May 2014 - mathfun@mailman.xmission.com

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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[math-fun] Alan Adler, of Aerobie fame,
by Bill Gosper 14 Sep '14

14 Sep '14

finds Euler's product over primes for zeta(s) somewhat magical, and wonders if we funsters know other really neat sums or products involving primes. Hence yesterday's "Minor twist". Otherwise, all I could remember was our Valentine's Day 2011 discussion of "Euler's crazy pi product". --rwg

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[math-fun] why can a line of slope 10^n determine n decimal digits of π?
by Gary Antonick 19 Jun '14

19 Jun '14

Just noticed this yesterday and can't figure out why it works. Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle. Now try again with a line of slope -100. The line will create an arc 1/314 of the circle. Starting to look a bit like pi. Am wondering why this is. In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.

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[math-fun] Subset-lex: did we miss an order?
by Joerg Arndt 10 Jun '14

10 Jun '14

Remember those combinatorial listings I used to annoy math-funsters with? Finally I managed to write things down: http://jjj.de/tmp-subset-lex/ If anyone wants to comment, I'd be grateful. Simple, loopfree, and fast algorithms for the generation of various combinatorial objects are given. Great care has been taken to keep things accessible as much as possible. Best, jj

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