June 2015 - mathfun@mailman.xmission.com

[math-fun] San Jose Math Decagon
by Bill Gosper 17 Feb '16

17 Feb '16

A couple of years ago Julian and I showed some animations to the San Jose Math Circle. gosper.org/decazoom.gif suggests why the area of a circle is the radius times half the circumference, and displays the continued fraction of a very lame approximation to π. --rwg

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[math-fun] xkcd
by rcs＠xmission.com 17 Dec '15

17 Dec '15

Today's XKCD cartoon has a nice collection of approximations. http://xkcd.com/1047/ There's the ancient expression 6 pi^5 for the mass ratio of the proton to the electron, and the more accurate (e^8 - 10)/phi. Robert Munafo & REIS are at least partially responsible. Rich

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[math-fun] Julian pimped my pump
by Bill Gosper 17 Dec '15

17 Dec '15

To run it <http://gosper.org/pumper_2.mp4> continuously, download with Miro, QuickTime-View-Loop. The pumped shapes split lengthwise resemble boats, so a grand entrance to a math museum would be a chain of these boats in an annular canal, passing through four rotors acting as an (approximate) air|water lock, with another four for the exit. Oh, the insurance. --rwg No manual entry for overboard

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[math-fun] Terminology: "Characteristic Generating Function"?
by Bill Gosper 23 Oct '15

23 Oct '15

"Indicator Generating Function"? "Predicate Generating Function"? What is the name for the G.F. that encodes an integer sequence in the exponents with coefficients in {0,1} ? E.g., the OGF for the squares is Sum[n^2*x^n, {n, ∞}] == -((x*(1 + x))/(-1 + x)^3), but the thing I'm looking for is Sum[x^n^2,{n,0,∞}]==(1/2)*(1 + EllipticTheta[3, 0, x]). (Years ago I sent some of you the GF for expressions for k as the sum of n (optionally distinct) members of such a set, in terms of the GF predicating membership in the set. It turned out to be a superspecial case of a formula in Generatingfunctionology that was so general I didn't even recognize it as the same problem.) The reason I ask is that Neil just came up with 1/2 (-1+EllipticTheta[2,0,x]/x^(1/4)+EllipticTheta[3,0,x]) predicating the quartersquares<http://en.wikipedia.org/wiki/Multiplication_algorithm#Quarter_square_multip…> A002620 <https://oeis.org/A002620>, from which he got (1/(2 (-1 + x)^2))((-1 + x) x^(3/4) EllipticTheta[2, 0, x] + x (3 - x + (-1 + x) EllipticTheta[3, 0, x])) generating what we think is the number of volumes you can pump with n of the recently discussed rotors, arranged optimally. --rwg

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Re: [math-fun] Another(?) stick length illusion
by Dan Asimov 25 Aug '15

25 Aug '15

Huh? At least to the nearest pixel, both lines are the same length. They are corresponding diagonals of congruent regular pentagons. I find this to be an impressive illusion. --Dan << << To me the stick that is in fact longer looks like it is in fact longer. If they're supposed to be the same length I think your image needs a bit of editing. >> On Thu, Aug 19, 2010 at 4:31 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > While hacking the pentagon fill, tutor Julian found > www.tweedledum.com/rwg/6pents.png, which seems to work backwards: > The stick larger than its context appears smaller than the > stick smaller than its context. >> _____________________________________________________________________ "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] New(?) elliptic K transformation
by Bill Gosper 20 Aug '15

20 Aug '15

On Mon, Nov 5, 2012 at 11:44 PM, Bill Gosper <billgosper(a)gmail.com> wrote: Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r] E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4] For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form. [Added 2012-11-16 07:03 Namely, EllipticK[1/2 + I/2] == Sqrt[-1 + Sqrt[2]] E^(I π/4) EllipticK[4 - 2 Sqrt[2]] And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I] ] None of the eight sign guesses works for Re(r)>1/2. If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg No one seemed to recognize it, so, grotesque though it is, NeilB and I decided it's time to write this up. So Neil bashed on it with Mathematica, and it got a simpler. Then I bashed on it so hard that it got unnervingly ungrotesque: (*) EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t] How could anything this simple be new? It looks like a routine 2F1 quadratic transformation. But which? As before, it makes the strange little results EllipticK[1/4] == E^(-((I \[Pi])/6)) EllipticK[Sqrt[3] E^((I \[Pi])/6)] and EllipticK[I] == (E^((I \[Pi])/16) EllipticK[1/2 - 1/4 Sqrt[7/2 - I/2]])/2^(1/8) Are these familiar? And somehow, (*) has miraculously lost its Re[t]<1/2 restriction:. E.g. for t=2, EllipticK[1/2] == (-1)^(1/4) EllipticK[2], as one might hope. For t=(-1)^(1/6), EllipticK[1/4 (2 + (-2 + (-1)^(1/6))/Sqrt[1 - (-1)^(1/6)])] == (2 - Sqrt[3])^(1/8) E^(-((5 I \[Pi])/48)) EllipticK[E^((I \[Pi])/6)] Angling for a K'/K identity, solve 1/4 (2 + (-2 + t)/Sqrt[1 - t]) == 1 - t : t -> 1/32 (31 + 3 I Sqrt[7]) So, Log[EllipticNomeQ[1/32 (31 - 3 I Sqrt[7])]] == -(1/2) Sqrt[1/2 (-3 I + Sqrt[7])] E^((I \[Pi])/4) \[Pi] along with its formal conjugate. Stop me if you've heard this one. --rwg (Is there a (dlmf?) page of special values of nome? Or just garden varieties:-)

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[math-fun] the 19th singular value
by Bill Gosper 16 Jul '15

16 Jul '15

ETA(%E^-(2*SQRT(19)*%PI)) = ((3*SQRT(57)+1)^(1/3)-8/(3*SQRT(57)+1)^(1/3))^(1/8)*(GAMMA(1/38)*GAMMA(5/38)*GAMMA(9/38)*GAMMA(11/38)*GAMMA(7/19)*GAMMA(17/38)*GAMMA(23/38)*GAMMA(25/38)/(38*GAMMA(13/38)*GAMMA(8/19)))^(1/4)/(2*2^(53/152)*%PI) where Corey's denester neatly whacks the 8th root: (-(8/(1 + 3*Sqrt[57])^(1/3)) + (1 + 3*Sqrt[57])^(1/3))^(1/8) == -(2^(3/8)/3) + ((-1 + 3*Sqrt[57])^(1/3)*(4*(-5 + Sqrt[57]) + (11 + Sqrt[57])*(-1 + 3*Sqrt[57])^(1/3)))/(96*2^(5/8)) . It also denests LambdaStar[19], but the result is badly undersimplified. Fullsimplify thuggishly reverts to the original Root object, even when given a ComplexityFunction that enormously penalizes _Root ! But it is possible to guess the general form from the denester mess and determine coefficients with PSLQ: LambdaStar[19]==(((-Sqrt[57] + Sqrt[19] + 7*Sqrt[3] + 11)*(3*Sqrt[57] + 1)^(1/3)/ 16) - ((Sqrt[57] + Sqrt[19] - 7*Sqrt[3] + 11)/(2*(3*Sqrt[57] + 1)^(1/3))) - ((Sqrt[19] - 1)/2))/(3*Sqrt[2]) I was unable to Google whether this denesting is known. Is there a table of singular values more extensive than MathWorld's? --rwg Joerg>Let's not forget Broadhurst's very neat expressions: Message-ID: <200803140250.m2E2oHi4002315(a)philter.princeton.idaccr.org> Date: Thu, 13 Mar 2008 22:50:17 -0400 From: David Broadhurst <D.Broadhurst(a)open.ac.uk> To: NMBRTHRY(a)LISTSERV.NODAK.EDU Reply-To: David Broadhurst <D.Broadhurst(a)open.ac.uk> Subject: Singular value for Euler's numerus idoneus N=1365 default(realprecision,1050); a_1 = 550*sqrt(13) + 318*sqrt(39) + 750*sqrt(7) + 433*sqrt(21); a_2 = 1986 + 1145*sqrt(3) + 208*sqrt(91) + 120*sqrt(273); b = ((a_1-a_2)^2 - 2)/2; k_1365 = sqrt(1/2 - b^2/4*sqrt(4-b^2) - (2-b^2)/4*sqrt(1-b^2)); --------------- The denester also reduced the depth of k_1365, but made a proverbial Very Large Expression, more than Mathematica could further chew.

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[math-fun] Definition of mathematics
by James Propp 06 Jul '15

06 Jul '15

A couple of weeks ago, in a gathering of mathematicians throwing out candidate definitions of mathematics, I half-seriously ventured the opinion that mathematics is the subset of philosophy consisting of those philosophical questions that actually have answers, along with the answers to those questions. But I don't think this is original. Whom am I quoting (or paraphrasing) here? Jim Propp

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[math-fun] New(?) way to tabulate pFq identities
by Bill Gosper 06 Jul '15

06 Jul '15

By a process not obvious how to automate, I extracted HypergeometricPFQ[{a, b, (2 a)/5 + (6 b)/5 + 1, 2 a + 2 b, a + 3 b}, {(2 a)/5 + (6 b)/5, a + (3 b)/2 + 1/2, a + (3 b)/2 + 1, 2 b + 1}, -(1/4)] == (4^-a Gamma[b + 1/2] Gamma[2 a + 3 b + 1])/( Gamma[a + b + 1/2] Gamma[a + 3 b + 1]) from a summation containing a messy, irreducible cubic factor, analogous to the sort used to discover recurrence relations via "creative telescopy". Punting the cubic cost two degrees of freedom, but there were hundreds of variants. The unpleasant choice seems to be between a useless messy cubic, and a nearly useless identity with only two degrees of freedom. Probably the identity you need was among the hundreds I threw away. Here's an interesting remedy. What you actually tabulate is, in this case, the fully general result in a somewhat intimidating form: ((-b - d) ((1/(-1 + b))(-1 + b - d) (-1 + a + b - d) (-1 + b + c - d) HypergeometricPFQ[{a, -1 + b, c, d, -1 - a + b + d, -1 + b - c + d}, {b/2 + d/2, 1/2 + b/2 + d/2, 1 - a + d, 1 - c + d, -a + b - c + d}, -(1/ 4)] - (19 + 5 b^2 - 7 c + b (-19 + 4 c - 11 d) + a (-7 + 4 b + c - 3 d) + 20 d - 3 c d + 5 d^2) HypergeometricPFQ[{a, b, c, d, -1 - a + b + d, -1 + b - c + d}, {b/2 + d/2, 1/2 + b/2 + d/2, 1 - a + d, 1 - c + d, -a + b - c + d}, -(1/ 4)] - 5 (-3 + b) (-2 + b) HypergeometricPFQ[{a, -1 + b, b, c, d, -1 - a + b + d, -1 + b - c + d}, {-3 + b, b/2 + d/2, 1/2 + b/2 + d/2, 1 - a + d, 1 - c + d, -a + b - c + d}, -(1/ 4)] + 3 (-2 + b) (-8 + a + 3 b + c - 3 d) HypergeometricPFQ[{a, -1 + b, b, c, d, -1 - a + b + d, -1 + b - c + d}, {-2 + b, b/2 + d/2, 1/2 + b/2 + d/2, 1 - a + d, 1 - c + d, -a + b - c + d}, -(1/ 4)]))/(a c (-1 + b + c) (1 - b + c - d) d) == -((Gamma[1 - a + d] Gamma[1 + b + d] Gamma[ 1 - c + d] Gamma[-a + b - c + d])/(a (-1 + b) c (-1 + b + c) Gamma[1 + d] Gamma[-1 - a + b + d] Gamma[ 1 - a - c + d] Gamma[b - c + d])) It's then fairly routine to determine whether your pFq[-1/4] is a special case. It's not pencil and paper, but many people now have access to computers. --rwg Count yourself lucky if you can read "Rauzy fractal" and not remember Them Moose Goosers.

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[math-fun] Draft of "The Lessons of a Square-Wheeled Trike"
by James Propp 03 Jul '15

03 Jul '15

I've finished a draft of installment #1 (as opposed to #0) of "Mathematical Enchantments", and I'd welcome comments. Jim Propp

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