November 2016 - mathfun@mailman.xmission.com

[math-fun] Julian has constructed a spacefill dense with sextuple points
by Bill Gosper 26 Jan '17

26 Jan '17

using a 25-way conditional. ClearAll[piecewiserecursivefractal]; piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := piecewiserecursivefractal[x, g_, which, iters, fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := Block[{y}, y /. Solve[f[y] == h[y], y]]); Union @@ ((fns[[#]] /@piecewiserecursivefractal[iters[[#]][x], Composition[f, fns[[#]]], which, iters, fns]) & /@which[x])); (I cannot praise this function enough.) In[363]:= trifil25[t_] := piecewiserecursivefractal[t, Identity, Piecewise[{{{1}, #1 == 0}, {{Ceiling[25*#1]}, Inequality[0, Less, #1, LessEqual, 1]}}, {}] & , {25*#1 + 0 & , -25*#1 + 2 & , 25*#1 - 2 & , 25*#1 - 3 & , 25*#1 - 4 & , 25*#1 - 5 & , 25*#1 - 6 & , 25*#1 - 7 & , 25*#1 - 8 & , -25*#1 + 10 & , -25*#1 + 11 & , 25*#1 - 11 & , -25*#1 + 13 & , -25*#1 + 14 & , 25*#1 - 14 & , -25*#1 + 16 & , 25*#1 - 16 & , -25*#1 + 18 & , 25*#1 - 18 & , 25*#1 - 19 & , -25*#1 + 21 & , -25*#1 + 22 & , 25*#1 - 22 & , -25*#1 + 24 & , 25*#1 - 24 & }, {#1/5 & , (1/10)*(3 + I*Sqrt[3] + (-1 - I*Sqrt[3])* #1) & , (1/10)*(3 + I*Sqrt[3] + #1 - I*Sqrt[3]*#1) & , (2 + #1)/5 & , (1/10)*(6 + #1 + I*Sqrt[3]*#1) & , (1/10)*(7 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (1/5)*(3 + I*Sqrt[3] - #1) & , (1/10)*(4 + 2*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , (1/10)*(3 + I*Sqrt[3] - 2*#1) & , (-(1/10))*I*(-I + Sqrt[3])*(-2 + #1) & , (1/5)*(2 + I*Sqrt[3] - #1) & , (1/5)*(2 + I*Sqrt[3] + #1) & , (1/10)*(5 + 3*I*Sqrt[3] + #1 - I*Sqrt[3]*#1) & , (1/10)*(4 + 2*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , (1/10)*(4 + 2*I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (-(1/10))*I*(-I + Sqrt[3])*(-4 + #1) & , (1/5)*(2 + 2*I*Sqrt[3] + #1) & , (1/10)*(5 + 3*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , (1/10)*(5 + 3*I*Sqrt[3] + 2*#1) & , (1/10)*(7 + 3*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , (1/5)*(4 + I*Sqrt[3] - #1) & , (1/10)*(9 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (1/10)*(9 + I*Sqrt[3] - 2*#1) & , (1/10)*(8 + I*(I + Sqrt[3])*#1) & , (4 + #1)/5 & }][[1]] E.g., In[361]:= trifil25[2/25 + 2/75] // ComplexExpand Out[361]= {1/2 + (I Sqrt[3])/10} In[367]:= Graphics[{Hue@#, Line[ReIm@{trifil25@#, trifil25[# + 5^-5]}]} & /@ Range[0, 1 - 5^-5, 5^-5]] gosper.org/trifil25.png (Those little white hexagons are sextuple points.) The inverse fractal, constructed more or less automatically: In[380]:= untrifil25[t_] := piecewiserecursivefractal[t, Identity, If[0 <= Re[#1] <= 1 && 0 <= Im[#1] <= Sqrt[3]/2, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}, {}] &, {ComplexExpand[5 #1] &, ComplexExpand[1/4 (6 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-4 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[-2 + 5 #1] &, ComplexExpand[1/4 (-6 + 6 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (4 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[3 + I Sqrt[3] - 5 #1] &, ComplexExpand[1/4 (10 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (3 + I Sqrt[3] - 10 #1)] &, ComplexExpand[1/4 (8 - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[2 + I Sqrt[3] - 5 #1] &, ComplexExpand[-2 - I Sqrt[3] + 5 #1] &, ComplexExpand[1/4 (4 - 8 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-10 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-2 + 6 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (16 - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[-2 - 2 I Sqrt[3] + 5 #1] &, ComplexExpand[1/4 (-14 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (-3 I Sqrt[3] + 5 (-1 + 2 #1))] &, ComplexExpand[1/4 (16 - 4 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[4 + I Sqrt[3] - 5 #1] &, ComplexExpand[1/4 (6 + 10 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (9 + I Sqrt[3] - 10 #1)] &, ComplexExpand[1/4 (8 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[-4 + 5 #1] &}, {#1/25 &, (2 - #1)/25 &, (2 + #1)/ 25 &, (3 + #1)/25 &, (4 + #1)/25 &, (5 + #1)/25 &, (6 + #1)/ 25 &, (7 + #1)/25 &, (8 + #1)/25 &, (10 - #1)/25 &, (11 - #1)/ 25 &, (11 + #1)/25 &, (13 - #1)/25 &, (14 - #1)/25 &, (14 + #1)/ 25 &, (16 - #1)/25 &, (16 + #1)/25 &, (18 - #1)/25 &, (18 + #1)/ 25 &, (19 + #1)/25 &, (21 - #1)/25 &, (22 - #1)/25 &, (22 + #1)/ 25 &, (24 - #1)/25 &, (24 + #1)/25 &}] Here it finds the preimages of the simplest sextuple point: In[381]:= untrifil25[1/2 + I Sqrt[3]/10] Out[381]= {8/75, 11/75, 14/75, 17/75, 4/15, 23/75} --rwg

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[math-fun] Trigonometric "ensemble"(?) sums (and products)
by Bill Gosper 02 Dec '16

02 Dec '16

I.e., equally spaced over a period. What's the term? (Special case of sum|product over a root system.) I once convinced myself that *all* of them, even such nasties as Sum[Cot[f + (j*Pi)/n]*Cot[g + (j*Pi)/n]*Cot[h + (j*Pi)/n], {j, 1, n}]/n == -Cot[f*n] - Cot[g*n] - Cot[h*n] - (1/2)*Csc[f - g]*Csc[f - h]*Csc[g - h]* ((-Cot[h*n])*Sin[2*f - 2*g] + Cot[g*n]*Sin[2*f - 2*h] - Cot[f*n]*Sin[2*g - 2*h]) come out in closed form as direct consequences of nothing more than x^n-1 == Product[x-I^(4*k/n), {k, 0,n -1}]. Computer algebra junkies may "enjoy" starting here and plowing through to Sum[1/(a + b*Cos[c + (2*k*Pi)/n]), {k, 0, -1 + n}] == ((b^(2*n) - (-a + Sqrt[a^2 - b^2])^(2*n))*n)/(Sqrt[a^2 - b^2]* (b^(2*n) + (-a + Sqrt[a^2 - b^2])^(2*n) - 2*b^n*(-a + Sqrt[a^2 - b^2])^n*Cos[c*n])) ("I ask a simple question; I get a pageant." --Stan Freberg) --rwg Note the messed up ordering of the nasty cot cot cot result. Is any Computer Algebra able to, e.g., convert Sin[x - y] - Sin[x - z] + Sin[y - z] to Sin[x - y] + Sin[y - z] + Sin[z - x] ? At least Macsyma had trigsign:false to partially inhibit the reverse transformation. But then the sum would be reversed. But it also had ordergreat and orderless, which aliased symbols to have different display orderings. It only accepted individual symbols, not relations between them, but it might barely suffice here. In Mathematica, you can Inactivate all the trigs and then run around manually flipping the anticyclic ones. I'll bet people have built such tools for themselves, but if anybody has a really good solution, WRI should have heard about it.

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[math-fun] We missed it.
by Bill Gosper 02 Dec '16

02 Dec '16

RKG turned 100 on Sep 30 ! --rwg

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Re: [math-fun] meth-fun
by Henry Baker 29 Nov '16

29 Nov '16

[This is a serious question.] Although it isn't much talked about (in some cases because those who took it weren't aware of what they were taking), meth was heavily used on both sides in WWII to attempt to accomplish 'impossible' tasks. Having seen "The Imitation Game", I have the question: did those working on Enigma et al use meth? At 09:02 AM 11/29/2016, James Propp wrote: >Are meth and math so different? Both can provide temporary escape from reality. :-) > >Jim Propp > >On Mon, Nov 28, 2016 at 8:01 PM, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > >> Serves 'em right --- they should have stuck to Matematica. WFL >> >> On 11/28/16, Bill Gosper <billgosper(a)gmail.com> wrote: >> > (via Carolyn Bickford) >> > >> > http://www.imediaethics.org/math-lab-bust-whoops-kentucky-tv-news-botches-o… >> > --rwg >> > And metaheadline writer botches oops into whoops.

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[math-fun] meth-fun
by Bill Gosper 29 Nov '16

29 Nov '16

(via Carolyn Bickford) http://www.imediaethics.org/math-lab-bust-whoops-kentucky-tv -news-botches-on-air-caption/ --rwg And metaheadline writer botches oops into whoops.

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[math-fun] Idea for a plaque
by Bill Gosper 29 Nov '16

29 Nov '16

Spelling lessons aren't quite as passé as spell-checkers might make them seam. --rwg

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Re: [math-fun] Exactly Distributed Digits in the Initial Digits of Pi
by Dan Asimov 26 Nov '16

26 Nov '16

Yes, that is a famous result of George Polya from 1921 that is explained in Feller vol. 1. I found it amazing at first that 1 and 2 dimensions should differ from all higher dimensions in that respect. This also has implications for heat flow in those dimensions: If n-space is a (homogenous and isotropic) heat-conducting medium with the unit ball kept at unit temperature and the rest of n-space beginning at temperature 0 at t = 0, then for n = 1 and n = 2, n-space asymptotically warms up to unit temperature but not for n >= 3 (cf. "There's no space like home", The Sciences, September-October 1995). —Dan >From: Hans Havermann <gladhobo(a)bell.net> >Sent: Nov 26, 2016 9:46 AM >I have been informed by Warren D. Smith that it is known that 1- and 2-dimensional random walks are recurrent, while 3-dimensional (and higher) such walks are transient. From that perspective, it strikes me that the sequence 15, 18, ... (third term > 8*10^9) [the number of initial digits of ternary pi where the digits 0, 1, and 2 appear exactly the same number of times] should be recurrent because it can be modeled by a 2-dimensional (presumably random) walk (just as I had done in my blog article). The implication would be that this sequence is infinite!

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Re: [math-fun] Exactly Distributed Digits in the Initial Digits of Pi
by Dan Asimov 26 Nov '16

26 Nov '16

Fascinating. (I'm not familiar with the equidistribution of pi digits in record-calculations, though.) I am somewhat familiar with the arcsine law for penny-matching, which implies that in the limit as the number N of coin flips approaches oo, the fraction of the N flips that are (say) heads that is *least* likely is 1/2 — just the opposite of what I would have guessed before taking a probability course in college that used Feller vol. 1 as the text. Now I'm wondering about something similar for a 3-sided coin. —Dan ----- Everyone is acquainted with record-calculations of pi and how their (base ten) digits are equally distributed. Closely, but not of course exactly. By contrast, in base two there are seemingly an infinite number of initial digits of pi that contain exactly the same number of zeros and ones: https://oeis.org/A039624 What a difference adding one digit/dimension can make. In base three, the random walk of digits allows the initial 15 and 18 digits of pi to be exactly equidistributed by sheer small-numbers chance [in base four, the first 4 digits (3.021); in base five, the first 75 digits]. I thought I had a shot at finding a third solution in base three but it was not to be: http://gladhoboexpress.blogspot.ca/2016/11/a-deep-walk-in-base-three-pi.html -----

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[math-fun] Exactly Distributed Digits in the Initial Digits of Pi
by Hans Havermann 26 Nov '16

26 Nov '16

Everyone is acquainted with record-calculations of pi and how their (base ten) digits are equally distributed. Closely, but not of course exactly. By contrast, in base two there are seemingly an infinite number of initial digits of pi that contain exactly the same number of zeros and ones: https://oeis.org/A039624 What a difference adding one digit/dimension can make. In base three, the random walk of digits allows the initial 15 and 18 digits of pi to be exactly equidistributed by sheer small-numbers chance [in base four, the first 4 digits (3.021); in base five, the first 75 digits]. I thought I had a shot at finding a third solution in base three but it was not to be: http://gladhoboexpress.blogspot.ca/2016/11/a-deep-walk-in-base-three-pi.html

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[math-fun] We missed it.
by Bill Gosper 25 Nov '16

25 Nov '16

[math-fun] Date: 2016-11-25 09:25 From: Hans Havermann <gladhobo(a)bell.net> RWG: "RKG turned 100 on Sep 30 !" So much for your professed (Sep 3) "distaste for decimal digits". ;) ------- 𝖨𝖨ché—I should've 𝚇cepted the integers! --rwg

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