April 2005 - mathfun@mailman.xmission.com

RE: [math-fun] elliptic mean
by Schroeppel, Richard 29 Apr '05

29 Apr '05

Must a Mean associate? -- Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of R. William Gosper Sent: Friday, April 29, 2005 11:29 AM To: math-fun(a)mailman.xmission.com Subject: Re: [math-fun] elliptic mean I blithered >(c562) circumference_ellipse(a,b)/2/%pi > 2 > b > 2 a elliptic_ec(1 - --) > 2 > a >(d562) ----------------------- > %pi >is clearly a legitimate mean(a,b), ... Guilt by nonassociation: (c138) em(a,b):=circumference_ellipse(a,b)/2/%pi$ (c139) dfloat(em(em(2,4),em(3,1))) (d139) 2.62751721836387d0 (c140) dfloat(em(em(2,1),em(3,4))); (d140) 2.62733046926582d0 Also, Department of Redundancy Department: >... discriminant of the nth Legendre polynomial discriminant. Sorry for the pollution. But thanks (Gary and Dick) for the solution. --rwg BLITHER HILBERT SPACE - PEACH BLISTER _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] elliptic mean
by R. William Gosper 29 Apr '05

29 Apr '05

(c562) circumference_ellipse(a,b)/2/%pi 2 b 2 a elliptic_ec(1 - --) 2 a (d562) ----------------------- %pi is clearly a legitimate mean(a,b), satisfying an identity: (c563) (circumference_ellipse(1-x,1+x)/2/%pi = rhs(d448)) 2 (1 - x) 2 (x + 1) elliptic_ec(1 - --------) 2 (x + 1) (d563) ----------------------------------- = %pi 2 2 2 (2 elliptic_ec(x ) - (1 - x) (x + 1) elliptic_kc(x )) ------------------------------------------------------- %pi (c564) taylor(%,x,0,12) 2 4 6 8 10 12 x x x 25 x 49 x 441 x (d564)/T/ 1 + -- + -- + --- + ----- + ------ + ------- + . . . = 4 64 256 16384 65536 1048576 2 4 6 8 10 12 x x x 25 x 49 x 441 x 1 + -- + -- + --- + ----- + ------ + ------- + . . . 4 64 256 16384 65536 1048576 (i.e. 2F1(-1/2,-1/2,1,x^2)) so it is >= arithmetic >= geometric means. Both the "aem" and the "egm" converge quadratically, like the agm, but to what? --rwg PS, John Brillhart challenged me to find the discriminant of the nth Legendre polynomial discriminant. I hope he doesn't mind my ducking and throwing open the question...

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[math-fun] Triangles question
by David Wilson 29 Apr '05

29 Apr '05

In the Mathworld article "Square Dissections" we find Guy (1989) asks if it is possible to triangulate a square with integer side lengths such that the resulting triangles have integer side lengths (Trott 2004, p. 104). There's more to this than I'm seeing right? If I interpret this to mean dissecting an square into integer-sided triangles, it doesn't seem very difficult. I can easily dissect a 12x12 square by tiling it with 3x4 rectangles and cutting the rectangles along random diagonals. I played with removing edges from those tilings and got a tiling of the 12x12 in 5 integer-sided triangles: 3-4-5, 5-5-6, 6-8-10, 10-10-12 and 9-12-15. But that solution dissection seems too simple that Guy would have overlooked it, so I'm thinking the original problem was more involved. So now the question is, can a square be dissected into fewer than 5 integer-sided triangles? 1 and 2 are clearly out. For 3, we cant use the square diagonal, so one of the triangles must have a square edge as a base and apex interior to the opposite side. I couldn't find integer solutions to this. For 4 triangles, I could not be sure I got all the possible configurations, but it might be an interesting investigation. - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"

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[math-fun] discriminant of Legendre polynomials
by Gary McGuire 28 Apr '05

28 Apr '05

Greg Fee wrote [>PS, John Brillhart challenged me to find the discriminant of the nth Legendre [>polynomial discriminant. I hope he doesn't mind my ducking and throwing open [>the question... The discriminants of many orthogonal polynomials can be found in the book "Orthogonal Polynomials" by Gabor Szego. (section 6.7) For the Legendre polynomials, if I've copied it right, you get 2^{-n(n-1)} product_{i=1}^n i^{3i-2n} (n+i)^{n-i}. One proves this using the formula for discriminants involving the derivative: D(f)=(-1)^{n(n-1)/2} a^{n-2} product_{i=1}^n f ' (alpha_i) where a=leading coefficient of f, alpha_i are the roots of f, f ' =derivative of f. Then use the differential equation to get f ' (alpha_i). Gary McGuire

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[math-fun] refractory telescopy
by R. William Gosper 27 Apr '05

27 Apr '05

I like to present convergent telescoping identities as infinite sums, e.g., inf ==== \ x x tan(x) = > (2 csc(------) - csc(--)), / n - 1 n ==== 2 2 n = 0 with a parameter (here x), two instances of which can be subtracted to give the sum over any range. But this example immediately suggests k - 1 ==== x \ x cot(--) - cot(x) = > csc(--), k / n 2 ==== 2 n = 0 which blows up in one direction and goes nuts in the other. Integrating, inf ==== | n | \ log(|cot(2 x)|) > ---------------- = - 2 (log(abs(sin(x))) + log(2)), / n ==== 2 n = 0 sort of reenabling the infinite series convention, but the lhs is undefined for x = dyadic rational * pi, a dense set! There may also be x for which it doesn't converge. Further massaging finally gives the very convergent product, 2 x %e coth(x) sqrt(coth(2 x)) sqrt(sqrt(coth(4 x))) . . . = ---------- . 2 4 sinh (x) --rwg

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[math-fun] A Malfatti Miracle.
by Richard Guy 25 Apr '05

25 Apr '05

It's well-known to those who well know it that there are 32 Malfatti configurations (of sets of 3 circles each touching 2 others & 2 sides of a triangle). Theorem. The 32 radical centres lie in pairs on 64 lines, 16 through each of the vertices of the triangle and 4 through each of 4 other, orthocentric, points, whose (common) 9-point centre is the orthocentre of the original triangle. I have a nice description, too long for the margin of this machine, which uses Conway's extraversion and nim-addition. Three of the last 4 points are in perspective with the original triangle. I suspect that there are centres (yes, genuine Kimberling ones) that may not be in the official collection. R.

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[math-fun] Legendre's relation empirically
by R. William Gosper 25 Apr '05

25 Apr '05

Just another nag/brag about TLINREL, the unreasonably effective, nine line Taylor-series analog of an integer relation-finder. Here it "discovers" Legendre's relation given all possible linear and quadratic combinations of K, E, K', and E', despite the messy logs and digammas introduced by shifting to 0 the point of expansion of K' and E'. (c72) (makelist((%pi*hyper_2f1(a/2,1/2,1,x)/2,hypersimp(%%)=%%),a,[1,-1]), append(%%,multthru(hyplin1mz(subst(1-x,x,%%))))) 1 1 %pi hyper_2f1(-, -, 1, x) 2 2 (d72) [elliptic_kc(x) = -------------------------, 2 1 1 %pi hyper_2f1(- -, -, 1, x) 2 2 elliptic_ec(x) = ---------------------------, 2 inf ==== 2 \ 1 n elliptic_kc(1 - x) = - ( > (-) x / 2 n ==== n = 0 1 2 (log(x) - 2 psi (n + 1) + 2 psi (n + -))/n! )/2, 0 0 2 inf ==== \ 1 3 n elliptic_ec(1 - x) = 1 - x ( > (-) (-) x / 2 n 2 n ==== n = 0 3 1 (log(x) - psi (n + 2) + psi (n + -) - psi (n + 1) + psi (n + -)) 0 0 2 0 0 2 /(n! (n + 1)!))/4] (c73) factor(tlinrel(makelist(lhs(e)=taylor(rhs(e),x,0,11),e, append(%,%^2,%*part(%,[2,3,4,1]),part(%,[1,2])*part(%,[3,4]))))) (d73) 98304 (2 elliptic_kc(1 - x) elliptic_kc(x) - 2 elliptic_ec(1 - x) elliptic_kc(x) - 2 elliptic_kc(1 - x) elliptic_ec(x) + %pi) 4 3 3 2 2 (54 log (x) - 864 log(2) log (x) + 242 log (x) + 5184 log (2) log (x) . . . 4 3 2 + 13824 log (2) - 15488 log (2) + 11168 log (2) - 3908 log(2) + 617) 4 4 3 3 /(%pi x (3432 log (x) - 54912 log(2) log (x) + 8362 log (x) . . . 3 2 - 535168 log (2) - 832096 log (2) + 161444 log(2) + 227688)) = 0 Expansion to 11th order was probably excessive, but higher order -> more confidence in the finding. Although the probability seems slim of TLINREL uncovering a functional relation that traditional analysts have somehow overlooked, the payoff would be large compared to the analogous detection of merely numerical relations. --rwg

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[math-fun] sum 1/p
by Richard Schroeppel 22 Apr '05

22 Apr '05

Richard Guy asked about a(4), the prime for which sum(1/p) exceeds 4. I responded: Rich> It would be straightforward to set up a distributed computation Rich> to sum 1/p for p < 1.8e18, and locate the specific P that bumps Rich> the sum over 4. In a few more years the resources will be Rich> available to many of us as individuals. I've thought about Rich> doing similar computations myself, but always get bogged down in Rich> "As long as you're touching all those primes, what else do you Rich> want to compute?". I also want sum 1/(p-1), product P, etc. ... Rich> Odlyzko & friends have pushed exact computations of pi(N) up to Rich> around N = 10^21 using a zeta function based formula, so perhaps Rich> sum(1/p) could be tackled similarly. This doesn't require Rich> touching every prime. which prompted victor(a)idaccr.org (Victor S. Miller) to elaborate: Ah, this piqued my attention. Since I was one of the "friends" involved in the combinatorial formula (which was the one used for the above quoted values -- the analytic formula seems much more difficult to get to work in a practical range), I can outline what needs to be done for an algorithm which will probably be O(x^(2/3) + epsilson) (but there's a serious issue of precision -- see below) to calculate sum_{p <= x} 1/p to sufficient accuracy. Once you have an algorithm for that, you can find the crossover point you want by binary search (or even better, interpolation search): for k a positive integer, let p_k denote the k'th prime, and set S(x,k) = sum { 1/n | 1 <= n <= x, and p_j does not divide n for j<=k }. Then, we have the recursion, for k>0: S(x,k) = S(x,k-1) - (1/p_k) S(x/p_k,k-1) (*) and S(x,0) = sum_{n<=x} 1/n Now S(N,sqrt(N)) = sum_{ sqrt(N) <= p <= N} 1/p [nit: ^^^^^^^ should be pi(sqrtN) ] (this is tricky this sum will be very close to log(2), so the "extra" amount is important). The key to the algorithm is to use the recursion (*) as long as k>=k_0 (for some convenient k_0, like 5) and x >= N^(2/3). We can then evaluate the values of S(x,a) that we have left by means for a tree data structure (see Lagarias, Miller and Odlyzko, Math. Comp. 1984 (I think)). Since the number of summands involved in all of the intermediate sums is <=N, it means that we need to know the initial values of sum_{1 <= i <= k} 1/k (for all k <= N^2/3) to at least an accuracy of (1/(2N)). However, the fact that the recursion (*) involves subtraction might well make this numerically unstable and force a lot more precision to be needed. Maybe someone will be motivated to make it work. I've included Victor's full message, since I had several comments. Re Future Prospects, for a(5), somewhere around 10^50. If we have an N^(2/3) algorithm, this would only be ~ 10^33. Although the sum total of the worlds computing resources today is perhaps only 10^25 cycles/year, we can't say 10^33 is forever out of range. Returning to more immediate possibilities: In the specific case of sum(1/p), the precision demands for Victor's recursion S(x,k) = S(x,k-1) - (1/p_k) S(x/p_k,k-1) (*) and base case S(x,0) = sum_{n<=x} 1/n seem manageable. Expanding the minuend recursively converts the recursion to S(x,k) = S(x,0) - sum (1/p_k) S(x/p_k,k-1). In this form, the depth of the recursion is at most log N, where N is the largest prime in our sum. The width is at most pi(sqrtN). Any given leaf at the bottom of the tree has an implicit maximum coefficient of (sqrtN)^logN in the final answer, even ignoring the 1/p_k reduction. This would mean an extra log(sqrtN^logN) bits of precision in the leaf calculations, around (logN)^2 bits. A more careful account might cut this to O(logN). Computing the base case harmonic numbers to moderate precision isn't difficult. Gosper has a nice accelerated series, and even Euler-Maclaurin summation should work. I mentioned wanting some other results, like sum 1/(p-1) and product P. The recursion only works for powers of P: we could use slight variations to sum P^2 or sqrt P, while redoing the error analysis a little. The required fixup for summing log P seems easy enough. We could also do sum 1/P^K for K = 2...100 or so, which should be enough to handle sum 1/(p-C) for small real and complex C. 1/(p-1) = 1/p + 1/p^2 + ..., and we can get stuff like sum 1/(p^2 + 2) using partial fractions. sum sqrt(P+3) can be done with another power series, 1/p^(K+1/2). I don't see how to do the sum of (logP)^2, or loglogP or 1/logP. Predicting future computing capability is hard: In the late 1800s, the mathmatician Jevons published a 20 digit number, the product of two 10 digit primes known only to himself. Anticipating RSA by almost a century, he confidently asserted that noone would ever know the factors. Cole exhibited the factors of M67 = 2^67-1, 21 digits, at a math conference in 1903. Rich rcs(a)cs.arizona.edu

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[math-fun] Re: math-fun Digest, Vol 26, Issue 7
by Gary McGuire 21 Apr '05

21 Apr '05

Date: Tue, 19 Apr 2005 10:41:01 -0400 >From: Bill Thurston <wpt4(a)cornell.edu> >Subject: Re: [math-fun] Density question >......... > >What's known about the chromatic number of the plane? I'm curious >whether enough is known to be able to prove there can't be a measurable >best coloring. > A google search came up with http://planetmath.org/encyclopedia/ChromaticNumberOfASpace.html which has a reference to the paper below, which proves that if a 4-colouring exists then one of the monochromatic sets is not Lebesgue measurable. K. J. Falconer. The realization of distances in measurable subsets covering R^n J Comb. Theory A 31, 184-189, 1981. Gary McGuire > >

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Re: [math-fun] Infinite epsilon-chains (set theory)
by Chris Landauer 21 Apr '05

21 Apr '05

hihi, all - the anti-foundation axiom and some of its consequences are explained in a book by Peter Aczel on non-well-founded sets, though that book is a little too much like the usual (non-naive) set theory for my taste: a good introduction to the axiom is in a book Jon Barwise, John Etchemendy, "The Liar: Essays on Truth and Circularity" i always thought that using sequences instead of sets, the same idea would make an excellent formal theory for the use of pointers and pointer arithmetic, iwht all its attendant dangers and difficulties more later, cal

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