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Re: [math-fun] Ferguson faces deadline
by Mike Stay 19 Dec '02

19 Dec '02

Please add my name to the letter of support. -- Mike Stay staym(a)clear.net.nz

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Re: [math-fun] Ferguson faces deadline
by asimovd＠aol.com 19 Dec '02

19 Dec '02

David Bailey forwarded this news story about mathematician and sculptor Helaman Ferguson: << Running Out Of Time, Space Math-Inspired Artist Under Eviction Threat ... ... He has worked since without a written agreement but with an unwritten pledge from one of the property owners that he could stay while completing his other large granite pieces. ... ... >> I sympathize tremendously with the plight that Ferguson, who is immensely talented, now finds himself in. I can't resist saying, however, that things like this are exactly why I am a strong believer in written contracts. --Dan

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RE: [math-fun] Inverse Symbolic Calculator and plouffe's inverter .
by Meeussen Wouter (bkarnd) 19 Dec '02

19 Dec '02

nice stuff, as Xmass present, I offer you 3.92065141231007653868232570898716939515935444313 which correctly Plouffe-inverts to Sum[(5/2*n^3 - 10*n^2 + 27/2*n - 4)/Binomial[2*n, n], {n, 1, Infinity}] aha, this equals Hypergeometric2F1[3, 4, 5/2, 1/4] which again equals 8/243*(81 + 7*Sqrt[3]*Pi) see? start anywhere, and you get Pi staring back at you. hmmm, can this be coincidence? ;-)) Wouter. -----Original Message----- From: Simon Plouffe [mailto:simon.plouffe@sympatico.ca] Sent: donderdag 19 december 2002 8:34 To: math-fun(a)mailman.xmission.com Subject: [math-fun] Inverse Symbolic Calculator and plouffe's inverter. hello, I agree the <reveng> engine is useless because of that java program which does not work for most people and platforms. (I did not programmed that junk!, I made the ISC only). The Inverse Symbolic Calculator still works there at the cecm, they did not unplugged my project there yet. http://www.cecm.sfu.ca/projects/ISC/ it uses the online (but limited) set of tests with 46 million constants. For lookups only I made the Inverter at http://pi.lacim.uqam.ca/eng/ and if you want a real test with 200 milllion constants + a real inverter which uses GFUN, Pari-GP with LLL you still can ask me to run a test on any number mail you number of any size to numbers(a)math.uqam.ca and I will run the tests on it. Of course : because of the exponential growth of CPU needed for solving anycase that has approx. 50 digits or more this last step is done interactively by me manually. I still could not find a way to automate that last step. Ah yes, the OEIS is embedded into the Inverter, I made a conversion of the integer sequences into real numbers using 18 different algorithms in 1999. I counted about 2 million entries that are made with the OEIS entries. Just in case that an unknown combination of integer sequence would lead to any interesting real number, and vice-versa. Simon Plouffe _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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[math-fun] the Inverse Symbolic Calculator
by Joerg Arndt 19 Dec '02

19 Dec '02

...was replaced by the utterly useless http://psg.cecm.sfu.ca/projects/revenge/client/RevEngClient.html Just in case somebody on the list has contact to the people maintaining this site: please tell those folks to get a clue about how to create a usable website. The page uses Java with no fallback and therefore is unusable for many people. The only text is included as jpeg, well well. Perfect to exclude the blind folks (screenreaders don't read images), not even an alt-tag there, jeeeesus... all the best, jj -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.

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Re: [math-fun] Puzzle: Find limiting angle as dimension -> oo
by Richard Guy 18 Dec '02

18 Dec '02

Sorry -- forgot to take the square root! I was looking at the scaled up (by a factor of n) version, and mistakenly though that that might be less easy for people to see. You're quite right. Limit is 1/2. R. On Wed, 18 Dec 2002 asimovd(a)aol.com wrote: > Hi, Richard. You wrote: > > The scalar product of (the numerators of) the > two vectors is > > 1*(n-1) + 2*(n-2) + ... + (n-1)*1 > > which is the (n-1)th tetrahedral number (see > Fig 2.34 on p.46 of the Book of Numbers, > where one sums a diagonal of the multiplication > table): (n+1)*n*(n-1)/6 so the scalar product > is (n+1)*(n-1)/6n > > The lengths are just the sum of the first (n-1) > squares, divided by n^2: n*(n-1)*(2n-1)/6n^2 > So cosine of angle twixt the two is > > (n+1)*(n-1)/6n / (n-1)*(n-1)*(2n-1)*(2n-1)/36n^2 > > = 6n*(n+1)/(n-1)*(2n-1)*(2n-1) which tends to 0. > Richard, since you've already multiplied the (equal) lengths of the integer vectors to get the sum of the first n-1 squares n*(n-1)*(2n-1)/6, there's no need to square this again in the denominator. And since (1,2,...,n-1) and (n-1,...,2,1) are just scaled versions of the vectors I asked about, they will have the same angle and we may as well just consider these. According to your dot product and length calculation, the cosine at the nth stage is (n+1)*n*(n-1)/6 divided by n*(n-1)*(2n-1)/6, and so the limit as n --> oo is 1/2. Regards, Dan Asimov

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Re: [math-fun] Puzzle: Find limiting angle as dimension -> oo
by asimovd＠aol.com 18 Dec '02

18 Dec '02

Richard Guy wrote: << I don't see why one shouldn't use very elementary summations I mentioned that only to make the problem more of a challenge. << The scalar product of (the numerators of) the two vectors is 1*(n-1) + 2*(n-2) + ... + (n-1)*1 which is the (n-1)th tetrahedral number (see Fig 2.34 on p.46 of the Book of Numbers, where one sums a diagonal of the multiplication table): (n+1)*n*(n-1)/6 so the scalar product is (n+1)*(n-1)/6n. The lengths are just the sum of the first (n-1) squares, divided by n^2: n*(n-1)*(2n-1)/6n^2. So cosine of angle twixt the two is (n+1)*(n-1)/6n / (n-1)*(n-1)*(2n-1)*(2n-1)/36n^2 = 6n*(n+1)/(n-1)*( 2n-1)*(2n-1) which tends to 0. >> Since the (equal) lengths of the integer vectors have been multiplied the to get the sum of the first n-1 squares n*(n-1)*(2n-1)/6, there's no need to square this again in the denominator. And since (1,2,...,n-1) and (n-1,...,2,1) are just scaled versions of the vectors I asked about, they will have the same angle and we may as well just consider these. According to the above dot product and length calculation, the cosine at the nth stage is ((n+1)*n*(n-1)/6) / (n*(n-1)*(2n-1)/6), and so the limit as n --> oo is 1/2, and the angle is pi/3. Dan Asimov

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[math-fun] Puzzle: Find limiting angle as dimension -> oo
by asimovd＠aol.com 18 Dec '02

18 Dec '02

Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n). Let w_n be the vector in R^(n-1) with the same coordinates in reverse order. Find the limit as n -> oo of the angle between v_n and w_n. (Try this without using numerical approximation or summation-of-powers formulae.) Is there a simple geometric reason for the answer? I don't know. --Dan

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Re: [math-fun] Puzzle: Find limiting angle as dimension --> oo
by asimovd＠aol.com 18 Dec '02

18 Dec '02

I asked: << Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n). Let w_n be the vector in R^(n-1) with the same coordinates in reverse order. Find the limit as n --> oo of the angle between v_n and w_n. (Try this without using numerical approximation or summation-of-powers formulae.) Is there a simple geometric reason for the answer? I don't know. >> Scott writes: << Ok, I cheated (numerical approximation) and found limiting angle pi/4 (45 degrees). >> That's not the answer I get. The cosine of the angle sought is the limit as n --> oo of the dot product f(n) = (v_n / ||v_n||) · (w_n / ||w_n||). Multiplying the numerator and denominator of f(n) each by 1/n, one sees that as n --> oo they approach integral_0^1 x(1-x) dx and integral_0^1 x^2 dx respectively, so the cosine is 1/2, and the angle is pi / 3. --Dan

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[math-fun] Reference book on voting systems, paradoxes, etc?
by Bernie Cosell 17 Dec '02

17 Dec '02

For some time now I've been putting off getting a book on Arrow's Theorem/paradox, Condorcet criteria, etc... and for no particularly good reason the topic has surfaced again for me and I thought I'd turn to you folks for recommendations for a good expository book on voting systems.. Thanks! /Bernie\ Bernie Cosell bernie(a)fantasyfarm.com

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[math-fun] René Thom
by asimovd＠aol.com 14 Dec '02

14 Dec '02

I just learned that French mathematician René Thom died on Nov. 17. He was awarded the Fields medal in 1958 for his work on cobordism theory, and is also well-known for originating catastrophe theory in the late 1960's. In case anyone's interested, the simplest case of cobordism theory defines an equivalence relation on all (compact and boundaryless) manifolds of a fixed dimension: two n-manifolds M,N are equivalent when there is an (n+1)-manifold W such that the boundary of W is M u N. With union as the additive operation, and cartesian product as the multiplication, the set of equivalence classes form a "graded algebra" over Z_2. Manifolds that are themselves the boundary of another manifold represent the zeros in this algebra, and it's over Z_2 since M u M is always the boundary of M x [0,1]. Thom's astonishingly elegant result is that this graded algebra over Z_2 has just one generator in each dimension *not* of the form 2^k - 1, for k >= 1. Put another way, the algebra is all polynomials over Z_2 formed by 1=x0 (represented by the point), x_2, x_4, x_5, x_6, x_8, x_9, etc., where the degree of any monomial is the sum of the indices of its factors. (The monomials of a given degree correspond to the equivalence classes among connected manifolds of that dimension.) Hence, among connected manifolds there are just two "cobordism" classes {0, x_2} in dimension 2; just one {0} in dimensions 1 or 3 (i.e. every manifold of dimension 1 or 3 is the boundary of another manifold), and e.g. three in dimension 9: {0, x_4 * x_5, x_9}. --Dan

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