December 2009 - mathfun@mailman.xmission.com

Re: [math-fun] Birthday paradox
by Henry Baker 12 Dec '09

12 Dec '09

Thirteen, as of last night. To save time, we may have to solve this problem in general -- perhaps as a generating function. At 10:59 AM 12/12/2009, Mike Stay wrote: >How many tigresses are there? > >On 12/12/09, Henry Baker <hbaker1(a)pipeline.com> wrote: >> [I know, this is a math email list.] >> >> One of the Tigresses complained that Tiger didn't remember her birthday. >> >> What is the probability that two of the Tigresses have the same birthday, >> thus making it easier to remember? >> >> http://en.wikipedia.org/wiki/Birthday_problem

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[math-fun] Birthday paradox
by Henry Baker 12 Dec '09

12 Dec '09

[I know, this is a math email list.] One of the Tigresses complained that Tiger didn't remember her birthday. What is the probability that two of the Tigresses have the same birthday, thus making it easier to remember? http://en.wikipedia.org/wiki/Birthday_problem

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Re: [math-fun] random rational numbers
by James Propp 12 Dec '09

12 Dec '09

Hint: It's not a countably-additive measure; it's finitely additive (every individual rational has measure zero, but every non-trivial interval of rationals has positive measure). If you prefer, you can think of it as a countably-additive measure on the reals that's associated with the rationals. Jim Propp

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[math-fun] question
by David Wilson 11 Dec '09

11 Dec '09

x == 1 (mod phi(x)) ==> x prime? (x >= 2)

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[math-fun] packing tetrahedrons at 83%
by rcs＠xmission.com 11 Dec '09

11 Dec '09

There's a newly discovered tetrahedron packing that achieves 83% fill. http://www.sciencedaily.com/releases/2009/12/091209134633.htm Rich

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Re: [math-fun] Divisibility sequences in OEIS.
by N. J. A. Sloane 11 Dec '09

11 Dec '09

Richard, rather than a new keyword, the way to handle this is to wait until the wiki is up and running and then to create a wiki category for divisibility sequences! Best regards Neil

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[math-fun] random rational numbers
by James Propp 11 Dec '09

11 Dec '09

I don't know the answer to Bill C.'s question about random positive integers, but I'll mention that there's a natural (and self-consistent) sense in which the average positive *rational* number is 3/2. As far as I'm aware, Bruce Reznick and I are the only people who know (or care about) this fact and the framework it comes from, but I've been curious about extending it further (e.g., what's the average positive rational between 0 and 1/2? what's the average of the squares of the positive rationals?), so maybe this forum is a good place to pursue this. I'll start by seeing if any of you can reconstruct the relevant notion of "average" given the hints that appear above (namely (a) the answer is 3/2 and (b) Bruze Reznick has worked on this). If nobody independently re-invents the theory over the weekend, I'll divulge more on Monday. Jim

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Re: [math-fun] N/b gives 'a' as remainder
by Eric Angelini 10 Dec '09

10 Dec '09

Results are now here (in French): http://www.cetteadressecomportecinquantesignes.com/Modulus.htm Best -- and many thanks to all, É.

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Re: [math-fun] characteristic polys of real symmetric matrices
by Henry Baker 09 Dec '09

09 Dec '09

I ran across the following beautiful paper which appears to completely answer my question in general. Quarez, Ronan. "Sturm and Sylvester algorithms revisited via tridiagonal determinantal representations". Nov. 14, 2008. http://hal.archives-ouvertes.fr/docs/00/33/89/25/PDF/prebubli_dual_bez_stur… In this paper, Ronan shows how to take a real polynomial p(x) with distinct roots and construct a tridiagonal matrix M which is symmetric iff p(x) has all real roots. In particular, he constructs a "polynomial remainder sequence" (Euclid's algorithm on polynomials for the rest of us) on p(x) and p'(x)/deg(p) and constructs the 3 diagonals of M from the sequences of quotients and remainders. If p(x) = x^2+2bx+c and q(x) = p'(x)/2 =x+b, then the tridiagonal matrix M is: [ -b e*sqrt(|b^2-c|) ] [ ] [ sqrt(|b^2-c|) -b ] where e = sgn(b^2-c) = +-1. (If y is real, then y = sgn(y)|y|.) If e = +1, then we have a symmetric matrix and 2 real roots. charpoly(M) = x^2+2bx+b^2-e*|b^2-c| = x^2+2bx+b^2-sgn(b^2-c)*|b^2-c| = x^2+2bx+b^2-b^2+c = x^2+2bx+c If p(x) = x^3-3px+2pq and q(x) = p'(x)/3 = x^2+p, then the tridiagonal matrix M is: [ q e1*sqrt(|p-q^2|) 0 ] [ ] [ sqrt(|p-q^2|) -q e2*sqrt(|2p|) ] [ ] [ 0 sqrt(|2p|) 0 ] where e1 = sgn(p-q^2) and e2 = sgn(2p). This matrix is symmetric when e1 = e2 = +1, in which case we have 3 real roots. The general matrix construction of this type fails if we have repeated roots, because gcd(p(x),p'(x)) /= 1. But this is good news, because we have constructed the polynomial gcd in the process, in which case we have a non-trivial divisor of p(x). Constructing matrices of this type involve only rational operations and _square roots_ of non-negative numbers. In particular, the "hard work" of cracking a polynomial remains ahead even after a tridiagonal matrix has been constructed. At 06:16 AM 11/16/2009, Henry Baker wrote: >We would like to represent the polynomial t^2+2*b*t-c as the characteristic polynomial of a 2x2 matrix. > >By translating the origin, we can get rid of b, which makes the following analysis more perspicuous. > >So we have p(t)=t^2-c as the polynomial we want to represent by the symmetric real matrix > >[x y] >[y z] = M. > >This matrix M satisfies its own characteristic polynomial, so we have the two equations: > >y*(x+z)=0 > >x^2+y^2-c=0 > >The first equation gives z=-x, while the second gives x=+- sqrt(c-y^2). > >So, M is > >[-sqrt(c-y^2) y] >[y sqrt(c-y^2)] > >Thus, so long as |y|<sqrt(c), M will represent p(t). > >In general, for 2x2 matrices, y^2 enters as an addition to the discriminant, so we have some freedom to choose y if the discriminant is positive. > >Unfortunately, we have effectively solved the polynomial p(t) already, prior to constructing M, so it would appear that constructing M is not easier than solving p(t).

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[math-fun] Vanishing triangle puzzle
by Dan Asimov 09 Dec '09

09 Dec '09

Let f: R -> R be a real analytic function with f(0) = f'(0) = f''(0) = 0. Let N_x denote the normal line at x to the graph of y = f(x), and let A(x) denote the area of the triangle formed by N_x and the x- and y-axes. What is the limit of A(x) as x -> 0 ? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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