September 2014 - mathfun@mailman.xmission.com

[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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[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] Fwd: harder 7x7x7?
by Bill Gosper 03 Jun '15

03 Jun '15

I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf ---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314(a)gmail.com> Cc: Julian Ziegler Hunts <julianj.zh(a)gmail.com>, Jon Ziegler < jonathan.zh(a)gmail.com>, Corey Ziegler Hunts <corwin.zh(a)gmail.com>, Michael Beeler <mikebeeler(a)verizon.net>, rcs(a)xmission.com, stan(a)isaacs.com Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill

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[math-fun] What is elliptic K and E closed form?
by Bill Gosper 25 May '15

25 May '15

I think the answer has changed in the last 50 yrs. A query from constantologist Steve Finch revived my interest in whether special value identities like EllipticK[(1 + Sqrt[2])^4] == ((1/4 - I/2)π^(3/2))/ ((2 + Sqrt[2]) Gamma[3/4]^2) and EllipticK[8 2^(1/4) (-1 + Sqrt[2])^3* E^(I ArcTan[(3 (3 + Sqrt[2]))/(4 2^(1/4))])] == (I (1 + Sqrt[2]) E^(-2 I ArcCot[2^(1/4)]) π^(3/2))/ (4 Sqrt[2] Gamma[3/4]^2) are already known, and where one might find tables much more extensive than, e.g., http://functions.wolfram.com/EllipticIntegrals/EllipticK/03/01/ or http://dlmf.nist.gov/19.6 . DLMF says "Exact values of K(k) and E(k) for various special values of k are given in Byrd and Friedman (1971, 111.10 and 111.11)" (1st English edition, 1954) Neil and his Mom were kind enough to fetch me this comprehensive (*Handbook* of *Elliptic* Integrals for Engineers and Physicists<http://www.amazon.com/Handbook-Elliptic-Integrals-Engineers-Physicists/dp/B…>), and to my surprise and disappointment, there was not a single identity of the above form, i.e. algebraics times Gammas(rationals). How could this be? My guess: In the 1950s K and E were considered closed forms while Gamma was not! It makes sense: K and E are of lower computational complexity. We Funsters once concluded that Gamma[k/24] was rapidly computable thanks to elliptic K identities. But now, in papers like "EVALUATION OF COMPLETE ELLIPTIC INTEGRALS OF THE FIRST KIND AT SINGULAR MODULI" (people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf*) "evaluation" means "determin[ation] explicitly in terms of Gamma values". Which is what I always thought. See, e.g., http://en.wikipedia.org/wiki/Elliptic_integral#Special_values --rwg

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Re: [math-fun] parametrizing the Koch snowflake curve
by Bill Gosper 24 Apr '15

24 Apr '15

Fourier epicycles thru nth: (d16) sum(((product((1 - (( - 1)^j * tan(((%pi * (2 * k + (1/3)))/(2^j)))/(sqrt(3)))),j,1,inf)) * (%e)^(%i * (6 * k + 1) * t)/((2 * k + (1/3))^2)),k, - n,n) See http://www.tweedledum.com/rwg/cog.htm Also http://gosper.org/fst.dvi Pix: http://gosper.org/fst1.png ,http://gosper.org/fst2.png , http://gosper.org/fst3.png ,http://gosper.org/fst4.png . (For parametric, take Re and Im.) Rich's base 4 recursion loops for rational arg. Find the fixed point of the pending stack ops to simulate infinite recursion, and get an exact val on the curve. Equate to this Fourier series for a *non-obvious* identity. --rwg > If you view the starting triangle as running from 0 to 3, > and write the fraction base 4, you can use the > fraction digits to select which part of the (sub)edge > you are on. The orientation of the edge can be calculated > from the 1s and 2s of the high order portion of the fraction, > and that should give a somewhat kludgy sum for the (x,y) > coordinates of the image point. > RWG likely has some nice FFT clock sum expression. > > Rich > > > -----Original Message----- > From: math-fun-bounces(a)mailman.xmission.com [mailto: math-fun-bounces(a)mailman.xmission.com] On > Behalf Of N. J. A. Sloane > Sent: Thursday, February 25, 2010 12:43 PM > To: math-fun(a)mailman.xmission.com > Cc: njas(a)research.att.com > Subject: Re: [math-fun] parametrizing the Koch snowflake curve > > Jim Propp asked: > >> Does anyone (Bill Gosper, maybe?) know an explicit parametrization > of the Koch snowflake curve? > > The Cantor set consists of the points 0 <= x <= 1 which have a base 3 "decimal" expansion that > does not contain a 1. Is there a similar characterization of the Koch snowflake? > > Neil Sloane > >

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[math-fun] That fractal pentagram again
by Bill Gosper 09 Apr '15

09 Apr '15

For some pictorial. mildly trick questions: http://gosper.org/foupent.pdf (1.6M) --rwg

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[math-fun] Microwave ovens
by James Propp 02 Feb '15

02 Feb '15

Why do microwave ovens alternate between rotating the food-carousel clockwise and rotating it counterclockwise (or, as the Brits would say, anticlockwise)? Jim Propp

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Re: [math-fun] analytic function for new Airbnb logo?
by Bill Gosper 14 Dec '14

14 Dec '14

Henry>I wanted an approximation that was entire over the complex plane, so the straight lines and discontinuities would have to be fudged. If I take the image, convert it to black-and-white, and erode the boundaries to a single pixel width, I will get a "continuous" curve that goes around the origin twice. If I consider this curve in (r,theta) coordinates, I can now consider the Fourier transform. I'd like to reset the origin in such a way that the "fundamental" is zero -- i.e., the lowest frequency (which takes 2 revolutions) has a coefficient of zero. (There are other possible locations for the origin: the balancing point, the center of gyration, etc.) That having been done, I could alternatively use (log(r),theta) coordinates, which have some nicer properties, because it is the actual complex log function. Fourier analysis can approximate the discontinuous changes in curvature; hopefully modifying these coefficients a little will remove these discontinuities with a minimum of ringing. The sides of your rounded edge squares are impressively flat. <HGB At 03:48 PM 7/30/2014, Bill Gosper wrote: HGB> My approach would be to take the Fourier transform of the polar plot, rwg>Defined as piecewise linear|circular? and then simplify to a small number of coefficients. Of course, it will take 2 circuits around the origin to complete the figure. The sides are perhaps a little too flat, which will blow up the number of coefficients required. I guess if you transform [0,4pi) to [0,1) you could use some sort of Cheby polynomial approximation. It would be nice to be able to transform it back to p(x,y)=0, where p is a polynomial in x,y. At 02:56 PM 7/29/2014, Whitfield Diffie wrote: HGB>Anyone game for an analytic function to approximate Airbnb's new logo? They should manufacture a paper clip and give it out at trade shows. Whithttp://www.underconsideration.com/brandnew/archives/airbnb_logo_detail.… rwg> You can get some impressively straight segments with theta fcns: gosper.org/halph.pdf (10MB?!) or see http://gosper.org/thetpak.html (scrolled 69%). (.3281... is âˆš(one ninth constant), which I claim should be named Halphen's. It's the largest q for which the curve is convex.)) As I mentioned here before, as 0<q<1, the curve continuously deforms from a tiny "circle" to a big "+" sign, sweeping out a fairly convincing spaceplane: http://gosper.org/sst.png --rwg But a problem with this logo is that it switches curvatures discontinuously. <rwg Another possibility: modulate the arcspeed (slow down for curves, speed up for straightaways) before taking the Fourier.

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