December 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Gamma(p/q) values
by Bill Gosper 29 Dec '11

29 Dec '11

WDS>Noticing 1 and 11 form a multiplicative subgroup mod 120, I tried Ries on (11/120)! * (1/120)! = 0.95023643380439452766545957659649988150532 and (11/120)! / (1/120)! = 0.95931292205678601103043910565064773096559 but it failed to find anything I liked. A certain proportion of Ries formulas seem to be "tasteless" and the sort of thing that are never going to happen, like using (e^pi) or 1/ln(pi) or e or sin(1) in an exponent... ---------------------------------------- In:= FindRoot[EllipticTheta[2, 0, x] == Gamma[1/4]/(2*Pi)^(3/4), {x, 0},WorkingPrecision -> 22] Out= {x -> 0.04321391826377225235322} billgosper$ ./ries .04321391826377225235322 Your target value: T = 0.0432139182637723 www.mrob.com/ries ln(x) = -pi (exact match) {49} pi,/x = 1/e for x = T + 2.08167e-17 {59} (Stopping now because best match is within 1e-15 of target value.) e = base of natural logarithms, 2.71828... A,/B = Ath root of B ln(x) = natural logarithm or log base e pi = 3.14159... --LHS-- --RHS-- -Total- max complexity: 36 25 61 dead-ends: 1357 550 1907 CPU time: 0.000 expressions: 41 44 85 distinct: 39 39 78 Memory: 64KiB Total equations tested: 1521 (1521) Also, a rhumb line from the south pole to the north pole of the unit sphere at a constant bearing a east of north has parametric (x,y,z) = (sech(w) * cos(tan(a) * w), sech(w) * sin(tan(a) * w), tanh(w)) -oo<w<oo. As w->oo, the distance from the north pole reduces by a factor approaching e^(2 π/tan(a)) for each circuit of the pole (2π change in longitude). For due northeast, a=π/4. (Since e^(2π) > 535, when you try to plot this spiral, you just see a curve that appears to reach the pole in about one orbit.) --rwg But things like sin(1) and cos(2) are exceptionally rare, so sin and cos should maybe be replaced with sin pi # and cos pi #. Irony: tan(1) is a bit more interesting due its continued fraction, but we'd need (sinpi 1/pi)/(cospi 1/pi) ! Also there's that pi,/4 limit at the beginning of http://www.tweedledum.com/rwg/idents.htm

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[math-fun] (1/3)!
by Bill Gosper 29 Dec '11

29 Dec '11

Seems about "twice as hard". (1/3)! == (2*2^(47/72)*((Pi*EllipticTheta[3, 0, E^(-((16*Pi)/Sqrt[3]))])/ (1/2^(3/16) + 1/(2^(7/16)*Sqrt[1 + Sqrt[3]]) + Sqrt[1 + Sqrt[3]]/(3 - 4*Sqrt[2] + 5*Sqrt[3])^(1/8)))^(2/3))/(3*3^(1/4)) Again just the terms through first order: 2 (1 + ----------------) Pi (16 Pi)/Sqrt[3] 47/72 E 2/3 2 2 (----------------------------------------------------------------) 1 1 Sqrt[1 + Sqrt[3]] ----- + ----------------------- + ------------------------------ 3/16 7/16 1/8 1 2 2 Sqrt[1 + Sqrt[3]] (3 - 4 Sqrt[2] + 5 Sqrt[3]) {(-)!, ------------------------------------------------------------------------------} 3 1/4 3 3 N[%, 51] {0.892979511569249211218564313658225881376229792652434, 0.892979511569249211218564313658225881376229792652429} --rwg

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[math-fun] Spot It card game
by Michael Kleber 29 Dec '11

29 Dec '11

I'm sorry, math-fun, I cheated on you. I recently played a cool card game called Spot It. It consists of a deck of 55 round cards; each card has eight assorted symbols on it. The deck has the property that every pair of cards shared exactly one common symbol, and they build a few games around the spot-the-shared-symbol mechanic. It's a lot of fun, and I've been playing it regularly with my 5-year-old and 12-year-old kids. The only annoying part? The any-two-cards-share-one-symbol thing is encoding point-line incidence in the projective plane over the field with 7 elements... in which there are fifty-SEVEN points. The manufacturers, for reasons I can't begin to guess, came up two cards short. (The cheating that I mentioned is because I wrote all this up, complete with pictures -- and posted it on Google Plus: https://plus.sandbox.google.com/114937925666302803969/posts/Fa9Znut5JmK. Go take a look -- I'm enjoying G+, and hey, my math-geeky post has been reshared 20 times now, with lots of non-math-focused people reading it. So there's something worthwhile going on in this slice of the social networking world.) So how does this generalize to n symbols per card? Any time n-1 is a prime power, the projective plane construction works, giving you n^2-n+1 cards (points) each showing n out of n^2-n+1 symbols (lines they're incident to). But when n-1 is not a prime power, e.g. when n=7, what can you get? That is, what's the largest finite geometry with a line between each pair of points, and at most n=7 lines incident to each point? Surely block design people must know the answer to this sort of question, but as far as I can tell, the OEIS doesn't have this. --Michael -- Forewarned is worth an octopus in the bush.

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[math-fun] ries
by Warren Smith 28 Dec '11

28 Dec '11

Cool! I tried ries on two of my favorite numbers (arising from theory of voting via some difficult integrals which might have a closed form but I used monte carlo) 0.2586333(3) = 0.2586335(4) and 0.1300873(4) = 0.1300876(2) [standard errors in last digit shown in parens] but it didn't find anything I liked. The correct formula (if there is one) may involve erf, e, pi, ln, and algebraic numbers. 0.1300876 = 1/(ln(2)*phi^5) [??] is suitably mysterious, I admit, and does match my accuracy. Then I tried Gosper's (1/4)! and I don't understand how Gosper got his formula. Actually, I practically never understand how Gosper gets his formulas. As I mentioned before, it'd be nice to find formulas for GAMMA(k/48) for example. Such formulas might exist in terms of algebraic numbers and the AGM.

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[math-fun] (1/4)!
by Bill Gosper 27 Dec '11

27 Dec '11

Neil fixed up my Mac with the next-to-latest version of Robert Munafo's ries searcher (www.mrob.com/ries) which led me to (1/4)! == Pi^(3/4)* EllipticTheta[3, 0, Exp[-4*Pi]]/(2^(1/4) + Sqrt[2]) which I had only known as a quotient of nested radicals, having neglected to attack it with Corey's denester. This also gives the simplifications {EllipticTheta[4, 0, E^(-4*Pi)] == (2*2^(1/16)*(1/4)!)/((-1 + Sqrt[2])^(1/4)* Pi^(3/4)), EllipticTheta[2, 0, E^(-4*Pi)] == ((-2^(1/4) + Sqrt[2])*(1/4)!)/Pi^(3/4)} Combining the theta[3] and theta[4], (1/4)! == ((2*Pi)^(3/4)*EllipticTheta[3, 0, E^(-16*Pi)])/ (1 + 2^(1/4) + 2^(13/16)/(-1 + Sqrt[2])^(1/4)) Taking the theta[3] terms just through first order, In[609]:= {(1/4)!, 2^(3/4)*(2/E^(16*Pi) + 1)* Pi^(3/4)/(2^(13/16)/(Sqrt[2] - 1)^(1/4) + 2^(1/4) + 1)} 2 3/4 (1 + ------) (2 Pi) 16 Pi 1 E {(-)!, ----------------------------} 4 13/16 1/4 2 1 + 2 + ----------------- 1/4 (-1 + Sqrt[2]) N[%, 88] {0.9064024770554770779826712889669180007487919207200163668583444998924798108846822804045900, 0.9064024770554770779826712889669180007487919207200163668583444998924798108846822804045892} --rwg Equal@@<the previous pair> tricks Mma into not saying False.

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[math-fun] Merry Christmath
by Dan Asimov 26 Dec '11

26 Dec '11

< http://www.youtube.com/watch?v=sxnX5_LbBDU&feature=youtu.be > ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Heron 4-simplex: proof of posability on the 4D "Hurwitz" lattice
by Warren Smith 25 Dec '11

25 Dec '11

A "Heron 4-simplex" is a 4D simplex with all edge lengths, triangle-areas, face-3D-volumes, and 4D-volume all integer. No Heron 4D simplex has ever been found and I suspect none exist. (Can any of you find one? Or prove nonexistence?) But if any do exist, then I now prove that it can be rotated and translated in 4-space so that all its vertices lie on the lattice of "Hurwitz integers" that is so that for each vertex EITHER all 4 coordinates are integers, OR all 4 coordinates are integers+1/2. We shall regard vertices as "Hurwitz quaternions" that is (A,B,C,D) is regarded as A+Bi+Cj+Dk. It is known that Hurwitz quaternion multiplication multiplies lengths, Hurwitzes have only integer lengths (see https://oeis.org/A004011 ) and that there is a GCD algorithm and unique factorization theorem, see book JH Conway & D Smith: On Quaternions and Octonions: Their Geometry, Arithmetic, and Symmetry, AK Peters, 2003 Specifically for any factorization F1 of |Q| into ordinary primes in any order, there exists a factorization F2 of the Hurwitz quaternion Q into Hurwitz primes so their lengths are the primes in F1 in the same order. PROOF OF HURWITZ-LATTICE POSABILITY: Use previous 3D posability proof to pose first 4 vertices on 3D integer lattice, then it is easy to see the 5th vertex has rational coordinates in 4D. Let L=LCM(denominators). Upscale your 4simplex by L so it is integer-posed but nonprimitive. All lengths now divisible by L. All vertices but one (namely E) now are on HurwitzLattice*L while E is on HurwitzLattice. E has length=integer hence |E|^2=squared integer, hence due to Hurwitz prime factorization theorem E is factorable within the Hurwitz quaternions E=A*B with |A|^2=|B|^2=|E|=integer. Apply rotation map X --> A~ X B to all 4simplex vertices X. (~ denotes quaternion conjugate, use quaternion multiplication.) Now still integer posed, and now newE = E' = A~ A B B = |A|^2 B^2 is on HurwitzLattice*|A|^2=HurwitzLattice*|E|. Meanwhile all other vertices lie on HurwitzLattice*L. Now we can safely de-scale by a factor of GCD(|E|, L) and still everything on HurwitzLattice. But |E| is divisible by L since |E|=integer*L. So the de-scaling factor is L. But that was precisely the upscaling factor in the first place. So the net effect has been a rotation without any rescaling, and it has made everything HurwitzLattice-posed. QED. NOTES: This theorem only needs to assume at least one edge has integer length and the other vertices (and there need not be 4 others, any number will do) all have integer 4D coordinates, and all the other edges have sqrt(integer) lengths. So it works for more general scenarios than merely Heron 4simplices. (Also my 3D proof worked for the more general scenario where one vertex E has rational xyz coordinates, any number of others have integer xyz coordinates, all edge lengths sqrt(integer) except at least one edge to E has integer length.) Fred Lunnon points out that Jan Fricke gave the example of a regular tetrahedron with edge-length 2 and integer vertices (0, 0, 0, 0), (1, 1, 1, 1), (2, 0, 0, 0), (1, 1, 1, -1). Dividing by 2 gives a set with all distances=1 which is easily seen not be posable on the 4D integer lattice, but it is posed on the HurwitzLattice. This suggests that my use of the HurwitzLattice in my proof (allowing some half-integer coordinates) was essential -- unlike in 3D and 2D where purely integer coordinates suffice for posability of Heron simplices. Finally I want to thank Fred Lunnon whose computer results stimulated all this and helped to restrict our wrong guesses. [It also is possible that a different 3D posability proof exists that is based on Fred Lunnon's computer program for finding poses -- by a semi-empirical process he found a fast pose-finding program that always seems to work, although it is currently not clear why.] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Heron tetrahedra: Proof of 3D integer-lattice posability
by Warren Smith 25 Dec '11

25 Dec '11

A "heron tetrahedron" is one with intger edge lengths, face-areas, and volume. Fred Lunnon and Jan Fricke both conjectured, both based on considerable computer evidence, that every Heron tetrahedron could be rotated & translated so that all its vertices had integer xyz coordinates. Indeed, Lunnon has now found at least one such "integer lattice pose" for every Heron tetrahedron with diameter<100,000. In this post, I will prove their conjecture. Idea behind proof: 3D integer lattice points will be regarded as "lipshitz quaternions" i.e. integer linear combinations of 1,i,j,k, where i*j=-j*i=k, j*k=-k*j=i, k*i=-i*k=j, i^2=j^2=k^2=-1. Fact: Any 3-dimensional (zero real part) Lipshitz quaternion E with integer length |E|, if scaled up by 2 times its length to get 2*|E|*E, becomes a square. Proof: Let R=|E|+E. Then R^2 = 2*|E|*E where E can be any pure-imaginary Lipshitz with integer length. QED Conjugation: R~ denotes quaternion conjugate: (A+Bi+Cj+Dk)~=(A-Bi-Cj-Dk). About R as a rotation map X --> R~ X R: this causes 2*|E|*E = R^2 to map to R~ R R R = |R|^2 R R = |R|^2 *2*|E|*E which lies on the integer 3D lattice times 2*|E|*|R|^2. Proof of integer pose existence theorem: Initially pose the tetrahedron so 3 of its vertices lie on the 2D integer lattice in the xy plane. (This can be done, as was prove in earlier posts using Gaussian integer GCD as a tool.) The remaining 4th vertex will be called E and is easily seen to have rational xyz coordinates (as a consequence of edge, area, and volume integrality). Let L=LCM(denominators). Upscale your tetrahedron by L so it is integer posed but nonprimitive. All edge lengths now divisible by L. All vertices but one (namely E) now are on lattice*L and B is on lattice. Compute R=|E|+E. Upscale tet again by 2*|E|. All vertices but one (namely newE=E') now are on lattice*2*|E|*L while E' has been put on lattice*2*|E|. Apply rotation map X --> R~ X R to all tet vertices X. Now still integer posed, and now E' is on lattice*2*|E|*|R|^2 and all other vertices on lattice*2*|E|*L. Now we can safely de-scale by a factor of 2*|E|*GCD(|R|^2, L) and still everything on integer lattice. But |R|^2=4*|E|^2 is divisible by L since |E|=integer*L. So the de-scaling factor is 2*|B|*L. But that was precisely the net upscaling factor in the first place. So the net effect has been a rotation without any rescaling, and it has made everything integer-lattice-posed. QED. In fact to do this I did not need to have a Heron tetrahedron. It sufficed if all vertices but one initially Lipshitz, and that one exception initially rational with length=integer. Notice we never even had to use quaternion-GCD anywhere and this proof works in 3D only. Finally, Lunnon & I have reason to suspect that eventually, for very large Heron tetrahedra (even primitive ones) there will often be many integer poses, i.e. arbitrarily high nonuniqueness. This contrasts with the 2D situation for Heron triangles. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Erich Friedman's game
by Warren Smith 24 Dec '11

24 Dec '11

As I daresay EF realizes, you can for any "one dimensional k-fold repeated game" solve by means of a recurrence where at each stage, the payoffs for the game are got from the solution to the next stage. [It also is possible to do this in more that one dimension, such as 2D array of games where after each game we move either North or East; this still is polynomial time to solve in any fixed dimension...] It might be interesting to write down the recurrences one can thus get: see if any are interesting+nice as recurrences and try to cook things up to get the simplest and most interesting recurrences you can. It seems to me EF's kind of scenario does arise in real life. Let me give some examples. In a multigame chess match (first to win 10 games win match). Each game you have option of playing a double-edged opening like sicilian defense, or a safer opening more likely to yield a draw. That choice is a meta-game analysable using EF-like approach. I thought in the Anand-Kasparov world champ match Anand was mis-playing this meta-game -- not that I'm really capable of disputing players of their calibre. In an american football game, again, each "down" is like an individual game and total yardage is like EF's points so far. In politics, Scum runs for the presidency versus Twit. Each week, another poll indicates who's currently ahead and each must decide what move to make that week, such as hacking the other guy's web site, making another incredibly stupid TV ad lying about opponent featuring mega-edited video, getting wife to come testify to cameras that you'd never betray her, arranging that voter registrations will be rejected in some state if they are not written on unavailable ultra-thick paper, etc. Some of these moves are riskier but have potentially higher payoff than others. Again, this is an iterated-game. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] DLMF fungous inflection
by Bill Gosper 24 Dec '11

24 Dec '11

Indubitably. But the author seemed to wish to qualitatively distinguish the the 2nd half of the interval. Spoiler lowermost. es>At K(k), the quarter period, the even derivatives of cn(x) vanish by symmetry. The special thing that happens at k = 1/sqrt(2) is that the third derivative vanishes. Thus the graph of cn(x) differs from a straight line by terms of 5-th order. -- Gene >________________________________> From: Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>>>To: math-fun(a)mailman.xmission.com <http://gosper.org/webmail/src/compose.php?send_to=math-fun%40mailman.xmissi…> >Sent: Friday, December 23, 2011 12:12 PM>Subject: [math-fun] DLMF fungous inflection> >(This was too easy for the kids, but it feels good>when you find the words an Authority missed.)>>See http://dlmf.nist.gov/22.3.F2.mag and http://dlmf.nist.gov/22.3.F3.mag>"For cn⁡(x,k) the curve for>k=1/sqrt2=0.70710⁢… is a boundary between the curves>that have an inflection point in the interval>0≤x≤2⁢K(⁡k), and its translates, and those>that do not;...">>a) why don't they just say "at x=K(k)" vs "in the interval 0≤x≤2⁢K(⁡k)"?>>b) Huh? cn⁡(x,k) always has an inflection point there! But *something*>happens at k=1/sqrt2. What should they have said?>--rwg>Caution: DLMF uses K(modulus) vs K(parameter), so if you plot this>with Macsyma or Mma, the magic k = 1/2, not 1/sqrt2. In either case,>K=4 (1/4)!^2/Sqrt[Pi] ~ 1.85407>(This K property is imperfectly analogous to the graphical definition of>Halphen's constant, about 69% into http://gosper.org/thetpak.html .)>_______________________________________________ For K > 4 (1/4)!^2/Sqrt[Pi], for x>K, the point of inflection of cn(x,k) splits into three! --rwg

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