September 2020 - mathfun@mailman.xmission.com

Re: [math-fun] A symmetrical rotation of Euclidean space
by Dan Asimov 21 Sep '20

21 Sep '20

----- I don't exactly know what you're asking for, but maybe you are trying to make a jump off Berlekamp's complex transformations by restricting to reals? I think the answer to your questions just falls out of basic representation theory of a finite cyclic group. ----- I thought it was clear I was asking for the angles theta_j. ----- In some choice of coordinates, your T commutes with a cyclic permutation of basis vectors, call that transformation T'. T' is a regular representation of the generator of a cyclic group, so we know its eigenbasis, which is shared with T. ----- The transformation T is *defined* as a cyclic permutation of (orthonormal) basis vectors. I don't know anything about representation theory, and didn't knowingly use it here. —Dan

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[math-fun] Two Teaching Points
by Brad Klee 21 Sep '20

21 Sep '20

No, Eigenvalues[H] is not broken, but buyer beware: https://arxiv.org/pdf/1809.08178.pdf S.M. Blinder's Mathematica solution of the Ammonia problem appears to suffer from "junk in, junk out", despite him having a canonical business address on the title page. One, the references seem to be out of date, and with too old data. Two, why choose a quartic potential? While "proof" checking, I compared quartic and sextic potentials. The sextic worked much better for extracting the barrier height from ground state splitting / first excitation. Generally I am interested in Blinder's demonstrations at: https://demonstrations.wolfram.com/author.html?author=S.+M.+Blinder Unfortunately, his tutorial on Ammonia appears to me, not right enough. Would anyone second this opinion, or is there dissent from my dissent? --Brad

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[math-fun] Torus tiling puzzle
by Dan Asimov 21 Sep '20

21 Sep '20

It's known that a certain torus T^2 (namely, the hexagonal torus obtained by identifying opposite edges of a regular hexagon) can be tiled by regular hexagons so that each hexagon shares an edge with each other one. Puzzle: For which N ≥ 1 does there exist a tiling of some torus (either the square or hexagonal one) by N regular polygons, any two of which share an edge? —Dan

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[math-fun] More iterations of triples
by Henry Baker 19 Sep '20

19 Sep '20

Here's one rationally reversible method to 'equilateralize' a triangle in the complex plane. This method is inspired by Gosper's continued fraction root extraction hack (perhaps in HAKMEM ??). Consider the cubic polynomial p(z) having the triangle vertices as roots. Translate the triangle centroid to the origin, then flip the polynomial end-for-end: z^3*p(1/z). (Of course, we need to remember this sequence of centroid translations to enable reversing.) Rinse & repeat. We are 'hollowing out' the internal coefficients of the polynomial, leaving only the z^3 and the z^0 terms in the limit. The result seems to converge to a 2-cycle of equilateral triangles, where the limit equilateral triangle inverts alternately inside and outside the unit circle. Below are 8 cycles starting from a 9-12-15 triangle; this iteration seems to converge pretty quickly. Q: The final result seems to remember two numbers, a 'size' and a 'phase' from the original triangle. I wonder what these quantities might be? BTW: This method is effectively Gosper's root extraction hack, because we can solve the limit equation z^3-C by inspection, and then reverse all of our steps. Note that unlike Gosper, we have produced *all three* of the roots of the original cubic equation. 3 2 2 (%o2) z - 9.0 %i z - 12.0 z + 108.0 %i z (%i3) forward(%); 3 2 (%o3) z + (0.12733403582719 - 0.045711275026344 %i) z - 0.002086406743941 %i - 0.0024868282402529 (%i4) forward(%); 3 2 (%o4) z + (0.23777047608189 - 1.844515505700051 %i) z + 208.5099028352554 %i - 222.3534715094485 (%i5) forward(%); 3 2 (%o5) z + (- 0.0031990745810353 %i - 0.0020077204240425) z - 0.0022424181810333 %i - 0.002388162593124 (%i6) forward(%); 3 2 (%o6) z + (0.0013849128634252 %i + 4.345783781893431E-4) z + 208.9519514637108 %i - 222.5322250905755 (%i7) forward(%); 3 2 (%o7) z + (- 3.3426031773526269E-10 %i - 2.2762130066155846E-9) z - 0.0022424186214955 %i - 0.0023881586265698 (%i8) forward(%); 3 2 (%o8) z + (4.8202318997299898E-16 - 2.4021268196176432E-16 %i) z + 208.9519514635722 %i - 222.5322250907546 (%i9) forward(%); 3 2 (%o9) z + (3.1212293864605986E-34 - 5.3805053878963621E-35 %i) z - 0.0022424186214955 %i - 0.0023881586265698 (%i10) forward(%); 3 2 (%o10) z + (4.6722777796490018E-66 - 9.0752194922195901E-66 %i) z + 208.9519514635722 %i - 222.5322250907546

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Re: [math-fun] Geometric mean of values of sin(x) between x=0 and x=π
by Dan Asimov 19 Sep '20

19 Sep '20

Very nice! Even if there may exist a more intuitive proof, this is very short and very clear, and it avoids actually doing any nasty definite integration. —Dan ----- On 19/09/2020 22:19, Dan Asimov wrote: > Wolfram alpha tells me that to 100 digits, the geometric mean > of the values of sin(x) for 0 ≤ x ≤ π is exactly 1/2. > > Can someone prove this? Even better, is there an intuitive > reason that it ought to be true? The following things are clearly equal; call them all M. - GM of sin x on (0,π). - GM of sin x on (0,π/2). - GM of cos x on (0,π/2). Second times third equals M^2 but also equals GM of 1/2 sin 2x on (0,π/2) equals GM of 1/2 sin x on (0,π) equals M/2. So M^2 = M/2 and so (modulo checking that M isn't zero) M = 1/2. This is a very short proof but I'm not sure it counts as an intuitive reason. Maybe there's some geometrical way to think about sin 2x = 2 sin x cos x that would bridge the gap? -----

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[math-fun] Geometric mean of values of sin(x) between x=0 and x=π
by Dan Asimov 19 Sep '20

19 Sep '20

Wolfram alpha tells me that to 100 digits, the geometric mean of the values of sin(x) for 0 ≤ x ≤ π is exactly 1/2. Can someone prove this? Even better, is there an intuitive reason that it ought to be true? —Dan

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[math-fun] AGM3 (was Re: Iterated averaging of triples)
by Keith F. Lynch 19 Sep '20

19 Sep '20

M F Hasler <mhasler(a)dsi972.fr> wrote: > Brad Klee <bradklee(a)gmail.com> wrote: >> a = (a+b+c)/3 >> b = [(a*b)^(1/2)+(b*c)^(1/2)+(c*d)^(1/2)]/3 >> c = (a*b*c)^(1/3) >> Is this the best generalization of the agm to three (or more) variables? >> Does it even converge? >> I checked that: agm3[1,2,3] = 1.8932121023002878019777... converges > I think the most natural generalisation is to take as "update" for > the k-th variable the k-th root of the k-th symmetric polynomial in > the variables, normalized by the number of terms. > For 3 that would mean > a = (a+b+c)/3 > b = [( a*b + b*c + c*a )/3 ]^(1/2) > c = (a*b*c)^(1/3) > It gives AGM(1,2,3) = 1.9099262335408153... Let's compromise by taking the AGM of Klee's 1.8932121023002878019777 and Hasler's 1.9099262335408153, which is about 1.9015599859028125. :-) How about a fully recursive definition with no visible addition, multiplication, powers, or roots: a = agm(agm(a,(b,c)) b = agm(agm(b,(a,c)) c = agm(agm(c,(a,b)) (As I assume Klee and Hasler did, I'm not replacing any of a,b, and c until all three assignments are done.) With 1,2,3 that converges to 1.9091504422214686... For four terms: a = agm(agm(a,b),agm(c,d)) b = agm(agm(a,c),agm(b,d)) c = agm(agm(a,d),agm(b,c)) then use my above agm3 on those three. (I thought they might immediately be equal, but no.) For 1,2,3,4 that converges to 2.3587709092584177... Is there a natural way to extend this approach to 5 or more terms? Of course it's required that it give the same answer if the terms are permuted.

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Re: [math-fun] Iterated averaging of triples
by Dan Asimov 19 Sep '20

19 Sep '20

Arclength along an ellipse is notorfiously hard to quantify. But if the ellipse is rescaled to be a circle, points at equal angles apart seem easier to deal with. —Dan Tom Karzes wrote: ----- This suggests the following question: For any three distinct, non-collinear points, does their exist an ellipse such that (1) the points all lie on the ellipse and (2) the distances between the points, along the ellipse, are equal? -----

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Re: [math-fun] More iterations of triples
by Dan Asimov 19 Sep '20

19 Sep '20

Of course, a polynomial with the vertices as its set of roots does not contain any information about the original polygon's cyclic order (or more accurately, dihedral order). I'm curious how the limiting ellipses differ for the same set of initial vertices with varying order. A priori, the ellipses might just represent the correlation coefficient of the original set of points. (Specifically, via a level curve of the best-approximating 2-dimensional normal distribution of having the initial vertices as sample points.) —Dan Henry Baker wrote: ----- Here's one rationally reversible method to 'equilateralize' a triangle in the complex plane. This method is inspired by Gosper's continued fraction root extraction hack (perhaps in HAKMEM ??). Consider the cubic polynomial p(z) having the triangle vertices as roots. ... ... -----

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[math-fun] Re Iterated averaging of triples
by Neil Sloane 18 Sep '20

18 Sep '20

This reminds me of a paper Elwyn Berlekamp wrote in (I think) the American Mathematical Monthly a long time ago.

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