(b),(c)]), the survival-pair condition is strong enough to imply integrality. Given the above-mentioned work in [5, Theorem, p. 38] and [1] on algebraic aspects of INC-pairs, it thus seems natural to pursue deeper connections between the survival-pair and integrality concepts. In this regard, Corollary 2.3 gives a new way in which the survival-pair concept can be used to characterize an integral ring extension. Moreover, Proposition 2.4 identifies a sense in which arbitrary survival-pairs (that is, LO-pairs) are closer to being integral than are arbitrary INC-pairs. Moving beyond the context of ring extensions to that of ring homomorphisms, recall for motivation that integrality and LO are universal properties (in the sense of [11], namely, being preserved by arbitrary base change) and that universally GU is equivalent to integrality. It seems natural to ask if/how these facts relate to the natural generalizations of the survival and survival-pair concepts to ring homomorphisms. The answers are given in Proposition 2.6(c) and Theorem 2.7: a ring homomorphism is integral (resp., satisfies LO) if and only if it is universally a survival-pair homomorphism (resp., universally a survival homomorphism). While these results frankly do not seem as deep to us as Theorem 2.2, they are pleasantly complete, they serve to place the survival-related concepts into a more appropriate categorical setting, and we would hope that readers find them of potential use in other studies. [...] REFERENCES [1] A. Ayache and A. Jaballah, Residually algebraic pairs of rings, Math. Z. 225 (1997), 49-65. http://www.springerlink.com/link.asp?id=5l1gb5r90glfbuta [2] A. Debremaeker and V. Van Lierde, Integral ring extensions and LO-pairs, Comm. Algebra 28 (2000), 2339-2341. http://www.wis.kuleuven.ac.be/algebra/artikels/rings2.ps [3] L. I. Dechene, Adjacent extensions of rings, dissertation, Univ. California, Riverside (1978). [4] M. Demazure and P. Gabriel, Introduction to Algebraic Geometry and Algebraic Groups, North Holland, Amsterdam, 1980. [5] D. E. Dobbs, On INC-extensions and polynomials with unit content, Can. Math. Bull. 23 (1980), 37-42. [6] D. E. Dobbs, Lying-over pairs of commutative rings, Can. J. Math. 33 (1981), 454-475. [7] D. E. Dobbs, Nearly integral homomorphisms of commutative rings, Bull. Austral. Math. Soc. 40 (1989), 1-12. [8] D. E. Dobbs, Prime ideals surviving in complete integral closures, Arch. Math. 69 (1997), 465-469. [9] M. Fontana, J. A. Huckaba, and I. J. Papick, Prufer Domains, Monographs and Textbooks Pure Appl. Math. 203,Marcel Dekker, New York, 1997. [10] R. Gilmer and J. F. Hoffmann, A characterization of Prufer domains in terms of polynomials, Pac. J. Math. 60 (1975), 81-85. [11] A. Grothendieck and J. A. Dieudonne, Elements de Geometrie Algebrique, Springer-Verlag, Berlin, 1971. [12] I. Kaplansky, Commutative Rings, rev. ed. Univ. Chicago Press, Chicago, 1974. [13] S. Visweswaran, Intermediate rings between D+I and K[y_1,...,y_t], Comm. Algebra 18 (1990), 309-345. 2000 Mathematics Subject Classification. Primary 13B24, 13B21, 13B99; Secondary 13A15, 13G05, 13B25, 14A15. Jim Coykendall Department of Mathematics North Dakota State University Fargo, ND. 58105-5075 E-mail: jim.coykendall@ndsu.nodak.edu David E. Dobbs Department of Mathematics University of Tennessee Knoxville, TN. 37996 E-mail: dobbs@math.utk.edu === Subject: Re: Uncle Al is Sadistic . >The performance of the American public school system, supports your >point. If pearls are cast before swine, they will still be swine. >>Well, what confuses lots of people is that we reached our present >>state following many centuries during which the majority of the >>population did not get educated due to luck of means, remaining dumb >>as the result. So, people got to equate stupidity with lack of means >>and to believe that if we'll only allocate sufficient resources, >>everybody will be educated and smart. This is not so. Availability >>of education may be a necessary situation but it is not sufficient. >True enough. The trouble is that some here appear to be arguing that >because it's not sufficient, it's not necessary. That's the connection >with which I have a problem. the issue becomes how much? >>And, beyond some level of expenditure, additional money seems to make >>little if any difference. When you look on the numbers, state by >>state (in the US) of per student expenditure and compare it with >>performance (as measured by SAT, CAT or any other measure), you find >>little if any correlation. >I think part of the problem is that we have such poor measures of what >it is to be educated. The College Board only claims that the SAT is >a very good predictor of first-year college grades; the average >correlation between SAT scores and _freshman_ grades is +0.52; the >correlation to being educated and smart is almost certainly quite a >bit smaller. Aha. The only problem is, we've (currently) no measure which gives a of deciding (in this field as in any other) that you won't use any measure until you've one which is perfect. >>Heck, the Washington DC school district >>has one of the highest per capita spending levels in the nation, and >>the results are dismaying. >The DC schools are a mess, but they do provide an example of why >looking at per capita spending levels may be overly simplistic. For >one thing, every school in DC borders on the ancient; the funding >required to keep them watertight, heated, and fire-resistant is much >more than it is in suburban school systems, where most of the >facilities are of much more recent vintage. (There's an annual >showdown each fall between the DC school board and a local judge about >finishing the maintenance that needs to get done every summer just to >open the schools safely). Aha. Do you've numbers regarding how much is being spent on maintenance, per year, relative to the suburb schools. >For another, the DC schools serve a much larger proportion of poor >children than any state. That accounts for some, but by no means all, >of the greater administrative costs; DC has more kids receiving free >meals at school, plus they have to hire more accountants to make sure >no one is stealing the free lunch money. I suspect that some per >capita spending totals include both the money for the free meals and >for the extra accountants at school headquarters, when it has more to >do with using schools as a convenient venue for delivering other >social services than it does with education. That's just one example >out of many. Some of it may. Again, what are the numbers? >Finally, although it may not be universally accepted, many people >accept that it costs more to educate someone who was brought up by a >single high-school dropout mother in a housing project than someone >living with two college graduate parents in the suburbs. This would've been much more credible, would there have been some results to show for the spending. While it is true to it costs money to solve problems, it is not automatically true that spending money solves problems. A fine distinction, but an important one. I could bring an exorcist to help me debug a faulty program, but this wouldn't do much good. So, then, I could bring an additional exorcist. Double the money spent, same results (or rather absence of). Mind you, I'm not saying that this is the case here. But I'm also not saying that it isn't. >Does this justify all the spending of the DC school board? I'm >skeptical, but I'm just as skeptical that a simple examination of >correlations between per capita spending levels and SAT scores is >going to give us a useful answer. These are not easy problems to >solve. For sure. But they are not going to be solved by hiding behind ohh, it is so complicated, we can't really measure what we're doing, so lets just keep spending and hope for the best. Anything that is fundamentally sound can only benefit from some serious questioning. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Uncle Al is Sadistic . >> You assume here, that everybody can be educated. There is little >> evidence to this effect. >I'm not a psychologist, so I have no idea about this one. But I would >think that if you spent lots and lots of time and money into this, at >least some of the scum would, eventually, be educated. >>We do spend lots of time and money on this. But some will be does >>not translate into everybody can be. And the issue here is one of >>proportion. When you reach the point at which you shortchange the >>education of those you know can be educated so as to spend more and >>more and those who maybe, possibly can be educated you're not acting >>reasonably. Resources are not unlimited and opportunity cost is not >>an empty phrase. >There may be little evidence that everybody can be educated. But if >one is to speak of opportunity cost the trick becomes how do we >distinguish those who can be educated from those who can not. Indeed. > The unstated premise of a number of previous posters to this thread is >that the race (or only slightly less perniciously, the socioeconomic >class) of one's parents provides that distinguishing factor. There are >sufficient counterexamples to indicate that it does not; there's an >opportunity cost to not providing sufficient resources to educate >talented kids being born in other than stable middle- or upper-class >families. Of course. I may not exactly be thrilled with the educational status >quo, but I'm appalled that one side on this thread has used phrases >like bet on the winners and [expletive] diversity, making it >explicit that we're basing all this on race and ethnic origin. Actually, it is the current notion of diversity which bases everything on race and ethnic origin. What opponents, by and large, advocate, is the removal of race and ethnic origin as consideration. This, in turn, is labeled in the PC Newspeak as racism. Interesting:-) >The other factor to consider when speaking of opportunity cost is that >we as a society take some pride in being a country of second >chances. There is an opportunity cost in not giving any second >chances, just as there is an opportunity cost in giving too many >second chances. At what point in a young person's life does the >opportunity cost rise to the point that no second chance should be >afforded? In principle, there should be a point of optimum >cost/benefit, but I strongly suspect it varies from person to person, >and I'm skeptical that we have the ability to identify that point with >any precision, all of which makes it a difficult basis for rational >policy. I fully agree. However, even if we cannot make a perfect determination, a determination needs to be made. When medical decision are being made, it is also rarely true that precise boundaries exist (as one of my MD friends uses to say, only an autopsy is for sure). Yet, again, decisions have to be made. In any diagnostic system you've non zero probabilities for false positives and false negatives. You should make your best effort to minimize these, but refraining from using the system until said probabilities are reduced to zero is hardly (if ever) an option. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Uncle Al is Sadistic . >> You assume here, that everybody can be educated. There is little >> evidence to this effect. >I'm not a psychologist, so I have no idea about this one. But I would >think that if you spent lots and lots of time and money into this, at >least some of the scum would, eventually, be educated. >>We do spend lots of time and money on this. But some will be does >>not translate into everybody can be. And the issue here is one of >>proportion. When you reach the point at which you shortchange the >>education of those you know can be educated so as to spend more and >>more and those who maybe, possibly can be educated you're not acting >>reasonably. Resources are not unlimited and opportunity cost is not >>an empty phrase. >There may be little evidence that everybody can be educated. But if >one is to speak of opportunity cost the trick becomes how do we >distinguish those who can be educated from those who can not. > Indeed. > The unstated premise of a number of previous posters to this thread is >that the race (or only slightly less perniciously, the socioeconomic >class) of one's parents provides that distinguishing factor. There are >sufficient counterexamples to indicate that it does not; there's an >opportunity cost to not providing sufficient resources to educate >talented kids being born in other than stable middle- or upper-class >families. > Of course. > I may not exactly be thrilled with the educational status >quo, but I'm appalled that one side on this thread has used phrases >like bet on the winners and [expletive] diversity, making it >explicit that we're basing all this on race and ethnic origin. > Actually, it is the current notion of diversity which bases > everything on race and ethnic origin. What opponents, by and large, > advocate, is the removal of race and ethnic origin as consideration. > This, in turn, is labeled in the PC Newspeak as racism. > Interesting:-) >The other factor to consider when speaking of opportunity cost is that >we as a society take some pride in being a country of second >chances. There is an opportunity cost in not giving any second >chances, just as there is an opportunity cost in giving too many >second chances. At what point in a young person's life does the >opportunity cost rise to the point that no second chance should be >afforded? In principle, there should be a point of optimum >cost/benefit, but I strongly suspect it varies from person to person, >and I'm skeptical that we have the ability to identify that point with >any precision, all of which makes it a difficult basis for rational >policy. > I fully agree. However, even if we cannot make a perfect > determination, a determination needs to be made. When medical > decision are being made, it is also rarely true that precise > boundaries exist (as one of my MD friends uses to say, only an > autopsy is for sure). Yet, again, decisions have to be made. In any > diagnostic system you've non zero probabilities for false positives > and false negatives. You should make your best effort to minimize > these, but refraining from using the system until said probabilities > are reduced to zero is hardly (if ever) an option. Education arising from finite resources requires triage. Better that the usable 80% be identified and 20% be erroneously discarded with 80% of the trash than embracing 100% of the dysfunctional and discarding everything useful in the name of compassion. Compassion for what, for garbage? Oh yeah... remember to reycle. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Uncle Al is Sadistic . >>The performance of the American public school system, supports your >>point. If pearls are cast before swine, they will still be swine. >Well, what confuses lots of people is that we reached our present >state following many centuries during which the majority of the >population did not get educated due to luck of means, remaining dumb >as the result. So, people got to equate stupidity with lack of means >and to believe that if we'll only allocate sufficient resources, >everybody will be educated and smart. This is not so. Availability >of education may be a necessary situation but it is not sufficient. True enough. The trouble is that some here appear to be arguing that because it's not sufficient, it's not necessary. That's the connection with which I have a problem. >And, beyond some level of expenditure, additional money seems to make >little if any difference. When you look on the numbers, state by >state (in the US) of per student expenditure and compare it with >performance (as measured by SAT, CAT or any other measure), you find >little if any correlation. I think part of the problem is that we have such poor measures of what it is to be educated. The College Board only claims that the SAT is a very good predictor of first-year college grades; the average correlation between SAT scores and _freshman_ grades is +0.52; the correlation to being educated and smart is almost certainly quite a bit smaller. >Heck, the Washington DC school district >has one of the highest per capita spending levels in the nation, and >the results are dismaying. The DC schools are a mess, but they do provide an example of why looking at per capita spending levels may be overly simplistic. For one thing, every school in DC borders on the ancient; the funding required to keep them watertight, heated, and fire-resistant is much more than it is in suburban school systems, where most of the facilities are of much more recent vintage. (There's an annual showdown each fall between the DC school board and a local judge about finishing the maintenance that needs to get done every summer just to open the schools safely). For another, the DC schools serve a much larger proportion of poor children than any state. That accounts for some, but by no means all, of the greater administrative costs; DC has more kids receiving free meals at school, plus they have to hire more accountants to make sure no one is stealing the free lunch money. I suspect that some per capita spending totals include both the money for the free meals and for the extra accountants at school headquarters, when it has more to do with using schools as a convenient venue for delivering other social services than it does with education. That's just one example out of many. Finally, although it may not be universally accepted, many people accept that it costs more to educate someone who was brought up by a single high-school dropout mother in a housing project than someone living with two college graduate parents in the suburbs. Does this justify all the spending of the DC school board? I'm skeptical, but I'm just as skeptical that a simple examination of correlations between per capita spending levels and SAT scores is going to give us a useful answer. These are not easy problems to solve. George ********************************************************************** Dr. George O. Bizzigotti Telephone: (703) 610-2115 Mitretek Systems, Inc. Fax: (703) 610-1558 3150 Fairview Park Drive South E-Mail: gbizzigo@mitretek.org Falls Church, Virginia, 22042-4519 ********************************************************************** === Subject: Re: Uncle Al is Sadistic . >The performance of the American public school system, supports your >point. If pearls are cast before swine, they will still be swine. >>Well, what confuses lots of people is that we reached our present >>state following many centuries during which the majority of the >>population did not get educated due to luck of means, remaining dumb >>as the result. So, people got to equate stupidity with lack of means >>and to believe that if we'll only allocate sufficient resources, >>everybody will be educated and smart. This is not so. Availability >>of education may be a necessary situation but it is not sufficient. > True enough. The trouble is that some here appear to be arguing that > because it's not sufficient, it's not necessary. That's the connection > with which I have a problem. We in the US, generally speaking, do not have a system of public education at all. Rather we have a system of public training. The stated purpose is to make good citizens. (See Dewey.) The institution of public schooling (in the US) was initially for the express purpose of preventing employers from using children to operate factory machines at pay scales lower than required and demanded by adults who they put out of work. If education had been a real goal it would not have taken till the mid-late 1800's to institute public schooling and it would not have had as short a school season as it initially did, just long enough per year to make hiring children into factory jobs not viable. === Subject: Re: Uncle Al is Sadistic . > You assume here, that everybody can be educated. There is little > evidence to this effect. >>I'm not a psychologist, so I have no idea about this one. But I would >>think that if you spent lots and lots of time and money into this, at >>least some of the scum would, eventually, be educated. >We do spend lots of time and money on this. But some will be does >not translate into everybody can be. And the issue here is one of >proportion. When you reach the point at which you shortchange the >education of those you know can be educated so as to spend more and >more and those who maybe, possibly can be educated you're not acting >reasonably. Resources are not unlimited and opportunity cost is not >an empty phrase. There may be little evidence that everybody can be educated. But if one is to speak of opportunity cost the trick becomes how do we distinguish those who can be educated from those who can not. The unstated premise of a number of previous posters to this thread is that the race (or only slightly less perniciously, the socioeconomic class) of one's parents provides that distinguishing factor. There are sufficient counterexamples to indicate that it does not; there's an opportunity cost to not providing sufficient resources to educate talented kids being born in other than stable middle- or upper-class families. I may not exactly be thrilled with the educational status quo, but I'm appalled that one side on this thread has used phrases like bet on the winners and [expletive] diversity, making it explicit that we're basing all this on race and ethnic origin. The other factor to consider when speaking of opportunity cost is that we as a society take some pride in being a country of second chances. There is an opportunity cost in not giving any second chances, just as there is an opportunity cost in giving too many second chances. At what point in a young person's life does the opportunity cost rise to the point that no second chance should be afforded? In principle, there should be a point of optimum cost/benefit, but I strongly suspect it varies from person to person, and I'm skeptical that we have the ability to identify that point with any precision, all of which makes it a difficult basis for rational policy. George ********************************************************************** Dr. George O. Bizzigotti Telephone: (703) 610-2115 Mitretek Systems, Inc. Fax: (703) 610-1558 3150 Fairview Park Drive South E-Mail: gbizzigo@mitretek.org Falls Church, Virginia, 22042-4519 ********************************************************************** === Subject: Re: Uncle Al is Sadistic . >Is there anything good about California? Anyone? I'd like to hear >something positive that I can hold onto, living here. >I was going to say the weather's nice. So nice that the entire state >goes up in flames when someone drops a cigarette. So nice that >when it does manage to rain, the mountain sides go sliding into the >valleys. So I guess you're right, there's nothing positive about >living in California. Well, it could be worse... > My friends that live in NY say things like, Well, it could be > worse, we could be living in California! oops. It was all foreseen in utter detail, even down to the words. === Planted in soc.singles; January 13, 1993; Subject: Could be worse >No matter how bad things are, take heart, it could be worse: >you could be in California. ... in reference to ... === Planted in sci.skeptic; November 13, 1991; Subject: Markadamus Quatrains >The best test of predictions is to simply wait and see them >emerge right before your very eyes. [...] >California will go bust by the end of the decade. ... and to the times following it in the early 21st century. === In reference to the sudden power shift in October >[...] receding back to male-majority right now (California in >2002 [...]) >Yes, Virginia. There is a Santa Claus. It is all really that >simple, after all. Humans don't drive history. Demographics >do. It's all just a numbers game; and humans (and their >movements) taking false credit for the inevitable consequences >of the natural historical evolutionary processes engendered >by numbers games. Pun intended. === Subject: Re: Uncle Al is Sadistic . ... lots of stuff ... >In the end, this is a lot of empirical evidence, but I'm not totally >convinced that it adds up to a proof. I've taught enough math at the college >level to know that student's aptititudes are in part, a reflection of their >upbringing as well as their heredity. Most probably so, but when you get the person, at age 18, with aptitudes which are significantly below par, the question of to what extent this is a result of heredity and to what extent this of early childhood upbringing, is somewhat academic. > If what you are saying is true, then >someone should have done some type of histological studies to try to discren >what these differences are in the brains of these different races at some >point, and I know of no such research. To the best of my knowledge the ability to to correlate observable aspects of a human brain with aptitudes is still way beyond the state of the art. Regardless of this, however, I would venture to predict that would somebody, in the current PC climate, attempt such study, then: 1) Funding for the proposal would not be approved. 2) If, by some miracle, the proposal made it through and the study yields any results but those officially approved, they won't be published. 3) If a journal brave enough to publish the results will be found, the results will meet with general derision and will be rejected (possibly terminating the academic career of the author, on the way). Consider the reception afforded the book The Bell Curve (mentioned by Unc). After all, the book primarily deals with compilation of data. Yet, it was broadly declared racist. Now, how can data be racist? All it can be is correct and incorrect. And if somebody thinks it is incorrect, a stronger argument that this just can't be so is needed. > As far as colleges being watered >down, well this would have happened with or without blacks: the average >American kid (of any color), puts in far too little work in h.s., and you >simply can't change the rules in college and expect them to change their >behavior like that. Indeed. The watering down starts at much earlier level than college. And it is attributable to multiple causes, not a single one. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Uncle Al is Sadistic . >> ... lots of stuff ... >>In the end, this is a lot of empirical evidence, but I'm not totally >>convinced that it adds up to a proof. I've taught enough math at the >college >>level to know that student's aptititudes are in part, a reflection of >their >>upbringing as well as their heredity. >> Most probably so, but when you get the person, at age 18, with >> aptitudes which are significantly below par, the question of to what >> extent this is a result of heredity and to what extent this of early >> childhood upbringing, is somewhat academic. >I agree with most of what you said except this part. It makes a huge >difference in that it colors people's perceptions and in fact leads to >racism. If you don't believe me, ask yourself if Uncle Al's data and his use >of racial epithets are not correlated? Maybe, maybe not. You've to go beyond first appearances, when you read Unc. >There are a large number of variables at work here including money and >privelege that are generally not discussed. I can tell you as the sone of a >blue-collared worker going to the University of Chicago, I learned all about >how academics in the higher circles is also in large part social also. I can tell you that any human activity (with the possible exception of hermits) is social. People are what they are. > If you were not raised in a certain environement, it is a lonely place and >success is more difficult to achieve. I can't even begin to imagine what it >would be like to have race piled on top of such a situtation. >> If what you are saying is true, then >>someone should have done some type of histological studies to try to >discren >>what these differences are in the brains of these different races at some >>point, and I know of no such research. >I don't think so: there are some very simple tests such as the density of >neurons, which could be done without much effort. I repeat, I know of no tests capable of discerning the intelligence level of the brain owner base on any physical or chemical observations (except for cases of gross deformities). Of course, this is not my field, so I may be mistaken. >> To the best of my knowledge the ability to to correlate observable >> aspects of a human brain with aptitudes is still way beyond the state >> of the art. Regardless of this, however, I would venture to predict >> that would somebody, in the current PC climate, attempt such study, >> then: >> 1) Funding for the proposal would not be approved. >> 2) If, by some miracle, the proposal made it through and the study >> yields any results but those officially approved, they won't be >> published. >> 3) If a journal brave enough to publish the results will be found, >> the results will meet with general derision and will be rejected >> (possibly terminating the academic career of the author, on the way). >Ah, but that's the beauty of science: every once in a while, you find the >person who's just after the truth. Fortunately, yes. > But... I'll conceded, it would be the >rare event. Maybe my question was rhetorical in nature: what it was intended >to do was to point out the fact that there is no real physical evidence that >some races are intellectually superior than others. Real physical evidence, of course not. Three aren't that many things related to the functioning of the organism, for which we've real physical evidence. More often than not we just have statistical evidence. Highly imperfect, but it doesn't necessarily translate to completely wrong. > On the basis of tests scores, some people have made some pretty strong >claims that really should be better substantiated. Quite possible. And lots of people made very strong claims in the opposite direction which are not substantiated at all. We've measures which, admittedly, are highly imperfect. Yet, when a huge amount of data, using numerous variations of existing measures, all shows a systematic correlation in the same direction, this cannot be dismissed just by it cannot be. Granted, the difference may be primarily cultural/environmental (and this can be studied as well). Bottom line, objectively, if we get to the point of judging individuals based on performance and nothing alse, all the above hardly matters. You'll not find an NBA team owner rejecting a 7'4 Chinese basketball player based on we all know that on the average the Chinese tend to be short. On the other hand, if somebody will claim that the NBA discriminates against orientals as proven by the fact that they're not represented there according to their weight in the population, we'll find this silly, won't we? Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Uncle Al is Sadistic . > ... lots of stuff ... >In the end, this is a lot of empirical evidence, but I'm not totally >convinced that it adds up to a proof. I've taught enough math at the college >level to know that student's aptititudes are in part, a reflection of their >upbringing as well as their heredity. > Most probably so, but when you get the person, at age 18, with > aptitudes which are significantly below par, the question of to what > extent this is a result of heredity and to what extent this of early > childhood upbringing, is somewhat academic. I agree with most of what you said except this part. It makes a huge difference in that it colors people's perceptions and in fact leads to racism. If you don't believe me, ask yourself if Uncle Al's data and his use of racial epithets are not correlated? There are a large number of variables at work here including money and privelege that are generally not discussed. I can tell you as the sone of a blue-collared worker going to the University of Chicago, I learned all about how academics in the higher circles is also in large part social also. If you were not raised in a certain environement, it is a lonely place and success is more difficult to achieve. I can't even begin to imagine what it would be like to have race piled on top of such a situtation. > If what you are saying is true, then >someone should have done some type of histological studies to try to discren >what these differences are in the brains of these different races at some >point, and I know of no such research. I don't think so: there are some very simple tests such as the density of neurons, which could be done without much effort. > To the best of my knowledge the ability to to correlate observable > aspects of a human brain with aptitudes is still way beyond the state > of the art. Regardless of this, however, I would venture to predict > that would somebody, in the current PC climate, attempt such study, > then: > 1) Funding for the proposal would not be approved. > 2) If, by some miracle, the proposal made it through and the study > yields any results but those officially approved, they won't be > published. > 3) If a journal brave enough to publish the results will be found, > the results will meet with general derision and will be rejected > (possibly terminating the academic career of the author, on the way). Ah, but that's the beauty of science: every once in a while, you find the person who's just after the truth. But... I'll conceded, it would be the rare event. Maybe my question was rhetorical in nature: what it was intended to do was to point out the fact that there is no real physical evidence that some races are intellectually superior than others. On the basis of tests scores, some people have made some pretty strong claims that really should be better substantiated. > Consider the reception afforded the book The Bell Curve (mentioned > by Unc). After all, the book primarily deals with compilation of > data. Yet, it was broadly declared racist. Now, how can data be > racist? All it can be is correct and incorrect. And if somebody > thinks it is incorrect, a stronger argument that this just can't be > so is needed. > As far as colleges being watered >down, well this would have happened with or without blacks: the average >American kid (of any color), puts in far too little work in h.s., and you >simply can't change the rules in college and expect them to change their >behavior like that. > Indeed. The watering down starts at much earlier level than college. > And it is attributable to multiple causes, not a single one. > Mati Meron | When you argue with a fool, > meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Uncle Al is Sadistic . > I agree with most of what you said except this part. It makes a huge > difference in that it colors people's perceptions and in fact leads to > racism. If you don't believe me, ask yourself if Uncle Al's data and his use > of racial epithets are not correlated? Schwartz is a lousy example for anything. He's never managed to climb out of his bankrupt Brooklyn background, a classic example of enjoying poor emotional health. Compare Mike Varney. > There are a large number of variables at work here including money and > privelege that are generally not discussed. I can tell you as the sone of a > blue-collared worker going to the University of Chicago, I learned all about > how academics in the higher circles is also in large part social also. If > you were not raised in a certain environement, it is a lonely place and > success is more difficult to achieve. I can't even begin to imagine what it > would be like to have race piled on top of such a situtation. It really has nothing to do with what environment one was actually raised in. What truly matters is how one fits into a given environment. Certainly it is a *lot* easier if one grew up in the targeted environment. Consider Pygmalion, the Kennedy family, and others who actually achieved that much sought after vertical mobility promised by the American dream. === Subject: Re: Uncle Al is Sadistic . > I'm curious as to when intelligence was consdered as a mutation. Is there > some firm scientific, genetic basis for this claim. I always thought that > intelligence, like height, was a trait that followed a Gaussian > distribution. > MB > IT'S BECAUSE YOU ONLY ADMITTED THE MUTANT SMARTEST ONES, FOOLS, AND > THEY PERFORMED TO SPEC. Now try applying that to the UC system as a > whole and see what you get - as opposed to what you've got. > The curve is not the same mean/median/mode for given populations. > Given, a European/American White bell curve centered at 100 with 15 > point standard deviation. American Blacks come in at 85, Hispanics at > 90, Jews and Asians around 110. That makes for a hugely > disproportionate difference for populations at the high end of > performance. Urban African Blacks given Black African IQ tests > written and administered by Black African psychologists come in around > 75 IQ compared to South African Boers (The Dutch East India Company > arrived at Table Bay in 1652). Boers are not the intellectual giants > of European-derived populations. Where is your information coming from? I seem to recall that an IQ range from 80-90 is termed dullness and 70-80 borderline deficiency. By your figures, you'd have to believe that over half the black population is at least dull. Do you believe this? > The Bell Curve, Herrnstein and Murray, p. 279. Every mass testing - > education, military, job, therapeutic - over almost 100 years of > looking is consistent. Some populations tend to be > blonde/brunette/redhead, tall/short, thin/fat, straight hair/curly > hair, smart/stupid, epicanthal fold present/absent. You can breed for > intelligence as you can breed for any other multi-gene trait. > New York City in the early 1970s was brutally assaulted by social > advocates for claimed racial discrimination on the basis of IQ and > other standardized achievement tests. Psychologists were called in to > create a Black IQ test normed at 100 IQ for the target population. > They did a credible job. When the tests were administered citywide, > Blacks and Browns clustered around 100 IQ. No prlbem. Psychology will > deliver any answer you seek to purchase. > Things would have gone along as usual if some pundit hadn't released > the rest of the results. In a test where Blacks and Browns scored 100 > IQ, Whites averaged around 120 IQ. The tests were valid objective > functional assessments of verbal and mathematical aptitudes, and > therefore racial discrimination. Social advocates screamed. > New York City Liberally removed all objective criteria for evaluation > (high school) and admission to public colleges and universities (now > only requiring a NYC high school diploma or GED). Starting in the > 1990s, a bachelor's diploma from any New York City public college or > university was toilet paper. It didn't even certify former high > school level proficiency. MS and PhD programs suitably scaled to have > proper racial quota outputs. Columbia University is no longer > respected as a whole. I _know_ this is not true nationwide. I can't count the number of stats graduate depts on both hands that are at _least_ 2/3 Chinese. > The SAT has been multiply revised to allow proper scores (SAT score/10 > = IQ) to be achieved by politically precious groups. Two salient > facts emerge. First, the test cannot be dumbed own enough to pull it > off. Anything involving reading, logical thought, or mathematics is > racial discrimination. Second, when Blacks and Browns nationwide can > score 1000 as a group, Whites as a group pull 1250. (Jocks are often > advised to make every answer c. Former instructions to fill in the > boxes with fast best guesses gave lower scores.) One then concludes > the test is overall a valid and objective skill-based assessment. It > must be changed. > In one way the demands of social advocacy have been met. Any NYC > diploma is crap (Bronx High School of Science excepted), regardless of > the color of the holder. The average has been lowered until everybody > is above it. Diversity admissions to universities nationwide are in > lock-step. Diversity does not include academic achievement - and it > doesn't. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > Quis custodiet ipsos custodes? The Net! In the end, this is a lot of empirical evidence, but I'm not totally convinced that it adds up to a proof. I've taught enough math at the college level to know that student's aptititudes are in part, a reflection of their upbringing as well as their heredity. If what you are saying is true, then someone should have done some type of histological studies to try to discren what these differences are in the brains of these different races at some point, and I know of no such research. As far as colleges being watered down, well this would have happened with or without blacks: the average American kid (of any color), puts in far too little work in h.s., and you simply can't change the rules in college and expect them to change their behavior like that. If you try to teach a math course for engineers the way it _should_ be taught, you'll find yourself in front of the dept head so fast it will make your head spin. I appreciate your response though. MB === Subject: Re: Uncle Al is Sadistic . In sci.math, jmfbahciv@aol.com : >Is there anything good about California? Anyone? I'd like to hear >something positive that I can hold onto, living here. >>I was going to say the weather's nice. So nice that the entire state >>goes up in flames when someone drops a cigarette. > Try allowing gleaning. Your greenies need to get some > education. >> .. So nice that >>when it does manage to rain, the mountain sides go sliding into the >>valleys. > Stop planting grass. Erm...dumb question: what should we do instead? :-) Mind you, grass is not native to California (I'm assuming you're referring to lawn grass of some sort [although one could be referring to that other grass, marijuana :-) ]; I'd have to look); planting drier landscapes would be preferable. >> ..So I guess you're right, there's nothing positive about >>living in California. > It's got an ocean. Drive a little bit, you can ski. Drive a little > bit, you can swim. Drive a little bit, you can die of thirst. Drive a little bit, one can choke on the smog. :-) > /BAH > Subtract a hundred and four for e-mail. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days Let's be serious for once. Consider an object being accelerated by a idealistic jet of water or a continuous 'stream of elastic ping pong balls'. What is its subsequent velocity pattern? accelerated beyond the operating speed of the accelerating fields, ie at 'c'. I hope it might also produce a relationship that is equivalent to mass 'appearing' to increase with velocity by gamma.) We have: S= -------m/s->Vo--------->M->v A stream of 'matter' emerges from source S at velocity Vo and m 'mass units' per second. It strikes unconstrained object (mass M), which subsequently accelerates away. Both energy and momentum must be conserved during this operation. The problem is to find the x/t relationship for three different situations. 1) nonelastic case where the jet ends up having the same velocity as the object after collision. 2) where the jet bounces back perfectly elastically. 3) (maybe) where the jet is absorbed into the object. WRT the source frame, the equation describing the nonelastic jet case is: M.dv/dt=m(Vo-v) or: (d2x/dt2)=K(Vo-dx/dt) Here, the water ends up traveling at M's velocity. In the elastic case, the water (or stream of ping pong balls) ends up moving backwards at Vo-2v, so the change in momentum per second is 2m(Vo-v) Do you all agree? I now wish to solve this equation. My maths is a bit rusty but I can probably work it out eventually. I get something starting with a factor e^-K (which is good). I'm not sure about the rest. Any offers? What are mathematicians for anyway? Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) > Let's be serious for once. Idiot. > Consider an object being accelerated by a idealistic jet of water or a > continuous 'stream of elastic ping pong balls'. What is its subsequent velocity > pattern? Do the integral including momentum transfer given center of mass angle vs. velocity vector. If you can't do calculus you should be mucking stalls instead of filling them with muck. > See why relativity is wrong: > http://www.users.bigpond.com/HeWn/index.htm Idiot. Experimental constraints on General Relativity. http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net!