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[math-fun] I was whimpering to Julian about being too stupid
by Bill Gosper 10 Dec '19

10 Dec '19

to use his tools to invert the semihexagons spacefiller. Unfortunately, Mac problems lost him access to Mathematica for a while. But now he's back with a vengeance. And I don't feel so stupid. The two semihexagons, closed loop & open curve, are defined in terms of each other in a simultaneous recursion. For which Julian's fractal inverter makes no provision. He somehow worked around this with an incredible contortion I will probably never understand. Here it is finding multiple points by inverting the semihexes at successive Wilf-Calkin fractions. (In[182]:= NestList[1/(#/(# - 1) + 2 Floor[1/(1 - #)]) &, 0, 11] Out[182]= {0, 1/2, 1/3, 2/3, 1/4, 3/5, 2/5, 3/4, 1/5, 4/7, 3/8, 5/7} ) foo = 0; Do[ If[Length[invsemi[semihexes[foo = nunu@foo][[1]]]] > 1, Print@{semihexes[foo][[1]], invsemi[Identity[semihexes@foo][[1]]]}], {999}] (Legend: {complex value, preimages}) {1/2+(I Sqrt[3])/2,{1/10,1/4,1/2,13/20}} {1/2+(I Sqrt[3])/2,{1/10,1/4,1/2,13/20}} {-1,{3/4,9/10}} {1+I Sqrt[3],{1/5,3/10}} {5/4+(I Sqrt[3])/4,{27/80,3/8,7/16,19/40}} {-(1/4)+(I Sqrt[3])/4,{47/80,5/8,11/16,29/40}} {-(1/2)+(I Sqrt[3])/2,{1/20,3/40,27/40,7/10,37/40,19/20}} [Sextuple!] . . . Here <http://gosper.org/hex&quad.png> is a triple point on a side of the hexagon, and a quadruple point interior. The interior corner of the little gold trapezoid is either triple(??), quad or hex. Three colors at a vertex guarantees at least a triple point, because spacefilling functions map closed intervals onto closed areas. Julian has shown that the preimages of all the sextuple points have denominators of the form 5⨉2ⁿ. Above, we saw that some quadruple point denominators are just 2ⁿ. But some double point denominators are more exotic: {2/3+(2 I)/Sqrt[3],{14/65,37/130}} {13/10+(I Sqrt[3])/5,{513/1360,37/85}} {23/14+(I Sqrt[3])/7,{41/105,1313/3360}} {17/10+(I Sqrt[3])/5,{33/85,1069/2720}} Uncountably many double points are transcendental, but I don't know how to exhibit any. It would probably involve implementing the semihexes as a finite-state machine, then finding a Thue-Morse-like preimage. . —rwg

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[math-fun] Most valuable moments
by Éric Angelini 10 Dec '19

10 Dec '19

Hello Math-Fun, say we could append a value to every moment of our life during a year; we would simply add the digits of the date (month and day) to the digits of a precise time (hours, minutes, seconds). For example, the second before 2020 would be worth (1+2) + (3+1) + (2+3+5+9) = 3+4+19 = 26. What moment of the year has the highest value? What is the most common value in a year? (midnight on the first of January = 2 = mid- night on October 10th) How many moments are prime in a year? Best, É. (sent at moment 27, Brussels time = 31 XII 19:55)

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[math-fun] π_q near q=1
by Bill Gosper 08 Dec '19

08 Dec '19

My idea for q-Trigonometry was simply to substitute QFactorial for Factorial in the identity z! (-z)! == π z/Sin[π z]. (z = -1/2 is where we get (-1/2)! = √π.) This determines the "q-extension" of π to be q^(1/4) QFactorial[-1/2,q^2]^2== π*(1+(q-1)+1/6 (q-1)^2-1/180 (q-1)^4+1/180 (q-1)^5-(37 (q-1)^6)/7560+4/945 (q-1)^7-(119 (q-1)^8)/32400+(9 (q-1)^9)/2800+O[q-1]^10) (Arguably, q^(1/8) QFactorial[-1/2,q]^2). This expansion (which should be in powers of 1-q) dates from an ancient Macsyma. I was unable to drag it out of Mathematica, which is woefully ignorant of expanding *q*-things near the all-important q➝1 limit. Likewise, other points on the unit circle. But Mathematica should be able to average π_q for us over the unit disk. For every triplet of distinct positive integers a,b,c, there is a polynomial identity relating π_q^a, π_q^b, and π_q^c, because π_q==(1-q^2) 𝜂(q^2)^4)/𝜂(q)^2, where 𝜂(q) := q^(1/24) Product[1-q^n,{n,∞}] = DedekindEta[Log@q/2/I/π] and 𝜂(q) satisfies an equivalent infinitude of three term identities. The simplest π_q identity is Out[95]= 4==-(1+q^4)^2 (π_q^2)^2/(π_q^4)^2+(1+q^2)^2 (1+q^4) (π_q)^2/π_q^2/π_q^4 (Converted to 𝜂 functions, this is Jacobi's Æquatio Identica Satis Abstrusa. But the *least* abstruse, it turns out!). Testing with our π_q expansion near 1, π_q:>π*(q+1/6 (-1+q)^2-1/180 (-1+q)^4+1/180 (-1+q)^5-(37 (-1+q)^6)/7560+4/945 (-1+q)^7-(119 (-1+q)^8)/32400+(9 (-1+q)^9)/2800) % /. q -> q^2; % /. q -> q^2; (My need to special case q^2 and q^4 affirms that the design of Series is seriously deficient.) %95 /. {%, %%, %%%} <Big mess> In[117]:= Series[Activate[%],{q,1,9}] Out[117]= 4==4+O[q-1]^10 Quod erat orandum. Why idolize a mere number? —rwg

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[math-fun] Raspberry Pi and the disappearing/reappearing Mathematica
by Hilarie Orman 08 Dec '19

08 Dec '19

>From what I've been able to determine, between July and September, there was no Mathematica included in the Raspberry Pi distribution. The latest version does have Mathematica. Hilarie

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[math-fun] Rasperry Pi 4 has no Mathematica
by Hilarie Orman 08 Dec '19

08 Dec '19

Until a few months ago, Raspberry Pis came with Mathematica already installed. That is no longer true. Don't buy a Pi with Mathematica in your eye. Hilarie

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Re: [math-fun] Blankets
by Henry Baker 08 Dec '19

08 Dec '19

Depends upon what the blanket is made of. E.g., I don't think that an ordinary wool blanket would commute with one of the silvery hi-tech blankets that first responders use. Those silvery blankets *reflect* (essentially all normal) wavelengths, while a wool blanket depends upon a quiet thickness of (hopefully dry) air; the wool is merely used to keep the thickness and reduce air movement (convection). I noticed that my down sleeping bag doesn't protect me from the cold ground -- I need a separate hard foam that keeps me a certain distance from the ground. On the other hand, those silvery first responder blankets also trap moisture, so you could end up with the inner surface covered with ice -- like uninsulated windows on a winter's day. At 07:19 AM 12/5/2019, James Propp wrote: >Do blankets commute? > >Not esthetically (of course), but thermally? > >Jim Propp

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[math-fun] Blankets
by James Propp 05 Dec '19

05 Dec '19

Do blankets commute? Not esthetically (of course), but thermally? Jim Propp

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[math-fun] Oh no! More Syzygies!
by Brad Klee 04 Dec '19

04 Dec '19

Previously I claimed that the null ring of NxN matrix invariants could be constructed by observing unitary transformations of the coefficients of the characteristic equation. This is an okay method, but not the only approach. Specht's theorem [1] provides an alternative using only traces of products of a square matrix A and it's complex conjugate A*. If we assume A*=A, then the only invariants are Tr(A^n) and should be finitely generated by T(A), T(A^2), T(A^3) up to a cutoff T(A^N), which depends on the dimension of A. For dim n=2, choose diagonal A=[L1,L2], then Tr(A)=L1+L2, and Tr(A^2)=L1^2+L2^2. Alternatively, det(A-Id2*L) = L^2 - Tr(A)*L+det(A). The syzygy is that det(A) = 1/2*(Tr(A)^2-Tr(A^2)). Similarly there are syzygies for dim(A)=3 to convert Tr(A), Tr(A^2), Tr(A^3) into the coefficients of the characteristic equation. Thus we can write a set of generators for the null ring (Identities: 0 + g = g, 0*g=g, g in NR ): NR={(g1+g2)*g3: g1,g2,g3 all in NR} U{ 0, Tr(A)-Tr(B), Tr(A^2)-Tr(B^2), Tr(A^3)-Tr(B^3) } with B unitarily equivalent to diagonal A. To solve the problem of changing symbols within R=K[[b_{i,j}, L1, L2, L3]] we look for identities of the form p2 = p1 + p3*pn, with p1,p1,p3 in R and pn in NR, and say p1=p2 mod 0. Let us not forget to cite Carl Pearcy. He also did some work on 3x3 matrices, and published in transactions of the AMS , 1962 on "A complete set of unitary invariants for 3x3 complex matrices" [2]. I want to reiterate again, or to re-reiterate--In my opinion Tao et al. did not cite the Sprecht theorem of 1940 and other works thereafter in an attempt to make their work seem more original and more "unexpected". --Brad https://en.wikipedia.org/wiki/Specht%27s_theorem https://www.ams.org/journals/tran/1962-104-03/S0002-9947-1962-0144911-4/

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[math-fun] Terence Tao's Blog: The Missing Proof.
by Brad Klee 04 Dec '19

04 Dec '19

https://terrytao.wordpress.com/2019/08/13/eigenvectors-from-eigenvalues/ Here is the proof I wrote, in Mathematica: ZS[n_] := M[n, n] (M[1, 1] + M[2, 2] + M[3, 3]) - M[n, n] (L1 + L2 + L3) ZS[n_, Li_] := Li (-M[n, n] - t @@ Complement[Range[3], {n}] + L1 + L2 + L3) ZS0 = Plus[M[1, 1] M[2, 2] - M[1, 2] M[2, 1], M[1, 1] M[3, 3] - M[1, 3] M[3, 1], M[2, 2] M[3, 3] - M[2, 3] M[3, 2], -(L1 L2 + L1 L3 + L2 L3)]; ZSComp[n_, Li_] := Subtract[ZS[n], ZS[n, Li] + ZS0] evsq[Li_] := With[{Ann = Complement[{L1, L2, L3}, {Li}]}, Part[Dot @@ ((Array[M, {3, 3}] - # IdentityMatrix[3]) & /@ Ann), #, #] & /@ Range[3]] Expand[evsq[L1] - (ZSComp[#, L1] & /@ Range[3])] // MatrixForm Expand[evsq[L2] - (ZSComp[#, L2] & /@ Range[3])] // MatrixForm Expand[evsq[L3] - (ZSComp[#, L3] & /@ Range[3])] // MatrixForm The so-called "new identity" is just the projector formula modulo a few zero-sums. The zero-sums are well known in the theory of matrix invariants, see for example: https://en.wikipedia.org/wiki/Invariants_of_tensors#Principal_invariants --Brad

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[math-fun] crab megastep
by Bill Gosper 03 Dec '19

03 Dec '19

2²⁰ steps of one of the six regions of a semihex grazing the point 1+i√3/2 (which I mistook for a vertex), crab megastep <http://gosper.org/crabmega.png>. This chain is the (closed loop) image of the (short) interval [13/40, 53/160]. Both endpoints map continuously to the tip of its tail. The other four tail tip preimages are 21/80, 43/160, 77/160, and 39/80. (Only three of the grazing regions have this shape.) I can still only find sextuple points by brute force guessing, until Julian mercifully creates his inversesemihex magic. Only the vertices of the texture represent visited points. The line segments just indicate order of visitation. But the "curve" solidly fills in the *all* the points in there, a dense subset of them six times over, where ever-smaller crab tails congregate. Such are the fairy tales logically coerced by belief in the continuum. —rwg

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