January 2021 - mathfun@mailman.xmission.com

[math-fun] More than you probably wanted to know about Riemann surfaces, part deux
by Dan Asimov 14 Jan '21

14 Jan '21

I posted to math-fun recently on how a complex power series f(z) = ∑ c_n z^n (where the summation is over n ≥ 0, and the c_n ∊ ℂ (where ℂ denotes the complex plane, and it is assumed convergent at least for all z in some neighborhood of the origin 0 ∊ ℂ), defines by analytic continuation a certain surface (called "the Riemann surface of the function f(z)"). A Riemann surface can be defined more abstractly, as a topological space X that is the union of some family {U_j} of open sets for j ∊ J, an arbitrary index set. Such that for each j ∊ J there is a homeomorphism h_j : U_j —> ℂ such that whenever U_j ∩ U_k ≠ ∅, the mapping h_j ∘ h_k^(-1) : h_k(U_j ∩ U_k) —> h_j(U_j ∩ U_k) is an analytic homeomorphism. Another more geometric way to say "an analytic homeomorphism" is to say that this map h_j ∘ h_k^(-1), whose domain and codomain are open subsets of ℂ, is *conformal*, that is, preserves angles and orientation. Then, the natural notion of when two Riemann surfaces are "equivalent" is when there is a conformal homeomorphism between them. This implies that a Riemann surface is the same as a smooth surface with a well-defined measure on the angles between any two curves that exit the same point. A simply connected subset of the complex plane ℂ is any subset in which every closed curve can be shrunk to a point. Simply connected subsets of ℂ can be ridiculously complicated. Imagine e.g. the "topologist's sine curve" with its endpoints connected by an ordinary path, making a closed curve. Now draw crazy hairs going inward from the boundary, and smaller hairs on hairs ad infinitum. The interior of this thing is simply connected. Yet the Riemann mapping theorem (not proved rigorously until the early 1900s) implies that all simply connected subsets of ℂ are homeomorphic. And if we exclude only ℂ itself, all the rest are even conformally equivalent to each other — homeomorphic by a homeomorphism that preserves angles. Also, the Koebe-Poincaré uniformization theorem implies that all smooth Riemann surfaces that are topologically a sphere S^2 are conformally equivalent to each other. But something different happens when considering Riemann surfaces that are topologically a torus, T^2. In this case it can be shown that the set of conformal equivalence classes of tori are in bijective correspondence with the points of the set X = {z ∊ ℂ | |z| > 1 and -1/2 ≤ Re(z) < 1/2}. (Note that its boundary consists of two semi-infinite lines and a 60º arc.) A finer tuning of this fact gives a geometric bijection between the set of conformal equivalence classes of tori with the *quotient* of the set X = {z ∊ ℂ | |z| > 1 and -1/2 ≤ Re(z) ≤ 1/2} (note that a less-than has become a less-than-or-equal-to) after any boundary point z ∊ ∂X and its reflection z' about the y-axis are identified with each other. That is, if z = x + iy then it is identified with z' = z - 2 Re(z) = -x + iy. The resulting shape is topologically a disk. But geometrically it has two special points, one of which has only 180º of angle around it and the other of which has only 120º of angle around it. These correspond to the square torus T_Q = ℂ/Z[i] and the hexagonal torus T_H = ℂ/Z[𝜔], (where 𝜔 = exp(2πi/3)), respectively. —Dan

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[math-fun] Sequence frameworks
by Hilarie Orman 14 Jan '21

14 Jan '21

I am interested in some general ways of defining sequences s[i]. 1. s[n+1] = f(s[n])*s[n-1] + f(s[n-1])*s[n] If f is the constant function "1", then this is the Fibonacci sequence. 2. s[n+1] = g(s[n]){s[n], s[n-1], ..., s[0]} Here, g is a function that has a number as an input and produces a multivariate function that is applied to previous sequence values. g could be, for example, a linear function with k terms using coefficients derived from s[n]. 3. s[n+1] = g(s[n]){t[n], t[n-1], ... t[0]} where t is a sequence defined on i=0...inf. As before, g is a multivariate function derived from the number s[n]. Is there some standard terminology for these methods that I could use to find prior work? The keywords just cause Google to latch onto well-known recursive sequences. Thanks, Hilarie

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[math-fun] Best way to handle large number of pdf files?
by Neil Sloane 09 Jan '21

09 Jan '21

Ever since it started in 2007 I've been using PAPERS ( https://en.wikipedia.org/wiki/Papers_(software) ) to manage pdf files on my Mac. But lately it has gotten very slow to respond, maybe because of the large number of pdf files I have. It often takes a minute to respond. I know there are newer programs (Mendeley, Benubird, Zotero), but even if they are faster at dealing with a large library of files, there would still be the problem of moving the files over from PAPERS. Anyone have any advice? I'm running Papers3 Version 3.4.20 (579) on an iMac running OS 10.13.6.

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[math-fun] Newton meets Cayley-Hamilton
by Henry Baker 06 Jan '21

06 Jan '21

The following has probably been proposed by many undergraduates learning matrix algebra for the first time, but I've never seen it mentioned before, and I couldn't find it with either Google or Scholar.Google. Newton's famous quadratically convergent method for finding roots of f(x) (so long as the root isn't a multiple root): x_(n+1) = x_n - f(x_n)/f'(x_n) Cayley-Hamilton: a square matrix M satisfies its own characteristic polynomial p(x): p(M)=0 Suppose now that we have a *diagonal matrix* E_n of *eigenvalue approximations* for M, where these approximations are 'sufficiently close' to the (distinct) eigenvalues of M. Then we can improve E_n with: E_(n+1) = E_n - p(E_n) (1/p'(E_n)) where (1/p'(E_n)) is the inverse of the matrix produced by substituting E_n into the derivative of the characteristic polynomial p(x). This works because Cayley-Hamilton doesn't care very much *which* matrix having this characteristic polynomial that you use. ----- I'm not claiming any great efficiency from this observation, but there may be some applications. ----- I'd love to find an analogous method which 'diagonalizes' M -> E_n, even if it is 'obviously' too expensive. (Expensive algorithms have a way of becoming cheaper due to Moore's Law, increasing core counts, and cleverer implementations.)

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[math-fun] Atonement
by Dan Asimov 06 Jan '21

06 Jan '21

After Maximilian Hasler pointed out my mistake, I want to set things right regarding unique and non-unique geodesics between a pair of points on the torus. Consider two points p ≠ q of T^2. Imagine p to lie in the center of a unit square Q = [0,1] x [0,1] in R^2 that is a fundamental domain for T^2. Then if dist(p,q) < 1/2, p and q must have a unique shortest geodesic between them. However, if 1/2 ≤ dist(p,q) < √(1/2 with one component (x or y) of the vector p-q having absolute value = 1/2 (i.e., so that q lies on the boundary ∂([0,1] x [0,1])) then there are exactly two distinct geodesics in T^2 that connect p and q. If dist(p,q) = √(1/2), the maximum possible distance on T^2, then there will be four distinct geodesic segments connecting p and q. Next case: the hexagonal T^2 = C / Z[𝜔], 𝜔 = exp(2πi/3). —Dan

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[math-fun] more groups than rings, surprise
by Neil Sloane 05 Jan '21

05 Jan '21

The Dec 2020 issue of the Amer Math Monthly has an article Desmond MacHale, Are there more finite rings than finite groups?, Amer. Math. Monthly, 127:10 (2020), 936-938. showing that there are infinitely many n such that there are more groups with n elements than rings with n elements. He gives one example, n = 36355, and asks if this is the smallest such n. This is a sequence not yet in the OEIS! Anyone care to investigate? These are the n such that A000001(n) > A027623(n).

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[math-fun] Puzzle and question about pairing up points on the torus with disjoint segments
by Dan Asimov 05 Jan '21

05 Jan '21

Let T^2 = R^2/Z^2 denote the square torus, i.e., a unit square with its opposite edges identified. Let X = {x_1, ..., x_n} be any set of n distinct points of T^2 where n = 2k is even. Puzzle: ------- Does there necessarily exist n/2 disjoint line segments in T^2 whose n endpoints form the entire set X ??? Proof or counterexample. (Note that on T^2 there are infinitely many line segments between points p and q.) Question: --------- When X is such that there does exist such a collection of n/2 disjoint line segments, can they always be chosen so as to be *shortest* line segments??? (If points p and q of T^2 are less than the maximum possible distance of √(1/2) apart, then there is a unique shortest segment connecting them. If their distance is √(1/2) then there are four such segments.) —Dan

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[math-fun] "A physicist's proof of the Basel problem"
by Dan Asimov 05 Jan '21

05 Jan '21

I thought this paper had an interesting (fully rigorous!) take on Euler's (not fully rigorous) proof that 1 + 1/2^2 + ... + 1/n^2 + ... = π^2/6: <https://arxiv.org/abs/1908.07518>. —Dan

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[math-fun] about some films made on a Tektronik equipment in 1976
by Simon Plouffe 05 Jan '21

05 Jan '21

Hello mr Gosper, I think you will find this of interest, https://vimeo.com/98285334 these are films made in or around 1976 with a tektronik screen, with dragon curves, triangles and such. They were made by a certain Eric Cornwell, and some other Cornwell in the same family I presume. also here : https://vimeo.com/channels/bkcfilms and here : https://vimeo.com/user543333/videos Best regards, Simon Plouffe ps : happy new year.

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[math-fun] Aware-like seq
by Éric Angelini 03 Jan '21

03 Jan '21

Hello Math-Fun (we need help to check, compute and submit this to the OEIS in good mathematical English — as we fear Dan's furrowed brow ;-D) This sequence is based on the Aware strategic game. We empty a “pit” and sow to the right its content – one seed per pit. When this is done, we continue from there, empty the next pit and sow again one seed per pit to the right. S0 = 1, 109, 2, 8, 27, 3, 26, 31, 5, 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … To start, we empty the first pit (1) and sow its unique seed to the right; 109 becomes 110 and S0 becomes S1 (the vertical stroke shows where we are in the process): S1 = _, 110, | 2, 8, 27, 3, 26, 31, 5, 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … We now empty the first “untouched” pit (2) and sow its content in the same way. S1 becomes: S2 = _, 110, _, 9, 28, | 3, 26, 31, 5, 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … We proceed like this, step by step, ad infinitum: S3 = _, 110, _, 9, 28, _, 27, 32, 6, | 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … S4 = _, 110, _, 9, 28, _, 27, 32, 6, _, 31, 54, 30, 53, | 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … S5 = _, 110, _, 9, 28, _, 27, 32, 6, _, 31, 54, 30, 53, _, 29, 52, 62, 8, 51, 61, | 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10, … S6 = _, 110, _, 9, 28, _, 27, 32, 6, _, 31, 54, 30, 53, _, 29, 52, 62, 8, 51, 61, _, 750, 60, 97, 49, 59, 96, 48, 58, 95, | 10, … If we now compare the end result Se to the starting sequence S0, we see that… S0 = 1, 109, 2, 8, 27, 3, 26, 31, 5, 4, 30, 53, 29, 52, 6, 28, 51, 61, 7, 50, 60, 9, 749, 59, 96, 48, 58, 95, 47, 57, 94, 10 … Se = _, 110, _, 9, 28, _, 27, 32, 6, _, 31, 54, 30, 53, _, 29, 52, 62, 8, 51, 61, _, 750, 60, 97, 49, 59, 96, 48, 58, 95, _, … … S0 and Se share the same digit succession. S0 should be the lexicographically earliest sequence of distinct positive integers with this property. This is not too tricky, I guess – still, I hope I didn't leave too many mistake above (and that this is not old hat). Best, É.

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