mm-2599 === Subject: continuity and uniform convergence Let f_n be a sequence of continuous functions that converge uniformly on each compact subset E of metric space X to a function f. Show f is continuous on X. Well, showing f is continuous on each compact subset E of X is a standard 3 epsilon proof. But, if I know that f is continuous on each compact subset of f can I conclude that f is continuous on X ? === Subject: Re: continuity and uniform convergence >Let f_n be a sequence of continuous functions that converge uniformly on each compact subset E of metric space X to a function f. Show f is continuous on X. >Well, showing f is continuous on each compact subset E of X is a standard 3 epsilon proof. But, if I know that f is continuous on each compact subset of f can I conclude that f is continuous on X ? Isn't every convergent sequence together with it's limit point a compact set? Make sure I'm right about that. If so, you should be able to use the version of continuity of the form f is continuous iff x_n --> x implies f(x_n) --> f(x) Hmm. Something about my idea must be wrong, since my plan only uses convergence on compact sets, not uniform convergence. But off hand, I don't see my error. quasi === Subject: Re: continuity and uniform convergence >>Let f_n be a sequence of continuous functions that converge uniformly on each compact subset E of metric space X to a function f. Show f is continuous on X. >>Well, showing f is continuous on each compact subset E of X is a standard 3 epsilon proof. But, if I know that f is continuous on each compact subset of f can I conclude that f is continuous on X ? >Isn't every convergent sequence together with it's limit point a >compact set? >Make sure I'm right about that. >If so, you should be able to use the version of continuity of the form >f is continuous iff x_n --> x implies f(x_n) --> f(x) >Hmm. Something about my idea must be wrong, since my plan only uses >convergence on compact sets, not uniform convergence. But off hand, I >don't see my error. >quasi No, I take it back -- the plan I outlined above should work. You'll need uniform convergence of the f_k's to get f(x_n) --> f(x). quasi === Subject: Re: Are Z-cutoffs transcendental? > I'm teaching statistics this semester [and I am appalled, once again, > at how desperately they want FORMULAS, and to avoid understanding], and > an unbidden thought came to my brain: are the Z-cutoff values such as > Z_0.025 = 1.96 (approx) algebraic or not? > Obviously a question of great interest to statisticians :-) > The cutoffs are defined by > Z_a = that c so that int_c^infty e^{-x^2/2}/sqrt(2 pi) dx = a. > OK, statisticians don't care. But from the point of view of > transcendental numbers, are there classes of integrals [or inverses of > integrals, as in this case] which are known to be transcendental? > One approach would be to show the continued-fraction development > converges too rapidly. But, alas, the CF development of Z_0.025 > appears to be something like > {1, 1, 23, 1, 43, 2, 6, 1, 2, 1, 1, 4, 2, 1, 1, 3, 7, 1, 3, 5, 2, 1, 3, > 1, 1, 1, 7, 1, 6, 7, 2, 4, 2, 1, 8, 13, 8, 1, 3, 18, 6, 1, 10, 2, 1, 1, > 1, 34, 1, 3, ...} > and is not at all very Liouville-like. I am too abysmally ignorant of > transcendental number theory to know of any other approach. Another > thought: erf satisfies an ODE. Is anything known about > transcendentality of solutions of ODE's (with, say, coefficients being > polynomials with integer coefficients) at particular points? (Or, as > in this case, their inverses?) Are there any famous conjectures? There is, in fact, quite a bit known about solutions to differential equations taking on transcendental values at rational arguments. Unfortunately, I am not one of the people who knows what is known. I think the key phrase to search for in the transcendence literature is G-functions. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Are Z-cutoffs transcendental? >I'm teaching statistics this semester [and I am appalled, once again, >at how desperately they want FORMULAS, and to avoid understanding], and >an unbidden thought came to my brain: are the Z-cutoff values such as >Z_0.025 = 1.96 (approx) algebraic or not? >Obviously a question of great interest to statisticians :-) >The cutoffs are defined by > Z_a = that c so that int_c^infty e^{-x^2/2}/sqrt(2 pi) dx = a. >OK, statisticians don't care. But from the point of view of >transcendental numbers, are there classes of integrals [or inverses of >integrals, as in this case] which are known to be transcendental? ... >and is not at all very Liouville-like. I am too abysmally ignorant of >transcendental number theory to know of any other approach. Another >thought: erf satisfies an ODE. Is anything known about >transcendentality of solutions of ODE's (with, say, coefficients being >polynomials with integer coefficients) at particular points? (Or, as >in this case, their inverses?) Are there any famous conjectures? I think the Siegel-Shidlovsky Theorems may be what you need, although I don't have a copy of Baker's Transcendental Number Theory handy. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Richardson's extrapolation Hello! Could anyone give an example how to apply Richardson's extrapolation to numerical integration? For example, given a function 1/(x*ln^2(x)), where ln - natural logarithm. I integrate it using rectangle method in interval 1.1 to 20 to a precision of 0.005. = 5120 (number of rectangle areas are calculated throughout whole interval). (Algorithm for checking precision calculates integral value using 10 rectangles and 20 rectangles, if precision is not reached (integral value approximated with 20 rectangles - value with 10 rectangles > precision) it increases rectangle count to 40, then to 80, etc. until precision is reached). How do I apply Richardson's extrapolation in this example? P.Krumins === Subject: Re: Richardson's extrapolation covers that topic in excruciating detail on pages 19-274. === Subject: Inverse Function Could somebody tell me how to find the inverse function for f(x) = (x^3)/(x^2+1) I'm having difficulties solving this eqn. === Subject: Re: Inverse Function >Could somebody tell me how to find the inverse function for f(x) = (x^3)/(x^2+1) >I'm having difficulties solving this eqn. Difficulties? That's no surprise for this problem. Forget trying to find f^(-1) explicitly. To solve explicitly for x requires solving an irreducible cubic equation. Of course you could always use the cubic formula but I suspect the result would be nightmarish. Instead, represent the inverse by an implicit equation. First, verify, that f is invertible. Taking the derivative, you get f'(x) =((x^2)*(x^2+3))/(x^2+1) Thus, f' is always >=0, and 0 at only one point, so f is increasing, hence f^(-1) exists. The equation for f is y=(x^3)/(x^2+1) This inverts easily -- just switch the variables. Then leave the equation implicit -- don't try to solve for y. Thus, you get the equation for f^(-1) as: x=(y^3)/(y^2+1) quasi === Subject: Re: Inverse Function >Could somebody tell me how to find the inverse function for f(x) = (x^3)/(x^2+1) >I'm having difficulties solving this eqn. ... over the real numbers, I presume. See Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: series converging >1) Find two series [sum](a_k) and [sum](b_k) >such that [sum](b_k) converges conditionally, >(a_k)/(b_k)->1 as k->oo, and [sum](a_k) diverges. Hint: take your favourite conditionally convergent series sum b_k, and let a_k = b_k + c_k where c_k > 0 and c_k -> 0 just a teeny bit faster than b_k. >2) Why is this impossible if the series [sum](b_k) >is absolutely convergent? Hint: what convergence tests do you know that work for absolutely convergent series? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: series converging 1) Find two series [sum](a_k) and [sum](b_k) >such that [sum](b_k) converges conditionally, >(a_k)/(b_k)->1 as k->oo, and [sum](a_k) diverges. Hint: take your favourite conditionally convergent series sum b_k, and let a_k = b_k + c_k where c_k > 0 and c_k -> 0 just a teeny bit faster than b_k. How about if I take (b_k)= ((-1)^n)/n. This series converges conditionally. And take (a_k)= |(b_k)|= 1/n, then [sum](a_k) diverges. But then (a_k)/(b_k)->1 as k->oo isn't true... grrrr... because (a_k)/(b_k)= 1/(-1)^n, which doesn't go to 1 as n->oo, it goes to 0. Please help! >2) Why is this impossible if the series [sum](b_k) >is absolutely convergent? Hint: what convergence tests do you know that work for absolutely convergent series? I don't know any specific tests... === Subject: Re: series converging >1) Find two series [sum](a_k) and [sum](b_k) >>such that [sum](b_k) converges conditionally, >>(a_k)/(b_k)->1 as k->oo, and [sum](a_k) diverges. >Hint: take your favourite conditionally convergent series sum b_k, >and let a_k = b_k + c_k where c_k > 0 and c_k -> 0 just a teeny >bit faster than b_k. >How about if I take (b_k)= ((-1)^n)/n. This series >converges conditionally. >And take (a_k)= |(b_k)|= 1/n, then [sum](a_k) diverges. >But then (a_k)/(b_k)->1 as k->oo isn't true... grrrr... >because (a_k)/(b_k)= 1/(-1)^n, which doesn't go >to 1 as n->oo, it goes to 0. >Please help! Assuming your favorite conditionally convergent series is oo --- k 1 > (-1) - --- k k=2 Show that the following series diverges: oo --- 1 > ------- --- n ln(n) k=2 Try to use these to make a divergent series with the proper condition. >>2) Why is this impossible if the series [sum](b_k) >>is absolutely convergent? >Hint: what convergence tests do you know that work for absolutely convergent series? >I don't know any specific tests... To work problems of this kind, knowledge of tests such as the comparison test are useful, if not necessary. Rob Johnson take out the trash before replying === Subject: Re: series converging >Show that the following series diverges: > oo > --- 1 > > ------- > --- n ln(n) > k=2 Instead, make that >Show that the following series diverges: > oo > --- 1 > > ------- > --- k ln(k) > k=2 Rob Johnson take out the trash before replying === Subject: Re: series converging > 1) Find two series [sum](a_k) and [sum](b_k) >>such that [sum](b_k) converges conditionally, >>(a_k)/(b_k)->1 as k->oo, and [sum](a_k) diverges. > Hint: take your favourite conditionally convergent series sum b_k, > and let a_k = b_k + c_k where c_k > 0 and c_k -> 0 just a teeny > bit faster than b_k. > How about if I take (b_k)= ((-1)^n)/n. This series > converges conditionally. > And take (a_k)= |(b_k)|= 1/n, then [sum](a_k) diverges. > But then (a_k)/(b_k)->1 as k->oo isn't true... grrrr... > because (a_k)/(b_k)= 1/(-1)^n, which doesn't go > to 1 as n->oo, it goes to 0. > Please help! >>2) Why is this impossible if the series [sum](b_k) >>is absolutely convergent? > Hint: what convergence tests do you know that work for absolutely > convergent series? > I don't know any specific tests... You either didn't read Robert's advice or you chose not to take it. What is c_k? Joe === Subject: Re: Inclusion-Exclusion > Hello everyone could someone tell me how to use the inclusion-exclusion > principle on this problem, tried a combinartorial proof but wish to see > it using the inclusion-exclusion principle. > Question: > Let S be a multiset with k distinct objects whose repetition numbers > are n1, n2, ..., nk, repsectively. Let r be a positive integer such > that there are at least one r-combination of S. Please clarify: what do you mean by an r-combination? Do you mean that r <= k and each object in the r-collection is different? (That would be my guess!) R.G. Vickson > Show that, in applying > the inclusion-exclusion principle to determine the number of > r-combinations of S, one has A1 n A2 n ... n Ak= the empty set. > n= and === Subject: Re: product r x c > Why do we always consider, as product of matrices, the rows by columns > product? What about if we consider one of the other three products (c x r, c > x c, r x r)? I think we won't have associativity anymore. For example I > tried this, column by column: > 1 0 0 1 0 1 1 0 0 0 0 0 > 0 0 0 0 0 0 = 0 0 0 1 = 0 0 > 1 0 0 1 0 1 0 1 0 1 0 0 > 0 0 0 0 0 0 = 0 0 0 0 = 0 1 > so we don't have associativity. > Regarding r x c product, is the preserving of associativity the only reason > why we prefer such a product? We prefer r x c because it actually means something. If the matrix A represents the linear transformation T from V to W (with respect to some particular bases for the vector spaces V and W) and the matrix B represents the linear transformation S from W to X (with respect to etc., etc.) then the matrix BA, defined in what you call the r x c way, represents the composition ST from V to X. That is, multiplication of matrices corresponds to composition of linear transformations. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: reference for automorphisms of nonabelian finite simple groups Hi. I need to cite something. Does anyone know of a reference for the automorphism groups of nonabelian finite simple groups? === Subject: Re: Constructing Complex Numbers Using Straightedge and Compass > can you help me a bit more? i have a question and i don't know its > answer (actually i know its answer but dont have a clue why it's > right). the question is the following: define z= e^(2*pi*i/13). which > of the following is true: > a. the number z+z^3+z^9 is constructable. > b. the number z+z^5+z^8+z^12 is constructable. > I'll be thankful if you could help me understand this stuff. > The answer is well-known. It is not hard to see that constructing a > primitive nth root of 1 is equivalent to constructing a regular n-gon. > For if you take the arc from 1 to that primite root and iterate it, > starting from the primitive root, you get all the roots and they are > the vertices of that regular n-gon. A regular n-gon is constructible > iff n is a power of 2 times a product distinct primes of the form > 2^{2^k} + 1, of which I think only five are known. IIRC, they are > called Fermat primes. At any rate, they are 3, 5, 17, 257, and 65,537 > (I think the last one is right; I don't have pencil and paper handy). > So, for example, a primitive 13th root is not constructible, but a > primitive 5140th is. (5140 = 2^2 x 5 x 257). Maybe so, but mostly irrelevant. Take either of your expressions and replace z, in turn, by z^2, z^3, ..., z^12, and see what you get - remembering always to use z^13 = 1 to reduce all exponents to the range 1, 2, ..., 12. For example, for z + z^3 + z^9, you get z^2 + z^6 + z^5, z^4 + z^12 + z^10, z^7 + z^8 + z^11, several times each. (You should check this) That says z + z^3 + z^9 has degree 4 over the rationals. If it's in a quadratic extension of a quadratic extension, then it's constructible. Have a look at (X - a) (X - b), where a = z + z^3 + z^9 and b = z^4 + z^12 + z^10. (By have a look, I mean, multiply it out, simplify, and think a bit. Maybe also have a look at (X - (a + b)) (X - (c + d)) where c = z^2 + z^6 + z^5 and d = z^7 + z^8 + z^11). -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: I need problems to solve for real analysis study... On 13 Nov 2005 12:16:43 -0800, comtech >I think you've hit the point. The whole point of real analysis is to >learn to write things correctly. I am a programmer and engineer. That's >why I want to learn this class well. I will try my best to get the TA's >help. But the TA usually did not go through all the problems himself so >requiring him to solve a problem that he has not seen before in real >time is both a torture to him and to me myself. Because if the TA does >not know the solution in advance, he begins to speak nonsense and waste >our time. >Anyway, I have to squeeze some time out to go to the instructor's >office hours... Why are you reluctant to post something that you have tried? I think the help is here; but you have to offer something to start with. === Subject: Re: I need problems to solve for real analysis study... Do you think posting is easy? It takes more than half an hour to describe a problem and write my whole proof out accurately and precisely in an good-old-ascii mode... This type of communication is not efficient... === Subject: Re: I need problems to solve for real analysis study... On 13 Nov 2005 23:36:31 -0800, comtech >Do you think posting is easy? >It takes more than half an hour to describe a problem and write my >whole proof out accurately and precisely in an good-old-ascii mode... >This type of communication is not efficient... I would have done something that long on the computer in the first place and then could just paste it into a message or something. If the problem is proof style, as it might seem to be, posting a piece of it would work. You win. I give up. I guess you are only up for easy stuff. Maybe if you give us your address, we will all fly out there, hold your hand, and do all the work for you. However, that is unlikely. Good day, sir! === Subject: Re: I need problems to solve for real analysis study... > Do you think posting is easy? You think anything worth while comes for free? > It takes more than half an hour to describe a problem and write my > whole proof out accurately and precisely in an good-old-ascii mode... And yet you expect people here to give you detailed help without you even showing what you've tried. So, is your time more valuable than ours, or what? > This type of communication is not efficient... You are, of course, free to go elsewhere for your free tuition. Alternatively, do what many others do: learn enough TeX to use it as an effective means of presenting mathematics via ASCII. === Subject: Re: I need problems to solve for real analysis study... > Do you think posting is easy? > It takes more than half an hour to describe a problem and write my > whole proof out accurately and precisely in an good-old-ascii mode... > This type of communication is not efficient... And it's made worse by not including context. === Subject: Re: Three Theorems of Godel I'm reading the Wikipedia page on Goedel's incompleteness theorem(s) here and it is some good reading, Aatu has done some good work on that. So, I am interested in the Goodstein theorem, that his confabulations go to zero, and am interested in a note about the Paris and Kirby result that apparently claims a statement about the naturals requires a nonstandard countable model of the natural integers. I'm interested in reading that, I should get a hold of a copy and check it out. I think that means that the infinite collection of natural integers is more than the sum of its parts. It's similar in a way to considering Cantor's first to show sequential reals, and the powerset result to show Ord = -1, or Goedel to demand no non-logical axioms. As I have said, I'm definitely not conventional when it comes to Goedel. Ross === Subject: Re: when convergence in distribution implies almost sure convergence? >Hello! >I know that almost sure convergence implies convergence in >distribution. >My question is: >Is there any special case under which convergence in distribution >implies almost sure convergence? By itself, only as stated below; there are two parts to the argument. Convergence in distribution implies convergence in probability (which is a topological type of convergence) if the limiting distribution is concentrated at one point, and not otherwise. The case of iid random variables shows that. To get the further almost sure convergence guaranteed, one would need that for every neighborhood of the limit point, the sum of the probabilities of the k-th random variable not being in the neighborhood is finite. These conditions are best possible; drop any one, and counterexamples exist. However, if S_n = sum^n X_i, and the X_i are independent random variables, the three types of convergence are equivalent. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Variance and moment >I can't understand the notion of moment in maths , particularly in >porbability theory. >The standard deviation is easy to understand but the variance , what is >the meaning of the average square of the difference between X and its >expected value. Consider the probability distribution as mass along the X axis. If this is pivoted about the mean, the variance is the moment of inertia. In fact, pivoted about 0, the second moment is the moment of inertia about that point. Now having the mean as the first moment, and the expected square as the second moment, the rest is a terminological generalization. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Variance and moment <23554206.1131844564503.JavaMail.jakarta@nitrogen.mathforum.org> 1) As David Kastrup mentioned: the variance of a sum is the sum of the variances (for independent random variables). 2) To compute the standard deviation, you first compute the variance, then take its square root. 3) If you have a Physics background, it may interest you to know that the mean is the center of gravity, while the variance is the moment of inertia (about the mean). === Subject: Re: Differentiation >I know that quaternions are not commutative and octonions are not associative. So it is not possible to define division in such a way that a/b makes sense. I also see that it is possible to define differentiation on real- or complex-valued functions. The question I have is, is it possible to define differentiation on quaternion- or octonion-valued functions? Are there any books or online resources I can check out? quaternions, using 2-sided division ends up with only functions ax+b, where a is real. For one-sided division, linear functions with multiplication on the other side are all that satisfy the condition. Octonions would only leave ax+b, a real. As you can see, this is of little use. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: illegal email and who is the person behind 234.98.227.217 > quoting the header of illegal email > > > I thought there was a law in place that email had to have a proper > return address so that unwanted email can be filtered. Of which the > above has no legal return address. > > Can you quote that law? > Netherlands probably does not have news of the USA law recently passed > on email. The new USA law forbids emails to have faked addresses. Which law? Is it a federal law? Anyway, it is *not* an international law. -- 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: illegal email and who is the person behind 234.98.227.217 > quoting the header of illegal email > > new. > > I thought there was a law in place that email had to have a > proper return address so that unwanted email can be > filtered. Of which the above has no legal return address. > > Can you quote that law? > > Netherlands probably does not have news of the USA law recently > passed on email. The new USA law forbids emails to have faked > addresses. > Which law? Is it a federal law? Anyway, it is *not* an > international law. Doesn't .nl have EC Directive 2002/58/EC Concerning the Processing of Personal Data and the Protection of Privacy in the Electronic Communications Sector enshrined in its own laws? properly. -- Rob Skedgell GnuPG/PGP: 7DA3 1579 C0DD 8748 C05A B984 E2A2 3234 D14B 6DD7 === Subject: Re: illegal email and who is the person behind 234.98.227.217 Which law? Is it a federal law? Anyway, it is *not* an international law. I remember the law discussed after it was passed by US Congress. It was an attempt to head off a similar law by a state government, I believe it was the state of Virginia who was going to pass a stiffer law I remember the federal law passed made it illegal to have a fake address on email because too many people were losing control of their computer email boxes. Those jerks that use fake addresses and constantly change the subject line were hard to filter out by most people. And so email became a burden and not a useful and pleasurable experience. I do not see the USA sending Marines into a foreign country chasing then another few seconds to filter out every email issued from that domain from entering into the USA. So I think it behooves all foreign countries to want to cooperate with these sensible and common sense laws of the USA about Internet use and Dik, around 1997 there was a website that listed the owners of domains such as 234.98.227.217. Is there still a website that can tell me who owns operates this domain of 234.98.227.217. It maybe a foreign outfit or it maybe in the USA. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: illegal email and who is the person behind 234.98.227.217 > Which law? Is it a federal law? Anyway, it is *not* an international > law. ... > I remember the federal law passed made it illegal to have a fake > address on email because too many people were losing control of their > computer email boxes. I found the act. It does *not* make it illegal to have a fake address on email. It makes it illegal to have a fake address on commercial email. Hoewever, I do not know how you think the following: > And there should be a Internet newsgroup that has resident lawyers, so > that we can post these violators and where legal action is taken on > them. News ADmin Net abuse newsgroups are too much of a slap on the > hand and do not do this, whereas we need a stiffer punishment for > those recurring abusers for they are really criminals. Their > harrassement of people and the time they waste of people makes them > criminals. should be organised and/or paid. -- 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: illegal email and who is the person behind 234.98.227.217 I found the act. It does *not* make it illegal to have a fake address on email. It makes it illegal to have a fake address on commercial email. Dik, you have troubles with the concept of definition in math and physics and it does not surprise me that you have trouble in recognizing a definition in common everyday living. If someone sends you an email talking about mathematics would you define that as a personal email, but if someone sends you a email trying to sell you something would that not be a commercial email. So any email that is unsolicited and selling something is what Dik? a commercial email is it not. Now if the commercial email has no valid address, well that makes it worse yet for it is illegal. If the above legal, because then filters can filter all future such crap. So the above email is illegal. And I wonder if that illegal email is coming from Europe because the last time someone tried to forge posts with my name onto the Internet was a culprit found to reside in Europe. So there is a good chance that this illegal email originates from some site in Europe. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: illegal email and who is the person behind 234.98.227.217 > I found the act. It does *not* make it illegal to have a fake > address on email. It makes it illegal to have a fake address on > commercial email. > Dik, you have troubles with the concept of definition in math and > physics and it does not surprise me that you have trouble in > recognizing a definition in common everyday living. You have trouble with reading. ... > So any email that is unsolicited and selling something is what Dik? a > commercial email is it not. Now if the commercial email has no valid > address, well that makes it worse yet for it is illegal. And, what did I write? But from what you presented it is *not* clear whether the mail was commercial. > If the above > legal, because then filters can filter all future such crap. Oh, possible, you will also filter mail that *is* legal. Considering the amount of bounce messages I receive for mail I did not send... The adress can be valid but still fake. But also see below. > So the above email is illegal. And I wonder if that illegal email is > coming from Europe because the last time someone tried to forge posts > with my name onto the Internet was a culprit found to reside in Europe. > So there is a good chance that this illegal email originates from some > site in Europe. Perhaps, but I think not. I suspect the mail originated in China or Taiwan. But let me see, from the headers: both appear to be valid e-mail addresses and not misleading. So, what is the problem? (Yes, 163.com does exist, as does dream918.com.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Algebra help I was given this practice problem from my professor when I asked for additional material. I believe I got the first part, but can someone provide a clean solution to parts b and c, or at least hints. And even if possible an outline for part a, since mine isn't so clean Let N be a normal subgroup of a finite group G and ler pi: G-->G/N be the natural projection. let S_p(G) be the set of p-sylow subgroups of G. a) assume p divides |N|. then P|-->PintN is a surjection from S_p(G) --> S_p(N). b) assume p divides |G/N|. then P |--> pi(P) is a surjection from S_p(G) --> S_p(G/N). c) suppose that the p-sylow subgroup of N is normal in N. show that it is a normal subgroup of G. === Subject: Re: Algebra help days. My association with the Department is that of an alumnus. >I was given this practice problem from my professor when I asked for >additional material. I believe I got the first part, but can someone >provide a clean solution to parts b and c, or at least hints. And even >if possible an outline for part a, since mine isn't so clean >Let N be a normal subgroup of a finite group G and ler pi: G-->G/N be >the natural projection. let S_p(G) be the set of p-sylow subgroups of >a) assume p divides |N|. then P|-->PintN is a surjection from S_p(G) >--> S_p(N). You want to show several things: (i) That if P is a p-sylow subgroup of G, then P/N is a p-Sylow subgroup of N. (ii) Every p-Sylow subgroup of N is equal to P/N for some p-Sylow subgroup P of G. The second part is the easiest: if Q is a p-Sylow subgroup of N, then is a subgroup of G; it need not be a p-Sylow subgroup of G (if the highest power of p that divides N is smaller than the highest power of P that divides G), but it will still be a subgroup of G of order p^k for some k. One of the parts of the Sylow Theorems should tell you that there is a p-Sylow subgroup P of G that contains Q. Now the only thing is to show that P/N = Q; it certainly contains Q; why can it not be any larger? (Shouldn't be too hard to do). First part is not too hard either: let P be any p-Sylow subgroup of call it P', such that P'/N is a p-sylow subgroup of N, call it Q. We also know, from the Sylow theorems, that P' is conjugate to P. You should be able to show then, using that N is normal in G, that P/N will be a conjugate of Q (conjugate in G, anyway), and thus contained in N and of the correct order, hence a p-Sylow subgroup of N. >b) assume p divides |G/N|. then P |--> pi(P) is a surjection from >S_p(G) --> S_p(G/N). Again, you need to do two things: (i) Show that P is a p-Sylow subgroup of G, then pi(P) (which should really by pi(P)N or pi(PN), by the by) is a p-Sylow subgroup of G/N; and (ii) Every p-Sylow subgroup of G/N can be obtained as pi(P)N of some p-Sylow subgroup P of G. For the first part, note that PN/N is isomorphic to P/(P/N), so the subgroup generated by pi(PN) is certainly a p-subgroup of G/N. By Sylow's theorems, there is a p-Sylow subgroup Q of G/N which contains PN/N. Lift Q to G to deduce that Q = PN/N. For the second part, take any p-Sylow Q of G/N, and lift it to G; then find a p-Sylow subgroup of the lift and check to see that one works. >c) suppose that the p-sylow subgroup of N is normal in N. show that it >is a normal subgroup of G. Prove the following: (*) If H is a group, and P is a normal p-Sylow subgroup of H, then P is characteristic: that is, for every ->automorphism<- f of H (not just the inner automorphisms), f(P)=P. Note that the image of a p-element under an automorphism must be a p-element. If P is normal, what can you say about the p-elements (elements whose order is a power of p)?. Then note that conjugation by g in G will map N to itself, and so will correspond to an automorphism (not necessarily inner) of N. So if the p-Sylow subgroup of N is normal in N, then it will be characteristic in N, so conjugation by g will map it to itself. -- 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: Simple iteration method for solving linear equation systems In the english literature the iteration is usually written as x_(k+1) = x_k + tau_k r_k where r_k = f - A x_k is the residual. That tau_k i s a scalar steplength. If tau_k = (r_k^T r_k)/(r_k^T A r_k), that is the steepest descent method, which often displays poor performance. A simple modification of this iteration gives the more effective conjugate gradient method. See the tutorial http://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf === Subject: Re: Simple iteration method for solving linear equation systems P.Krumins === Subject: Re: Simple iteration method for solving linear equation systems In the english literature the iteration is usually written as x_(k+1) = x_k + tau_k r_k where r_k = f - A x_k is the residual. That tau_k looks like a steplength. If tau_k = 1 for any k that is the steepest descent method, which often displays poor performance. A simple modification of this iteration gives the more effective conjugate gradient method. See http://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf === Subject: Re: Simple iteration method for solving linear equation systems In the english literature the iteration is usually written as x_(k+1) = x_k + tau_k r_k where r_k = f - A x_k is the residual. That tau_k looks like a steplength. If tau_k = 1 for all k it is called the steepest descent method, which BTW is generally a very poor scheme. === Subject: Re: Simple iteration method for solving linear equation systems >In the english literature the iteration is usually written as > x_(k+1) = x_k + tau_k r_k >where r_k = f - A x_k is the residual. That tau_k looks like a >steplength. If tau_k = 1 for all k it is called the steepest descent >method, which BTW is generally a very poor scheme. No, that's not steepest descent, or at least, not either of the two very different things I've ever seen called steepest descent: 1) a method of minimizing a nonlinear function F(x) by the iteration x_{k+1} = x_k - tau_k (grad F)(x_k) choosing the scalar tau_k to (at least approximately) minimize F(x_{k+1}) 2) a method of finding asymptotics of a function of a complex variable defined as an integral on a curve. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Simple iteration method for solving linear equation systems That is the name I always used in my courses, at least to warn students to stay away from. It agrees with the common terminology in the english numerical analysis literature, for example in Shewchuk's tutorial http://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf === Subject: Re: Simple iteration method for solving linear equation systems >In the english literature the iteration is usually written as > x_(k+1) = x_k + tau_k r_k >where r_k = f - A x_k is the residual. That tau_k looks like a >steplength. If tau_k = 1 for all k it is called the steepest descent >method, which BTW is generally a very poor scheme. > No, that's not steepest descent, or at least, not either of the two > very different things I've ever seen called steepest descent: > 1) a method of minimizing a nonlinear function F(x) by the iteration > x_{k+1} = x_k - tau_k (grad F)(x_k) > choosing the scalar tau_k to (at least approximately) minimize F(x_{k+1}) Actually, I think Carlos is saying that this is steepest descent (or gradient descent) of that type for the least squares problem, where F = || f - A x||^2. And I believe he's correct. But you're right also, that step size in most modern algorithms is not 1 but is chosen by some sort of line search algorithm along the descent direction. Usually convergence is pretty good if instead of searching for an optimum you search for a sufficiently small step where the new iteration of F is sufficiently smaller. It's early in the morning and I'm on travel, so I'm feeling to brain dead to decide if that step direction is gradient descent (-grad F), or is a Newton step -inv[H(x)]'grad(F) where H is the Hessian (second derivative matrix) of F. Since the least-squares objective function is a convex quadratic function of x, Newton's method should converge very well. - Randy === Subject: Re: Simple iteration method for solving linear equation systems > The text it is mentioned in just contains single (recursive) equation > and I do not understand it. > Here it is: > (x_(k+1) - x_k)/tau_k + A.x_k = f, Also known as: x_(k+1) = tau_k (f - A. x_k) + x_k > for solving A.x = f equation system. Solution x_0 is chosen > arbitrarily. > I have no idea what tau_k is, the text says k-th iteration > parameter. > Does anyone recognize this iterative method? If so what is its name in > English? I don't, but it seems to me that any good textbook on numerical analysis will contain iterative techniques for solving systems of linear equation. For example, the second edition of Burden, Faires, and Reynolds mentions the Jacobi Iterative Algorithm, the Gauss-Seidel Algorithm, and methods of Successive Over Relaxation. Your algorithm doesn't seem to be any of these, though. I think any such algorithm needs a convergence condition, e.g. that some vector norm of (x_(k+1) - x_k) falls below some specified small value epsilon. -- Mostly economics: r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Can an immortal solve most integer-related problems? >However, the Germans are always causing problems. >This time it was some guy called Goedel. > Austrian. > Wasn't there some other Austrian guy contemporary with Goedel > who also caused a lot of trouble? Yes, that Wittgenstein was a real pain. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Pricing Formula Question I'm trying to come up with a formula I can use for determining the final price for a product to assure a certain percentage profit. My problem is that I have to factor in the commision I will be charged for selling that product. The commision I am charged is 8% of the first $25 then 5% of the rest. So what I am looking for is... what final price must I charge in order to come out with 15% profit at the end of the transaction. Final Price = Cost + Markup + Commision And I am trying to final a formula to give me the markup to get 15% profit. Say it cost me $60 to purchase a certain item, and I want to sell it to make 15% profit, taking into account the commision charged on the final total, how much markup must I use. I'd appreciate any help I can get. --Mike === Subject: Re: Pricing Formula Question On 13 Nov 2005 18:07:18 -0800, Michael Shnitzer >I'm trying to come up with a formula I can use for determining the >final price for a product to assure a certain percentage profit. My >problem is that I have to factor in the commision I will be charged for >selling that product. >The commision I am charged is 8% of the first $25 then 5% of the rest. >So what I am looking for is... what final price must I charge in order >to come out with 15% profit at the end of the transaction. >Final Price = Cost + Markup + Commision >And I am trying to final a formula to give me the markup to get 15% >profit. >Say it cost me $60 to purchase a certain item, and I want to sell it to >make 15% profit, taking into account the commision charged on the final >total, how much markup must I use. >I'd appreciate any help I can get. >--Mike Your problem isn't stated clearly. When you say you want a 15% profit, does that mean that your profit is to be 15% of the final price, which includes the markup itself? I am going to assume the answer to that is yes, and, at least for now, assume the cost is greater than 25. Let the final price F be 1.25c+.9375 where c is the cost. Then in your example, if c = $60 we get with rounding: F = $75.94 Commission = $4.55 which is 8% of $25 + 5% of the rest Markup = 15% of F = $11.39 Cost = $60 Check: 60 + 11.39 + 4.55 = 75.94 --Lynn === Subject: Re: Pricing Formula Question You are correct Lynn. I didn't even think of that until I saw Arturo's response. I should be including the commission fees as part of my cost. --Mike === Subject: Re: Pricing Formula Question days. My association with the Department is that of an alumnus. >You are correct Lynn. I didn't even think of that until I saw Arturo's >response. I should be including the commission fees as part of my >cost. So, if F is your final price, C the cost, and M the commission, you want F-(C+M) to be 15% of F? I.e., F-(C+M) = .15F, or .85F = C+M. And the problem is of course that M is not constant, but a function of F. If the cost is $19.25 or less, then your final price will be 25 or less, so the commission will stay at the 8% of final price level. In that case, you have M = .08F, so .85F = C + 0.8F, or .77F = C, so F = C/(0.77). Check: if F = C/(0.77) = 100C/77, with C<= 19.25, then F<=25; so the commission is .08F = (.08/.77)*C = (8/77)C. Thus your total cost is C + M = C + .08F = C + (8/77)C = (85/77)C, so your profit is F - (C+M) = (100/77)C - (85/77)C = (15/77)C, which is exactly 15% of 100/77C, which was your final price. If the cost is $19.25 or more, then after the markup the final price will be 25 or more, so then your commission will be given by 8% on the first 25 and 5% on everything after that, so you can use Lynn Kurtz's formula. So we have: / (100/77)*C if C<= 19.25. F = { 1.25C + .9375 if C>= 19.25 -- 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: Pricing Formula Question days. My association with the Department is that of an alumnus. >I'm trying to come up with a formula I can use for determining the >final price for a product to assure a certain percentage profit. My >problem is that I have to factor in the commision I will be charged for >selling that product. You posted almost this exact same question back on May 9th. I answered it then. What is the problem? Apparently that the percentages are different this time. Was it really so hard to understand the solution you were given six months ago? Or is that six months ago you wanted a fixed profit and now you want a percentage profit? >The commision I am charged is 8% of the first $25 then 5% of the rest. >So what I am looking for is... what final price must I charge in order >to come out with 15% profit at the end of the transaction. >Final Price = Cost + Markup + Commision >And I am trying to final a formula to give me the markup to get 15% >profit. Let M be the comission: M = min(F,25)*.08 + max(0,(F-25))*.05 The easiest way for the max function is: max(0,F-25)= (1/2)*[(F-25) + |F-25|]. If F>= 25, then |F-25| (the absolute value) is equal to F-25, so you get (1/2)[F-25 + F - 25] = F-25. If F<25, then |F-25| = 25-F, so you get (1/2)*[(F-25)+(25-F)] = 0. For the min function, there is a similar process: (1/2)*[(F+25) - |25-F|] which will give 25 if F>=25, and F if F<25. So M = (1/2)*[ (F+25) - |25-F|]*.08 + (1/2)*[(F-25)+|F-25|]*.05 Since you want a 15% profit, that means that you want F - (M+C) = .15(C+M) which means F = 1.15C + 1.15M. That gives F = 1.15C + 1.15*[(1/2)*[ (F+25) - |25-F|]*.08 + (1/2)*[(F-25)+|F-25|]*.05] This will be difficult to solve for F unless you know already if F will be more than 25 or less than 25. That can be a problem if your cost is around 21 or so, because then the comission may or may not push you over the magic line of 25. But if you already know the final price is going to be 25 then you have F = 1.15C + 1.15*25*.08 + 1.15*(F-25)*.05 F = 1.15C + 1.15*25*.03 + 1.15*.05*F F = 1.15C + 1.15*.75 + .0575F .9425 F = 1.15C + 0.8625 F = (1.15C + 0.8625)/.9425 >Say it cost me $60 to purchase a certain item, and I want to sell it to >make 15% profit, taking into account the commision charged on the final >total, how much markup must I use. If C= 60, then plugging into the above I get $74.13 (rounded up). Let's see: the comission will be 8% on the first 25 (or for the layman, $2), and 5% for the remaining $49.13 (which gives $2.46, rounded up). So you are paying $4.46 in commission, plus 60 of cost, for a total of 64.46. 15% of 64.46 is $9.67, and your profit was 74.13 - 64.46 = 9.67 exactly. -- 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: integrating cardinal-valued functions Let S be a set and let f be a cardinal-valued function defined on f. Can we define the concept of the integral of f on S? We use the notation int(f,S) to denote the integral of f on S. The requirements are as follows ... (1) For all sets S and all cardinal-valued functions f on S, int(f,S) exists and is a cardinal number. (2) int(1,S)=card(S) (3) int(a*f,S)=a*int(f,S) where a is any constant cardinal. (4) int(f+g,S)=int(f,S)+int(g,S). This is just finite additivity with respect to the integrand, but if possible, we would like a more general additivity, perhaps arbitrary. (5) If S1, S2 are disjoint sets, whose union is S, then int(f,S)=int(f,S1)+int(f,S2). This is just finite additivity with respect to a partition of the domain. Here too, we would prefer a more general additivity, perhaps arbitrary. (6) if S is empty, then int(f,S)=0. Is this condition redundant? If not then I'm pretty sure it would be redundant if condition (5) was strengthened to arbitrary additivity. Ok, so here are some questions ... Question (1): Are the above conditions consistent? In other words, are the conditions satisfiable by some definition of int? Without thinking about it too hard, I'm pretty sure the answer is yes, but if the answer is no, relax some of the requirements and try again. Question (2): Assuming that there is such an int, is it unique? If not, can conditions (4) and/or (5) be replaced by stronger versions of additivity to force uniqueness? Question (3): Assuming a unique int, is there some concept of derivative that corresponds to int? If so, is there also a concept of antiderivative? If so, can we get an analog of the Fundamental Theorem of Calculus? Since I just made these questions up out of nothing, and I've spent no time trying to answer them, it wouldn't surprise me if the answers are both immediate and trivial. If so, it's probably because cardinals are just too coarse to get anything but blunt answers (so it's not my fault). quasi === Subject: Re: integrating cardinal-valued functions > Let S be a set and let f be a cardinal-valued function defined on f. > Can we define the concept of the integral of f on S? > We use the notation int(f,S) to denote the integral of f on S. > The requirements are as follows ... > (1) For all sets S and all cardinal-valued functions f on S, int(f,S) > exists and is a cardinal number. > (2) int(1,S)=card(S) > (3) int(a*f,S)=a*int(f,S) where a is any constant cardinal. > (4) int(f+g,S)=int(f,S)+int(g,S). This is just finite additivity with > respect to the integrand, but if possible, we would like a more > general additivity, perhaps arbitrary. > (5) If S1, S2 are disjoint sets, whose union is S, then > int(f,S)=int(f,S1)+int(f,S2). This is just finite additivity with > respect to a partition of the domain. Here too, we would prefer a > more general additivity, perhaps arbitrary. It seems to me that what you are defining is int(f,S) = sum_{s in S} f(s) which could be defined to be the cardinality of the union of {s} times A_s, where A_s is a typical set of cardinality f(s). (Am I missing something?) > (6) if S is empty, then int(f,S)=0. Is this condition redundant? If > not then I'm pretty sure it would be redundant if condition (5) was > strengthened to arbitrary additivity. I don't think this is redundant as subtraction of cardinalities is not well defined. > Ok, so here are some questions ... > Question (1): Are the above conditions consistent? In other words, are > the conditions satisfiable by some definition of int? Without thinking > about it too hard, I'm pretty sure the answer is yes, but if the > answer is no, relax some of the requirements and try again. Since I think my definition satisfies your requirements, I think the answer is yes. > Question (2): Assuming that there is such an int, is it unique? If > not, can conditions (4) and/or (5) be replaced by stronger versions of > additivity to force uniqueness? I don't think that it is unique, but now I think about it, my proposed counterexample probably fails (4). > Question (3): Assuming a unique int, is there some concept of > derivative that corresponds to int? If so, is there also a concept of > antiderivative? If so, can we get an analog of the Fundamental Theorem > of Calculus? Derivatives are really differences, and subtraction is not well defined with cardinalities. > Since I just made these questions up out of nothing, and I've spent no > time trying to answer them, it wouldn't surprise me if the answers are > both immediate and trivial. If so, it's probably because cardinals are > just too coarse to get anything but blunt answers (so it's not my > fault). > quasi === Subject: Re: integrating cardinal-valued functions > (5) If S1, S2 are disjoint sets, whose union is S, then > int(f,S)=int(f,S1)+int(f,S2). This is just finite additivity with > respect to a partition of the domain. Here too, we would prefer a > more general additivity, perhaps arbitrary. So you want ALL subsets, and ALL functions, and not just some collection of measurable sets or functions? If so, with countable additiveity you are probably out of luck, even for functions with values 0 and 1. Unless there are measurable cardinals, then the only possibility will be like int(f,S) = f(x), where x is a fixed element of S. Certainly this defeats uniqueness. In the Hilbert space theory of Spectral Multiplicity there is some discussion of cardinal-valued functions, and integration. ref: Halmos, Paul R., Introduction to Hilbert Space and the theory of Spectral Multiplicity. Chelsea Publishing Company, New York, N. Y., 1951. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: integrating cardinal-valued functions >Let S be a set and let f be a cardinal-valued function defined on f. That must be a typo. The poster probably meant: Let S be a set and let f be a cardinal-valued function defined on S. Either that or f is a truly recursive function. quasi === Subject: degree of certain maps on S^3 First, is it true that the map S^3 to S^3 defined by g mapsto g^n has degree n in H^n(S^3) (or pi_n(S^3) )? Second, if the answer is yes does anyone have a reference to a proof? ptk === Subject: Re: degree of certain maps on S^3 Sorry I'm a bad typist - the dimension of the cohomology and homotopy groups should be 3. To be clear: 1. Does deg(g --> g^n) = n in H^3(S^3) (or pi_3(S^3)) 2. proof/reference requested === Subject: Re: degree of certain maps on S^3 define the vector field g---->g^n,the origin is the unique singularity which index is n,since (g-e1)(g-e2)(g-e3)....(g-en) is a perturbation of g^n and this perturbed vector field has n difeerent simple singul;arity with index 1 further the index of an isolated siungularity of X has the following interpretation: The Degree of X as a map from small sphere to another small sphere,it is independednt of choosing sphere soi you can choose S^3 you can find ,perhaps,in Hirsh Diff Topology book, It would be intersting to consider the same question for arbitrary compact lie group > Sorry I'm a bad typist - the dimension of the > cohomology and homotopy > groups should be 3. To be clear: > 1. Does deg(g --> g^n) = n in H^3(S^3) (or pi_3(S^3)) > 2. proof/reference requested === Subject: Re: degree of certain maps on S^3 <7668602.1131995324510.JavaMail.jakarta@nitrogen.mathforum.org> > define the vector field g---->g^n,the origin is the unique singularity which index is n,since > (g-e1)(g-e2)(g-e3)....(g-en) is a perturbation of g^n > and this perturbed vector field has n difeerent simple singul;arity with index 1 > further the index of an isolated siungularity of X has the following interpretation: > The Degree of X as a map from small sphere to another small sphere,it is independednt of choosing sphere soi you can choose S^3 > you can find ,perhaps,in Hirsh Diff Topology book, > It would be intersting to consider the same question for arbitrary compact lie group Very early in Spanier's 'Algebraic Topology', there's a relevant discussion of H-spaces and H-cogroups. Among other results, there is a proof of the following: Theorem. Let X and Y be objects in a category and let * and *' be two laws of composition in hom(X,Y) such that (a) * and *' have a common two-sided identity element (b) * and *' are mutually distributive Then * and *' are equal, and each is commutative and associative. There's a corollary that answers the OP's question: Corollary. If P is an H space and Q is any H cogroup, then [Q;P] is an abelian group and the group structure is defined by the multiplication map in P. The OP's situation is P = Q = S^3 (which is a H-space since S^3 is a topological group) and an H cogroup since S^3 is a suspension. But ... this also applies to the more general question that's raised in this post: the group operation on pi_n(G) with G a topological group is induced by the group operation on G ... > Sorry I'm a bad typist - the dimension of the > cohomology and homotopy > groups should be 3. To be clear: > 1. Does deg(g --> g^n) = n in H^3(S^3) (or pi_3(S^3)) > 2. proof/reference requested > === Subject: Principal square roots Hi Terminology/notation question. I am confused about whether the term principal square root can be applied to negative, imaginary and complex numbers, and whether, for example, expressions like sqrt(-9), sqrt(i), sqrt(-i), sqrt(i+1) etc. are well-defined (in terms of distinguishing which root is meant, once i has been assigned to one of the square roots of -1). For example, http://college.hmco.com/mathematics/larson/calculus_analytic/7e/shared/downl oads/clc7eap0f01.pdf says If a is a positive number, the principal square root of the negative number -a is sqrt(a)*i, and http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam; action=display;num=1114282118 says that The complex Principal Square Root sqrt(z) of a complex number z is the square root (one of two if z != 0) whose Real Part is nonnegative. However, http://mathworld.wolfram.com/PrincipalSquareRoot.html says The concept of principal square root cannot be extended to real negative numbers... Can anyone shed any light on this? TIA === Subject: Re: Principal square roots >Terminology/notation question. >I am confused about whether the term principal square root can be >applied to negative, imaginary and complex numbers, and whether, for >example, expressions like sqrt(-9), sqrt(i), sqrt(-i), sqrt(i+1) etc. >are well-defined (in terms of distinguishing which root is meant, once >i has been assigned to one of the square roots of -1). If z = r exp(it) with r >= 0 and -pi < t <= pi, the principal square root of z is r^(1/2) exp(it/2). Equivalently, the principal square root of z is the square root that is either in the open right half-plane or on the nonnegative imaginary axis. This is perfectly well-defined for any complex number. However, you should keep in mind that when talking about a square root, people don't always mean the principal square root. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Principal square roots > If z = r exp(it) with r >= 0 and -pi < t <= pi, the principal square > root of z is r^(1/2) exp(it/2). Equivalently, the principal square > root of z is the square root that is either in the open right half-plane > or on the nonnegative imaginary axis. > This is perfectly well-defined for any complex number. However, you > should keep in mind that when talking about a square root, people don't > always mean the principal square root. But then we have: 1=(-1)*(-1)=[(-1)*(-1)]^{1/2}=(-1)^{1/2}*(-1)^{1/2}=i*i=-1 How to deal with this? Ciao Karl === Subject: Re: Principal square roots <4378476a$0$21954$9b4e6d93@newsread2.arcor-online.net> > If z = r exp(it) with r >= 0 and -pi < t <= pi, the principal square > root of z is r^(1/2) exp(it/2). Equivalently, the principal square > root of z is the square root that is either in the open right half-plane > or on the nonnegative imaginary axis. > This is perfectly well-defined for any complex number. However, you > should keep in mind that when talking about a square root, people don't > always mean the principal square root. > But then we have: > 1=(-1)*(-1)=[(-1)*(-1)]^{1/2}=(-1)^{1/2}*(-1)^{1/2}=i*i=-1 > How to deal with this? The formula (a*b)^(1/2) = a^(1/2)*b^(1/2) is wrong. What is true, if you're using principal values, is that (a*b)^(1/2) = (+ or -) a^(1/2) * b^(1/2) where + or - is chosen to make -pi/2 < arg((a*b)^(1/2)) <= pi/2. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Principal square roots >> If z = r exp(it) with r >= 0 and -pi < t <= pi, the principal square >> root of z is r^(1/2) exp(it/2). Equivalently, the principal square >> root of z is the square root that is either in the open right half-plane >> or on the nonnegative imaginary axis. >> This is perfectly well-defined for any complex number. However, you >> should keep in mind that when talking about a square root, people >> don't always mean the principal square root. > But then we have: > 1=(-1)*(-1)=[(-1)*(-1)]^{1/2}=(-1)^{1/2}*(-1)^{1/2}=i*i=-1 > How to deal with this? > Ciao > Karl I cannot, since I don't see where you know the third equation from as you are talking about *principal* roots only. ;) Best wishes, J. === Subject: Re: Principal square roots >> If z = r exp(it) with r >= 0 and -pi < t <= pi, the principal square >> root of z is r^(1/2) exp(it/2). Equivalently, the principal square >> root of z is the square root that is either in the open right half-plane >> or on the nonnegative imaginary axis. >> This is perfectly well-defined for any complex number. However, you >> should keep in mind that when talking about a square root, people don't >> always mean the principal square root. >But then we have: >1=(-1)*(-1)=[(-1)*(-1)]^{1/2}=(-1)^{1/2}*(-1)^{1/2}=i*i=-1 >How to deal with this? Just accept it -- 1 and -1 are equal. Welcome to Z_2 === Subject: Re: Principal square roots > Hi > Terminology/notation question. > I am confused about whether the term principal square root can be > applied to negative, imaginary and complex numbers, and whether, for > example, expressions like sqrt(-9), sqrt(i), sqrt(-i), sqrt(i+1) etc. > are well-defined (in terms of distinguishing which root is meant, once > i has been assigned to one of the square roots of -1). The general consensus is that the square rooting operation on numbers other than non-negative reals cannot be usefully constrained to give a unique result. Thus the notion of principle square root is not usefully extendable to operation on other than non-negative reals in most people's opinion. Some people try to do it anyway. === Subject: rigorous physics textbook sought hi all can anybody recommend a book on physics (especially mechanics) that builds solidly from the ground up? i enjoy pure mathematics, and i know that physics must make sense, but i can't seem to make sense out of any of the physics textbooks i've looked at. too many loose ends and unsound inferences. please help. peace, stm === Subject: Re: rigorous physics textbook sought The Feynmann Lecture Series is pretty good, once you are in 3rd year, wondering what it was all about. You also might try the following, since not only does the math have to make sense but so does the physics: Matter and Interactions I : Modern Mechanics (Paperback) by Ruth W. Chabay, Bruce A. Sherwood and Classical Mechanics (3rd Edition) (Hardcover) by Herbert Goldstein, Charles P. Poole, John L. Safko -- Casey === Subject: Re: rigorous physics textbook sought <6k5hn1hdnhujv8utl1vkqunbt8ackd88vl@4ax.com> >> can anybody recommend a book on physics (especially >> mechanics) that builds solidly from the ground up? >> i enjoy pure mathematics, and i know that physics >> must make sense, but i can't seem to make sense out >> of any of the physics textbooks i've looked at. >> too many loose ends and unsound inferences. please help. > The Feynmann Lecture Series is pretty good, > once you are in 3rd year, wondering what > it was all about. I think you can't do much better than the Feynmann series. It's intended for a first college course (in theory, at any rate), so the mathematics isn't at a very high level. But the _cognitive_reasoning_ level is high. Moreover, it should appeal to most anyone with a mathematical bent, since Feynmann is very careful in being explicit with saying what implies what, what suggests (but not necessarily implies) what, and what is at best just an analogy. > Classical Mechanics (3rd Edition) (Hardcover) > by Herbert Goldstein, Charles P. Poole, John L. Safko It's interesting that you mentioned Goldstein, because when I was an undergraduate in the late 1970's I was always hearing the math graduate students and a few of the more mathematically oriented physics graduate students say how cook-book'ish and non-rigorous Goldstein was. Arnold's Mathematical Methods of Classical Mechanics was the recommended book, to be followed by Abraham & Marsden's Foundations of Classical Mechanics, although this isn't the route someone planning to be a physicist should take. Dave L. Renfro === Subject: Re: rigorous physics textbook sought > hi all > can anybody recommend a book on physics (especially mechanics) that > builds solidly from the ground up? i enjoy pure mathematics, and i know > that physics must make sense, but i can't seem to make sense out of any > of the physics textbooks i've looked at. too many loose ends and > unsound inferences. please help. > peace, > stm I note that you have not simultaneously posted this to sci,physics. David Ames === Subject: Re: rigorous physics textbook sought ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt hi all > can anybody recommend a book on physics (especially mechanics) that > builds solidly from the ground up? i enjoy pure mathematics, and i know > that physics must make sense, but i can't seem to make sense out of any > of the physics textbooks i've looked at. too many loose ends and > unsound inferences. please help. The Feynman lectures on physics I-III are pretty good. The first volume has probably what you seek. Your university library should also have an exercise companion text to it, which is important for getting practice in doing problems. === Subject: Re: Well Ordering the Reals > I don't care which putative well-ordering of the reals you think there > is. Via extension of Cantor's first, it's a proof that there are > adjacent points in the normal ordering of the reals, or where sets are > well-orderable, the reals are not a set, or not the complete ordered > field. > Great news!!! You only have to say this > 845,678,340,236,742,488,956,354,675,864,532,467,047,090,223,423,147,920 > more times, and it will be true. Keep at it Ross!! > Brian Chandler > http://imaginatorium.org shift the addend.... Yep, not funny. That falls beneath the threshold of something funny, but does give you an excuse for nervous giggles as I suggest well-ordering the reals, and dire, for your professed theory, consequences thereupon. Levity aside, what was that again? You're saying that after repeating it some n number of times it's true? Your puny number, some 8.5 * 10^53, say Brian's number, is very much less than infinity. While larger than any number of discrete items on Earth not counting functions between them, it's but an infinitesimal to infinity. It's piffling. So, well-order the reals and see that in extension Cantor's first is a proof of their adjacency in the contiguous sequence that is the continuous reals. Infinite sets are equivalent. Ross === Subject: Re: Well Ordering the Reals > I don't care which putative well-ordering of the reals you think there > is. Via extension of Cantor's first, it's a proof that there are > adjacent points in the normal ordering of the reals, or where sets are > well-orderable, the reals are not a set, or not the complete ordered > field. > Great news!!! You only have to say this > 845,678,340,236,742,488,956,354,675,864,532,467,047,090,223,423,147,920 > more times, and it will be true. Keep at it Ross!! > Brian Chandler > http://imaginatorium.org > Your puny number, some 8.5 * 10^53, say Brian's number, is very much > less than infinity. While larger than any number of discrete items on > Earth not counting functions between them, it's but an infinitesimal to > infinity. It's piffling. > Ross And Ross is the acknowledged expert at piffling! === Subject: Re: Well Ordering the Reals > > I don't care which putative well-ordering of the reals you think there > is. Via extension of Cantor's first, it's a proof that there are > adjacent points in the normal ordering of the reals, or where sets are > well-orderable, the reals are not a set, or not the complete ordered > field. > > Great news!!! You only have to say this > 845,678,340,236,742,488,956,354,675,864,532,467,047,090,223,423,147,920 > more times, and it will be true. Keep at it Ross!! > > Brian Chandler > http://imaginatorium.org > Your puny number, some 8.5 * 10^53, say Brian's number, is very much > less than infinity. While larger than any number of discrete items on > Earth not counting functions between them, it's but an infinitesimal to > infinity. It's piffling. > Ross > And Ross is the acknowledged expert at piffling! I am not concerned of Brian's number, that was mockery of it. Brian may well attempt to correct this interpretation: what he says there is that it's true. Virgil breaks his silence to squawk. How profound. What a troll. That's ridiculous. Humor is appreciated, but irrelevant. Let's see, last time we were discussing this and Virgil ventured an opinion, what was that again, Virgil? What's your analysis of this well-ordering of the reals and extension of Cantor's first, again? Well-order the real numbers. Are there thus only countably many nested intervals? Stop being a hypocrite. Why don't you answer the last dozen posts that I've written directly to you? This one wasn't. That's not very considerate. So, I encourage you to go on at length concerning the issue at hand, well-ordering the reals in application to the described extension of nested intervals. Ross === Subject: Re: Well Ordering the Reals > Your puny number, some 8.5 * 10^53, say Brian's number, is very much > less than infinity. While larger than any number of discrete items on > Earth not counting functions between them, it's but an infinitesimal to > infinity. It's piffling. > > > Ross > And Ross is the acknowledged expert at piffling! > I am not concerned of Brian's number, that was mockery of it. Brian > may well attempt to correct this interpretation: what he says there is > that it's true. > Virgil breaks his silence to squawk. How profound. What a troll. > That's ridiculous. Humor is appreciated, but irrelevant. > Let's see, last time we were discussing this and Virgil ventured an > opinion, what was that again, Virgil? What's your analysis of this > well-ordering of the reals and extension of Cantor's first, again? > Well-order the real numbers. Are there thus only countably many nested > intervals? > Stop being a hypocrite. Why don't you answer the last dozen posts that > I've written directly to you? This one wasn't. That's not very > considerate. I accidentally les my killfine on Ross run out, and see what happened! > So, I encourage you to go on at length concerning the issue at hand, > well-ordering the reals in application to the described extension of > nested intervals. > Ross === Subject: Re: Well Ordering the Reals > Your puny number, some 8.5 * 10^53, say Brian's number, is very much > less than infinity. While larger than any number of discrete items on > Earth not counting functions between them, it's but an infinitesimal to > infinity. It's piffling. > > > Ross > > And Ross is the acknowledged expert at piffling! > I am not concerned of Brian's number, that was mockery of it. Brian > may well attempt to correct this interpretation: what he says there is > that it's true. > Virgil breaks his silence to squawk. How profound. What a troll. > That's ridiculous. Humor is appreciated, but irrelevant. > Let's see, last time we were discussing this and Virgil ventured an > opinion, what was that again, Virgil? What's your analysis of this > well-ordering of the reals and extension of Cantor's first, again? > Well-order the real numbers. Are there thus only countably many nested > intervals? > Stop being a hypocrite. Why don't you answer the last dozen posts that > I've written directly to you? This one wasn't. That's not very > considerate. > I accidentally les my killfine on Ross run out, and see what happened! > > So, I encourage you to go on at length concerning the issue at hand, > well-ordering the reals in application to the described extension of > nested intervals. > Ross Heh heh, Virgil's disparate Plonk. Please don't let it happen again. So what? He's running away from the question. Now, running away is often a good solution, but this particular problem will _always_ be here and _never_ go away, and Virgil eventually _will_ have to confront it. I don't care, he reads the posts anyways, if he doesn't present a refutation to something I say, it can be presumed he can't, because otherwise he would revel in it, as he does in mock indignation over the joke of Brian's number. great word. I was thinking it meant trifling, you know, minimal, nominal, insignificant, arbitrarily small, that kind of thing, irrelevant. Then, in researching the definition of the word, I found it also had other applicable meanings, eg distractory. Whatever it is, it's definitely not addressal of the issue: well-ordering the reals leads to a form where they are not shown uncountable by Cantor's first. So... the same goes for the rest of you all, and us, including myself, in well-ordering the reals, via Cantor's first, if there are not adjacent points in the reals then there is no well-ordering of the set of the real numbers, the complete ordered field. One notion is that the reals are not well-orderable. Yet, then it would not be possible to quantify over them, and in general the well-ordering principle holds, via axiomatization in the fantasy land of transfinite cardinals. The reals are a set, or, they aren't objects of a set theory, so, if they're a set and sets are well-orderable then they're well-orderable. Then, in their well-ordering, there's a least element, and, disjoint of that a different least element, and so on and so forth. Using these values, construct nested intervals a la Cantor's first, more expansively known in general as Cantor's first proof of the uncountability of the reals, while being shown here Cantor's first proof of adjacent points in the reals. Construct the nested intervals, as many as you care, there are countably many nested intervals, and the well-ordering would not at that point be complete, except in that case the reals are countable. Then, with some higher cardinal assigned to the reals, as in these theories of cardinals with no applications except those by having transfinite cardinals imposter themselves as segments of the continua, with some higher cardinal there are uncountably many intervals. The rationals are thus uncountable, another contradiction. The well-ordering is not yet complete, because until there are two adjacent points in the reals addressed, that complete ordered field's elements will be divisible. That's deductive: the necessary existence of the result of infinite divisbility, and basically a non-zero result. This notion of contiguous or adjacent points in the reals is not new. It's basically what Leibniz, and, Newton, used to invent the integral calculus, or as they knew it the infinitesimal analysis. It's what precaculus instructors warn about because it's not rigorous, and leads to dangerous intuitions about the real numbers. Instead, with reinterpreting Cantor's first to mean a more applicably truthful thing, it _is_ rigorous, there's a proof of it, the existence of contiguous points. That's in seeming contradiction to that the reals are the complete ordered field, that any element of the integers divided by any other, except for zero and that's a relaxable condition, is an element. The definition continues that as is each value converged upon by a sequence of those values of the closure of the integers to division, the rationals. The notion of having contiguous points in the reals is that there are yet more, or not, with the information from one notion of addressal, the limit of a convergent sequence, lost to the notion of the other, an element on the continuous, contiguous and sequential, real line, and each is still the other. Infinite sets are equivalent. Ross === Subject: Re: Well Ordering the Reals Ross A. Finlayson says... >Whatever it is, it's definitely not addressal of the issue: >well-ordering the reals leads to a form where they are not shown >uncountable by Cantor's first. There is no connection between Cantor's first and well-ordering of the reals. Cantor's proofs show that the reals cannot be enumerated using the natural numbers, but that doesn't mean that they can't be well-ordered. -- Daryl McCullough Ithaca, NY === Subject: Re: Well Ordering the Reals > Ross A. Finlayson says... >Whatever it is, it's definitely not addressal of the issue: >well-ordering the reals leads to a form where they are not shown >uncountable by Cantor's first. > There is no connection between Cantor's first and well-ordering > of the reals. Cantor's proofs show that the reals cannot be > enumerated using the natural numbers, but that doesn't mean > that they can't be well-ordered. > -- > Daryl McCullough > Ithaca, NY The reals are ordering-sensitive. Consider: http://www.cs.nyu.edu/pipermail/fom/2005-November/009330.html . Infinite sets are equivalent. Ross === Subject: Re: Well Ordering the Reals Ross A. Finlayson says... >> There is no connection between Cantor's first and well-ordering >> of the reals. Cantor's proofs show that the reals cannot be >> enumerated using the natural numbers, but that doesn't mean >> that they can't be well-ordered. >The reals are ordering-sensitive. >Consider: http://www.cs.nyu.edu/pipermail/fom/2005-November/009330.html What does that URL have to do with Cantor's proof, or with well-orderings of the reals? -- Daryl McCullough Ithaca, NY === Subject: Re: Well Ordering the Reals > Ross A. Finlayson says... >> There is no connection between Cantor's first and well-ordering >> of the reals. Cantor's proofs show that the reals cannot be >> enumerated using the natural numbers, but that doesn't mean >> that they can't be well-ordered. >The reals are ordering-sensitive. >Consider: http://www.cs.nyu.edu/pipermail/fom/2005-November/009330.html > What does that URL have to do with Cantor's proof, or with well-orderings > of the reals? > -- > Daryl McCullough > Ithaca, NY That's what the large-cardinalists are trying to use to well-order the reals, trying to find utility in large cardinals, from the people who brought epsilon chains of length fifteen, no wait, sixteen. It also represents the most very recent post to that discussion group, and you measurability of sets, of numbers. If the cardinality of the continuum c is consistently equivalent to Aleph_1 and Aleph_2 in ZFC, then aren't those consistently equivalent? Hausdorff once noted that an uncountable set was a countable collection. Skolemize, your model is countable, and the set in the generic extension contains no new elements. Infinite sets are equivalent. Ross === Subject: Re: Well Ordering the Reals Ross A. Finlayson says... >>Consider: http://www.cs.nyu.edu/pipermail/fom/2005-November/009330.html >> What does that URL have to do with Cantor's proof, or with well-orderings >> of the reals? >That's what the large-cardinalists are trying to use to well-order the >reals, trying to find utility in large cardinals, from the people who >brought epsilon chains of length fifteen, no wait, sixteen. It also >represents the most very recent post to that discussion group, and you >measurability of sets, of numbers. What does any of that have to do with Cantor's proof? -- Daryl McCullough Ithaca, NY === Subject: Re: Well Ordering the Reals > Ross A. Finlayson says... >>Consider: http://www.cs.nyu.edu/pipermail/fom/2005-November/009330.html >> >> What does that URL have to do with Cantor's proof, or with well-orderings >> of the reals? >That's what the large-cardinalists are trying to use to well-order the >reals, trying to find utility in large cardinals, from the people who >brought epsilon chains of length fifteen, no wait, sixteen. It also >represents the most very recent post to that discussion group, and you >measurability of sets, of numbers. > What does any of that have to do with Cantor's proof? > -- > Daryl McCullough > Ithaca, NY Well, ignore it, it's just kind of interesting in the broader context. It's relevant because the grancardinalistas are very much interested in application to justify their results, as it is arguable that there is none. This is where as time has progressed they've noticed that the cardinality of the continuum has inflated beyond all known cardinals, stopping at each along the way. Compared to that, the unit interval is a microcosm where it is seen that to well-order the reals there will be adjacent points in the normal ordering, and as that's denied there will not be resolution in the eternal and futile neverending buck-passing to the next yet-to-be-discovered larger cardinal. So, well-order the reals. Ross === Subject: Re: Well Ordering the Reals > The reals are a set, or, they aren't objects of a set theory, so, if > they're a set and sets are well-orderable then they're well-orderable. > Then, in their well-ordering, there's a least element, and, disjoint of > that a different least element, and so on and so forth. Using these > values, construct nested intervals a la Cantor's first, more > expansively known in general as Cantor's first proof of the > uncountability of the reals, while being shown here Cantor's first > proof of adjacent points in the reals. Construct the nested intervals, > as many as you care, there are countably many nested intervals, and the > well-ordering would not at that point be complete, except in that case > the reals are countable. Then, with some higher cardinal assigned to > the reals, as in these theories of cardinals with no applications > except those by having transfinite cardinals imposter themselves as > segments of the continua, with some higher cardinal there are > uncountably many intervals. The rationals are thus uncountable, > another contradiction. The well-ordering is not yet complete, because > until there are two adjacent points in the reals addressed, that > complete ordered field's elements will be divisible. Your prose is hard to follow, but I think you are assuming that some well-ordering of the reals must imply a well-ordering under their normal order. It is my understanding that Cantor' first proof of the uncountability of the reals starts with the assumption that the reals are countable and uses countably infinite nested intervals to prove that those intervals do not include all reals, thus proving by contradiction that the reals are uncountable. But Cantor's proof appears to operate only within the normal ordering of the reals. So I fail to see how a different ordering (one that is well-ordered) is subject to the same contraints (and thus the same conclusion) as the normal ordering. === Subject: Re: Well Ordering the Reals > The reals are a set, or, they aren't objects of a set theory, so, if > they're a set and sets are well-orderable then they're well-orderable. > Then, in their well-ordering, there's a least element, and, disjoint of > that a different least element, and so on and so forth. Using these > values, construct nested intervals a la Cantor's first, more > expansively known in general as Cantor's first proof of the > uncountability of the reals, while being shown here Cantor's first > proof of adjacent points in the reals. Construct the nested intervals, > as many as you care, there are countably many nested intervals, and the > well-ordering would not at that point be complete, except in that case > the reals are countable. Then, with some higher cardinal assigned to > the reals, as in these theories of cardinals with no applications > except those by having transfinite cardinals imposter themselves as > segments of the continua, with some higher cardinal there are > uncountably many intervals. The rationals are thus uncountable, > another contradiction. The well-ordering is not yet complete, because > until there are two adjacent points in the reals addressed, that > complete ordered field's elements will be divisible. > Your prose is hard to follow, but I think you are assuming that some > well-ordering of the reals must imply a well-ordering under their > normal order. > It is my understanding that Cantor' first proof of the uncountability > of the reals starts with the assumption that the reals are countable > and uses countably infinite nested intervals to prove that those > intervals do not include all reals, thus proving by contradiction that > the reals are uncountable. Two nits to pick: (1) The first proof is a *direct* proof that any mapping from the naturals to the reals omits some reals, just as is the second(diagonal) proof, and therefore proving there are no surjections from the naturals to the reals. > But Cantor's proof appears to operate only within the normal ordering > of the reals. So I fail to see how a different ordering (one that is > well-ordered) is subject to the same contraints (and thus the same > conclusion) as the normal ordering. Two nits to pick: (1) The first Cantor proof is a *direct* proof that any mapping from the naturals to the reals omits some reals, just as is the second (diagonal) proof, proving there are no surjections from the naturals to the reals. (2) A different ordering of the reals does not destroy the properties of the standard ordering, so does not destroy the validity of a proof based on the standard ordering. And such a proof for the standard reals with the standard ordering will also be valid for any set that the reals can be injected into, regardless of how it is ordered. > === Subject: Re: Well Ordering the Reals >> The reals are a set, or, they aren't objects of a set theory, so, if >> they're a set and sets are well-orderable then they're well-orderable. >> Your prose is hard to follow, but I think you are assuming that some >> well-ordering of the reals must imply a well-ordering under their >> normal order. >> It is my understanding that Cantor' first proof of the uncountability >> of the reals starts with the assumption that the reals are countable >> and uses countably infinite nested intervals to prove that those >> intervals do not include all reals, thus proving by contradiction that >> the reals are uncountable. >> But Cantor's proof appears to operate only within the normal ordering >> of the reals. So I fail to see how a different ordering (one that is >> well-ordered) is subject to the same contraints (and thus the same >> conclusion) as the normal ordering. > Two nits to pick: > (1) The first proof is a *direct* proof that any mapping from the > naturals to the reals omits some reals, just as is the second(diagonal) > proof, and therefore proving there are no surjections from the naturals > to the reals. > (2) A different ordering of the reals does not destroy the properties of > the standard ordering, so does not destroy the validity of a proof based > on the standard ordering. And such a proof for the standard reals with > the standard ordering will also be valid for any set that the reals can > be injected into, regardless of how it is ordered. Corrections noted. I was confused. If Cantor's first proof shows that the reals are uncountable under their normal ordering, then it can also show that they are uncountable under any other ordering. This is _obviously_ true. So even if the reals can be well-ordered, they are still uncountable. Or as, Daryl says: > There is no connection between Cantor's first and well-ordering > of the reals. Cantor's proofs show that the reals cannot be > enumerated using the natural numbers, but that doesn't mean > that they can't be well-ordered. But Ross appears to be arguing for the opposite, that well-ordering the reals would render them countable (although it's not always clear to me what he is arguing). === Subject: Re: Well Ordering the Reals >> The reals are a set, or, they aren't objects of a set theory, so, if >> they're a set and sets are well-orderable then they're well-orderable. >> Your prose is hard to follow, but I think you are assuming that some >> well-ordering of the reals must imply a well-ordering under their >> normal order. >> >> It is my understanding that Cantor' first proof of the uncountability >> of the reals starts with the assumption that the reals are countable >> and uses countably infinite nested intervals to prove that those >> intervals do not include all reals, thus proving by contradiction that >> the reals are uncountable. >> >> But Cantor's proof appears to operate only within the normal ordering >> of the reals. So I fail to see how a different ordering (one that is >> well-ordered) is subject to the same contraints (and thus the same >> conclusion) as the normal ordering. > Two nits to pick: > (1) The first proof is a *direct* proof that any mapping from the > naturals to the reals omits some reals, just as is the second(diagonal) > proof, and therefore proving there are no surjections from the naturals > to the reals. > (2) A different ordering of the reals does not destroy the properties of > the standard ordering, so does not destroy the validity of a proof based > on the standard ordering. And such a proof for the standard reals with > the standard ordering will also be valid for any set that the reals can > be injected into, regardless of how it is ordered. > Corrections noted. I was confused. > If Cantor's first proof shows that the reals are uncountable under > their normal ordering, then it can also show that they are uncountable > under any other ordering. This is _obviously_ true. > So even if the reals can be well-ordered, they are still uncountable. > Or as, Daryl says: > There is no connection between Cantor's first and well-ordering > of the reals. Cantor's proofs show that the reals cannot be > enumerated using the natural numbers, but that doesn't mean > that they can't be well-ordered. > But Ross appears to be arguing for the opposite, that well-ordering > the reals would render them countable (although it's not always clear > to me what he is arguing). No, determining that the normal ordering is the well-ordering is the result, not the assumption. A well-ordering of the reals will have a least element, like an enumeration has a least element. Then, the disjoint of that element from that set, the remaining piece of the set, will have a least element. Those will form an interval that basically represents a pair (a_0, b_0), the initial elements of the sequences a and b in Cantor's first. Then, in the progression of the well-ordering, elements that fall within that interval will form another, (a_1, b_1). Then, the set of those endpoint pairs is well-ordered. Then, for whatever cardinal, where each cardinal is well-ordered, it being an ordinal, for whatever cardinal the reals may have, then unless there is (a_n, b_n) with those being adjacent points in the reals, there is no well-ordering of the reals. The argument, Cantor's first, is based upon the completeness of the reals, the complete ordered field, the gaplessness property of the reals. That does not change regardless of what enumeration, or well-ordering, or perhaps well-ordering's initial segment, is used, unless there are adjacent points in the reals. For example, if you can conceive of the reals as integral iota-multiples, basically half-point or one-sided point widths, where the (non-negative) reals could be addressed as points in the normal ordering, ie zero, 1 iota, 2 iota, 3 iota, ..., then Cantor's first does not preclude that from being a well-ordering or even an enumeration. There is obviously the problem with that that the reals are the complete ordered field, so you want to divide 1 iota by two or average 1 and 2 iota, the result is undefined. That's a problem, I agree. While that is so, the alternative leads to a variety of conclusions that would be contradictory. These are basically notions of a result of infinite divisibility that is not zero, Zeno in reverse. As has been noted, the notion of these reals as points on a line is not new. Spinoza for exampled compared the natural integers to the continuum. The natural integers are an interesting set, not {0, 1, 2} except in extension but having the properties of compactness and so on that their compactification is trivial, implicit, and a variety of extensions of decidability related results may lead to the notion that N E N, in discussion of a variety of number-theoretic and metatheoretical related issues. Anyways, Leibniz for example has that they're infinitesimals, and that lead to centuries of consternation and the constraints of limit, which are useful but not necessarily unrestricted enough in comprehension. Then, many modern analysts have heard of varieties of nonstandard analysis of the reals, most prominently Robinso(h)n with his hyperreals, Nelson's IST, and less mainstream Schmieden and Laugwitz and other various constructions, including deductive constructions, of the real numbers. Well-order the reals, unless there is a least real greater than zero, that is not a well-ordering of the reals. V = L. Ross === Subject: Re: Well Ordering the Reals >> The reals are a set, or, they aren't objects of a set theory, so, if >> they're a set and sets are well-orderable then they're well-orderable. >> If Cantor's first proof shows that the reals are uncountable under >> their normal ordering, then it can also show that they are uncountable >> under any other ordering. This is _obviously_ true. >> So even if the reals can be well-ordered, they are still uncountable. >> But Ross appears to be arguing for the opposite, that well-ordering >> the reals would render them countable (although it's not always clear >> to me what he is arguing). > No, determining that the normal ordering is the well-ordering is the > result, not the assumption. > A well-ordering of the reals will have a least element, like an > enumeration has a least element. Then, the disjoint of that element > from that set, the remaining piece of the set, will have a least > element. Those will form an interval that basically represents a pair > (a_0, b_0), the initial elements of the sequences a and b in Cantor's > first. Then, in the progression of the well-ordering, elements that > fall within that interval will form another, (a_1, b_1). Then, the set > of those endpoint pairs is well-ordered. Then, for whatever cardinal, > where each cardinal is well-ordered, it being an ordinal, for whatever > cardinal the reals may have, then unless there is (a_n, b_n) with those > being adjacent points in the reals, there is no well-ordering of the > reals. But you are assuming that the well-ordering produces sequential elements in the normal ordering, which is probably impossible. Let's say a well-ordering F of the reals exists, and produces (1, 27) as your (a_0, b_0). Since it's a well-ordering, 27 is the successor to 1 in that ordering. Let's also say that (3, 1/7) is your (a_1, b_1). Likewise, 3 is the successor to 27, and 1/17 is the successor to 3. But the fact that the intervals (1, 27) and (3, 1/17) contain other reals under the normal ordering of R has no bearing on ordering F. As long as F exhaustively orders all of the members of R (an uncountable set), it can indeed be a well-ordering. But that fact has no bearing on the normal ordering of the reals, which is not well-ordered. The fact that the normal ordering of the reals is not a well-ordering does not mean that no orderings of the reals are well-ordered. > The argument, Cantor's first, is based upon the completeness of the > reals, the complete ordered field, the gaplessness property of the > reals. That does not change regardless of what enumeration, or > well-ordering, or perhaps well-ordering's initial segment, is used, > unless there are adjacent points in the reals. There can be adjacent reals (actually, successor reals) under ordering F but not necessarily under the normal ordering of R. As long as F is exhaustive, there is no problem. > For example, if you can conceive of the reals as integral > iota-multiples, basically half-point or one-sided point widths, where > the (non-negative) reals could be addressed as points in the normal > ordering, ie zero, 1 iota, 2 iota, 3 iota, ..., then Cantor's first > does not preclude that from being a well-ordering or even an > enumeration. If your iotas defined a well-ordering, then there would be no points between any two adjacent points in that ordering. But if points exist between any two iotas, then the iotas do not define a well-ordering. But any well-ordering of the reals must exhaustively order an uncountable number of reals. (Which means that the ordering cannot be a countable sequence.) > There is obviously the problem with that that the reals are the > complete ordered field, so you want to divide 1 iota by two or average > 1 and 2 iota, the result is undefined. That's a problem, I agree. > While that is so, the alternative leads to a variety of conclusions > that would be contradictory. If (a_i + a_j)/2 is defined for j=i+1, then a_n is not a well-ordering. > These are basically notions of a result of infinite divisibility that > is not zero, Zeno in reverse. > Well-order the reals, unless there is a least real greater than zero, > that is not a well-ordering of the reals. Under the normal ordering of the reals. But not necessarily under ALL orderings. === Subject: Re: Well Ordering the Reals >> The reals are a set, or, they aren't objects of a set theory, so, if >> they're a set and sets are well-orderable then they're well-orderable. >> Your prose is hard to follow, but I think you are assuming that some >> well-ordering of the reals must imply a well-ordering under their >> normal order. >> >> It is my understanding that Cantor' first proof of the uncountability >> of the reals starts with the assumption that the reals are countable >> and uses countably infinite nested intervals to prove that those >> intervals do not include all reals, thus proving by contradiction that >> the reals are uncountable. >> >> But Cantor's proof appears to operate only within the normal ordering >> of the reals. So I fail to see how a different ordering (one that is >> well-ordered) is subject to the same contraints (and thus the same >> conclusion) as the normal ordering. > Two nits to pick: > (1) The first proof is a *direct* proof that any mapping from the > naturals to the reals omits some reals, just as is the second(diagonal) > proof, and therefore proving there are no surjections from the naturals > to the reals. > (2) A different ordering of the reals does not destroy the properties of > the standard ordering, so does not destroy the validity of a proof based > on the standard ordering. And such a proof for the standard reals with > the standard ordering will also be valid for any set that the reals can > be injected into, regardless of how it is ordered. > Corrections noted. I was confused. > If Cantor's first proof shows that the reals are uncountable under > their normal ordering, then it can also show that they are uncountable > under any other ordering. This is _obviously_ true. > So even if the reals can be well-ordered, they are still uncountable. > Or as, Daryl says: > There is no connection between Cantor's first and well-ordering > of the reals. Cantor's proofs show that the reals cannot be > enumerated using the natural numbers, but that doesn't mean > that they can't be well-ordered. > But Ross appears to be arguing for the opposite, that well-ordering > the reals would render them countable (although it's not always clear > to me what he is arguing). It is not always clear to Ross what he is arguing. === Subject: Re: Well Ordering the Reals > I thought it was > 845,678,340,236,742,488,956,354,675,864,532,467,047,090,223,423,147,921 > times. Not possible. That's one greater than the largest natural number. MoeBlee === Subject: Re: Well Ordering the Reals > I thought it was > 845,678,340,236,742,488,956,354,675,864,532,467,047,090,223,423,147,921 > times. > Not possible. That's one greater than the largest natural number. > MoeBlee Does that mean that 845,678,340,236,742,488,956,354,675,864,532,467,047,090,223,423,147,921 is unnatural, or that it is not a number? === Subject: Re: Well Ordering the Reals >> I thought it was >> 845,678,340,236,742,488,956,354,675,864,532,467,047,090,223,423,147,921 >> times. >> Not possible. That's one greater than the largest natural number. > Does that mean that > 845,678,340,236,742,488,956,354,675,864,532,467,047,090,223,423,147,921 > is unnatural, or that it is not a number? Perhaps it is a TO-infinite (Orlovian) number? === Subject: Re: Well Ordering the Reals MoeBlee === Subject: Re: Rest In Peace: Eddie Guerrero (1967-2005) >It has just been learned that Eddie Guerrero has passed away. RIP. After Googling this name (I never heard of him), I have no idea what he has to do with ec.arts.sf.written, sci.math,comp.os.linux.misc, or rec.gambling.poker. === Subject: Re: Rest In Peace: Eddie Guerrero (1967-2005) Stop trying to take credit for your mama's work. Irish Mike > You guys won't believe this but I ed him one time. === Subject: =?iso-8859-1?B?QWRvYmUgQ3JlYXRpdmUgU3VpdGUgKDUgQ0QpIC0gNzUlIE9GRg==?= type=multipart/alternative; boundary=----=_NextPart_000_0000_BCD87264.064A74E3 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id jAE3Eg911938 by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id jAE3E3F4017704 --------------------------------------------------------------------- boundary=----=_NextPart_001_0001_98055D7C.B133E3EB --------------------------------------------------------------------- Get access to all the popular software possible for extremely low prices! We sell software 2-6 times cheaper than retail price. A few examples: $79.95 Windows XP Professional (Including: Service Pack 2) $99.95 Adobe Photoshop 8.0/CS (Including: ImageReady CS) $149.95 Adobe Creative Suite Premium (5 CD) $79.95 Adobe Acrobat 6.0 Professional Special offers: $89.95 Windows XP Pro + Office XP Pro $129.95 Photoshop 7 + Premiere 7 + Illustrator 10 All main products from Microsoft, Adobe, Macromedia, Corel, etc. And many other... To view full list of products go: http://www.alldisks.net Best, Jennifer Clark ________________________________ To stop further mailings, go here ________________________________ === Subject: the infinite earring Let C_n be the circle of radius 1/n in R^2 with center at the point (1/n,0). Let X be the subspace of R^2 that is the union of these circles; Let S_n be the circle of radius n in R^2 with center at the point (n,0). Let Y be the subspace of R^2 that is the union of these circles. I don't know the difference between these two. And why the second Pi_1(Y.p) is a free group with {[f_n]} as a system of free generators, where f_n is a loop representing a generator of pi_1(S_n,p) but the first one can't? === Subject: Re: the infinite earring ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt, > Let C_n be the circle of radius 1/n in R^2 with center at the point (1/n,0). > Let X be the subspace of R^2 that is the union of these circles; > Let S_n be the circle of radius n in R^2 with center at the point (n,0). Let > Y be the subspace of R^2 that is the union of these circles. > I don't know the difference between these two. > And why the second Pi_1(Y.p) is a free group with {[f_n]} as a system of free > generators, where f_n is a loop representing a generator of pi_1(S_n,p) > but the first one can't? One is actually a wedge of circles, as you might expect from the description of its fundamental group. The other which you call the infinite earring seems more commonly in my experience known as the Hawaiian earring. Its group is bigger since there are loops that go around every ring. http://en.wikipedia.org/wiki/Hawaiian_earring === Subject: Re: the infinite earring Yes,I know that Y is not homeomorphic to X, but I can't tell whether or not Y is normal === Subject: Re: the infinite earring >Yes,I know that Y is not homeomorphic to X, >but I can't tell whether or not Y is normal Since X,Y are subsets of R^2, they are both metrizable, hence they are both normal since every metric space is normal. === Subject: Re: the infinite earring >Let C_n be the circle of radius 1/n in R^2 with center at the point (1/n,0). Let X be the subspace of R^2 that is the union of these circles; >Let S_n be the circle of radius n in R^2 with center at the point (n,0). Let Y be the subspace of R^2 that is the union of these circles. >I don't know the difference between these two. >And why the second Pi_1(Y.p) is a free group with {[f_n]} as a system of free generators, where f_n is a loop representing a generator of pi_1(S_n,p) >but the first one can't? Y is a closed set, while X is not. But that doesn't affect the topology -- they look homeomorphic to me. Maybe you stated the problem incorrectly? quasi === Subject: Re: the infinite earring >>Let C_n be the circle of radius 1/n in R^2 with center at the point (1/n,0). Let X be the subspace of R^2 that is the union of these circles; >>Let S_n be the circle of radius n in R^2 with center at the point (n,0). Let Y be the subspace of R^2 that is the union of these circles. >>I don't know the difference between these two. >>And why the second Pi_1(Y.p) is a free group with {[f_n]} as a system of free generators, where f_n is a loop representing a generator of pi_1(S_n,p) >>but the first one can't? >Y is a closed set, while X is not. >But that doesn't affect the topology -- they look homeomorphic to me. >Maybe you stated the problem incorrectly? >quasi Whoops -- I forgot to put on my 2D glasses. X and Y are both closed, but X is compact, while Y is not. So they're not homeomorphic. quasi === Subject: Re: the infinite earring >Let C_n be the circle of radius 1/n in R^2 with center at the point (1/n,0). Let X be the subspace of R^2 that is the union of these circles; >Let S_n be the circle of radius n in R^2 with center at the point (n,0). Let Y be the subspace of R^2 that is the union of these circles. >I don't know the difference between these two. >And why the second Pi_1(Y.p) is a free group with {[f_n]} as a system of free generators, where f_n is a loop representing a generator of pi_1(S_n,p) >but the first one can't? >>Y is a closed set, while X is not. >>But that doesn't affect the topology -- they look homeomorphic to me. >>Maybe you stated the problem incorrectly? >>quasi >Whoops -- I forgot to put on my 2D glasses. >X and Y are both closed, but X is compact, while Y is not. >So they're not homeomorphic. >quasi Well, now that I look more closely, Y isn't even closed. === Subject: Re: the infinite earring >>Let C_n be the circle of radius 1/n in R^2 with center at the point (1/n,0). Let X be the subspace of R^2 that is the union of these circles; >>Let S_n be the circle of radius n in R^2 with center at the point (n,0). Let Y be the subspace of R^2 that is the union of these circles. >>I don't know the difference between these two. >>And why the second Pi_1(Y.p) is a free group with {[f_n]} as a system of free generators, where f_n is a loop representing a generator of pi_1(S_n,p) >>but the first one can't? >Y is a closed set, while X is not. >But that doesn't affect the topology -- they look homeomorphic to me. >Maybe you stated the problem incorrectly? >quasi >>Whoops -- I forgot to put on my 2D glasses. >>X and Y are both closed, but X is compact, while Y is not. >>So they're not homeomorphic. >>quasi >Well, now that I look more closely, Y isn't even closed. Now that I have the sets X,Y in focus, I can see a loop in X which has no analog in Y. Letting p=(0,0), we can glue all the circles of X together, to form a single loop. Just define a loop [0,1]->X as follows: loop around C_1 once for 0<=x<=1/2 loop around C_2 once for 1/2<=x<=3/4 ... loop around C_n once for 1-1/2^n<=x<=1-1/2^(n+1) ... and finally map x=1 to (0,0) where all the above loops start and end at (0,0). Since Y is not compact, there's no way to do the same thing -- if you try the same construction, it will fail when you try to map x=1. quasi === Subject: Curve - identify which points are significant Suppose I have a set of points (x,y) which forms a curve on a 2D plot. What I want to find out is which points are significant to the shape of the curve and which aren't. For example, if y = x, then the two end points would be the most significant. Is there a branch of mathematics which study this? Any feedback would be appreciated. === Subject: Re: Curve - identify which points are significant research form the internet, I found something called sampling theory which could be useful. However, the original version of the theory works only when the data points are regularly spaced but this doesn't happen in my case. At the moment, I am still looking for a better way to do the thing that I want to do. > Suppose I have a set of points (x,y) which forms a curve on a 2D plot. > What I want to find out is which points are significant to the shape of > the curve and which aren't. For example, if y = x, then the two end points > would be the most significant. Is there a branch of mathematics which > study this? Any feedback would be appreciated. === Subject: Re: Curve - identify which points are significant Consider a third or forth degree polynomial and find best fit by least squares. Leave out points beyond an assigned margin of error. === Subject: Re: Curve - identify which points are significant > Suppose I have a set of points (x,y) which forms a curve on a 2D plot. What > I want to find out is which points are significant to the shape of the curve > and which aren't. For example, if y = x, then the two end points would be > the most significant. Is there a branch of mathematics which study this? Any > feedback would be appreciated. In my calculus class, when we did curve sketching, we used to have to remember M2I2ACID. This stood for: Minima Maxima Points of Inflection Asymptotes Curvature Increasing intervals Decreasing intervals === Subject: Re: Curve - identify which points are significant > Suppose I have a set of points (x,y) which forms a curve on a 2D plot. > What I want to find out is which points are significant to the shape of > the curve and which aren't. For example, if y = x, then the two end points > would be the most significant. Is there a branch of mathematics which > study this? Any feedback would be appreciated. Your question is entirely dependent upon what you mean by curve. === Subject: Re: Curve - identify which points are significant I am in fact looking for a general method which can be applied to different types of not too irregular curves. What I really want to achieve here is to eliminate those insignificant data points from a set of sample. I somehow believe that this problem must have been studied because I can think of so many applications of it (e.g. image compression, shape recognition). >> Suppose I have a set of points (x,y) which forms a curve on a 2D plot. >> What I want to find out is which points are significant to the shape of >> the curve and which aren't. For example, if y = x, then the two end >> points would be the most significant. Is there a branch of mathematics >> which study this? Any feedback would be appreciated. > Your question is entirely dependent upon what you mean by curve. === Subject: Re: Curve - identify which points are significant <43781380$0$73336$892e7fe2@authen.yellow.readfreenews.net> <43781b76$0$9821$c30e37c6@ken-reader.news.telstra.net> First, look for symmetry, and tendencies toward infinity, if any. Do this by substitution. Like, determine if subtituting two linear combinations of x and y produce the same result for the former, and subtitute the infinities for x, y, or both and also look to see if a linear combination of x and y has a limit when x and y approach the infinities to determine asymptotic behaviour. Then, determine the curvature, and locate where its absolute value is maximum and locate where its absolute value is minimum. Do this by differentiating the curvature formula. By looking at the curvature and seeing where it is undefined you can determine any sharp angles, and by looking at where it is zero and changes from positive to negative, you can determine any points of inflection. There is also a method, though difficult to use, of finding the smallest circle in which the shape can be inscribed. That is, if there are maximum values of x, y, and linear combinations of x and y, there will be a smallest circle. If you don't know how to do this, just use an approximate method. Also, look for lines that can never be crossed, that will dictate the boundary of the curve. Just about everything else is completely relative to the axis of rotation. For example, relative minimums and relative maximums will vary depending on the axis of rotation you use for your curve. I am sure there are other examples, like the pure acceleration of one variable with respect to the other, and the third and fourth derivatives with respect to eachother, and so on.use for your curve. I am sure there are other examples, like the pure acceleration of one variable with respect to the other, and so on. === Subject: Re: Curve - identify which points are significant > Suppose I have a set of points (x,y) which forms a curve on a 2D plot. What > I want to find out is which points are significant to the shape of the curve > and which aren't. For example, if y = x, then the two end points would be > the most significant. Is there a branch of mathematics which study this? Any > feedback would be appreciated. If the curve has one or two endpoints, these would be significant points. If the curve has points at which there is more than one line of tangency, these points would be important, either as points at which the curve intersects itself or points at which the tangents lines are not turning continuously. If the curve has a curvature defined as a function of some directed distance along the curve, then points at which that curvature changes sign, or is undefined, would be points of interest. === Subject: Regularity of Minimizers I want to show that minimizers of $F$ are more regular. Minimize F where F(u) = int_Omega( 1/2 |grad u|^2 + W(u) ) dx over $X$ where $X$ denotes the zero average subspace of $ H^1(Omega)$. Here $Omega $ is open , bounded , smooth subset of $ R^3 $. W(u) = 1/6 u^6 -1/2 u^2 +1/3. So we can show $u$ solves - Laplacian u = u - u^5 -C in Omega with Neummann boundary conditions and where $C$ denotes average of (u-u^5) over $Omega$. Since H^1 continuously imbedded in L^6 we see f in L^{6/5} and so by L^p regularity theory we have u in W^{1,2} intersect W^{2 , 6/5 } . This isn't enough regularity for me to start the bootsrapping process. But I can carry out some formal calculations to get Laplacian u in L^2 and this gives me enough to start bootstrapping process. Any help would be greatly appreciated. craig === Subject: Planck Era Emergent Gravity & Leptoquark Masses Note two meanings of inertia. Gravity as a force is locally indistinguishable from an inertial force in a non-inertial local frame that is either rotating or on a time-like nongeodesic path or both with respect to the Levi-Civita connection. Inertia as the rest mass of quarks, leptons and W-mesons. Our universe is thought to have formed in a sequence of spontaneous symmetry breakings of the physical vacuum. Assuming the old Heisenberg uncertainty principle Planck scale breaking ~ 10^19 Gev ~ 10^-33 cm GUT scale breaking ~ 10^15 Gev ~ 10^-29 cm Electroweak breaking ~ 250 Gev ~ 10^-16 cm In more detail Time (s) T (Kelvin) Reference Event CBR THE QUANTUM GRAVITY ERA 1x10^-43 1x10^32 KT 72 Quantum limit of classical general relativity [My comment 1: I postulate, as there is no fundamental theory of this as yet, the simplest possibility G(Planck)---> H(Planck) V (Planck)= G(Planck)/H(Planck) ~ S1(Planck) from first homotopy group PI1(S1) = Z (integer winding numbers) i.e. line defects i.e. 1-dim strings Vacuum coherent order parameter is a single complex scalar field PSI(Planck) = |PSI|e^iargPSI Planck-scale Higgs amplitude is |PSI|, Planck-scale Goldstone phase is argPSI. argPSI is a 0-form Curved space-time tetrad 1-form is B = Lp'd'argPSI ds^2 = (1^a + B^a)(Minkowski)ab(1^b + B^b) is EEP B is from local gauging of SPACE-TIME SYMMETRY T4 to Diff(4) AND ALSO from spontaneous breaking of INTERNAL SYMMETRY G(Planck) -> H(Planck) to S1 Post-Planck Vacuum Manifold PSI is the INFLATION FIELD Diff(4) covariant exterior derivative is D = d + W/ W^ac = T4 spin-connection The vanishing torsion 2-form in 1915 GR model is T^a = De^a = de^a + W^ac/(1 + B)^c = dB^a + W^ac/(1 + B)^c = 0 Invert this equation to get the 1-form spin connection W^a^c = -*[dB^a/(1 - B)^c] * is Hodge dual based on Minkowski metric in tangent fiber space. The geodesic deviation curvature 2-form is R^ab = DW^ab = dW^ab + W^acW^cb The Bianchi identity is the 3-form equation (analog to dF(EM) = 0 3-form) DR^ab = 0 The Einstein-Hilbert vacuum Lagrangian density is the 0-form *[R^ab/e^c/e^d] The Einstein curvature 4-form is G^ab = D*W^ab The Einstein 4-form source equation (analog to d*F(EM) = *j(electric) 3-form) is G^ab = *J(T4)^ab Local conservation of stress-energy source current densities is the vanishing 5-form DG^ab = D*J(T4)^ab = 0 5-form in 4-space is identically zero. G^ab = *J(T4)^ab is 4-form because of equivalence principle where even zero point energy with w = -1 gravitates absolutely and not relatively.] THE INFLATION AND SYMMETRY BREAKING ERA 1x10^-38 1x10^29 KT 72 Limit of perturbative interaction thermalization of universe 1x10^-35 1x10^28 KT 72 Grand unification spontaneous symmetry breaking [My Comment 2 OK, here H(Planck) ---> H(GUT) = U(1)hyperchargeSU(2)weak SU(3)quark coupling constant splits off from U(1)hyperchargeSU(2)weak with same coupling constant We do not know the topology of the new vacuum manifold H(Planck)/U(1)hyperchargeSU(2)weak = V(GUT) = ? Note the sequence: G(Planck) ---> H(Planck) ---> H(GUT) = U(1)hyperchargeSU(2)weak G/(Planck)/H(Planck) ---> H(Planck)/H(GUT) i.e. V(Planck) ---> V(GUT) But what does this mean in terms of the vacuum order parameter? V(Planck) does not disappear. It is robust from Goldstone phase rigidity and it is the Hercules supporting The World Geometrodynamic Field V(Planck) Indeed the total vacuum manifold has grown more complex into the TENSOR PRODUCT V(Planck)xV(GUT) 1x10^-34 1x10^27 KT 274 Approximate start of inflation 1x10^-32 1x10^27 KT 274 Approximate start of reheating and end of inflation 1x10^-11 3x10^15 KT 72 Electroweak unification spontaneous symmetry breaking [My Comment 3 Here H(GUT) = U(1)hyperchargeSU(2)weak --3x3 adjoint irrep--> U(1)em V(Electroweak) = U(1)hyperchargeSU(2)weak/U(1)em ~ S^2 with point defects. W+,W0, W- massive A(photon) massless, 1 Higgs & masses of leptons & quarks from Yukawa couplings. Prior to that 250 Gev scale all quantum fields are massless. There is gravity already emergent from Planck era. It does not have to be put in by hand adhoc anymore. You can have curved spacetime without massive principle. Now the vacuum manifold is V(Planck)xV(GUT)xV(Electroweak) Although the short-scale vacuum order parameter may be a single complex scalar field with an S1 vacuum manifold, there is reason to believe that the larger-scale vacuum order parameter after the electroweak splitting of the weak SU(2) charge from the U(1) electromagnetic charge is essentially of the character of isovector, i.e. 3x3 adjoint irrep of SU(2) with effective vacuum manifold S2 hence geometrodynamic point defects not to be confused with magnetic monopoles. Having geometrodynamic point defects instantly gives the Hawking-Bekenstein area quantization because of the second homotopy group PI2(S2) = Z giving stable 2D wrapping numbers instead of the stable 1D winding numbers. The 3-adjoint irrep of SU(2) is 3 real scalar fields of which in one Higgs model only one scalar field develops a non-vanishing VEV Higgs amplitude. We can imagine its conjugate Goldstone phase as the rotation about it as an axis in V(Electroweak) ~ S2 i.e. the azimuthal angle about the z axis chosen so that the order parameter is electrically neutral. We then must assume that this is the same Goldstone phase that survived from the Planck era. These speculations are of course very new.] THE QUARK-LEPTON ERA 2x10^-7 2x10^13 HA 353 Tauon anti-tauon annihilation 1x10^-5 2x10^12 KT 72 Formation of hadrons from quarks 7x10^-5 1x10^12 HA 353 Muon anti-muon annihilation 5x10^-4 4x10^11 KT 159, 281 By this time the universe has a baryon-antibaryon asymmetry which results from post-inflationary B,C,CP violating processes 1x10^-1 3x10^10 BS Neutral current weak interactions become too slow and neutrinos decouple 1x10^0 1x10^10 BS Charged current weak interactions become too slow and neutron to proton ratio freezes out 1x10^1 5x10^9 BS Electron positron annihilation THE RADIATION ERA 1x10^2 1x10^9 BS Typical photon energies drop below the deuteron binding energy and nucleosynthesis begins barrier energies and nucleosynthesis ends 4x10^10 6x10^4 KT 77 Matter density becomes equal to radiation density THE MATTER ERA 4x10^12 3.5x10^3 KT 78 Electrons and protons recombine into hydrogen atoms 6x10^12 3.0x10^3 KT 80 Photon decoupling 2x10^16 20 HA 349 Formation of galaxies (X-1.4x10^17) --- KT 12 Formation of the solar system (X-9.9x10^16) --- HA 390 Emergence of life on Earth X ~ 4.7x10^17 2.726 KT 12 Today These values are rough estimates only. I have taken Omega_0=1, Lambda_0=0, and h=1. http://www.talkorigins.org/origins/cosmo_timeline.html === Subject: CONTACT THE PAYING BANK by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id jAE5DC921224 by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id jAE5CTF4018203 Applecrest & Grant Lotteries, 1022 Kingsway Ave., Osborne, Western Australia, 6018. Head office: Maliebaan 38-40, 3581CR, Utrecht, The Netherlands. YOU WON: $1,000,000.00! With utmost joy, we are pleased to inform you of the result of the Apple crest & Grant Lotteries International programs held on the 14th, of November 2005 drew cash Prize of $1,000,000.00. Your e-mail address attached to Code number 370011672387-290-02 with claim Number 12093782/42 drew these lucky numbers 894-4563-001-764-64-2911, which consequently won in the 1st category. You have been approved to receive a cash prize of $1,000,000.00 (One Million USD). Be advised that due to incessant mix up in cash prize transfer to previous winners, we advise that you Keep your winning information confidential until your claims is been processed and your funds ($1,000,000.00) Transferred to you. This security measure is important; its protocol meant to avoid double claims, transfer and unwarranted abuse of this program by other participants. METHOD OF SELECTING WINNERS: Participants were selected anonymously through a private ballot system (computer ballot) from over 35,000 Companies and 70,000 individual e-mail addresses all over the world. At the final draw your e-mail address Emerged as the winner of our last jackpot.This is promotional program takes place bi-annually, and it is promoted and sponsored Orient software Corporation (Orient Networks). It is met to reward some individuals and corporations who have devote Their time and resources surfing the world wide web (www) with their email addresses. Remember, all winning must be claimed not later than 5 working days from today. After the expiration of that time frame, if your funds are unclaimed, it will be refunded to us and included in our next draws.We hope you will use part of your cash prize to participate in our next $18million draws in March 2006. Lastly, in order to avoid unnecessary delays/complications with the transfer of your funds to you, please Always remember to quote your Code number in all correspondence with our paying bank (Deutsche Bank-Amsterdam). Furthermore, should there be any change of your present address, please notify us Immediately so to make the appropriate adjustment. Congratulations from the management and staffs of Applecrest & Grant Lotteries. ----------------------------------------------------- Kindly, immediately, contact our Paying Bank (Deutsche Bank-Amsterdam) for your payment advice. Send them following e-mail: I?????(your names)??.., with e-mail address???..(your e-mail address)?.., contact address???..(postal address)??.and code number:???.(quote your code number)??.is the winner of the last Applecrest & Grant Lottery draws, with a cash prize of $1,000,000.00. Kindly advise me with details of the process of the transfer of my cash prize within the next 5-working days. Contact details of our Paying Bank: Contact person: Mr. Larry Martin. Department of Private Banking, Deutsche Bank, Amsterdam, the Netherlands. E-mail: custservicess@netscape.net Tel: 31 647-210-426 Fax 31-847-235-420 Congratulations. Yours in services, Scott Frederic.(Lottery Coordinator) Note: Participants below 21 years are automatically disqualified. All e-mail transmission/correspondence are 100% protected by our Secure Socket Layer (SSL) server. === Subject: Re: Standard Deviation of PISA : > Why are Korean and Gaelic the same language? More than 3,000 years >> ago, Hebrew was spoken all the way from Ireland to Phoenicia to Judaea >> to Korea, and the odds are good that several of the hundred plus >> Indian languages are Hebrew. > Indian as in feather or dot? Because if it's > feather, then Lakota obviously isn't. >> Indian as in India. >> Even American Indians know this isn't India. >> To be here as long as they were, American Indians have to be REAL >> stupid to never develop a written language. Even the Indians of India, >> whose average earnings are $100 per month, were smart enough to develop >> several written languages (or at least smart enough to inherit written >> languages from the original developers and maintain it). > Which part of Mayan seems to be giving you trouble? > -- cary ScatBoy is sponsored today by the letter K and the number eleven. ++ gray === Subject: Re: Standard Deviation of PISA <0001HW.BF9D882F00DDD321F0284550@news.giganews.com> joo Schlochkley, Could it be that the following is a reference to you?: Genetic predisposition to violence, in that, to a dim mind, this is the easiest answer. This isn't unique to the U.S., but follows their race wherever they live globally. Desolation has also followed their race throughout all of recorded history Actually, he's referring to niggers--but you've succeeded in convincing us that there's not much difference between the jew mind and the nigger mind. Identifying information has been removed, the poll is at http://christianparty.net/pollblacks.htm At this point, regarding the $200k option [read: exiling niggers with a $200,000 payment], only 17.4% Strongly Disagree and 9.5% Disagree, for a total of 26.9%. BUT, 53% who Strongly Agree either agree or don't know about the other two options, and 72% of those who Disagree either agree or don't know about the other two options. This leaves a whopping 10.9% who oppose exiling niggers with a $200k payment. Done deal. The only remaining issue is the amount. $200,000 each? $35,000 each? ten cents each? The niggers repay us for doubling their life expectancy, increasing their household incomes 30 FOLD, and reducing their odds of getting AIDS by 10X? If niggers pay US, how much do you think they owe us for this, Schlochkley? John Knight race: White sex: Man age: 58 === Subject: race SubjectOther: Username: Dave UserEmail: private UserTel: private UserFAX: n/a ContactRequested: question1: Strongly Agree question2: Agree onepercent: Strongly Agree zimbabwe: Strongly Agree la: Agree justicestats: Strongly Agree hispanics: Strongly Agree nd: Strongly Agree povertyb: Strongly Agree povertyh: Strongly Agree twentiethcentury: Strongly Agree twentyfirst: Strongly Agree clearance: Strongly Agree drugs: Strongly Agree wardrugs: Agree guncontrol: Strongly Agree second: Strongly Agree stp: Strongly Agree taxes: Agree taxesh: Agree savings: Strongly Agree slaves: Agree prison: Strongly Agree potential: Strongly Agree desire: Strongly Agree loss: Strongly Agree aids: Strongly Agree life: Agree income: Strongly Agree math: Strongly Agree dc: Strongly Agree education: Strongly Agree ancestor: solomon: Strongly Agree sin: Strongly Agree beasts: Strongly Agree liberia: Strongly Agree ps: Strongly Agree cuttaxes: reducehomicide: Agree increaseclearance: Strongly Agree raisescores: Strongly Agree speech: Strongly Agree repealgunlaws: Strongly Agree reducedrugs: Agree autotheft: Strongly Agree rape: Strongly Agree assault: Strongly Agree robbery: drugabuse: Strongly Agree dui: Strongly Agree theft: Agree lawyers: Agree divorce: Agree illegitimacy: Agree illegitmacy: inmates: Strongly Agree garden: discriminated: repatriation: Strongly Agree betterenvironment: twohundredk: Agree thirtyfivek: Agree exile: Strongly Agree Time: 19:15:08 Remote Name: .........162.9 Remote User: HTTP User Agent: Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; FunWebProducts; SV1) Comments: Genetic predisposition to violence, in that, to a dim mind, this is the easiest answer. This isn't unique to the U.S., but follows their race wherever they live globally. Desolation has also followed their race throughout all of recorded history === Subject: Re: Standard Deviation of PISA <0001HW.BF9D882F00DDD321F0284550@news.giganews.com> It's a stretch to call the Mayan pictographs a written language, even though it's the ONLY Indian language which could represent spoken language to the same degree as the written language of Europe. All of the written languages of Europe are derived from Hebrew, with 95% of English words being of Hebrew origin from 5,000 to 6,000 years ago. The Hebrew that the Masoretic jews developed prior to the translation of the King James Version of the Holy Bible wasn't even as similar to paleo Hebrew as English is. http://www.crystalinks.com/mayanwriting.html Furthermore, nobody but royalty could read this language, making it just about worthless. They had no Shakespeare, and where they had enough math skills to pile rocks on top of each other which survived several centuries, these structures are nowhere close to the quality of the grand pyramid in Giza, the Parthenon in Greece, the Roman Amphitheatre, Stonehenge (built by Israelites), and Newgrange in Ireland, tomb of King Zedekiah's grandson Angus. You'd have to ask, though, why the descendants of people like the Mayans in Peru are so much stupider now than the rocks their putative ancestors piled on top of each other. Peru's one of the grandest contenders for the Lowest Math Score in the World, POSSIBLY even lower than Washingonian niggers (though even Peru probably scored higher on PISA than niggerized DC would have scored had they dared to participate in PISA). Girls in Peruvian public schools scored 266, which is a whopping 353 points lower than boys in private schools in the Netherlands of 579, an indication of the difference in intellect between a bunch of Indians whose grandest accomplishment in millennia was pictographs, and the Dutch who ACTUALLY did the original research which LYING PLAGIARIZING jew [though I repeat myself] Einstein STOLE, then claimed to be his own work. http://christianparty.net/einstein.htm John Knight === Subject: Re: Standard Deviation of PISA <0001HW.BF9D882F00DDD321F0284550@news.giganews.com> > It's a stretch to call the Mayan pictographs a written language, even > though it's the ONLY Indian language which could represent spoken > language to the same degree as the written language of Europe. All of > the written languages of Europe are derived from Hebrew, with 95% of > English words being of Hebrew origin from 5,000 to 6,000 years ago. Evidence would be nice. > superficial resemblance to the Egyptian writing, to which it is not > related) was a combination of phonetic symbols and ideograms. It is the > only writing system of the Pre-Columbian New World that can completely > represent spoken language to the same degree as the written language of > the old world.> > http://www.crystalinks.com/mayanwriting.html crystalinks is NOT a reliable site. And if I mentioned other written languages, would you hate me? > Furthermore, nobody but royalty could read this language, making it > just about worthless. Citation? > They had no Shakespeare, and where they had > enough math skills to pile rocks on top of each other which survived > several centuries, these structures are nowhere close to the quality of > the grand pyramid in Giza, the Parthenon in Greece, the Roman > Amphitheatre, Stonehenge (built by Israelites), and Newgrange in > Ireland, tomb of King Zedekiah's grandson Angus. I think that Mayan pyramids are at LEAST as grandiose as a pile of rocks in a circle. > You'd have to ask, though, why the descendants of people like the > Mayans in Peru ^^^^^^ ^^^^ Mayans in Peru. LOL > are so much stupider now than the rocks their putative > ancestors piled on top of each other. Peru's one of the grandest > contenders for the Lowest Math Score in the World, POSSIBLY even > lower than Washingonian niggers (though even Peru probably scored > higher on PISA than niggerized DC would have scored had they dared to > participate in PISA). Girls in Peruvian public schools scored 266, > which is a whopping 353 points lower than boys in private schools in > the Netherlands of 579, an indication of the difference in intellect > between a bunch of Indians whose grandest accomplishment in millennia > was pictographs, Actually, I'd say their grandest accomplishments were in the fields of medicine and astronomy. > and the Dutch who ACTUALLY did the original research > which LYING PLAGIARIZING jew [though I repeat myself] Einstein STOLE, > then claimed to be his own work. > http://christianparty.net/einstein.htm > John Knight === Subject: Re: Standard Deviation of PISA <0001HW.BF9D882F00DDD321F0284550@news.giganews.com> <<< I'd have to agree with that. Medicine was their grandest accomplishment. That's why they're all dead. John Knight === Subject: Re: Standard Deviation of PISA > Actually, I'd say their grandest accomplishments were in the > fields of medicine and astronomy. > and the Dutch who ACTUALLY did the original research > which LYING PLAGIARIZING jew [though I repeat myself] Einstein STOLE, > then claimed to be his own work. > http://christianparty.net/einstein.htm Don't bother with the link; by the Dutch, I'm sure John is talking about poor old Henrik Lorentz, from whom Einstein stole Special Relativity. In John-world, that is. Last time we went through this, I put up excerpts a popular-science Courant, Novermber 1919, just after the first experimantal proof of Einstein's General theory had been reported. You can see how violently Lorentz hates Big Al for the theft, how venom just drips from every word: THE EINSTEIN THEORY OF RELATIVITY A Concise Statement by Prof. H. A. Lorentz, of the University of Leyden [all comments in brackets are my own] The total eclipse of the sun of May 29, resulted in a striking confirmation of the new theory of the universal attractive power of gravitation developed by Albert Einstein, and thus reinforced the conviction that the defining of this theory is one of the most important steps ever taken in the domain of natural science. In response to a request by the editor, I will attempt to contribute something to its general appreciation in the following lines . . . Einstein has put an end to this isolation [that gravity seemed to stand apart from electromagnetic phonomena]; it is now well established that gravitation affects not only matter, but also light. Thus strengthened in the faith that his theory already has inspired, we may assume with him that there is not a single physical or chemical phenomenon--which does not feel, although very probably in an unnoticeable degree, the influence of gravitation, and that, on the other side, the attraction exercised by a body is limited in the first place by the quantity of matter it contains and also, to some degree, by motion and by the physical and chemical condition in which it moves. It is comprehensible that a person could not have arrived at such a far- reaching change of view by continuing to follow the old beaten paths, but only by introducing some sort of new idea. Indeed, Einstein arrived at his theory through a train of thought of great originality. Let me try to restate it in concise terms . . . Every attempt along that line [to detect the aether], however, has remained fruitless; all the phenomena examined were evidently independent of the motion of the earth. That this is the way they do function was brought to the front by Einstein in his first or special theory of relativity. For him the ether does not function and in the sketch that he draws of natural phenomena there is no mention of that intermediate matter . . . [Lorentz disagrees that Einstein has disproved the existence of the aether. nonetheless:] Nevertheless, even without the color and clearness that the ether theories and the other models may be able to give, and even, we can feel it this way, just because of the soberness induced by their absence, Einstein's work, we may now positively expect, will remain a monument of science; his theory entirely fulfills the first and principal demand that we may make, that of deducing the course of phenomena from certain principles exactly and to the smallest details. It was certainly fortunate that he himself put the ether in the background; if he had not done so, he probably would never have come upon the idea that has been the foundation of all his examinations. great difficulties to overcome in his attempts, Einstein has attained the results, which I have tried to sketch, while still young; he is now 45 years old. He completed his first investigations in Switzerland, where he first was engaged in the Patent Bureau at Berne and later as a professor at the Polytechnic in Zurich. After having been a professor for a short time at the University of Prague, he settled in Berlin, where the Kaiser Wilhelm Institute afforded him the opportunity to devote himself exclusively to his scientific work. He repeatedly visited our country and made his Netherland colleagues, among whom he counts many good friends, partners in his studies and his results. He attended the last meeting of the department of natural philosophy of the Royal Academy of Sciences, and the members then had the privilege of hearing him explain, in his own fascinating, clear and simple way, his interpretations of the fundamental questions to which his theory gives rise. My god, could Lorentz have condemned Einstein any more thoroughly? -- cary === Subject: Re: Standard Deviation of PISA > It's a stretch to call the Mayan pictographs a written language, even > though it's the ONLY Indian language which could represent spoken > language to the same degree as the written language of Europe. All of > the written languages of Europe are derived from Hebrew, with 95% of > English words being of Hebrew origin from 5,000 to 6,000 years ago. > Evidence would be nice. > superficial resemblance to the Egyptian writing, to which it is not > related) was a combination of phonetic symbols and ideograms. It is the > only writing system of the Pre-Columbian New World that can completely > represent spoken language to the same degree as the written language of > the old world.> > http://www.crystalinks.com/mayanwriting.html > crystalinks is NOT a reliable site. And if I mentioned other > written languages, would you hate me? Hey, GOD hates you, since you're clearly no Israelite. Of course, nobody's an Isrealite when you come to think of it, so God, John's version of, pretty much hates everybody. > Furthermore, nobody but royalty could read this language, making it > just about worthless. > Citation? There are more readers of Mayan today than there were at the time of the Empire. (my source: Ronald Wright's Massey Lectures). Good thing, too, since there's a ton of documents yet to be translated. And of course John's flea-ridden, furs-wearing, village-raiding Germanic and Nordic Israelites weren't exactly noted for their high rates of literacy either. I suppose this makes his imaginary paleo-phoneo-hebraeo-anglo proto-language just about worthless too. > They had no Shakespeare, and where they had > enough math skills to pile rocks on top of each other which survived > several centuries, these structures are nowhere close to the quality of > the grand pyramid in Giza, the Parthenon in Greece, the Roman > Amphitheatre, Stonehenge (built by Israelites), and Newgrange in > Ireland, tomb of King Zedekiah's grandson Angus. > I think that Mayan pyramids are at LEAST as grandiose as a pile > of rocks in a circle. > You'd have to ask, though, why the descendants of people like the > Mayans in Peru > ^^^^^^ ^^^^ > Mayans in Peru. LOL Hey, as errors in John's post go, Mayans in Peru doesn't even poke out of the general noise. Besides, there WERE Mayans in Peru. It says so right there in the Celestine Prophecies. > are so much stupider now than the rocks their putative > ancestors piled on top of each other. Peru's one of the grandest > contenders for the Lowest Math Score in the World, POSSIBLY even > lower than Washingonian niggers (though even Peru probably scored > higher on PISA than niggerized DC would have scored had they dared to > participate in PISA). Girls in Peruvian public schools scored 266, > which is a whopping 353 points lower than boys in private schools in > the Netherlands of 579, an indication of the difference in intellect > between a bunch of Indians whose grandest accomplishment in millennia > was pictographs, > Actually, I'd say their grandest accomplishments were in the > fields of medicine and astronomy. -- cary === Subject: Re: Standard Deviation of PISA > You'd have to ask, though, why the descendants of people like the > Mayans in Peru > ^^^^^^ ^^^^ > Mayans in Peru. LOL Hey, as errors in John's post go, Mayans in Peru doesn't even poke out of the general noise. Besides, there WERE Mayans in Peru. It says so right there in the Celestine Prophecies. Why would you think there were no Mayans in Peru, if they were so smart? National Geographic had a picture of a Mayan girl who'd been buried on a Peruvian mountain top in a sitting position, proof enough that Mayans were in Peru. What they missed was even more important than what they discovered. The shoes the Mayan girl was wearing were unmistakably Korean shoes, a design which pre-dated the Mayans. Does this prove that the Mayans were Koreans? No. But unless Mayans did their shopping in Korea, it's likely that the Koreans at least travelled across the ocean from Korea to South America and had at least enough influence to sell them shoes. You keep forgetting to answer the question, though, which is if the Mayans were so advanced thousands of years ago, then why are they so backward today? [read: why did nobody in the world score lower on PISA than Peruvian girls in Peruvian public schools?]. How did they drop from the foreskin of technology, all the way to the rear, in such a short spell? Did they mate with niggers as DNA suggests? Is the offspring of a Mayan who mates with a nigger even stupider than the nigger? Did their dumb offspring suddenly find that God's Law is ah violashun of mah snivel rahts, like today's niggers think? Are they waiting around fo da whiat man [read: da man, or da honkey] to rebuild their temples for them the way niggers in France and LA and New Orleans are waiting fo honkies da rebuild the cities the niggers destroyed? John Knight ps--http://christianparty.net/pollblacks.htm === Subject: Re: Standard Deviation of PISA > It's a stretch to call the Mayan pictographs a written language, Pictographs? Only if you regard Chinese and Japanese as pictographs. > even > though it's the ONLY Indian language which could represent spoken > language to the same degree as the written language of Europe. Well, then, there you go. Out of your own mouth. > All of the written languages of Europe are derived from Hebrew, mummmm heh heh heh... > with 95% of > English words being of Hebrew origin > from 5,000 to 6,000 years ago. heh heh ha ha Ha Ha Ha HA HA.... > The Hebrew that the Masoretic jews developed prior to the translation > of the King James Version of the Holy Bible wasn't even as similar to > paleo Hebrew as English is. BWAAAAAAAHAHAHAHAHAHAHAHAHAHAHaHaHahahahah...gigglesmurflesnortle blork!!! Ah, man. Don't ever change John, don't ever change. Tafsik lezyen bamoah basehel. As we say in paleo-English. (incidentally, with relation to the math sneer I snipped, the Mayan calendar was slightly more accurate than the Gregorian) -- cary === Subject: Re: Standard Deviation of PISA <0001HW.BF9D882F00DDD321F0284550@news.giganews.com> > (incidentally, with relation to the math sneer I snipped, > the Mayan calendar was slightly more accurate than the Gregorian) And the Mayans had some holidays in relation to the Venerian year, something Johnny knows nothing about because the people he claims are his ancestors believed in a flat Earth with a dome around it to keep rain out. === Subject: Re: Standard Deviation of PISA <<> Lookie what I found on the net, which must be true, since we live in a democracy according to Bush, and which If I Found It On The Web, It Must Be True according to cary nation. This is a profile of the replies to the Poll To Exile Blacks which as you now already know can be found at http://christianparty.net/pollblacks.htm EXILE WITH NO REPARATIONS Of the 13% who Strongly Disagree with exile no reparations: 13% of them agree or don't know about $35k 9% of them don't know or agree with $200k NET = 10% Of the 6% who Disagree with $0k: 60% Don't Know, Agree or Strongly Agree with $35k or $200k NET = 2.4% TOTAL WHO DISAGREE = 12.4% TOTAL WHO DON'T OBJECT [READ: Strongly Agree, Agree, or Don't Know] = 87.6% Of the 11.5% who were blacks, 25% favored their own exile, either with $200k or with $35k in reparations. More White women than black men Disagree or Strongly Disagree with exiling blacks. Women of all races who Strongly Agree that the 24,000 gun control laws are unconstitutional also Strongly Agree with exiling blacks with no reparations. PROFILE OF THE RESPONDENTS 66.7% Agree or Strongly Agree ... the war on drugs was necessary because of blacks and Hispanics 68.2% Agree or Strongly Agree that the American White man must work at least 11 days each year just to earn the money which is paid just in taxes which funds just the welfare which is paid just to American Hispanics 68.4% Agree or Strongly Agree that ... each of these 23,000 gun control laws to be a violation of the spirit and intent of the Second Amendment 69% Agree or Strongly Agree that the primary reason American White men have a negative personal savings rate, compared to more than 25% in most European nations and more than 25% in the US just half a century ago, is the extraordinary cost of supporting American blacks 70.1% Agree or Strongly Agree that the American White man must work at least 42 days each year just to earn the money which is paid just in taxes which funds just the welfare which is paid just to American blacks 74.2% social transfer payments [read: welfare] to American blacks exceeds their contribution to GDP by more than $850 billion, almost twice the entire economic output of 750 million blacks on the Continent of Africa The questions to which the highest percentage of responses were Don't Know were: 20.2% I believe Solomon sinned by marrying foreign or non-Israelite wives like Egyptians, and perhaps blacks 21.4% I believe the 400,000 black slaves of the 17th Century were never more productive than their cousins in Africa who now earn less than $50 per month. 22% I believe that exiling all blacks would reduce the percentage of the lawyers in the world who are in the US from 70% to at most 40% the first year 34% I believe it was Hispanics and not blacks who caused Los Angeles to once hold the title of Murder Capitol of the World. 35% I believe Hispanics commit homicide at a rate almost half that of blacks. 40% I believe that exiling all blacks would reduce the illegitimacy rate by at least 50% within 5 years The questions to which the lowest percentage of responses were Don't Know were: 3.4% I believe drug wars and gang related violence are related primarily to blacks and Hispanics 3.5% I believe that the primary reason American blacks have a life expectancy of 64, much lower than Europeans but almost twice that of African blacks, is primarily the influence and accomplishments of American Whites 4.3% I believe that 23,000 gun control laws were ineffective because blacks and Hispanics cannot be trusted with guns 4.5% I believe the Holy Bible and our Founding Forefathers who considered blacks to be beasts and animals were correct and that modern political correctness is in serious error 6.7% I believe the reason the murder rate in North Dakota is 1/350th that of Washington, DC, is because there are so few blacks and Hispanics in North Dakota This is an unprecedented and unexpected result. No poll we've ever conducted on the internet has had such a high percentage of support for such a key moral issue. More Americans want to exile blacks than want to exile jews (only 83%), want creation to be taught in public schools (only 81%), want spoken Christian prayer in public schools (only 70%), blame Israel for the attack on the WTC and Pentagon (68%), believe that abortion is murder (57%), voted to ban affirmative action in California with Proposition 209 (54%). http://christianparty.net/surveys.htm President Bush claims we're a democracy, and that this is what he's making the world safe for. In a democracy, the majority rules. This is not a mere majority, though. This is not a mere two thirds. This is not a mere three quarters, the requirement for a Constitutional Amendment. This is not a mere five sixths, a level of support no president has ever held for long if ever. This is almost 9 out of every ten Americans who want, no demand, the immediate restoration of the founding Christian principles our Founding Forefathers fought and shed blood and died for, including a significant proportion of the niggers themselves. Bear in mind that the 1.2 million niggers in prison don't have access to this poll, but do you need to guess what their reply would be? implement it here at home and quit making it so safe for the hundreds of thousands of Iraqi women and children you've cluster bombed into modernism. You're duty bound to uphold your oath of office, requiring you to protect the US Constitution from foreign *AND DOMESTIC* enemies, and we hereby declare blacks to be public enemy number one. http://christianparty.net/homicide.htm IMPLEMENT THE PLAN TO EXILE THEM NOW. No more sound bytes. No more debates. No more discussions. No more excuses. No more blackening of our TalmudVision sets, and no more niggerization of America. Each American black has proven himself to be a million times more dangerous to our safety than each Muslim throughout the entire Arab world. John Knight ps--note that the percentages above are based on the number of people who answered each question, rather than the total number of respondents. F'rinstance, if 5% of all the respondents Strongly Agree with Point A, but only half the respondents answered that one question, then 10% of those who responded to the question were counted as Strongly Agree. All of the replies are in a data base which you will soon be able to query on line. === Subject: could you please recommend some good topology and set theory reference/problems books? HI all, I am struggling with my real analysis study. We are using Mardsen's book Elementary Classical Analysis... It took me a while to figure out that the first a few chapters are really set theory and topology: Chapter 1. The Real Line and Euclidean Space Chapter 2. The Topology of Euclidean Space Chapter 3. Compact and Connected Sets Chapter 4. Continuous Mappings ... Thus I think I really should find some more problems to solve in point set theory and basic topology... I should have taken these two classes first then I come to the real analysis class... I did not know that since I am a engineer... Now knowing these facts, please could you recommend some good topology and set theory reference books and problems books? It would be great if the books have many examples and precise proofs to show how to write math in a correct way? === Subject: Re: could you please recommend some good topology and set theory reference/problems books? > Now knowing these facts, please could you recommend some good topology > and set theory reference books and problems books? It would be great if > the books have many examples and precise proofs to show how to write > math in a correct way? R. Engelking, General Topology - a basic reference book in topology with hundreds of problems, from easy exercises to graduate-level questions. Pawel Gladki === Subject: Re: could you please recommend some good topology and set theory reference/problems books? >> I am struggling with my real analysis study. >> We are using Mardsen's book Elementary Classical Analysis... >> Now knowing these facts, please could you recommend >> some good topology and set theory reference books and >> problems books? It would be great if the books have many >> examples and precise proofs to show how to write math >> in a correct way? > R. Engelking, General Topology - a basic reference book > in topology with hundreds of problems, from easy exercises > to graduate-level questions. And completely inappropriate for anyone who is struggling with Mardsen's text, as much as I like Engelking's book. Dave L. Renfro === Subject: Re: could you please recommend some good topology and set theory reference/problems books? > And completely inappropriate for anyone who is struggling > with Mardsen's text, as much as I like Engelking's book. I'd be hard put to think of a text at the same level as Marsden's which does a better job of presenting worked examples and showing just how to write proofs. The first several chapters *are* a good introductory text on metric spaces and R^n in particular. (And it even has worked solutions to a lot of the exercises in the back.) === Subject: Re: could you please recommend some good topology and set theory reference/problems books? > HI all, > I am struggling with my real analysis study. > We are using Mardsen's book Elementary Classical Analysis... > It took me a while to figure out that the first a few chapters are > really set theory and topology: > Chapter 1. The Real Line and Euclidean Space > Chapter 2. The Topology of Euclidean Space > Chapter 3. Compact and Connected Sets > Chapter 4. Continuous Mappings > ... > Thus I think I really should find some more problems to solve in point > set theory and basic topology... I should have taken these two classes > first then I come to the real analysis class... I did not know that > since I am a engineer... > Now knowing these facts, please could you recommend some good topology > and set theory reference books and problems books? It would be great if > the books have many examples and precise proofs to show how to write > math in a correct way? If I remember correctly, the book How to read and do proofs contains many elementary proof exercises in set-theory with detailed explanation of how one discover such proofs. === Subject: Re: finite presentation of free prduct >BTW, is the debut for the complete and finished theorem with all >queasiness about (iii) removed, to be presented this weekend? > Yes -- last night I posted a complete version (of the construction of > a commutative ring such that all elements except 1 are zero divisors). Glich, gleech didn't show up or got missed. Was in new thread or old? Please repost as new thread or email. What was subject? === Subject: Re: finite presentation of free prduct >>BTW, is the debut for the complete and finished theorem with all >>queasiness about (iii) removed, to be presented this weekend? >> Yes -- last night I posted a complete version (of the construction of >> a commutative ring such that all elements except 1 are zero divisors). >Glich, gleech didn't show up or got missed. Was in new >thread or old? Please repost as new thread or email. >What was subject? Same subject as before -- same thread: === Subject: Re: commutative rings such that all elements except 1 are zero divisors It shows on Google -- here's a link: quasi === Subject: Re: Set Theory-Axiom of Congruence. >> Robert Low > As soon as you show how adding an axiom to an inconsistent > system can make it consistent. >> Euh... do you happen to know the Hilbert axioms for Euclidean geometry? >> They become inconsistent when you remove the Archimedian axiom. > I don't know that framework. But I find it hard to understand > your claim. If you can derive a contradiction from all but one > of the axioms, surely you can derive a contradiction when > the axioms include the Archimedian one, because any proof > of the original system is still a proof---it just doesn't happen > to use the Archimedian axiom anywhere. Nope. A simpler variant is the following: Suppose we want to axiomatize the real number system. So we say: 1. R is a totally ordered field 2. the set {1,1+1,...} doesn't have an upper bound (Archimedian axiom) 3. R is maximal [wrt these assumptions] As is well known, there exists one and only one model of these axioms, hence, as it stands, the axiom set is consistent. However, if we leave out 2), then the upward Skolem-Loewenheim theorem shows that there no longer exists any model of 1) + 3), as there are no maximal totally ordered fields. (Some hints in case you're confused about this: note that 1) is expressible in FOL, while 2) and 3) are not. Also note that if we leave out 2), the contents or 'meaning' of assumption 3) changes. 3) 'refers' to the other assumptions, so to say. That explains why this can happen.) You might call this cheating, but that would not be entirely fair. There is no a-priori reason why axiom system should always be first order. -- Herman Jurjus === Subject: Re: Set Theory-Axiom of Congruence. > (Some hints in case you're confused about this: note that 1) is > expressible in FOL, while 2) and 3) are not. Also note that if we leave > out 2), the contents or 'meaning' of assumption 3) changes. 3) 'refers' > to the other assumptions, so to say. That explains why this can happen.) > You might call this cheating, but that would not be entirely fair. There > is no a-priori reason why axiom system should always be first order. If necessary, then: OK, it's not cheating, but I'm talking about first order systems. (That said, I'm not convinced that the assertion that it's cheating is entirely unfair, either.) === Subject: Re: Set Theory-Axiom of Congruence. > (That said, I'm not convinced that the assertion > that it's cheating is entirely unfair, either.) It's no more cheating than the following example. We start with axioms 1. P 2. Q 3. P <-> ~P v Q This is consistent. We now remove axiom 2 - naturally also adjusting axiom 3 to P <-> ~P - and get an inconsistent theory. === Subject: Re: Set Theory-Axiom of Congruence. >>(That said, I'm not convinced that the assertion >>that it's cheating is entirely unfair, either.) > It's no more cheating than the following example. We start with axioms > 1. P > 2. Q > 3. P <-> ~P v Q > This is consistent. We now remove axiom 2 - naturally also adjusting > axiom 3 to P <-> ~P - and get an inconsistent theory. I can't find any cause to fault this analogy. === Subject: Re: Set Theory-Axiom of Congruence. > (Some hints in case you're confused about this: note that 1) is > expressible in FOL, while 2) and 3) are not. Also note that if we leave > out 2), the contents or 'meaning' of assumption 3) changes. 3) 'refers' > to the other assumptions, so to say. That explains why this can > happen.) Other aspects of your terminology apart, since you've not only removed one axiom, but changed another axiom, the observation that the revised theory may have no model although the original theory does is not on the face of it relevant to showing how a consistent theory can be made inconsistent by removing an axiom. === Subject: Re: Field extensions > Let p be prime q=p^r > Any field F_q^2 is an extension of field F_q and can be obtained by adjoining a root of the irriducible polynomial x^2-a=0, a in F_q. False, if p=2, true otherwise. > For which q, can the field F_q^3 be an extension of F_q, by adjoining the root of x^3-a=0, a in F_q. This works, if and only if you can locate an element in F_q that has no cubic root in F_q. Such an element exists, if and only if 3 divides q-1. Both are immediate consequences of the fact that the multiplicative group of a finite field is cyclic. Jyrki === Subject: Field extensions How would one show that for a field F_q, (q=p^n), there always exists an extension of degree r, where r is prime, for all q. If r divides q-1, then we can use the polynomial x^r-a=0, for some a in F_q. But what about the case where r does not divide q-1? Maybe an induction argument of some sort? gillian === Subject: Re: Field extensions >How would one show that for a field F_q, (q=p^n), there always exists an extension of degree r, where r is prime, for all q. Let K be the splitting field over F_q of f = x^(q^r) - x. Show (using derivative) that the roots of f are distinct in K. Then show that the roots of f form a subfield of K that containg F_q, which must be equal to K. Since any extension of F_q of degree r must be the splitting field of f, this also shows uniqueness of the extension. Derek Holt. >If r divides q-1, then we can use the polynomial >x^r-a=0, for some a in F_q. >But what about the case where r does not divide q-1? >Maybe an induction argument of some sort? >gillian === Subject: Re: Field extensions >Let K be the splitting field over F_q of f = x^(q^r) - x. >Show (using derivative) that the roots of f are distinct in K. >Then show that the roots of f form a subfield of K that containg F_q, >which must be equal to K. I'm lost here, how would you show the roots of f form a subfield of K, containing F_q, and that it equals to K? >Since any extension of F_q of degree r must be the splitting field of f, >this also shows uniqueness of the extension. Um why? And is K=F_q^r ? Aren't we looking for a degree r extension? Is K a degree r extension? === Subject: Re: Field extensions >>Let K be the splitting field over F_q of f = x^(q^r) - x. >>Show (using derivative) that the roots of f are distinct in K. >>Then show that the roots of f form a subfield of K that containg F_q, >>which must be equal to K. >I'm lost here, how would you show the roots of f form a subfield of K, containing F_q, and that it equals to K? You just need to check that, if a and b are roots of f, then so are a+b, -a, ab, and b^-1, which is routine. It contains K, because every element of K is a root of x. Since the splitting field of f is by definition the smallest field containing all of the roots of f, and the roots of f form a subfield, that subfield must be equal to K. >>Since any extension of F_q of degree r must be the splitting field of f, >>this also shows uniqueness of the extension. >Um why? That follows from the uniqueness os splitting fields. >And is K=F_q^r ? Aren't we looking for a degree r extension? Is K a degree r extension? Yes, K is F_q^r, which is a degree r extension of F_q. But you should not really start using the notation F_q at all until you have proved that there is a unique finite field of each prime power order. Derek Holt. === Subject: Open subset in the algebra B(X) Let B(X) be the Banach algebra of linear and bounded operators on the Banach space X. Is the set S(X) of all surjective operators in B(X) an open set in B(X)? I know that the result is true when X is a Hilbert space, but what is the general situation? Mate === Subject: Re: Open subset in the algebra B(X) >Let B(X) be the Banach algebra of linear and bounded operators on the >Banach space X. >Is the set S(X) of all surjective operators in B(X) an open set in >B(X)? I believe so. >I know that the result is true when X is a Hilbert space, but what is >the general situation? What's a Hilbert-space proof that does not work in a Banach space? Anyway, say T:X->X is surjective. Choose r > 0 such that for every y in X there exists x in X with Tx = y and ||x|| <= r ||x||. Seems to me that if ||S|| = e and er < 1 then T+S is surjective: Say y in X with ||y|| = 1. Choose x_0 in X with ||x_0|| < r and Tx_0 = y_0. Then (T+S)x_0 = y_0 + Sx_0 and ||Sx_0|| < er. Choose x_1 in X with ||x_1|| < er^2 and Tx_1 = -Sx_0. Then (T+S)(x_0+x_1) = y + Sx_1, and ||Sx_1|| < e^2r. Choose x_2 with ||x_2|| < (er)^2 and Tx_2 = - Sx_1. Etc. Then sum x_j converges and T(sum x_j) converges to y. >Mate ************************ David C. Ullrich === Subject: Re: Open subset in the algebra B(X) I had some problems following your argument due to some typos. My main objection is that you considered that if T in B(X) is surjective then there exists r>0 such that for each y in X, there is x in X satisfying Tx = y and ||x|| < r ||y||. As far as I know, this is not true in general (the Bartle-Graves result says less; it is however true when X is Hilbert, but then your proof simplifies a bit; it is also true when T is a compact perturbation of the identity). Mate === Subject: Re: Open subset in the algebra B(X) >I had some problems following your argument due to some typos. >My main objection is that you considered that if T in B(X) is >surjective then there exists r>0 such that for each y in X, > there is x in X satisfying Tx = y and ||x|| < r ||y||. >As far as I know, this is not true in general It is true. It's just the Open Mapping Theorem. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Open subset in the algebra B(X) It is the open mapping theorem only if T is bijective. When T is only surjective one has to use a Bartle-Graves type result and it does guarantee such a r (the right inverse is not necessarily linear!). Mate === Subject: Re: Open subset in the algebra B(X) > It is the open mapping theorem only if T is bijective. > When T is only surjective one has to use a Bartle-Graves type result > and it does guarantee such a r (the right inverse is not necessarily > linear!). No, it's immediate from the Open Mapping Theorem: T(B(0,1)) contains B(0,r) for some r > 0, etc. === Subject: Re: Open subset in the algebra B(X) >It is the open mapping theorem only if T is bijective. >When T is only surjective one has to use a Bartle-Graves type result >and it does guarantee such a r (the right inverse is not necessarily >linear!). I don't know what you mean by Bartle-Graves, but the versions of the OMT that I've seen don't require bijective. Reed & Simon, Functional Analysis, Theorem III.10 (open mapping theorem): Let T: X -> Y be a bounded linear transformation of one Banach space onto another Banach space Y. Then if M is an open set in X, T[M] is open in Y. So take M to be the open unit ball in X, and r such that the ball of radius 1/r about 0 is in T[M]. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Open subset in the algebra B(X) You are right, Robert. My problem was that I was looking for a _continuous_ mapping y |--> x such that ||x|| <= r ||y||. For such situations the Bartle-Graves theorem applies, but it seems that the continuity is not needed for the initial problem. Mate === Subject: New mathematics/physical sciences positions at http://jobs.phds.org, November 14, 2005 New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Portfolio Analytics/ Software Engineers: Martingale International Search Inc., New York. JOB DESCRIPTION: Portfolio Analytics team of the major Investment Bank is seeking experienced and detail oriented C++ software engineers to research, design, develop, and test Portfolio analytics software applications and components. Applicant will be... http://jobs.phds.org/jobs/dinkak/listing_2005_11_07 * Assistant Professor of Mathematics, Southwestern College: Academic Careers Online, Winfield, KS. Teach a broad range of undergraduate courses in both mathematics and statistics, supporting a small major but primarily servicing majors in science, business, education, and computer science. 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JOB DESCRIPTION: RMS developes stochastic models for assessment of the financial risk due to natural catastrophic events such as windstorms, and earthquakes, as well as man-made risks such as fire and terrorism, working in collaboration with a group consisting... http://jobs.phds.org/jobs/recruiter/listing_2005_11_10 * Senior Risk Manager- Hedge Fund, London : Financial Executive Search, London. This well established global hedge fund is seeking a senior risk manager responsible for overseeing their cross asset range of products. Due to continued, successful years of trading and new product generation they require a focused individual to move into this... http://jobs.phds.org/jobs/dietrichyap/listing_2005_11_11 * PhD Statistics Graduate for Quantitative Finance Role: Huxley Associates, New York. The head of Quantitative Trading Research for a top-tier European Bank is looking to hire a PhD statistician to work in his New York team. You will contribute to the development of quantitative trading strategies for the prop trading groups. This role will be fairly... http://jobs.phds.org/jobs/huxley/listing_2005_11_07 * Senior Stat Arb Trader, London Role!: Financial Executive Search, London. Leading US investment bank seeks a senior statistical arbitrage trader to join their highly rated proprietary trading group. You will join a global team of experienced traders who have a natural aptitude for developing quantitative and systematic market neutral... http://jobs.phds.org/jobs/dietrichyap/listing_2005_11_08 === Subject: Re: Matrix rank problem > No. Consider > M = [ x 0 1 ] > [ 1 y 2 ] > [ 1 3 z ] > Then r = 1, since there are no 2 x 2 submatrices with only > given entries, but it's easy to see that the actual rank > must be at least 2. Yup, that solves my general problem - always took too few unknowns in my experiments. Although this again leaves half a dozen new questions, e.g. a m12 m13 m14 m21 b m23 m24 m31 m32 c m34 m41 m42 m43 d a,b,c,d are unknown, m_ij are given such that r=1 (e.g. fix m31=1,m32=2,m41=4,m42=8 and the rest of the m_ij follow modulo a common denominator which factors from the determinant) - then a,b,c,d can be chosen such that R=1. So: In which cases is R=r? -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Manifolds and Their Boundaries let M and N be two compact 1-codimensional smooth submanifolds of R^n. Then M and N bound two n-dimensional smooth submanifolds of R^n, P and Q. Let us suppose that M and N are diffeomorphic. The question is: are P and Q homeomorphic? The study of the differentiable version of the Schoenflies Problem (see Milnor, Ann. Math. 1957) excludes that P and Q are generally diffeomorphic. What do you know about this problem? Maury === Subject: Re: Manifolds and Their Boundaries No. There is a knot in S^3 whose knot group is non-abelian. Take an open tubular neighborhood of this knot in S^3 and let P denote its complement. Then Bd P = S^1 x S^1 and P lies in R^3. Take Q to be S^1 x D^2 so that Bd Q = S^1 x S^1. Both P and Q lie in R^3 and have diffeomorphic boundaries. Note that the fundamental group of P is non-abelian, but that of Q is abelian, so P is not diffeomorphic (or homeomorphic) to Q. === Subject: Aha! Found out about manifold stripings - Foliations A month or so ago I posted a question in a sci.math thread titled What manifolds allow an everywhere parallel vector field? I recently found that the idea I was trying to express is called Foliation Theory, and is currently an active geometry and 3-manifolds by James Otterson at: http://math.berkeley.edu/~alanw/242papers02/otterson.pdf on symplectic geometry. The idea of foliation theory is to break down a manifold M into immersed [*] submanifolds that fit together nicely; i.e. locally they look like a piled ream of paper. It is hoped to get new insights about the manifold by looking at such assembled subpieces. [*] Dammit - I just spent £40 on a book on Riemannian sub mersions, thinking they might be relevant! This extract (on page 1) is followed by a diagram showing a torus surface comprising a stack of congruent circles in rotational symmetry about its principle axis. Just shows that often half the problem asking experts for information is to know the correct buzzwords, as they often can't or won't attempt to read between the lines however one struggles to express an idea otherwise. Foliations of manifolds are also defined in Wikipedia: http://en.wikipedia.org/wiki/Foliation === Subject: A graph theoretic problem What is the minimal number n, such that for any graph G with n vertices, either G or Complement(G) is non-planar? We think that the answer is 9, but we're unable to prove it. There are example with 8 vertices such that both G and Complement(G) are planar. === Subject: Re: A graph theoretic problem G must be planar. Then G can have up to 3*n-6 edges. C(G) can have a maximum of ((n^2)*n)/2 -3*n +6 edges. To guarantee that C(G) is non planar, it must have more than 3*n-6 edges. n=11 is the smallest n for which this is possible.. If n = 10, C(G) may have no more than 21 edges and therefore, can be planar If n = 11, C(G) must have at least 28 edges and therefore, must be non planar. === Subject: Re: A graph theoretic problem >G must be planar. Then G can have up to 3*n-6 edges. > C(G) can have a maximum of ((n^2)*n)/2 -3*n +6 edges. > To guarantee that C(G) is non planar, it must have more than 3*n-6 > edges. > n=11 is the smallest n for which this is possible.. > If n = 10, C(G) may have no more than 21 edges and therefore, can be > planar > If n = 11, C(G) must have at least 28 edges and therefore, must be non > planar. Yes, we know this already, so the answer must be less or equal to 11. This is a simple corollary of Euler's formula. === Subject: Re: A graph theoretic problem You didn't include the context, bill! Don't use Google Groups's Reply option! Click on show options, and then Reply in _the menu that opens up_. The OP was: > What is the minimal number n, such that for any graph G with n vertices, > either G or Complement(G) is non-planar? > We think that the answer is 9, but we're unable to prove it. > There are example with 8 vertices such that both G and Complement(G) are > planar. > G must be planar. Then G can have up to 3*n-6 edges. > C(G) can have a maximum of ((n^2)*n)/2 -3*n +6 edges. I think you mis-typed something there. If G has at most 3*n-6 edges, then C(G) has at least n(n-1)/2 - 3*n + 6 edges. So you're looking for what values of n have 3*n - 6 < n(n-1)/2 - 3*n + 6, which should be true if n <= 8 but not true for n >= 9.. --- Christopher Heckman > To guarantee that C(G) is non planar, it must have more than 3*n-6 > edges. > n=11 is the smallest n for which this is possible.. > If n = 10, C(G) may have no more than 21 edges and therefore, can be > planar > If n = 11, C(G) must have at least 28 edges and therefore, must be non > planar. === Subject: Re: A graph theoretic problem > You didn't include the context, bill! Don't use Google Groups's Reply > option! Click on show options, and then Reply in _the menu that > opens up_. > The OP was: > What is the minimal number n, such that for any graph G with n vertices, > either G or Complement(G) is non-planar? > We think that the answer is 9, but we're unable to prove it. > There are example with 8 vertices such that both G and Complement(G) are > planar. > G must be planar. Then G can have up to 3*n-6 edges. > C(G) can have a maximum of ((n^2)*n)/2 -3*n +6 edges. > I think you mis-typed something there. If G has at most 3*n-6 edges, > then C(G) has at least > n(n-1)/2 - 3*n + 6 edges. So you're looking for what values of n have > 3*n - 6 < n(n-1)/2 - 3*n + 6, > which should be true if n <= 8 but not true for n >= 9. If n=9, then 3*9 - 6 =21 9*(9-1)/2 - 3*9 + 6 = 72/2 - 27 + 6 = 36 - 27 + 6 = 15 21 is not less than 15. Another approach: G can have 3*n-6 edges, C(G) must have >= 3*n-5 edges to be non planar. Therefore the complete graph must have >= 6*n - 11 edges. Then ((n-1)*n)/2 >= 6*n - 11 n= 11 is the smallest value for which the inequality is true! Bill > --- Christopher Heckman > To guarantee that C(G) is non planar, it must have more than 3*n-6 > edges. > n=11 is the smallest n for which this is possible.. > If n = 10, C(G) may have no more than 21 edges and therefore, can be > planar > If n = 11, C(G) must have at least 28 edges and therefore, must be non > planar. === Subject: Re: A graph theoretic problem >What is the minimal number n, such that for any graph G with n vertices, >either G or Complement(G) is non-planar? >We think that the answer is 9, but we're unable to prove it. >There are example with 8 vertices such that both G and Complement(G) are >planar. Here's an example with vertices 1 to 9, constructed with Maple: edges of G: {3, 8}, {1, 4}, {2, 3}, {2, 6}, {3, 6}, {3, 7}, {3, 9}, {4, 7}, {4, 8}, {5, 8}, {7, 9}, {3, 5}, {4, 5}, {1, 7}, {1, 9}, {2, 4}, {2, 7}, {4, 6}, {8, 9} edges of complement(G): {2, 9}, {1, 3}, {4, 9}, {6, 7}, {1, 2}, {1, 5}, {1, 8}, {2, 5}, {5, 6}, {6, 9}, {3, 4}, {1, 6}, {2, 8}, {5, 9}, {5, 7}, {6, 8}, {7, 8} Both are planar according to isplanar: I haven't verified directly, but I assume it's right. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: A graph theoretic problem > edges of complement(G): > {2, 9}, {1, 3}, {4, 9}, {6, 7}, {1, 2}, {1, 5}, {1, 8}, {2, 5}, > {5, 6}, {6, 9}, {3, 4}, {1, 6}, {2, 8}, {5, 9}, {5, 7}, {6, 8}, {7, 8} No, this is not planar. Checked by Mathematica. Though G is indeed planar. === Subject: definition of Riemann sum (double integral) I have a question about a definition of Riemann (double) sum in some real analysis textbook *. There's some notation I can't understand when one choose a point from each cell of a partition of a rectangle. The definition starts as follows: (AXB means the cartesian product of two sets A and B) Let f be a bounded, real-valued function defined on a rectangle R = [a_1,b_1]X[a_2,b_2]. Let pi_1 be a partition of [a_1,b_1] and let pi_2 be a partition of [a_2,b_2], that is, pi_1 : a_1 = x_0 < x_1 < ... < x_p = b_1 pi_2 : a_2 = y_0 < y_1 < ... < y_q = b_2 We call [x_(i-1),x_i]X[y_(j-1),y_j] the (i,j) cell and denote it by C_ij. Now, here comes what I can't understand: For each i and j, choose a point (s_i, t_j) in C_ij and form the sum ... That phrase looks strange. I think the expression (s_i, t_j) should be replaced with (s_ij,t_ij). This notation (s_i,t_j) keeps appearing again and again throughout the corresponding chapter of the book. Maybe I am being stupid. But I think the only restriction on choice of points when forming a Riemann sum should be that each point be in its corresponding cell. What do you think? * Introduction to Mathematical Analysis - Steven A. Douglass === Subject: Re: definition of Riemann sum (double integral) > The definition starts as follows: > (AXB means the cartesian product of two sets A and B) > Let f be a bounded, real-valued function defined on a rectangle R = > [a_1,b_1]X[a_2,b_2]. > Let pi_1 be a partition of [a_1,b_1] and let pi_2 be a partition of > [a_2,b_2], that is, > pi_1 : a_1 = x_0 < x_1 < ... < x_p = b_1 > pi_2 : a_2 = y_0 < y_1 < ... < y_q = b_2 > We call [x_(i-1),x_i]X[y_(j-1),y_j] the (i,j) cell and denote it by C_ij. > Now, here comes what I can't understand: > For each i and j, choose a point (s_i, t_j) in C_ij and form the sum ... > That phrase looks strange. I think the expression (s_i, t_j) should be > replaced with (s_ij,t_ij). This notation (s_i,t_j) keeps appearing again > and again throughout the corresponding chapter of the book. > Maybe I am being stupid. But I think the only restriction on choice of > points when forming a Riemann sum should be that each point be in its > corresponding cell. What do you think? You are right. But... on the other hand, a Riemann sum is independent on a choice of sample points, so we might as well assume that all are the same - and similarly for y-coordinates of sample points in cells C_1j, C_2j, ..., C_pj. This simplifies the notation in your book a lot. Pawel Gladki === Subject: Re: definition of Riemann sum (double integral) > You are right. But... on the other hand, a Riemann sum is independent on > a choice of sample points, so we might as well assume that all > x-coordinates of sample points located in cells C_i1, C_i2, ..., C_iq > are the same - and similarly for y-coordinates of sample points in cells > C_1j, C_2j, ..., C_pj. This simplifies the notation in your book a lot. Interesting question. Is there an example of a function so that the Riemann sums converge when tags of the form (s_i, t_j) are used, but does not converge when arbitrary tags in the cells are allowed? That is, a function not integrable according to the standard definition, but integrable according to the Douglass definition. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: definition of Riemann sum (double integral) > The definition starts as follows: > > (AXB means the cartesian product of two sets A and B) > Let f be a bounded, real-valued function defined on a rectangle R = > [a_1,b_1]X[a_2,b_2]. > Let pi_1 be a partition of [a_1,b_1] and let pi_2 be a partition of > [a_2,b_2], that is, > pi_1 : a_1 = x_0 < x_1 < ... < x_p = b_1 > pi_2 : a_2 = y_0 < y_1 < ... < y_q = b_2 > We call [x_(i-1),x_i]X[y_(j-1),y_j] the (i,j) cell and denote it by C_ij. > > Now, here comes what I can't understand: > For each i and j, choose a point (s_i, t_j) in C_ij and form the sum ... > > That phrase looks strange. I think the expression (s_i, t_j) should be > replaced with (s_ij,t_ij). This notation (s_i,t_j) keeps appearing > again > and again throughout the corresponding chapter of the book. > Maybe I am being stupid. But I think the only restriction on choice of > points when forming a Riemann sum should be that each point be in its > corresponding cell. What do you think? > You are right. But... on the other hand, a Riemann sum is independent on > a choice of sample points No, it certainly is not. >, so we might as well assume that all > x-coordinates of sample points located in cells C_i1, C_i2, ..., C_iq > are the same - and similarly for y-coordinates of sample points in cells > C_1j, C_2j, ..., C_pj. This simplifies the notation in your book a lot. It simplifies the notation, but it's wrong. > Pawel Gladki === Subject: Re: definition of Riemann sum (double integral) >(AXB means the cartesian product of two sets A and B) >Let f be a bounded, real-valued function defined on a rectangle R = >[a_1,b_1]X[a_2,b_2]. >Let pi_1 be a partition of [a_1,b_1] and let pi_2 be a partition of >[a_2,b_2], that is, >pi_1 : a_1 = x_0 < x_1 < ... < x_p = b_1 >pi_2 : a_2 = y_0 < y_1 < ... < y_q = b_2 >We call [x_(i-1),x_i]X[y_(j-1),y_j] the (i,j) cell and denote it by C_ij. >Now, here comes what I can't understand: >For each i and j, choose a point (s_i, t_j) in C_ij and form the sum ... >That phrase looks strange. I think the expression (s_i, t_j) should be >replaced with (s_ij,t_ij). This notation (s_i,t_j) keeps appearing >again >and again throughout the corresponding chapter of the book. >Maybe I am being stupid. But I think the only restriction on choice of >points when forming a Riemann sum should be that each point be in its >corresponding cell. What do you think? >>You are right. But... on the other hand, a Riemann sum is independent on >>a choice of sample points > No, it certainly is not. If the integral exists, I should have added. Otherwise, I don't know what you mean. Pawel Gladki === Subject: Re: definition of Riemann sum (double integral) > You are right. But... on the other hand, a Riemann sum is independent > on a choice of sample points >> No, it certainly is not. > If the integral exists, I should have added. Otherwise, I don't know > what you mean. Ah, of course... a limit of Riemann sums is independent on a choice of you, I should be more careful :) Pawel Gladki === Subject: Re: Convergence of a function sequence in L^infty(R) >>Huh? What other definition >>of the sum would give a different result? (For a given x all >>but one term vanishes, so it's hard to see how we're going to >>come up with some other sort of sum that doesn't give the >>same answer here...) >Of course, every reasonable definition of sum will produce the same >result *here*, but perhaps that the OP is not aware of the fact that, in >general, the meaning of the expression sum_{n in Z} a_n is very much >context-dependent. >> Could be. I suppose that every time you see the expression >> sum_{-infty}^infty in a book or a paper you write the >> author informing him of this fact? >Every time I see that expression in a book or a paper I expect >to see somewhere an explanation of its meaning. And I think that >I've never failed to find it (when I've looked for it...). Even in cases where only finitely many terms are non-zero? (Or more generally when it's clear the series converges absolutely?) >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: Convergence of a function sequence in L^infty(R) >Huh? What other definition >of the sum would give a different result? (For a given x all >but one term vanishes, so it's hard to see how we're going to >come up with some other sort of sum that doesn't give the >same answer here...) >> >>Of course, every reasonable definition of sum will produce the same >>result *here*, but perhaps that the OP is not aware of the fact that, in >>general, the meaning of the expression sum_{n in Z} a_n is very much >>context-dependent. >Could be. I suppose that every time you see the expression >sum_{-infty}^infty in a book or a paper you write the >author informing him of this fact? >>Every time I see that expression in a book or a paper I expect >>to see somewhere an explanation of its meaning. And I think that >>I've never failed to find it (when I've looked for it...). > Even in cases where only finitely many terms are non-zero? > (Or more generally when it's clear the series converges > absolutely?) No. However, like I said before, perhaps that the OP is not aware of the fact that, in general, the meaning of the expression sum_{n in Z} a_n is very much context-dependent. We've had much more interesting discussions in the past, did you know? :-) Jose Carlos Santos === Subject: closure of equicontinuous family is equicontinuous Let F be a family of equicontinuous functions and F' be the same family with all pointwise limits added in i.e. f in F' if there is a sequence {f_n} in F such that f = lim f_n. Then F' is equicontinuous. how does one attack this problem? === Subject: Re: closure of equicontinuous family is equicontinuous > Let F be a family of equicontinuous functions and F' be the same family > with all pointwise limits added in i.e. f in F' if there is a sequence > {f_n} in F such that f = lim f_n. Then F' is equicontinuous. > how does one attack this problem? With epsilons and deltas. Jose Carlos Santos === Subject: Re: local measure & local cardinality >[...] >Terminology: local cardinality >Let E be a subset of R. For p in E, define the local cardinality of p >in E, denoted card_loc(p), as the limit as t-->0^+ of card(b(p,t) >intersect E). >Actually, I'm not sure if that definition makes sense. Can cardinals >even have a limit? If not, use the locally constant version ... As Robert Israel suggested, this is a decreasing family of cardinals, so its infiumum deserves to be called its limit. There is no strictly decreasing sequence of cardinals, so in fact the inf is the minimum. >Let E be a subset of R. For p in E, define the locally constant >cardinality of p in E, denoted card_loc_const(p), as a cardinal number >b (what letters are usually used for cardinals?), provided such a >cardinal exists, such that for some t0>0 and for all t with 0card(b(p,t) intersect E)=b. And hence card_loc = card_loc_const. >For the questions below, the terminology almost all will mean almost >all in the sense of cardinality -- that is, all but a set of lesser >cardinality than the full set. >Question (1'): >If E is a subset of R, must card_loc(p) exist for all p in E? Yes, just because any family of cardinals has a least element. >If not, >then for almost all p in E? If still not, then for at least one p in >Question (2'): >Assuming the appropriate yes answers to Question (1'), we have the >following analog of Question (2). >If E is a subset of R, must card_loc(p)=card(E) for almost all p in E? >If not, then for at least one p in E? E = {1,2} is a counterexample. If E is infinite I suspect that the answer is yes for almost all p in E... Yes, it's yes. [no, wait, not quite...] Say E is contained in [0,1]. Let E_0 be the intersection of E and [0,1/2] and E_1 the intersection of E with [1/2,1]. If card(E_j) < card(E) then call E_j part of A, if not then subdivide E_j into two halves E_j0 and E_j1. Continue until done. Let B be the set of all points of E that were not added to A at some point. Then card_loc = card(E) at every point of B. And A, the set of points that got discarded, is a countable union of sets of cardinality less than card(E), and hence has cardinality less than card(E). Oops, that's wrong. A countable union of sets of cardinality less then card(E) can have cardinality card(E). In fact E = {1, 1/2, 1/3,...} is an obvious counterexample. If you assume that card(E) = c and you _also_ assume the continuum hypothesis then a countable union of sets of cardinality less than card(E) must have cardinality less than card(E), and the answer becomes yes. >Question (3'): >Let E be a subset of R. Suppose card_loc(p) exists at p. Must the >defining limit for card_loc(p) be locally constant, Clearly not. > that is, if >card_loc(p) exists, must card_loc_const(p) also exist? This is not the same thing at all. >If so, it's >clear that we must have card_loc(p)=card_loc_const(p). >Question (4'): >Does there exist a subset E of R, with more than 1 point, such that >for all p in E, card_loc(p) exists and is different for each point p >of E? >Of course, for question (4'), if a yes answer is possible, then it's >necessary to abandon the Continuum Hypothesis. So no problem, you have >permission to put CH aside while answering this question, but please >put it back as it was when you're done. >quasi ************************ David C. Ullrich === Subject: Re: local measure & local cardinality > If you assume that card(E) = c and you _also_ assume the > continuum hypothesis then a countable union of sets of > cardinality less than card(E) must have cardinality less > than card(E), and the answer becomes yes. Actually, you don't need to assume the continuum hypothesis for this result. In ZFC the cardinal c has uncountable cofinality, a result that can be proved by the Konig cardinality inequality for arbitrary products and sums. (The links below are for others who might be interested, because I know you know this and just overlooked it.) sci.math thread: How bad can 2^(Aleph_0) be? history question Dave L. Renfro === Subject: Re: local measure & local cardinality >A few seemingly related questions have suggested themselves to me and >although I explained to the questions that I'm not the right person to >provide answers, the questions wouldn't go away. >So I'll pose the questions to sci.math. >First, a disclaimer: >With regard to measure theory, I've forgotten all of whatever I once >knew, and with regard to set theory, I never really knew very much in >the first place. Thus, my terminology may be non-standard or maybe not >even meaningful. I'll gladly accept corrections. >Terminology: local measure >Let E be a measurable subset of R. For p in E, define the local >measure of p in E, denoted m_loc(p), as the limit as t-->0^+ of >m(b(p,t) intersect E)/(2*t) where b(p,t)=(p-t,p+t). >Question (1): >If E is a measurable subset of R, must m_loc(p) exist for all p in E? >If not, then for almost all p in E? If still not, then for at least >one p in E? No and yes. For the no, let E be the union of {0} and the intervals [2^{-2n},2^{1-2n}] for positive integers n, and take p = 0. For t = 2^{1-2n} the quotient is 1/3, for t = 2^{-2n} it is 1/6. For the yes, see Lebesgue Density Theorem. >Question (2): >If E is a bounded, measurable subset of R such that m_loc(p) exists >for all (or almost all) p in E, must the function m_loc(p) be >integrable? If so, must the integral of m_loc(p) over E = m(E)? Yes and yes: Lebesgue Density Theorem again. >Question (3): >Let E be a measurable subset of R. Suppose m_loc(p) exists at p. Must >the defining limit for m_loc(p) be locally constant, that is, must >there exist t0>0 such that for all t with 0E)/(2*t)=m_loc(p)? No. Counterexamples are easy. >Question (4): >Does there exist a measurable subset E of R, with more than 1 point, >such that for all p in E, m_loc(p) exists and is different for each >point p of E? No. Lebesgue Density Theorem again. >Terminology: local cardinality >Let E be a subset of R. For p in E, define the local cardinality of p >in E, denoted card_loc(p), as the limit as t-->0^+ of card(b(p,t) >intersect E). >Actually, I'm not sure if that definition makes sense. Can cardinals >even have a limit? If not, use the locally constant version ... Call it infimum instead of limit. Cardinals have that. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: local measure & local cardinality >>A few seemingly related questions have suggested themselves to me and >>although I explained to the questions that I'm not the right person to >>provide answers, the questions wouldn't go away. >>So I'll pose the questions to sci.math. >>First, a disclaimer: >>With regard to measure theory, I've forgotten all of whatever I once >>knew, and with regard to set theory, I never really knew very much in >>the first place. Thus, my terminology may be non-standard or maybe not >>even meaningful. I'll gladly accept corrections. >>Terminology: local measure >>Let E be a measurable subset of R. For p in E, define the local >>measure of p in E, denoted m_loc(p), as the limit as t-->0^+ of >>m(b(p,t) intersect E)/(2*t) where b(p,t)=(p-t,p+t). >>Question (1): >>If E is a measurable subset of R, must m_loc(p) exist for all p in E? >>If not, then for almost all p in E? If still not, then for at least >>one p in E? >No and yes. >For the no, let E be the union of {0} and the intervals [2^{-2n},2^{1-2n}] >for positive integers n, and take p = 0. For t = 2^{1-2n} the quotient >is 1/3, for t = 2^{-2n} it is 1/6. >For the yes, see Lebesgue Density Theorem. >>Question (2): >>If E is a bounded, measurable subset of R such that m_loc(p) exists >>for all (or almost all) p in E, must the function m_loc(p) be >>integrable? If so, must the integral of m_loc(p) over E = m(E)? >Yes and yes: Lebesgue Density Theorem again. >>Question (3): >>Let E be a measurable subset of R. Suppose m_loc(p) exists at p. Must >>the defining limit for m_loc(p) be locally constant, that is, must >>there exist t0>0 such that for all t with 0>E)/(2*t)=m_loc(p)? >No. Counterexamples are easy. >>Question (4): >>Does there exist a measurable subset E of R, with more than 1 point, >>such that for all p in E, m_loc(p) exists and is different for each >>point p of E? >No. Lebesgue Density Theorem again. I just looked the Lebesgue Density Theorem, and I recognize it as something I once knew. It's a fascinating theorem. Apparently the Lebesgue Density Theorem pretty much clinches all the questions I asked about what I called local measure and which I now see should have been called density. So a natural question is -- is there an analog of the Lebesgue density theorem for the concept of cardinality? If so, it would probably clinch the answers to the questions I asked about local cardinality. quasi === Subject: Re: local measure & local cardinality > So a natural question is -- is there an analog of the > Lebesgue density theorem for the concept of cardinality? > If so, it would probably clinch the answers to the > questions I asked about local cardinality. There is an interesting result by Maurice Machover that I thought I'd posted about at one time or another, but I can't I did learn that it's available on the internet (for everyone, and not just those with certain university access). Maurice Machover, An independent statement about metric spaces, Notre Dame Journal of Formal Logic 17 (1976), 131-134. [MR 55 #5442; Zbl 305.02070] http://www.emis.de/cgi-bin/MATH-item?0305.02070 http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.ndjfl/1093887431 In a metric space can the points near some point x pack close to each other with ever-increasing density (in the sense of cardinality or power) as x is approached, or must it always be the case that this density reaches a maximum at a certain distance from x and does not increase for smaller distances? We give a precise definition of this density concept and show that the former case can happen (for a space of cardinality Aleph_omega) but that the question as to whether it can happen in a space of power less than or equal to that of the continuum cannot be answered. Our results are based on some of the recent independence results in set theory. In what follows I've changed notation slightly for posting purposes and as a result of my notational tastes. Let (X,d) be a metric space, E be a subset of X, and x in X. Given delta > 0, B(x, delta) denotes the delta-ball of all points in X that are a distance less than delta from x. Less than or equal to is denoted by leq and the set difference operation will be denoted by a minus sign (-). The cardinality of a set A is denoted by card(A). Definition: The E-packing power near x is P_E(x) = sup{b leq card(X): there exists epsilon > 0 such that for all 0 < delta_1 < delta_2 < epsilon we have card[(B(x, delta_2) - B(x, delta_1)) intersect E] is greater than or equal to b}. The idea is to look at the maximum cardinality in shells (or generalized annuli) as you approach x. As Machover says, We remark that P_E(x) measures how close (in the sense of cardinality) to _each_other_ the points near x are packed, not how closely they pack about x itself. Dave L. Renfro === Subject: Error estimation of Gauss-Kronrod method for double integrals I need to implement the method of Gauss-Kronrod for computation of double integral with a predefined tolerance. But I have no info about the error of the numerical integration of double integrals and conditions the integrand should satisfy for these error estimations (continuity of a function and/or derivatives along X and Y etc). They say that there were no such investigations, but I suspect that this is an old-dated information... === Subject: 2-dimensional subspaces of a 5-dimensional vector space over F_2. Need a help in the following problem: Let V be a 5-dimensional vector space over finite field F_2={0,1}, with standerd basis {e_1,e_2,..,e_5}, where e_i=(0,0,..,1,..,0) , 1 is at ith position. I want the list of all 2-dimensional subspaces of V over F_2. I know that this number is 155 (using Gaussian binomial ,we get this no). Arun === Subject: Re: 2-dimensional subspaces of a 5-dimensional vector space over F_2. > Need a help in the following problem: > Let V be a 5-dimensional vector space over finite field F_2={0,1}, with > standerd basis {e_1,e_2,..,e_5}, where e_i=(0,0,..,1,..,0) , 1 is at > ith position. I want the list of all 2-dimensional subspaces of V over > F_2. I know that this number is 155 (using Gaussian binomial ,we get > this no). > Arun All the 2-d subspaces appear as row spaces of 2 by 5 matrices with entries in F_2. As any matrix shares its row space of a unique matrix in the reduced row echelon form, you will get the spaces without replication simply by listing all the binary matrices that are in reduced row echelon form. There are ('*' will denote an unknown binary entry): A) 64 matrices of the form (1 0 * * *) (0 1 * * *), B) 32 matrices of the form (1 * 0 * *) (0 0 1 * *), C) 16 matrices of the form (1 * * 0 *) (0 0 0 1 *), ... J) 1 matrix of the form (0 0 0 1 0) (0 0 0 0 1). Why don't you fill in the missing parts! In this case the row space of a matrix consists simply of the all zero vector, the two rows, and their sum. Jyrki === Subject: A triangle center family The Torricelli point T has ATB=pi/3,BTC=pi/3,CTA=pi/3. The orthocenter H has AHB=pi-y,BHC=pi-@,CHA=pi-¤. The incenter I has AIB=(pi+y)/2,BHC=(pi+@)/2,CHA=(pi+¤)/2. The circumcenter U has AUB=2y,BUC=2@,CUA=2¤. The general center of this family G evidently has AGB=(2-l)*pi/3+l*y,... with l a parameter. What's the locus of all G(l)? (And since the above centers are l=0,-1,1/2,2, do l=-1/2,1,-2 have names? Kimberling number suffices :-) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: I don't know what the question asks...(express exp(tA) in polynomial in matrix.. HI there, I am confused about this, Express e^(tA) in a polynomial in A if 1 0 3 A=(8 1 -1) 5 1 -1 Obviousely, we can easily express exp(tA) from Taylor series, e^(tA) = I + tA/1 + (tA)^2/2! + (tA)^3/3! + .... But I am confused why the question give you munbers of matrix A. I do this is because the question asks us express that in a polynomial in A, So, is that correct answer? === Subject: Re: I don't know what the question asks...(express exp(tA) in polynomial in matrix.. > HI there, I am confused about this, > Express e^(tA) in a polynomial in A if > 1 0 3 > A=(8 1 -1) > 5 1 -1 > Obviousely, we can easily express exp(tA) from Taylor series, > e^(tA) = I + tA/1 + (tA)^2/2! + (tA)^3/3! + .... the key is that the matrix satisfies its own characteristic equation (that's the one you write to find eigenvalues). in this case, it means that A^3 can be written as a linear combination of A^0 = I, A, and A^2. then A^4 = A A^3 can be written as a combination of powers up to A^3, which can be rewritten as powers up to A^2. the infinite series, then, collapses to 3 terms: a0 I + a1 A + a2 A^2, and they want you to find a0, a1, a2. i hope this helps. vale, rip -- NB eddress is r i p 1 AT c o m c a s t DOT n e t === Subject: Express exp(tA) as a polynomial in A ...... HI there, I am confused about this, Express e^(tA) as a polynomial in A if 1 0 3 A=(8 1 -1) 5 1 -1 Obviousely, we can easily express exp(tA) from Taylor series, e^(tA) = I + tA/1 + (tA)^2/2! + (tA)^3/3! + .... But I am confused why the question give you munbers of matrix A. I do this is because the question asks us express that as a polynomial in A, So, is that correct answer? === Subject: Re: Express exp(tA) as a polynomial in A ...... > HI there, I am confused about this, > Express e^(tA) as a polynomial in A if > 1 0 3 > A=(8 1 -1) > 5 1 -1 > Obviousely, we can easily express exp(tA) from Taylor series, > e^(tA) = I + tA/1 + (tA)^2/2! + (tA)^3/3! + .... Do you know about the characteristic polynomial of a matrix? If you don't, you can Google it to find out, or look in a book. Anyway, the characteristic polynomial P(x) of an nxn matrix is of degree n, with leading term x^n. It is a theorem that any matrix satisfies its own characteristic equation, which means that if you substitute the matrix x in place of the variable x, you get P(A) = A^n - c_1 A^(n-1) - ... - c_n I = 0, where I = identity matrix. Therefore, in your problem, you can write A^3 = c_1 A^2 + c_2 A + c_3 I, with easily-computed coefficients c_i. Then A^4 = A(A^3) = c_1 A^3 + c_2 A^2 + c_3 A = c_1 (c_1 A^2 + c_2 A + c_3 I) + c_2 A^2 + c_3 A = polynomial of degree 2 in A, etc. In this way, the whole series can, in principle, be expressed as a 2nd degree polynomial in A. However, expressing the coefficients of A^2, A and I will be difficult. R.G. Vickson > But I am confused why the question give you munbers of matrix A. > I do this is because the question asks us express that as a polynomial > in A, > So, is that correct answer? === Subject: Re: Express exp(tA) as a polynomial in A ...... > HI there, I am confused about this, > Express e^(tA) as a polynomial in A if > 1 0 3 > A=(8 1 -1) > 5 1 -1 > Obviousely, we can easily express exp(tA) from Taylor series, > e^(tA) = I + tA/1 + (tA)^2/2! + (tA)^3/3! + .... > But I am confused why the question give you munbers of matrix A. > I do this is because the question asks us express that as a polynomial > in A, > So, is that correct answer? No. A polynomial in A is a finite linear combination of powers of A, not an infinite linear combination as you have. === Subject: Re: Express exp(tA) as a polynomial in A ...... > HI there, I am confused about this, > > > Express e^(tA) as a polynomial in A if > > > 1 0 3 > A=(8 1 -1) > 5 1 -1 > Obviousely, we can easily express exp(tA) from Taylor series, > > > e^(tA) = I + tA/1 + (tA)^2/2! + (tA)^3/3! + .... > > > But I am confused why the question give you munbers of matrix A. > If A is diagonizable (for example if it has all distinct real eigenvalues) then A = inv(B)*D*B for some invertable B and diagonal D, with D = [[d_ij]], d_ii are eigenvalues and d_ij = 0 for i <> j. . Then e^(t*A) = inv(B)*e^(t*D)*B and e^(t*D) is a diagonal matrix with diagonal entries e^(t^d_ii), and other etnries zero. If A is not diagonizable, one can still do something with the Jordan decomposition of the matrix A, q.v. > > I do this is because the question asks us express that as a polynomial > > in A, > > > So, is that correct answer? > No. A polynomial in A is a finite linear combination of > powers of A, not an infinite linear combination > as you have. === Subject: Re: Express exp(tA) as a polynomial in A ...... >HI there, I am confused about this, >Express e^(tA) as a polynomial in A if > 1 0 3 >A=(8 1 -1) > 5 1 -1 Hint: The answer is a quadratic, because A satisfies a cubic. Interpolate: P(r) = exp(tr) for each eigenvalue r of A. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Express exp(tA) as a polynomial in A ...... On 14 Nov 2005 08:03:37 -0800, Harry901.tw >HI there, I am confused about this, >Express e^(tA) as a polynomial in A if > 1 0 3 >A=(8 1 -1) > 5 1 -1 >Obviousely, we can easily express exp(tA) from Taylor series, >e^(tA) = I + tA/1 + (tA)^2/2! + (tA)^3/3! + .... >But I am confused why the question give you munbers of matrix A. >I do this is because the question asks us express that as a polynomial >in A, >So, is that correct answer? The following is true: Let A be an n x n matrix over R. Then every convergent power series in A converges to a polynomial in A. proof: Let A be an n x n matrix over R where R denotes the field of real numbers. Suppose there is a power series in A which converges to the matrix B. We want to show that B is a polynomial in A. Let M be the set of all n x n matrices over R, and let R[A] be the set of polynomials in the matrix A. As a topological space, M is isomorphic to R^2n. Any vector subspace of M is a closed subset of M, hence, since R[A] is vector subspace of M, R[A] is a closed subset of M. The partial sums of the power series are in R[A], and they converge to B, hence, since R[A] is closed, B must be in R[A], as claimed. I haven't shown you how to actually find the polynomial in A that equals B -- I don't have time right now -- but at least you know that such a polynomial exists. quasi === Subject: Geometrodynamic monopole makes Bekenstein-Hawking BITS On the other hand, if the initial Planck scale vacuum phase transition has a degenerate vacuum manifold V(Planck) = G(False Vacuum)/H(GUT) ~ S2 rather has 2D point monopole defects where the quantized 1D winding numbers are replaced by 2D wrapping numbers. That is, we have 2 alternative models for the initial emergence of Einstein's gravity at 10^-43 sec. I. V(Planck) ~ S1 circle Here there is only ONE Goldstone phase THETA with 1D string defects where the Higgs magnitude of the vacuum order parameter vanishes along the string and the Goldstone phase THETA is undefined. This is the usual easy to visualize Mexican Hat Potential for renormalizable quantum field theories that spontaneously break internal symmetries. Here the local virtual vacuum order parameter is a single complex scalar field like in superfluid helium and the BCS superconductor for real bosons on-mass-shell. Stable quantized circulation comes from the first homotopy group being non-trivial, i.e. PI1(S1) = Z (integers) B(curved) = Lp'd'THETA 1-form curved tetrad of Einstein's GR The geometrodynamic vorticity flux dB(curved) is quantized through areas including the string (vortex) defect. This is an example of Bohm-Aharonov Flux without flux because dB = 0 on the loop L in physical space that surrounds the singularity with the Higgs magnitude vanishes and the Goldstone phase is undefined. Going around the loop one in physical space means going around the S1 circle that is the vacuum manifold N times. N is the number of trapped geometrodynamics flux quanta. This is not to be confounded with magnetic flux. II. V(Planck) ~ S2 spherical surface Here we obviously have TWO Goldstone phases THETA and PHI with point defects. How do we construct B(curved)? Here the local vacuum order parameter is 3 real scalar fields like the 3x3 adjoint irreducible representation of the SU(2) group corresponding to spin 1 rather than spin 1/2. However, the real non-vanishing vacuum order parameter can be chosen to lie completely along the arbitrarily chosen polar axis of S2 = G(False Vacuum)/H(GUT) so that the complementary Goldstone phase is the azimuthal angle PHI in the equatorial plane in this fiber space of degenerate order parameters beyond space-time. i.e. B(curved) = Lp'd'PHI 1-form Here its the second homotopy group that is non-trivial PI2(S2) = Z The curvature 2-form with physical dimension 1/(area) is R = DW where W is the spin connection 1-form W = -*[dB/(1 - B)] Therefore, the dimensionless DeRham integral of R over a surrounding S2 surface in physical space enclosing the point defect is quantized even though the Bianchi identity DR = 0 everywhere other than the point defect singularity. That is closed surface integral of the curvature 2-form R ~ Z The wrapping number around V = G(False Vacuum)/H(GUT) = S2 is the number of Bekenstein bits equal to the number of curvature quanta through the closed surface. This argument implies that there are no Bekenstein bits in globally flat spacetime. Note that the curvature 2-form maps to the 4th rank Riemann-Christoffel curvature tensor i.e. R^ab = R^abuvdx^udx^v R^wluv = R^abuvea^we^bl That is, the Bekenstein BITs require B(curved) =/= 0. Therefore, quantization of area has the same status as quantization of circulation in a superfluid. Both point to an underlying ground state order parameter. Geometrodynamics is an emergent local gauge theory of T4 ---> Diff(4) from the spontaneous breaking of an internal symmetry at the Planck era. We only need ONE point defect geometrodynamic monopole in the universe to produce the Bekenstein BIT exactly like we only need 1 magnetic monopole in the universe at the end of the Dirac string to produce quantization of electric charge. Note two meanings of inertia. Gravity as a force is locally indistinguishable from an inertial force in a non-inertial local frame that is either rotating or on a time-like nongeodesic path or both with respect to the Levi-Civita connection. Inertia as the rest mass of quarks, leptons and W-mesons. Our universe is thought to have formed in a sequence of spontaneous symmetry breakings of the physical vacuum. Assuming the old Heisenberg uncertainty principle Planck scale breaking ~ 10^19 Gev ~ 10^-33 cm GUT scale breaking ~ 10^15 Gev ~ 10^-29 cm Electroweak breaking ~ 250 Gev ~ 10^-16 cm In more detail Time (s) T (Kelvin) Reference Event CBR THE QUANTUM GRAVITY ERA 1x10^-43 1x10^32 KT 72 Quantum limit of classical general relativity [My comment 1: I postulate, as there is no fundamental theory of this as yet, the simplest possibility G(Planck)---> H(Planck) V (Planck)= G(Planck)/H(Planck) ~ S1(Planck) from first homotopy group PI1(S1) = Z (integer winding numbers) i.e. line defects i.e. 1-dim strings Vacuum coherent order parameter is a single complex scalar field PSI(Planck) = |PSI|e^iargPSI Planck-scale Higgs amplitude is |PSI|, Planck-scale Goldstone phase is argPSI. argPSI is a 0-form Curved space-time tetrad 1-form is B = Lp'd'argPSI ds^2 = (1^a + B^a)(Minkowski)ab(1^b + B^b) is EEP B is from local gauging of SPACE-TIME SYMMETRY T4 to Diff(4) AND ALSO from spontaneous breaking of INTERNAL SYMMETRY G(Planck) -> H(Planck) to S1 Post-Planck Vacuum Manifold PSI is the INFLATION FIELD Diff(4) covariant exterior derivative is D = d + W/ W^ac = T4 spin-connection The vanishing torsion 2-form in 1915 GR model is T^a = De^a = de^a + W^ac/(1 + B)^c = dB^a + W^ac/(1 + B)^c = 0 Invert this equation to get the 1-form spin connection W^a^c = -*[dB^a/(1 - B)^c] * is Hodge dual based on Minkowski metric in tangent fiber space. The geodesic deviation curvature 2-form is R^ab = DW^ab = dW^ab + W^acW^cb The Bianchi identity is the 3-form equation (analog to dF(EM) = 0 3-form) DR^ab = 0 The Einstein-Hilbert vacuum Lagrangian density is the 0-form *[R^ab/e^c/e^d] The Einstein curvature 4-form is G^ab = D*W^ab The Einstein 4-form source equation (analog to d*F(EM) = *j(electric) 3-form) is G^ab = *J(T4)^ab Local conservation of stress-energy source current densities is the vanishing 5-form DG^ab = D*J(T4)^ab = 0 5-form in 4-space is identically zero. G^ab = *J(T4)^ab is 4-form because of equivalence principle where even zero point energy with w = -1 gravitates absolutely and not relatively.] THE INFLATION AND SYMMETRY BREAKING ERA 1x10^-38 1x10^29 KT 72 Limit of perturbative interaction thermalization of universe 1x10^-35 1x10^28 KT 72 Grand unification spontaneous symmetry breaking [My Comment 2 OK, here H(Planck) ---> H(GUT) = U(1)hyperchargeSU(2)weak SU(3)quark coupling constant splits off from U(1)hyperchargeSU(2)weak with same coupling constant We do not know the topology of the new vacuum manifold H(Planck)/U(1)hyperchargeSU(2)weak = V(GUT) = ? Note the sequence: G(Planck) ---> H(Planck) ---> H(GUT) = U(1)hyperchargeSU(2)weak G/(Planck)/H(Planck) ---> H(Planck)/H(GUT) i.e. V(Planck) ---> V(GUT) But what does this mean in terms of the vacuum order parameter? V(Planck) does not disappear. It is robust from Goldstone phase rigidity and it is the Hercules supporting The World Geometrodynamic Field V(Planck) Indeed the total vacuum manifold has grown more complex into the TENSOR PRODUCT V(Planck)xV(GUT) 1x10^-34 1x10^27 KT 274 Approximate start of inflation 1x10^-32 1x10^27 KT 274 Approximate start of reheating and end of inflation 1x10^-11 3x10^15 KT 72 Electroweak unification spontaneous symmetry breaking [My Comment 3 Here H(GUT) = U(1)hyperchargeSU(2)weak --3x3 adjoint irrep--> U(1)em V(Electroweak) = U(1)hyperchargeSU(2)weak/U(1)em ~ S^2 with point defects. W+,W0, W- massive A(photon) massless, 1 Higgs & masses of leptons & quarks from Yukawa couplings. Prior to that 250 Gev scale all quantum fields are massless. There is gravity already emergent from Planck era. It does not have to be put in by hand adhoc anymore. You can have curved spacetime without massive principle. Now the vacuum manifold is V(Planck)xV(GUT)xV(Electroweak) Although the short-scale vacuum order parameter may be a single complex scalar field with an S1 vacuum manifold, there is reason to believe that the larger-scale vacuum order parameter after the electroweak splitting of the weak SU(2) charge from the U(1) electromagnetic charge is essentially of the character of isovector, i.e. 3x3 adjoint irrep of SU(2) with effective vacuum manifold S2 hence geometrodynamic point defects not to be confused with magnetic monopoles. Having geometrodynamic point defects instantly gives the Hawking-Bekenstein area quantization because of the second homotopy group PI2(S2) = Z giving stable 2D wrapping numbers instead of the stable 1D winding numbers. The 3-adjoint irrep of SU(2) is 3 real scalar fields of which in one Higgs model only one scalar field develops a non-vanishing VEV Higgs amplitude. We can imagine its conjugate Goldstone phase as the rotation about it as an axis in V(Electroweak) ~ S2 i.e. the azimuthal angle about the z axis chosen so that the order parameter is electrically neutral. We then must assume that this is the same Goldstone phase that survived from the Planck era. These speculations are of course very new.] THE QUARK-LEPTON ERA 2x10^-7 2x10^13 HA 353 Tauon anti-tauon annihilation 1x10^-5 2x10^12 KT 72 Formation of hadrons from quarks 7x10^-5 1x10^12 HA 353 Muon anti-muon annihilation 5x10^-4 4x10^11 KT 159, 281 By this time the universe has a baryon-antibaryon asymmetry which results from post-inflationary B,C,CP violating processes 1x10^-1 3x10^10 BS Neutral current weak interactions become too slow and neutrinos decouple 1x10^0 1x10^10 BS Charged current weak interactions become too slow and neutron to proton ratio freezes out 1x10^1 5x10^9 BS Electron positron annihilation THE RADIATION ERA 1x10^2 1x10^9 BS Typical photon energies drop below the deuteron binding energy and nucleosynthesis begins barrier energies and nucleosynthesis ends 4x10^10 6x10^4 KT 77 Matter density becomes equal to radiation density THE MATTER ERA 4x10^12 3.5x10^3 KT 78 Electrons and protons recombine into hydrogen atoms 6x10^12 3.0x10^3 KT 80 Photon decoupling 2x10^16 20 HA 349 Formation of galaxies (X-1.4x10^17) --- KT 12 Formation of the solar system (X-9.9x10^16) --- HA 390 Emergence of life on Earth X ~ 4.7x10^17 2.726 KT 12 Today These values are rough estimates only. I have taken Omega_0=1, Lambda_0=0, and h=1. http://www.talkorigins.org/origins/cosmo_timeline.html === Subject: Re: Finding areas of eight sectors within a circle >I want to find the area of eight sectors within a circle. >Imagine that I place a point randomly within the circle. The point is >not placed at the center of the circle. I then draw four lines through >the point, dividing the circle into eight unequal sections, each one >with an angle of 45 degrees. >I am trying to find the area of these sectors. I realize that a true >sector is based at the center of the circle. I am not sure what the >correct name for these pieces is. >Can someone help me find the areas, and possibly suggest a name for the >sections? This problem is ordinarily described using the term pizza slices, where the chef can't seem to get the cuts to go through the center but he somehow manages to make them all concurrent at some other point, with all angles exactly equal. Maybe he's using a cookie cutter to cut the pizza. http://www.math.hmc.edu/funfacts/ffiles/10001.2.shtml Pizza slice doesn't sound very mathematical, though. The shape is a union of a triangle and a segment when the segment's chord is one side of the triangle... but that's too wordy. If I were writing about it formally, I'd probably define the term off-center sector and use that. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Finding areas of eight sectors within a circle A = Area of any eccentric sector s = arc length between two points of the circle. a = radius of circle P is any fixed point in the circle r = radius vector length ( from P to any point on circle ) ecc = eccentricity between circle center and P To get A, integrate from: dA / ds = ( 2 * r ^ 2 - ecc ^ 2 )/( 4 * a ) === Subject: Re: Finding areas of eight sectors within a circle > A = Area of any eccentric sector > s = arc length between two points of the circle. > a = radius of circle > P is any fixed point in the circle > r = radius vector length ( from P to any point on circle ) > ecc = eccentricity between circle center and P > To get A, integrate from: > dA / ds = ( 2 * r ^ 2 - ecc ^ 2 )/( 4 * a ) ;-) first you have to find r as a function of s. then the limits s0, s1 for integration, s1 as a function of angle *from P* = 45 deg and initial vector (s0). I wonder if you could succeed in this process... If you really want to calculate A, the best is (using your notations, but ecc=p to simplify writing) let t = angle (OP, OM) of the vector from origin law of cosine gives : a^2 = r^2 + p^2 - 2*r*p*cos(t) solve for r = f(t) then A = 1/2 int(t=t0, t=t0+45Á, f(t)^2 dt) But this is not analytically solvable with elementary functions, needs at least elliptic ones... Luckily, the purpose of Diana was not to calculate each area individually but to proove pizza theorem, hence was just interested in *sum* of some areas... -- philippe mail : chephip at free dot fr site : http://chephip.free.fr/ === Subject: Re: Another word for 'intersection of two circles'? <111120050629097149%edgar@math.ohio-state.edu.invalid> <3tjjjdFsuhm5U1@individual.net> I remember an incident from High School. The annual MAA exam, given to high school students asked the question: I am surprised by the lack of responses. What? No flames??? No You were rude to a teacher comments? There was some more interested conversation that took place. I will relate it if there is any interest... === Subject: Re: Another word for 'intersection of two circles'? >I remember an incident from High School. The annual MAA exam, given >high school students asked the question: > >I am surprised by the lack of responses. What? No flames??? >No You were rude to a teacher comments? OK, if you insist: it is rude, and generally counter-productive, to call a teacher an idiot who had no business teaching mathematics, even if that happens to be true. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Another word for 'intersection of two circles'? <3tjjjdFsuhm5U1@individual.net> OK, if you insist: it is rude, and generally counter-productive, to call a teacher an idiot who had no business teaching mathematics, even if that happens to be true. I agree. However, I had asked the teacher to check with the head of the department. He refused, saying that he had been given authority to grade the exams and he did not need to check with anyone else. This teacher was new, and did not know me personally. However, the fact that this was the only wrong answer, it might have suggested that perhaps I knew what I was saying...... How does one deal with such knee-jerk authoritarians??? By refusing to consider what I had to say, and refusing to check with anyone else, the *teacher* was also being rude. === Subject: Google Inc! Protest for humanity By Google archiving usenet, Google turns the international territory of usenet into official business ground. Google suggests: *Be careful how you do business on usenet*. The logic works for businessmen in general. But what that translates to for the genral public, is really what Ceaucescu wanted with his street cameras: a business ground, the core formula of dictatorships, or in other words: unilateralism and crimes against humanity at large in the 21st century. === Subject: x^3-x = 6y^3 On Sept 30 the following question was posted to Sci. Math Ricardo Alfaro Sep 30, 8:14 am author === Subject: x^3-x=6*y^3 Has anyone reference to this Diophantine equation? or a proof that it has only the trivial solutions (0,0),(1,0),(-1,0)? I shall give a fairly complete proof that the only integer solutions of this equation are (x,y) = (-2,-1), (-1,0), (0,0), (1,0) and (2,1). Before I start let me state that this problem is solved in a paper by Michael Bennett (J. Austral. Math. Soc. (Series A2), vol. 62(1997), pp. 342 and 343.) I shall give the essentials of Bennett's proof here and then present an alternate solution, using algebraic number theory. I. Reduction to binomial cubic Thue equations. Write the equation x^3 -x = 6y^3 (1) as x^3 - 6y^3 = x Clearly, (0,0) is a solution here. We will find all the other integer solutions. Let (x, y) = d, x = da, y = db, (a,b) = 1. Then d^2(a^3 - 6b^3) = a Now let a^3 - 6b^3 = c Then (b,c) = 1, a = cd^2 and c^3 d^6 - c = 6b^3. Since (b,c) = 1, this implies c | 6. So the problem reduces to solving a^3 - 6b^3 = pm 1, pm 2, pm 3 and pm 6. (2) II. Bennett's solution If b = 0, then the only solutions of (2) are (-1,0) and (1,0). Suppose next that b = pm 1. By direct substitution we find that the only solutions of (2) in this case are (a,b) = (-2, -1), (0, -1), (0,1) and (2,1). So it suffices to show there are no solutions with |b| > 1 Next, the system of equations (2) is equivalent to the Diophantine inequality | a^3 - 6b^3| <= 6 (3) with a^3 - 6b^3 ne pm 4 or pm 5. Let t = 6^(1.3) and factor the LHS of 3. Then |(t - a/b)| |(a^2/b^2) + a/b * t + t^2| <=6/ |b| ^3. (4) Now since |b| >=2, the rhs of (4) is less than 1, while the second factor of the lhs is at least 3. Thus |t - a/b| < 6/ |b|^3 But Bennett proves, using Pade approximations and the continued fraction expansion of t, that |t - a/b| > b^{-2.35}. provided |a| ne 465 and |b| ne 257. (This does not hold here since then |a^3- 6b^3| would equal 5.) Thus |b|^ 0.65 < 6, which implies |b| <= 15. Checking the range 2 <= | b| <= 15, we find no further solutions. III. Solution, using algebraic number theory. Let's tackle the cases 2 at a time. First, a^3 - 6b^3 = pm 3 Implies a^3 - 6b^3 = pm 3(mod 7), which is impossible. For a^3 - 6b^3 = pm 1, we can use Nagell's theorem: The Diophantine equation Ax^3 + By^3 =1 has at most one solution in nonzero integers. If A = 1, this solution must come from the fundamental unit of the field Q(B ^ (1/3)) or its square. The latter happens only when B = 19, 20 or 28. Since the fundamental unit of Q(t) is eps = 1 - 6t + 3t^2, the only solutions in this case are (-1, 0) and (1,0). Now let a^3 - 6b^3 = pm 2. Then a is even so we let a = 2u and get 4u^3 - 3b^3 = pm 1, Again by Nagell's theorem, the only solutions of these equations are (-1, -1) and (1,1), so the only solutions in this case are (-2, 1) and (2,1). There remains the case a^3 - 6b^3 = pm 6. Here a = 6u, then 36u^3 - b^3 = pm 1 Now we can use the following theorem of Hemer: Let K(t) be a cubic field over Q and let alpha = At^2 + Bt + C be an integer in the ring (1, t, t^2). Suppose A = B = 0(mod p^k), where p is an odd rational prime and (alpha, p) = 1. Further, suppose that tA + sB is not congruent to 0 mod (p^{2k}), where t, s, k are rational integers and k > 0. Then if alpha ^ n = A_nt^2 + B_n t + C_n, tA_n + sB_n is never 0 for any n ne 0. In the case at hand,we have to solve eps^n = a_n + c_n t^2, i.e., b_n = 0. But b_n = 3(mod 9), so the only solution occurs for n = 0. Thus the only solutions in this case are (a,b) = (0, -1) and (0,1). Finally, working back through the various changes of variable, we find that (1) has exactly the solutions stated. Sorry if any of this seems a bit awkward. Ray Steiner === Subject: Re: question about non-uniform sampling? > Hi all, > Can non-uniform sampled signal be used to perfectly reconstruct the > original continuous time signal? > What is the Nyquist sampling rate in the non-uniform case? > -L This was a popular problem in the 1960's. A fairly thorough presentation is made in the book Discrete Time Systems - An Introduction to the Theory by Herbert Freeman, Wiley and Sons 1965; section 4.8 - Nonuniform Sampling. The main result for this issue is : a function f(t), bandlimited to -B/2<=w<=B/2 (where B=2pi/T) can be uniquely reconstructed from a set of samples which are nonuniformly spaced but satisfy the condition that there be precisely N distinct samples to every interval of length NT, where N is some finite integer . The actual reconstruction is shown in the book and is straight forward; but really ugly. This may be a more severe condition that it seems at first; especially the precisely N distinct ... . There are probably better results since this book was published forty years ago. MzF === Subject: Re: question about non-uniform sampling? > Can non-uniform sampled signal be used to perfectly reconstruct the > original continuous time signal? Interesting homework question old man. The answer is probably no. All the sample points could end up at the same point in time without further specification on the non-uniformity. > What is the Nyquist sampling rate in the non-uniform case? DC or infinity? YMMV. -- rhn === Subject: Re: question about non-uniform sampling? A specific, and limited, example is when you do have a reference that indicates where in its cycle the signal is, regardless of time. For instance with rotating machinery, you might have a reference signal that tells you each time one of the shafts rotates to a given position. Then, you can use that to resample the (time) signals you measured, so that they are evenly spaced with respect to the rotation (usually within one cycle of some part of the machinery). This then lets you enforce that your samples always happen to sample complete cycles of the rotation, and that gives you the happy effect that the signal is then ideally sampled. (This corresponds to a case that I describe on our web site: http://www.bores.com/courses/intro/freq/3_exact.htm ) This technique was called 'order processing' and was developed and publicised by Hewlett Packard some years ago - I don't know if references are still available or if the method is widely used still. Chris Chris Bore BORES Signal Processing www.bores.com === Subject: Re: question about non-uniform sampling? > A specific, and limited, example is when you do have a reference that > indicates where in its cycle the signal is, regardless of time. > For instance with rotating machinery, you might have a reference signal > that tells you each time one of the shafts rotates to a given position. > Then, you can use that to resample the (time) signals you measured, so > that they are evenly spaced with respect to the rotation (usually > within one cycle of some part of the machinery). This then lets you > enforce that your samples always happen to sample complete cycles of > the rotation, and that gives you the happy effect that the signal is > then ideally sampled. (This corresponds to a case that I describe on > our web site: http://www.bores.com/courses/intro/freq/3_exact.htm ) > This technique was called 'order processing' and was developed and > publicised by Hewlett Packard some years ago - I don't know if > references are still available or if the method is widely used still. That amounts to sampling that's uniform in space, rather than in time. The results have spatial significance, rather than temporal. It's so standard in image processing that we don't even think about it. Jerry -- Engineering is the art of making what you want from things you can get. żżżżż żżż[OS lash]żżżż[DownQuest ion]żżż żżżżż żżż[OSl ash]żżżż[DownQuesti on]żżż żżżżż żżż[OSl ash]żżżż[DownQuesti on]żżż żżżżż żżż[OSl ash]żżżż[DownQuesti on]żżż żżż === Subject: a series, convergent, divergent, oscillating A series can be convergent; summation is finite a series can be divergent, summation tends to either infinity (is this a correct definition for divergence) a series is oscillating; (-1)^n, the sum is neither convergent or divergent. and is there any other form a series can take besides these three... ta === Subject: Re: a series, convergent, divergent, oscillating > A series can be convergent; summation is finite > a series can be divergent, summation tends to either infinity (is this > a correct definition for divergence) > a series is oscillating; (-1)^n, the sum is neither convergent or > divergent. > and is there any other form a series can take besides these three... > ta I can't locate the book right now, but in The Fractal Geometry of Nature I recall Mandelbrot using the expression converges to infinity. That's the only author I've seen do so, however. Wish I could find his book, to verify the memory. Other than that one instance, a series diverges if it does not converge. [Yeah, I asked my instructors your same question, long ago, expecting a (converges, diverges, other} answer as you seem wanting.] Lynn Killingbeck === Subject: Re: a series, convergent, divergent, oscillating >> A series can be convergent; summation is finite >> a series can be divergent, summation tends to either infinity (is this >> a correct definition for divergence) >> a series is oscillating; (-1)^n, the sum is neither convergent or >> divergent. >> and is there any other form a series can take besides these three... >> ta > I can't locate the book right now, but in The Fractal Geometry > of Nature I recall Mandelbrot using the expression converges to > infinity. That's the only author I've seen do so, however. Wish I > could find his book, to verify the memory. Other than that one > instance, a series diverges if it does not converge. [Yeah, I > asked my instructors your same question, long ago, expecting a > (converges, diverges, other} answer as you seem wanting.] > Lynn Killingbeck Lynn, Here is a link to the Mandelbrot page that says converges to infinity Joe === Subject: Re: a series, convergent, divergent, oscillating > a series is oscillating; (-1)^n, the sum is > neither convergent or divergent. > and is there any other form a series can > take besides these three... my earlier post. Convergence and divergence is like an animal being alive and dead, with oscillating being what kind of animal we're looking at (or how much it weighs, or some other aspect of the animal that we might be interested in which both alive and dead animals can have). So there are just two main categories, convergence and divergence. Dave L. Renfro === Subject: Re: a series, convergent, divergent, oscillating > A series can be convergent; summation is finite > a series can be divergent, summation tends to > either infinity (is this a correct definition > for divergence) > a series is oscillating; (-1)^n, the sum is > neither convergent or divergent. > and is there any other form a series can > take besides these three... Well, a series can oscillate towards every single real number, or, more generally, towards every number in any specified closed set of real numbers. But if you're just learning about series, this is a good start. However, keep in mind that a series can be oscillating in other ways than +-+-+-+ ... For example, the terms might have other sign patterns, such as ++--++--++ ..., +--+--+--+-- ..., +-++-+++-++++- ..., etc. Also, for each of these possibilities a series may or may not converge. For example, 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... converges (by a result often called the n'th term alternating series convergence theorem). Dave L. Renfro === Subject: Re: a series, convergent, divergent, oscillating >A series can be convergent; summation is finite > a series can be divergent, summation tends to either infinity (is this a > correct definition for divergence) > a series is oscillating; (-1)^n, the sum is neither convergent or > divergent. > and is there any other form a series can take besides these three... > ta The terminology is strictly this.. A series that does not converge is divergent. There is no middle ground so Sum (-1)^n is a divergent series becasue it does not converge. Joe === Subject: Re: a series, convergent, divergent, oscillating > A series can be convergent; summation is finite > a series can be divergent, summation tends to either infinity (is this a > correct definition for divergence) No,divergence just means not convergent and includes oscillating > a series is oscillating; (-1)^n, the sum is neither convergent or divergent. if the nth term is (-1)^n b_n with b_n>0 the series is called alternating. Such a series may be convergent or divergent (either to +oo or -oo or have no limit for the partial sums. > and is there any other form a series can take besides these three... > ta Consult your local calculus book for more detail smn === Subject: Re: a series, convergent, divergent, oscillating >A series can be convergent; summation is finite > a series can be divergent, summation tends to either infinity (is this a > correct definition for divergence) No. A series that does not converge diverges, like Sum((-1)^n), which divereges but is bounded. > a series is oscillating; (-1)^n, the sum is neither convergent or > divergent. wrong. > and is there any other form a series can take besides these three... Some series converge conditionally. === Subject: monic polynomials Hi Can someone please tell if this argument is sound? Q: How many monic irreducible polynomials of degree 4 over F_p (p prime) are there? A: If f is monic irreducible, then F_p^4=F_p/(f(x)). So the question how many elements in F_p^4 are of degree 4 over F_p. So, if b in F_p^4, then the only subfields of F_p^4 are F_p^2 and F_p, but F_p is in F_p^2. So, if b is in F_p^2 then b is of degree 2 over over F_p. The only other possiblity is that b is not in F_p^2, then b is of degree 4 over F_p. Since an irreducible polynomial has distinct roots (in F_p^4), and there are 4 roots in a polynomial of degree 4, so the number of monic irreducible polynomials is: (p^4 - p^2)/4 === Subject: Graph theory I have a little question: I am learnig a basics of graph theory, and i really don't have any idea how can i count how many graphs exists which have n vertex and m egdes. And something similiar: how can i count how many subgrpahs (which has n vertex) can i create from graph with n vertex and m edges? I know how to count all different graphs with n vertex: 2^(n(n-1)/2) but i don't know what to do with this m egdes. Any idea? === Subject: Re: Graph theory n(n-1)/2) is the total number of edges in a graph with 'n' vertices. Let the total number of edges = x. Then you want the number of combinations of 'x' things taken 'm' at a time! . === Subject: Re: Graph theory > I have a little question: I am learnig a basics of graph theory, and i > really don't have any idea how can i count how many graphs exists which > have n vertex and m egdes. If you're counting labeled graphs, then you have a set of possible edges (C(n,2), if your graph is simple and loopless), and you want to choose m of them. If you're counting unlabeled graphs, the problem is much harder. > And something similiar: how can i count how many subgraphs (which has n > vertex) can i create from graph [G] with n vertex and m edges? Count how many possible edges there are for your subgraph, and choose m of them. (Hint: The answer depends on the number of edges of the larger graph G and not on the number of vertices.) > I know how to count all different graphs with n vertex: 2^(n(n-1)/2) > but i don't know what to do with this m egdes. > Any idea? You should also do some reading about combinatorics. --- Christopher Heckman === Subject: Lie algebras : Definition of direct sum If g is a Lie algebra and g_1, ..., g_n are Lie subalgebras of g, what does it mean for g to be the direct sum of g_1, ..., g_n? I know what it means for a vector space to be the direct sum of subspaces (i.e. if V = U (+) W and U,W are subspaces of V, then either U / W = {0}, or equivalently every element in V can be uniquely expressed as u + w, where u in U, w in W). What does it mean for Lie algebras? Also, this probably will follow, but if I know that a Lie algebra g has Lie subalgebras g_1, ..., g_n, and I also know that g = g_1 (+) g_2 (+) ... (+) g_n as vector spaces, what would I need in order for this direct sum to also be a direct sum as Lie algebras? Finally, I ask these questions because of the following : Let g be a Lie algebra, and g_1, ..., g_n be ideals of g (i.e. complex subalgebras such that [g, g_i] is in g_i for all i). Suppose I know that g = g_1 (+) ... (+) g_n as vector spaces. Then g_i are ideals, so in particular they are Lie subalgebras of g. But from the proof of a theorem I am reading in a book, this is not enough for g to be the Lie algebra direct sum of the g_i. We only know that it is a direct sum as vector spaces. I appreciate any of your thoughts, James === Subject: Re: Lie algebras : Definition of direct sum > If g is a Lie algebra and g_1, ..., g_n are Lie subalgebras of g, what does > it mean for g to be the direct sum of g_1, ..., g_n? I know what it means > for a vector space to be the direct sum of subspaces (i.e. if V = U (+) W > and U,W are subspaces of V, then either U / W = {0}, or equivalently every > element in V can be uniquely expressed as u + w, where u in U, w in W). > What does it mean for Lie algebras? It can be shown that the underlying vector space of the direct sum of two Lie algebras (in the category of Lie algebras) is the direct sum as vector spaces. By the universal property of the direct sum, the projection maps on the first and second factor, resp., are Lie algebra homomorphisms. Use this to show that for two given Lie algebras g1 und g2, identified with their images in the direct sum g1 (+) g2, it holds [g1,g2] = 0 (i.e. g1 and g2 are perpendicular to each other). With this in mind, your question could be easily answered. > Also, this probably will follow, but if I know that a Lie algebra g has Lie > subalgebras g_1, ..., g_n, and I also know that g = g_1 (+) g_2 (+) ... (+) > g_n as vector spaces, what would I need in order for this direct sum to > also be a direct sum as Lie algebras? Please see my remarks above. > Finally, I ask these questions because of the following : > Let g be a Lie algebra, and g_1, ..., g_n be ideals of g (i.e. complex > subalgebras such that [g, g_i] is in g_i for all i). Suppose I know that g > = g_1 (+) ... (+) g_n as vector spaces. Then g_i are ideals, so in > particular they are Lie subalgebras of g. But from the proof of a theorem I > am reading in a book, this is not enough for g to be the Lie algebra direct > sum of the g_i. We only know that it is a direct sum as vector spaces. > I appreciate any of your thoughts, > James === Subject: Re: PLEASE SOMEBODY HELP!!! Actually,Oakland CA has approximately 700 officers. You must be thinking of a different Oakland? But then no other Oakland has the slightest chance of playing in the Super Bowl. Did you post this question just to take a dig at the fans of the Oakland Raiders? === Subject: Limit Point Compactness I just wanted to see if I've done this homework exercise correctly. Some notation; Let R_l be R with the lower limit topology. That is to say R_l has basis elements of the form [a, b) = {x : a <= x < b}. Also, a space X is said to be limit point compact if every infinite subset of X has a limit point. The exercise I have is as follows: Show that [0, 1] is not limit point compact as a subspace of R_l. PROOF. First note that {1} is open in the subspace [0, 1] since for any b > 1 the set [1, b) is a basis element for R_l and thus [0, 1] / [1, b) = {1} is open. Now consider S = {n / (n + 1) : n is in N} as an infinite subset of the subspace [0, 1]. Clearly for any point p in S we may choose e > 0 such that [p - e, p + e) / S is empty. Moreover, the point 1 is *not* a limit point of S since {1} is open and S / {1} is empty. Thus, [0, 1] is not limit point compact under the lower limit subspace topology. [] Kyle === Subject: uniformly continuous, vanishing Assume x(t) and y(t) are the solutions of the following linear differential equation: dx/dt = a x + b y + d1 dy/dt = c x +d2 (I) where a,b,c,d1 and d2 are bounded for all t>0. Moreover we know that int |x(t)| dt over [t0,+oo) (II) is bounded. (t0>0). Is x(t) bounded? is it uniformly continuous? If so we can immediately concluded that x(t)-->0 as t-->+oo. (Using Barbalat's Lemma) My thought: I could prove that if x(t) is the solution of dx/dt = a x + d1 (III) and integral (II) is bounded, then x(t) is uniformly continuous. As follows: By contradiction suppose that x(t) is not bounded. It can only go unbounded as t-->oo. Therefore there is a sequence {tn} such that { |x(tn)| } is monotonically increasing and lim |x(t_n)| = oo as n-->oo. Assume t(i+1)-t(i)>e>0 and t1>M1 for some large enough M1>0. (if not simply choose a subsequence that satisfies these conditions). Then int_oo |x(t)| > sum_n int_[tn,tn+e] |x(t)| dt. Using (III), one can show that min_[tn,tn+e] |x(t)| > M2, for some M2>0 if M1 is chosen large enough (some function of d1 and a). Consequently, right hand side summation goes unbounded which contradicts the assumption of therefore x(t) is uniformly continuous. But I could not do the same using (I). Does anybody have a suggestion === Subject: YES, and...YAAAAAAAAY! : D Hi there! My name is YOU, and I just dunked my head into a huge wet bucket of my own feces. Anyone care to comment? === Subject: A geometry question: transformation Let f be a one-to-one map from E^2 ONTO E^2, where E^2 is the Euclidean plane. If f maps collinear points into collinear points, ( i.e. if A, B, C are collinear, then so are f(A), f(B), f(C)), must f be continuous? === Subject: important by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id jAF0pf904445 by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id jAF0p3F4025459 ] helo=mail.nitros6.org) by .striker.ottawa.on.ca with esmtp ( 3.35 #1 ()) id 133nlc-0098MM-00 Sensational revolution in medicine! Enlarge your penis up to 10 cm or up to 4 inches! It's herbal solution what hasn't side effect, but has 100% guaranteed results! Don't loose your chance and but know wihtout doubts, you will be impressed with results! Click here: http://cpomobilityproducts.info === Subject: Re: Nearing breakthrough...need magic :-) >>So, what's the breakthrough you're nearing? I'm trying to come up with a better way to do integer division in computer hardware. Currently integer division is more computationally expensive than even floating point multiplication and division, and is done by finding the binary representation of the reciprical of the divisor using variations on newtonian iteration, and then performing integer multiplication against the numerator. The newtonian iteration is of course the expensive part, so I'm trying to trying to find a better way to find a reciprocal in binary. === Subject: Re: Nearing breakthrough...need magic :-) <436ba300$0$63086$ed2e19e4@ptn-nntp-reader04.plus.net> >>How about: >> e = (n-1)! >> m = (2^e - 1) / n === Subject: Re: Smoothness of time series data > I wonder if anyone can help me? I am writing a windows application which > displays and analyzes sensor data. What I need is a function which > returns the smoothness of the data in a specific time period. The > sensor data is roughly acquired in 10 minute intervals and can be > anything from temperature, flow rate, rotational speed etc. > What I want to do is to highlight to the user data which is erratic. > What I imagine is a smoothness function which analyzes the data in a > time period and returns a value based upon its smoothness. A > smoothness of 0 could indicate a non changing sensor reading. A value > of 1 could indicate a sensor reading that increases or decreases at a > constant rate. Values greater than 1 would indicate how unsmooth the > data was. > Anyone know how to achieve this? > Is there some appropriate mathemtical theory I could study? > Graham === Subject: Magic numbers Hi all We have a developing post at http://www.menkaura.com/Forum/index.php?topic=294 if anyone is interested. We'd love your insight if possible. === Subject: basic statistics query (and sanity check!) My wife is doing a college course part of which is a GCSE in maths, whiuch historically has been a sticky subkect for her. Me, on the other hand, always found maths my easiest subject, and I ended up with s staticstics degree! That degree however WAS 21 years ago now however and I accept that I may now be totally and utterly confused today... so I turn to your good selves for some feedback. So... here is the question verbatim from her assigment sheet: 2 a) 1250 tickets are sold in a raffle. They are individually placed in a big hat. If you only draw one ticket what is the probability that your number will be drawn first? [My wife answered 1/1250, and was marked correct on this. I agree!] b)A fair die is rolled 2 times and each time it shows a 4. What is the probability that the die will show a 4 again on the third roll? well??? (scroll down) My wife answered 1/6. That is also my answer on the basis that is it a FAIR die (as stated in the question) and that each roll is therefore a random event, unaffected by anything that has happened previously. However... her LECTURER told her it was NOT 1/6 but..... ONE! ie 1/1! Because , he says, HISTORICAL evidence shows that it MUST be a 4! Surely that cannot be right? Surely it IS 1/6? HEELLLPPPP! My sanity is at stake! didds2 === Subject: Re: basic statistics query (and sanity check!) >My wife is doing a college course part of which is a GCSE in maths, >whiuch historically has been a sticky subkect for her. Me, on the >other hand, always found maths my easiest subject, and I ended up with >s staticstics degree! >That degree however WAS 21 years ago now however and I accept that I >may now be totally and utterly confused today... so I turn to your >good selves for some feedback. >So... here is the question verbatim from her assigment sheet: >a) 1250 tickets are sold in a raffle. >They are individually placed in a big hat. If you only draw one ticket >what is the probability that your number will be drawn first? >[My wife answered 1/1250, and was marked correct on this. I agree!] >b)A fair die is rolled 2 times and each time it shows a 4. What is the >probability that the die will show a 4 again on the third roll? >well??? >My wife answered 1/6. That is also my answer on the basis that is it a >FAIR die (as stated in the question) and that each roll is therefore a >random event, unaffected by anything that has happened previously. >However... her LECTURER told her it was NOT 1/6 but..... ONE! ie >1/1! >Because , he says, HISTORICAL evidence shows that it MUST be a 4! >Surely that cannot be right? Surely it IS 1/6? >HEELLLPPPP! My sanity is at stake! >didds2 Nothing wrong with your sanity. The only question here is what her lecturer has been smoking. --Lynn