January 2010 - mathfun@mailman.xmission.com

Re: [math-fun] Closed curve puzzle (SPOILER)
by Dan Asimov 11 Jan '10

11 Jan '10

I wrote: << In the plane: Let C be a C^oo simple closed curve. Let a "double-normal" be a line segment whose endpoints lie on C and which is normal to C at each of them. C must have a double-normal. (Proof: Consider the longest segment from C to C). Question: Let a "simple" double-normal be one that intersects C only at its endpoints. Must C have at least one simple double-normal? Prove or find a counterexample. >> Consider an equilateral triangle T in the plane, and on each side build a 15-30-135 triangle, in the same sense, pointing into T. (Each side of T is the longest side of one 15-30-135 triangle.) Now remove the three original sides of T, leaving the 6 remaining sides of the obtuse triangles, which make a starlike hexagon H with 3-fold rotational symmetry. It's easy to check that if H is C^1-approximated by a smooth curve, then every double-normal segment (there are three) intersects the curve in its interior. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] new pi computation, 2700 billion digits... on a PC...!
by Simon Plouffe 08 Jan '10

08 Jan '10

http://bellard.org/pi/pi2700e9/announce.html was carried out by Fabrice Bellard in base 10 and 16. The computation was checked using the BBP formula for the base 16. It took more than 100 days to compute everything, maybe much more of elapsed time. This is the world record, the previous record was 2576 billion digits. This is an historic moment since it is the first time an ordinary PC does better than a super-computer (and a lot of determination). Congratulations to Fabrice Bellard! simon plouffe

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[math-fun] More on the solutions of a real cubic
by Henry Baker 07 Jan '10

07 Jan '10

Given a monic cubic x^3+bx^2+cx+d=0, with real b,c,d, we can translate the origin to produce x^3+3px+2q=0, where if p=0, we have a trivial cube root solution. If p/=0, we can scale x to produce one of the following 3 exclusive cases, with the parameter a real: x^3-3x-2cos(3a)=0, [sqrt(discriminant)=sqrt(108)isin(3a)] /* 3 real roots */ x^3-3x-2cosh(3a)=0, or [sqrt(discriminant)=sqrt(108)sinh(3a)] x^3+3x-2sinh(3a)=0. [sqrt(discriminant)=sqrt(108)cosh(3a)] These equations factor as follows: x^3-3x-2cos(3a) = (x-2cos(a))(x-2cos(a+2pi/3))(x-2cos(a-2pi/3)) x^3-3x-2cosh(3a) = (x-2cosh(a))(x+cosh(a)+isqrt(3)sinh(a))(x+cosh(a)-isqrt(3)sinh(a)) x^3+3x-2sinh(3a) = (x-2sinh(a))(x+sinh(a)+isqrt(3)cosh(a))(x+sinh(a)-isqrt(3)cosh(a)) We can represent the case with 3 real roots as the characteristic polynomial of the following tridiagonal symmetric matrix: [-cos(3a) sin(3a) 0 ] [ sin(3a) cos(3a) sqrt(2)] [ 0 sqrt(2) 0 ] The eigenvectors corresponding to these real eigenvalues are the columns of this orthogonal matrix: [ sin(a) sin(a+2pi/3) sin(a-2pi/3)] [ cos(a) cos(a+2pi/3) cos(a-2pi/3)]*sqrt(2/3) [1/sqrt(2) 1/sqrt(2) 1/sqrt(2) ]

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Re: [math-fun] Closed curve puzzle
by Dan Asimov 06 Jan '10

06 Jan '10

Clarification: I wrote: << In the plane: Let C be a C^oo simple closed curve. Let a "double-normal" be a line segment whose endpoints lie on C and which is normal to C at each of them. C must have a double-normal. (Proof: Consider the longest segment from C to C). Question: Let a "simple" double-normal be one that intersects C only at its endpoints. Must C have at least one simple double-normal? Prove or find a counterexample. >> Veit's counterexample (of a smoothed {7/3} star polygon) is a brilliant example and maybe even the simplest one possible, if the question had not required that the closed curve be simple, i.e., non-self-intersecting. But it does, so the original puzzle is still open for solutions. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Closed curve puzzle
by Dan Asimov 06 Jan '10

06 Jan '10

In the plane: Let C be a C^oo simple closed curve. Let a "double-normal" be a line segment whose endpoints lie on C and which is normal to C at each of them. C must have a double-normal. (Proof: Consider the longest segment from C to C). Question: Let a "simple" double-normal be one that intersects C only at its endpoints. Must C have at least one simple double-normal? Prove or find a counterexample. --Dan _____________________________________________________________ * Thanks to Evan O'Dorney for this quick proof and for asking this question. _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Elementary (?) matrix question
by Dan Asimov 05 Jan '10

05 Jan '10

Jean writes: << The non-negative square roots, sigma_1, ..., sigma_n, of the eigenvalues of M.M' are indeed known as the singular values of M. The Singular Value DecompositionTheorem (SVD) says that M = U Sigma V', for some unitary matrices, U , V , and where Sigma = diag(sigma_1, ..., sigma_n) (I assume M is an n X n matrix). >> The Singular Value Decomposition has an interesting interpretation over the reals. Suppose L_1 and L_2 are linear n-dimensional subspaces of R^k. If n = 1 their relative position in R^k is described by one number, the angle between these two lines (or its cosine). Let {u_1,...u_n} and {v_1,...v_n} be orthonormal bases for L_1 respectively L_2 in R^k. Let U and V be the k x n matrices whose columns are the u_j's respectively the v_j's. Then the n x n matrix A := U^t V has its ij^th entry just <u_i, v_j> (dot product). The singular value decomposition of A is SVD = P^t A Q, where SVD is a diagonal matrix of nonnegative entries in descending order, and P and Q are orthogonal. This defines SVD uniquely. Back to the geometry of L_1 and L_2: Among all the unit vectors x in L_1 and y in L_2, there is an x and y that maximize the dot product <x,y>, i.e., minimize their angle. (It may happen that the angle is 0.) This dot product will be the first entry of the SVD matrix S. Now let L_1' and L_2' be the orthogonal complements of x respectively y in L_1 and L_2. Again let x' and y' be the vectors of L_1' and L_2' respectively having the least angle. Then <x',y'> is the second entry of SVD. Etc. The sequence of vectors x, x', x'',... is orthonormal, as is y, y', y'',.... The x's are given by the columns of the n x k matrix VQ, and likewise for the y's and UP. The sequence of (nonnegative) cosines, i.e., the singular values, on the diagonal of SVD completely characterize the relative positions of L_1 and L_2 up to an isometry of R^k. The angles (inverse cosines of the singular values) are called the principal angles of L_1 and L_2. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Elementary (?) matrix question
by Jean Gallier 05 Jan '10

05 Jan '10

The non-negative square roots, sigma_1, ..., sigma_n, of the eigenvalues of M.M' are indeed known as the singular values of M. The Singular Value DecompositionTheorem (SVD) says that M = U Sigma V', for some unitary matrices, U , V , and where Sigma = diag(sigma_1, ..., sigma_n) (I assume M is an n X n matrix). A non-trivial relationship between the eigenvalues, lambda_1, ..., lambda_n, of M, and its singular values, sigma_1, .., sigma_n, assuming that | lambda_1 | \geq | lambda_2 | \geq ... \geq | lambda_n | and sigma_1 \geq sigma_2 \geg ... \geq sigma_n, are the Weyl Inequalities (1949): | lambda_1 | ... | lambda_k | \leq sigma_1 ... sigma_k, for k = 1, .., n - 1 and | lambda_1 | ... | lambda_n | = sigma_1 ... sigma_n. See Horn and Johnson, Topics in Matrix Analysis, page 171. Best, -- Jean Gallier

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[math-fun] Elementary (?) matrix question
by Henry Baker 04 Jan '10

04 Jan '10

I hate to be so dense about linear algebra, but here's another matrix question: Given an _arbitrary_ square matrix M with complex entries, we can compute the matrix product M.M', where M' is the conjugate transpose of M. Clearly, M.M' is Hermitian, so all of its eigenvalues are real. Question: Is there any interesting relationship between the eigenvalues of M.M' and those of M ? There's an obvious one from the fact det(M.M')=det(M)det(M'). Also, trace(M.M') is the sum of the squares of the absolute values of the entries of M.

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[math-fun] Third Millennium Passage
by rtwainwright＠optonline.net 01 Jan '10

01 Jan '10

Hello everyone. As we pass through the third millennium, HAPPY: SUM (n^3 - n/3) n = 0...9 R. Wainwright

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[math-fun] Twisted Life
by Paul Reiners 31 Dec '09

31 Dec '09

Here is a video game I wrote using something that is sort of a CA: http://bit.ly/7xkjiH Constructive criticism is welcome.

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