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[math-fun] Re: galois fields, 2-var polynomials
by Mike Stay 12 Jan '03

12 Jan '03

I think what I described is a commutative ring. Are there irreducible polynomials in two variables that would give a representation of GF(p^nm)? Or are rings the only thing you can get out of a construct like that? -- Mike Stay staym(a)clear.net.nz

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[math-fun] galois fields, 2-var polynomials?
by Mike Stay 12 Jan '03

12 Jan '03

Galois fields talk about polynomials in one variable modulo some irreducible polynomial in that variable. When we go up to two variables, what kinds of structures emerge? Do they map onto larger fields somehow, or rings, or what? I'm looking at something like a polynomial that's nth order in x, mth order in y, with coefficients modulo a prime. If you hold one variable or the other constant, you can treat it like a galois field and reduce it modulo some irreducible polynomial in that variable. -- Mike Stay staym(a)clear.net.nz

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[math-fun] "On Quaterions and Octonions" finally available.
by asimovd＠aol.com 10 Jan '03

10 Jan '03

At long last, AK Peters has informed me that "On Quaternions and Octonions", by our own John Conway, and Derek Smith, has been published and will be available at Amazon next week. The blurb: << On Quaternions and Octonions John Conway/Derek Smith This book investigates the geometry of the quaternion and octonion algebras. Following a comprehensive historical introduction, the special properties of 3- and 4-dimensional Euclidean spaces are illuminated using quaternions, leading to enumerations of the corresponding finite groups of symmetries. The second half of the book discusses the less familiar octonion algebra, concentrating on its remarkable "triality symmetry" after an appropriate study of Moufang loops. The arithmetics of the quaternions and octonions are also described, and the book concludes with a new theory of octonion factorization. Topics covered include: history; the geometry of complex numbers; quaternions and 3-dimensional groups; quaternions and 4-dimensional groups; the Hurwitz integral quaternions; the composition algebras, Moufang loops; octonions and 8-dimensional geometry; integral octonions; and the octonion projective plane. Year: 2003 ISBN: 1-56881-134-9 Price: $29 160 pages. Hardcover. >> --Dan

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[math-fun] galois field in exponent
by Mike Stay 09 Jan '03

09 Jan '03

I thought this up today. Has anyone ever seen anything like this? It's reminiscent of eliptic curves, but it's not the same as far as I can tell. Take a_n, b_n as elements of a prime-order multiplicative subgroup of a field. Let g be the generator of the subgroup, and lg be discrete log base g. For example, nonzero elements of GF(2^3), g=x. Then we can form a kind of "super-polynomial" A(y) = (a_n)^(y^n) (a_(n-1))^(y^(n-1)) ... (a_0) which we can rewrite as A(y) = g^( (y^n) lg a_n + (y^(n-1)) lg a_(n-1) + ... + lg a_0 ) Since the logs are all mod 7, the exponent is a polynomial in GF(7) and can be given an irreducible modulus with degree d yeilding a representation of GF(7^d). Multiplication of elements in the field is done by exponentiation: A(y) (X) B(y) = A(y)^(lg B(y)) = B(y)^(lg A(y)) Since each of the a_n are polynomials in x over GF(2^3), A(y) is really A(x,y), so it's more fun than just GF(7^d). Since we can stack two of these, we can stack N of them; pick some prime-order subgroup of 7^d and let the log of those elements be the coefficients of a polynomial in z. What happens when you add these things? -- Mike Stay staym(a)datawest.net staym(a)clear.net.nz

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[math-fun] Analytic function with Fibonacci coefficients
by asimovd＠aol.com 09 Jan '03

09 Jan '03

Just for fun, find f(z) = sum from k = 0 to oo of (F_n z^n), and its radius of convergence where F_n is the nth Fibonacci number (F_0 = 0, F_1 = 1,...). --Dan

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[math-fun] 1.24 trillion digits of Pi
by Simon Plouffe 09 Jan '03

09 Jan '03

hello, there is more, Kanada used Machin formula and another one(?) to verify the computation apparently and made basic statistics about the distribution of digits (0 to 9), that's about it. see the article of Peterson in : http://www.sciencenews.org/20021214/mathtrek.asp see also : http://www.super-computing.org/ I find it surprising that he took that formula which is by far not the best. Simon Plouffe

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[math-fun] 1.24 Trillion digits of pi
by Richard Schroeppel 09 Jan '03

09 Jan '03

http://www.guardian.co.uk/worldlatest/story/0,1280,-2223732,00.html Pi Value Calculation Earns Record Friday December 6, 2002 7:20 AM TOKYO (AP) - A team of researchers at a leading national university have set a world record by calculating the value of pi to 1.24 trillion places, one of the researchers said Friday. Professor Yasumasa Kanada and nine other researchers at the Information Technology Center at Tokyo University calculated the value for pi with a Hitachi supercomputer over 400 hours in September, project team member Makoto Kudo said. The new calculation is more than six times the number of places in the record currently recognized by Guinness World Records - 206.158 billion places - which Kanada also helped calculate in 1999. Kanada's team spent five years designing the program used in the September experiment, Kudo said. The Hitachi supercomputer is capable of 2 trillion calculations per second, or twice as fast as the one used for the current Guinness record calculation.

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[math-fun] (no subject)
by Helger Lipmaa 09 Jan '03

09 Jan '03

Simon Plouffe was wondering that some people are calculating pi by using imperfect algorithms. Here is an example of some people doing factoring by using imperfect algorithms. Helger ----------------------------------------------------- Date: Tue, 7 Jan 2003 13:00:30 +0200 (EET) From: Markku-Juhani O. Saarinen <mjos(a)tcs.hut.fi> To: Phil Carmody <fatphil(a)asdf.org> Cc: Helger Lipmaa <helger(a)tcs.hut.fi>, Johan Wallen <johan(a)tcs.hut.fi> Subject: 2000+ complete idiots doing factoring Hi, This is from "The Neo Project", http://www.theneoproject.com, which recently got to the news: http://news.com.com/2100-1040-979301.html They've got a lot of people working on this: LIVE RSA-576 Project Statistics Live: Tuesday January 07, 2003 05:39:28 EST (GMT -5) Total Users: 2,069 Total Nodes: 6,457 Total Teams: 1648 Total Packets Complete: 36,878,155 Total Keys Checked: .14633251904E+15 Total CPU Years: 160.3000 Now.. the good bit. Check out their method (from the FAQ's [sic]): How are we going to crack the RSA 576 challenge? [..] You can basically get a bunch of computers together and start sieving, but that could take months, years, decades, even centuries. So we decided to take the random approach. We do this by to generating a random number less than or equal to the square root of the challenge number and add 1 digit until its a possible prime number, as per Miller's Primality Test (while keeping the number the same length as the square root of the challenge) and see if the challenge number is evenly divisible by the "altered" randomly generated number. If it is... it is figured out. We are looking in to incorporating a few known methods, but keeping it a little random! Cheers, - mjos Markku-Juhani O. Saarinen <mjos(a)tcs.hut.fi> Helsinki University of Technology

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Re: [math-fun] Fibonacci power triangle
by Richard Schroeppel 08 Jan '03

08 Jan '03

Date: Tue, 7 Jan 2003 17:12:06 -0800 (PST) From: "R. William Gosper" <rwg(a)spnet.com> Rich said >This can probably be connected with q-binomial coefficients. Bingo. But to make matters worse, there are q-Fibonacci numbers. >Some other curiosities: >(-1 2 2 -1) also works for F*F' It works for Fib(n+a)*Fib(n+b), a, b arbitrary integers. >(-1 -3 6 3 -1) works for F*F'*F'' Likewise. If I load("c:\\rwg\\climax\\share\\differ6.mac") Warning - you are redefining the MACSYMA function DIFFERENCEQ (which is why it doesn't load automatically) (c17) makerecurrence(f(n),fib(n-12)*fib(n-9)*fib(n+5)) (d17) f(n) - 3 f(n - 1) - 6 f(n - 2) + 3 f(n - 3) + f(n - 4) However, trying to recover fib(n)^3 by imposing initial conditions, (c18) differenceq(%,f(n),f(0)=0,f(1)=1,f(2)=1,f(3)=8) we get instead the alternative form fib(3 n) + 3 fib(- n) (d18) f(n) = --------------------- 5 Without the initial conditions, (c19) differenceq(d17,f(n)) (d19) f(n) = %r15 fib(3 n + 1) + %r16 fib(3 n) + %r14 fib(- n) + %r13 fib(1 - n) I.e., for any integers a,b,c, there are constant values of %r13,...,%r16 to make f(n) = fib(n+a)*fib(n+b)*fib(n+c). Similarly for fib(n)^2: (c21) makerecurrence(g(k),fib(k)^2) (d21) g(k) - 2 g(k - 1) - 2 g(k - 2) + g(k - 3) (c22) differenceq(%,g(k),g(0)=0,g(1)=1,g(2)=1) k 2 fib(2 k + 1) - fib(2 k) - 2 (- 1) (d22) g(k) = ------------------------------------ 5 a peculiar expression for fib(k)^2. To those of you running my Macsyma enhancements, I apologize upon discovering that the MAKERECURRENCE in \rwg\climax\share\differ6.mac returns recurrences of needlessly high order, which I just corrected in differ6 by exhaustively trying all "factors" of the difference operator, in spite of a so-far successful conjecture as to which factor to try. Also, since the even power annihilators contain the difference operator as a factor, there is a lower order operator that renders fib(k)^(2*n) a non-zero constant. --rwg

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Re: [math-fun] Re: Krantz review of Wolfram Book
by asimovd＠aol.com 08 Jan '03

08 Jan '03

I strongly urge anyone who feels the Bulletin carried an ill-considered review to write to the Editor about it (Donald Saari, bams(a)ams.org) (I was very disappointed in some of the "survey" articles in the Jan. Bulletin, due to their utter inaccessibility -- at least to this mathematician. I think several were given a pass because of the fame of their authors and because they are just replicas of talks given at the L.A. Millennial meeting in 2000. But I don't think any of this is a justification for printing impenetrable survey articles.) --Dan

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