December 2019 - mathfun@mailman.xmission.com

[math-fun] OEIS A016825 has 28(?) characterizations,
by Bill Gosper 16 Dec '19

16 Dec '19

e.g., In[214]:= ContinuedFraction@Coth@.5 Out[214]= {2, 6, 10, 14, 18, 22, 26} e.g., the oddly even numbers. A 29th: Polar rose (r = cos(k t)) unachievable petal counts. (https://en.wikipedia.org/wiki/Polar_coordinate_system#Polar_rose) —rwg

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[math-fun] Shew that PolarPlot[1/(√2 - Cos@u), {u, 0, 2 π}]
by Bill Gosper 15 Dec '19

15 Dec '19

is an ellipse with focus at the origin. —rwg

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[math-fun] Check your color vision.
by Bill Gosper 15 Dec '19

15 Dec '19

http://www.tweedledum.com/rwg/squares.htm problem 2, can you see the color boundary between "18" and "14"? Maximal zoom. Slide "10" and "4" off the left edge to show only the middle right portions of "14" and "18". Slide up and down. Suspicion: Girls can see it better. Color vision is on the X chromosome. —rwg ("8" and "10" are identical shades of light blue. Proof: TextEdit>Format>Font>Show Colors> <eyedropper>.) I used to grumble at color-enhanced astronomy photos. Until I realized I am colorblind compared to most space aliens.

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[math-fun] Does this define sin(z)?
by Bill Gosper 14 Dec '19

14 Dec '19

1) sin(0) := 0 and 2) |sin(x) - sin(y)| ≤ |x - y| and 3) sin(x) = sin(x/3) (3 - 4 sin(x/3)^2). Note that 2) isn't even true, e.g. x=0, y=i. If this means "No", then what about for real z? —rwg

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Re: [math-fun] Does this define sin(z)?
by Keith F. Lynch 14 Dec '19

14 Dec '19

Bill Gosper <billgosper(a)gmail.com> wrote: > 2) |sin(x) - sin(y)| ? |x - y| and The digest option of this list turns all non-ASCII characters into question marks. I have no idea what was intended to go there, so I can make no sense of the question.

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[math-fun] A problem about chords of Jordan curves
by James Propp 13 Dec '19

13 Dec '19

Given a rectifiable simple closed curve, can one always find points P and Q on the curve that cut the curve into two arcs of equal length, neither of which crosses the chord joining P and Q? Here’s a proof that works if the curve is starlike from an interior point O: Every line through O determines a chord that cuts the curve into two pieces, one longer than the other (unless they’re the same length in which case we’re done). As you rotate the line by 180 degrees, the short piece becomes the long piece and vice versa. Since the lengths vary continuously, there must be some intermediate position of the line that makes the two pieces have equal length. For general Jordan curves, I can find either a proof nor a counterexample. (Seems like the kind of problem that Polish mathematicians in the 1920s would have looked at.) Jim Propp

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[math-fun] "Self-avoiding spacefilling curve" is an oxymoron.
by Bill Gosper 11 Dec '19

11 Dec '19

Besides completely covering the closure of its range, a spacefilling function doubly covers an uncountable subset of it, and triple-covers a dense subset of that. (Can we define a finite "fill factor" to describe *how* dense are these subsets?) The useful idea that we're trying to express is perhaps "noncrossing", meaning that there are arbitrarily fine polygonal samplings that do self-avoid <http://gosper.org/hexpicules.png>. Could we say "self avoidable"? —rwg Another one of those "Hey, wait a minute" math posters. Julian: If you feel like it, please explain how on Earth these two brain hemispheres are born horizontally separated by exactly 10 units. Talk about agenesis corpus callosum.

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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 09 Dec '19

09 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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