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Re: [math-fun] September 10000, 1993
by Keith F. Lynch 17 Jan '21

17 Jan '21

Allan Wechsler <acwacw(a)gmail.com> wrote: > I'm guessing that anybody that has dived into calendar math for any > time at all can figure this one out. > I don't want to give spoilers, so I will digress a bit, with > apologies to Keith. A vaguely related curiosity is that given dates > of the year fall on the seven days of the week with *almost* equal > probability -- almost but not exactly equal. Right. Most notoriously, the 13th is slightly more likely to land on a Friday than to be any one of the other six days of the week. > This is because the central postulate of the Gregorian calendar is > that 400 years equals exactly 146,097 days. Unfortunately, this > period in days is exactly divisible by seven, so the day-of-the-week > pattern in our calendar repeats exactly every four centuries. Right. Isaac Asimov got this wrong, saying that the calendar repeats every 2800 years. As you said, it repeats every 400. (Except for the dates of Easter, which repeat every 5,700,000 years.) But the calendar repetition period has nothing to do with the solution to my problem, which would be the same even if the calendar never repeated. (An example of a non-repeating calendar is one in which a day is added to the year for every odd number the year number is divisible by and subtracted from the year for every even number the year number is divisible by. The average year length would then be a transcendental number, hence the calendar would never repeat. (There would also be zero-length and negative-length years, but those would be extremely rare.)) > Calendar math is full of weirdness like this. I *still* haven't > wrapped my head around the Hebrew calendar. The Islamic calendar > is far simpler in principle, and (intentionally) full of mystery > in practice. I thought the Islamic calendar was unpredictable, as it depends on when, after each new moon, the moon is first observed by someone in authority, which in turn depends on the weather. The solution to a related but more difficult calendar puzzle *does* depend on the details of our calendar. What about power-of-ten dates of *all* months? Is there any day which is *not* such a day? (Using the proleptic Gregorian calendar.) The 1st and the 10th of each month are of course the 0th and 1st powers of 10 of that month. But what about the other 26, 27, 28, or 29 days of that month? About once a month is the 100th day of some previous month, and about once a month is the 1000th day of some other previous month, and about once a month is the 10,000th day of some other previous month, etc. So most likely yes. But with an average of about 30.4 days per month, we'd have to go to at least 30th powers of 10, meaning most such previous months would be long before the Big Bang. Probably a lot more than 30th powers, since (presumably) nothing keeps such days from landing on the same dates, leaving other dates empty. Since there's no shortage of negative-numbered years (given that we've abandoned the constraint that the Gregorian calendar, the Earth, or even the universe, needs to have existed at the time) is it inevitable that every day has this property? Not necessarily, since the calendar does repeat. Successive powers of 10 number of days are 2.7379070069885... 27.379070069885..., 273.79070069885... years ago, etc. Since that number (400/146097) is rational, the decimal expansion repeats, so there are only a finite number of distinct fractional numbers of years ago, no matter how many powers of 10 you try. I don't know the period of repetition of 400/146097, but since 146097 is 3^3 * 7 * 773, I think it's at most 6*772=4632 decimal digits. Also, since the calendar does repeat, if we find solutions for 146097 consecutive days, we'll have found a solution for all days. A simpler version is what if every month had the same length? If so, I'm pretty sure that there would be no solutions for some days unless that month length was relatively prime to 10.

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Re: [math-fun] Conjecture about the distribution of nontrivial powers.
by Keith F. Lynch 17 Jan '21

17 Jan '21

Mike Speciner <ms(a)alum.mit.edu> wrote: > Some of the cubes and fifth powers are also squares. And some of > the cubes and fifth powers are the same, as they are fifteenth > powers. In the text you quoted in that very message, Allan Wechsler said: >> ... but there is some ambiguity here, exemplified by 16 = 4^2 = >> 2^4. Let us, for the moment, resolve the ambiguity by preferring, >> for this purpose, the representation with the smallest exponent.

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[math-fun] Regular (until they're flexed) dodecahedra are slightly flexible?
by Bill Gosper 16 Jan '21

16 Jan '21

But only this much <http://gosper.org/dodecasquish.gif>? (No stretch; some squish? I don't trust this.) —rwg

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[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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