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[math-fun] Fun series
by Dan Asimov 13 Dec '18

13 Dec '18

Pick an integer p in {2, 3, 4, ...} = Z+ - {1}. If you add up the curves in the complex plane C given by f_n(t) = (1/p^n) exp(p^n * it), (so we don't have to worry about convergence), we get the closed plane curve f(t) = Sum_{0 <= n < oo} (1/p^n) exp(p^n * it), 0 <= t <= 2π in C. From plotting this, I know there are a few parts of the curve where f is not one-to-one. It is also clear that the curve contains a *simple* closed curve that bounds a unique minimal open set of positive area. Some questions: --------------- 1. Is the image of f locally simply connected anywhere at all?* 2. Does the image of f contain a differentiable arc anywhere at all? 3. Do the answers depend on the choice of integer p >= 2 ? —Dan ————— * A space X is locally simply connected if for each x in X, every neighborhood of x contains an neighborhood U of x such that U is simply connected: All closed loops in U can be continuously shrunk to a point in U.

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[math-fun] Messages in pi
by James Propp 13 Dec '18

13 Dec '18

Keith Lynch writes: “Carl Sagan’s 1985 novel _Contact_ is a classic, though not very original, SETI novel. Unoriginal except for a subplot about the aliens having discovered messages hidden in the base-eleven digits of pi, which was unfortunately left out of the 1997 film of the same name. I find it an interesting question whether evidence for such a message in pi can ever overcome the possibilities that it's either an astonishing coincidence or someone hacking the computer that's doing the calculation. Presumably if there really was such a message, it would have to be from God, rather than from advanced aliens.” I’m not sure if we’ve discussed this before on math-fun, but I’ve never found this aspect of Sagan’s novel compelling or even coherent. I can imagine a God who intervenes in ways that locally violate the laws of physics, or a God who chooses the laws of physics, but not a God who chooses the laws of mathematics. Is there anything that this might mean? My guess is that, even though Sagan acknowledges in the novel that pi is a mathematical constant rather than a physical one, he didn’t really understand what this entails. Hiding a message in the digits of the base ten expansion of pi is no different from hiding a message in the digits of the base ten expansion of seventeen. (“Mathematicians will tell you that it’s all zeros after the decimal point, but how far out how they really checked?”) I know that the mathematician and science-fiction novelist Greg Egan plays thought experiments with the mutability of math, but I always get the sense with him that at least a fraction of his tongue is in his cheek when he does this, whereas I think Sagan was invoking mathematical ideas in an attempt to convey a sense of transcendence, much as Arthur C. Clarke had done in _2001: A Space Odyssey_ (when the newly transhuman Bowman chides his recently human self for failing to imagine that the numerical pattern governing the proportions of the monolith were limited to three dimensions). Trying to convey something beyond human comprehension is a tricky business; it’s akin to designing good technobabble, but harder. Jim Propp

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[math-fun] Can anyone source this mean joke?
by James Propp 13 Dec '18

13 Dec '18

“I had a weak doctoral student once. I didn’t mind telling him what problem to work on. I didn’t mind solving it for him. I didn’t even mind writing up the proof. But I really resented having to EXPLAIN it to him.” (heard at UC Berkeley in the 1980s?) Jim Propp

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Re: [math-fun] spacefilling surface
by Dan Asimov 13 Dec '18

13 Dec '18

Still, I wonder if there exist space-filling surfaces that are limits of surfaces S_n in R^3 as n —> oo, such that ... if for some eps > 0, the epsilon neighborhood of S_n (meaning {x in R^3 | ||x-p|| < eps for some p in S_n}) is denoted by N_eps(S_n), then *hypothetically*: N_eps(S_n) contains N_(1/eps)(0) which we require must hold for all n, for some sequence of eps = eps_n > 0 that decreases to 0 as n —> oo. And of course we require that for some sequence h_n of embeddings h_n : R^2 —> R^3 of the plane into R^3 with h_n(R^2) = S_n, we have that the limit lim_{n—>oo} h_n exists and is continuous. It's space-filling by the hypothetical construction. —Dan Andy Latto wrote ----- Aren't there even simpler solutions? If f:R -> R^3 is space-filling, define g:R^2 -> R^3 by g(x,y) = f(x). >On Thu, Dec 13, 2018 at 4:34 PM Bill Gosper <billgosper(a)gmail.com> wrote: >> >> Rohan requested a continuous map from a 2D surface onto a non-porous patch >> of >> positive 3D volume.

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[math-fun] spacefilling surface
by Bill Gosper 13 Dec '18

13 Dec '18

Rohan requested a continuous map from a 2D surface onto a non-porous patch of positive 3D volume. There's one under our noses, but maybe not where you think. We certainly have continuous maps from 1D onto 3D, so it is tempting to compose such a volume-filling function with some sort of (necessarily multivalued) 1D inverse of a planefilling function. But such inverses are wildly discontinuous, no matter which values you choose as the definition. Although 1/2+i/2 is "merely" triple point, the Hilbert "curve" is dense with quadruple points. The preimage of a sufficiently small neighborhood of such a point will comprise at least four disjoint pieces. Could the 1D to 3D map possibly smooth everything out? Maybe, but there's a much cleaner solution. SPOILER: gosper.org/3dsno.png gives a continuous mapping of an equilateral triangle onto a standard 3 simplex. —rwg

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[math-fun] a Christmas carpet
by Fred Lunnon 13 Dec '18

13 Dec '18

Glad tilings of great joy: a Christmas carpet. Herewith the only greeting card I am ever likely to get around to sending, folks. Make the most of it --- but don't spy this at phone! In connection with Penrose tilings, `empire' denotes the subset of tile placements forced by some given subset of compatible placements; `country' denotes any connected region in such an empire. Modulo geometric congruence and de/inflation, there are just 7 distinct classes of country, identifiable by their simple polygonal boundary shapes. The graphic illustrates the empire of a `free wheel', or cartwheel for which a location is specified, but not an orientation. It somehow appeals to this supercilious misanthropist as vaguely festive, even to the extent of omitting any coffin for Tiny Tim: https://www.dropbox.com/s/ixfdnjax12kn7id/glad_tiling.png Amongst others, the recent paper by Fang Fang, Dugan Hammock, Klee Irwin, "Methods for Calculating Empires in Quasicrystals" (2017) https://www.mdpi.com/2073-4352/7/10/304/htm provides more concrete illustrations of 3 of the 7 home-country empires, populated with "Penrose rhomb" slings-and-arrows tiles, rather than visually preferable kites-and-darts. Figs. 9a, 9b, 9f illustrate instances of single worms (sandwiches) with increasing sizes; 9c, 8b, 9e cartwheels (coffins); 9d, 8a pentagonal stars. By the way, I am in the process of providing more material on this topic, available to anybody sufficiently reckless to enquire further. A Dickens of a lot more ... Bah! Humbug! WFL

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[math-fun] Curious parametrized family of loci
by Dan Asimov 12 Dec '18

12 Dec '18

Let z be a complex variable, t real. Then the locus L(t) given by L(t) = {z in C | |z-1|^(1/|z+1|) = t} goes through an interesting bifurcation as t passes through 1.260522 . I doubt I would have predicted it. —Dan

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Re: [math-fun] Egyptian Fractions (was Re: Messages in pi (cont'd))
by Dan Asimov 12 Dec '18

12 Dec '18

There is the greedy-harmonic-series-with-signs method: Let C be the irrational real number to approximate; say C > 0. Then let p1 be the least integer for which 1 + 1/2 + 1/3 + ... + 1/p1 > C. Now let n1 be the least integer > p1 for which 1 + 1/2 + 1/3 + ... + 1/p1 - (1/(p1+1) + 1/(p1+2) + ... + 1/n1) < C Continue this way, alternately adding and subtracting groups of consecutive terms, resulting in a signed harmonic series that converges to C. —Dan Keith Lynch wrote: ----- ..... What are some exotic ways to approximate real numbers to any desired precision? Extra points if you can come up with something completely original. -----

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[math-fun] How AB-normal can numbers get?
by Henry Baker 12 Dec '18

12 Dec '18

This talk of pi being 'normal' got me to thinking: "How bad can an ab-normal number get?" I'm not talking about numbers like 1.000000...; I'm talking about numbers which "look" random to the unaided eye, but which are quite abnormal -- e.g., they might be quite compressible by something as simple as zip or gzip. In particular, a gzip'd number might be 1/100'th the size of the un-gzip'd number. Are there easy constructions for such numbers -- hopefully constructions which don't involve individual bits or Turing Machines ? Perhaps using only "calculator" functions ?

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Re: [math-fun] Egyptian Fractions (was Re: Messages in pi (cont'd))
by Dan Asimov 12 Dec '18

12 Dec '18

Yes, any divergent series whose terms approach 0. Say we're choosing the sign of each term. Then all possible sign choices amount to the countable product K (cartesian power) of {-,+}, which has the size of 2^aleph_0 = the continuum. The product K naturally has the topology of the Cantor set. So for each point of the Cantor set K — i.e., for each choice of signs S : Z+ —> {-,+} — this S gives rise to the real number Sum_{n in Z+} S(n)/n. So assuming the probability of convergence is 1 (which it is), we get a map K —> R from the Cantor set onto the real numbers. This should by rights be continuous, so it's a funny map — a continuous surjection from the totally disconnected Cantor set onto the real numbers. But basic theorems in topology rule this out (the image must be compact). Therefore, mathematics is inconsistent. Another related thing is to look at the distribution of values obtained when choosing signs randomly. Or for more fun, use points on the unit circle S^1 in the complex plane, chosen independently at random from S^1, as the coefficient for each term of the harmonic series. What does the distribution of sums in the plane look like? —Dan Cris Moore wrote: ----- This applies to any divergent series, right? And the fun thing is that any C has a series with any initial subsequence. I suppose this means any C has countably many sequences, although it’s not clear that taking a finite initial subsequence and then turning greedy covers all of them… > On Dec 10, 2018, at 3:43 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > > There is the greedy-harmonic-series-with-signs method: ... -----

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