March 2019 - mathfun@mailman.xmission.com

Re: [math-fun] Statistics for a class of random solids
by Dan Asimov 10 Mar '19

10 Mar '19

You can certainly use the "rejection" method for picking points uniformly distributed on a sphere by first choosing n+1 real numbers x_0, ..., x_n in [-1, 1] and using only those sets of points with x_0^2 + ... + x_n^2 <= 1. But as the dimension increases, the fraction rejected becomes very high very fast. A better method is based on a smart way to choose points from the standard normal distribution in R^2: 1. Pick x, y from the uniform distribution on [0, 1]. 2. Let A = sqrt(-2 ln(x)) cos(2πy) B = sqrt(-2 ln(x)) sin(2πy). Then A and B are independent standard normal variates.* 3. Repeating this floor[n/2] times gives the coordinates of a standard normal point in R^n. Dividing by its length gives a uniformly distributed point on the unit sphere S^(n-1). —Dan ————— * For more on this see <https://en.wikipedia.org/wiki/Box-Muller_transform> Allan Wechsler: ----- That means that, to answer another question somebody asked here recently, you can produce points uniformly distributed on a sphere by generating N Gaussian coordinates and normalizing the resulting vector. -----

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Re: [math-fun] Statistics for a class of random solids
by Keith F. Lynch 10 Mar '19

10 Mar '19

Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > Assuming "randomness" defined in a spherically symmetric manner, the > resulting polyhedron will presumably resemble a lumpy sphere with > occasional mountains. On the plains in between, most polygons will > be triangles at 6-valent vertices. Whether mountains dominate > plains would depend on the distribution tail --- I'd guess not for > any reasonable extension of normal. To me, "mountain" implies concavity (at the base of the mountain, not at its peak). So by definition a convex hull would have no mountains. That raises the question of how far above the horizontal a feature could be on such a "planet," where horizontal is defined in reference to the local gravity. And where the local gravity is defined as what you would get if the interior has some fixed density and the exterior has zero density. Which raises the question of efficient ways to calculate the surface gravity on a uniform polyhedron. > Computing the "convex hull" of a point set is a typical example of > "geometric sorting" problems, to which computer graphics buffs have > devoted considerable attention. Googling (ugh!) those terms should > locate plenty of references. Thanks. I see that key words include "Graham scan," "Chan's algorithm," "Quickhull," "gift wrapping," "monotone chain," and "Jarvis march."

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Re: [math-fun] Statistics for a class of random solids
by Keith F. Lynch 09 Mar '19

09 Mar '19

Allan Wechsler <acwacw(a)gmail.com> wrote: > Suppose we have a random point generator, that generates points in > R^3 with all three coordinates normally distributed around 0, say > with a standard deviation of 1, though it doesn't really matter. Do you mean that the values of the three coordinates are independently normally distributed, or that the distance from the origin is? It's not obvious to me that these would give the same distribution. I suspect that the former wouldn't be isotropic. I could be wrong. > Take N random points in this manner, and form the convex closure. > This will be a polyhedron. What are the expected numbers of faces, > vertices, and edges, in terms of N? Do they approach finite limits > as N increases? I imagine that the surface area grows without > bound. What about the expected number of faces that have 3, 4, > 5, 6 edges, and so on? I would expect that all the faces would be triangles, since the odds of any four or more of the N random points being coplanar is zero. > I can think of many more questions of this sort, and am wondering if > anybody here knows what's known. I have no idea, but I've been thinking about how to simulate it, so as to count the faces. Decades ago, I was tasked to simulate two random noise sources with a desired correlation. I decided to generate points whose distance from the origin was normally distributed and whose direction from the origin was uniformly distributed. The X coordinates of successive points would be one of the noise sources. To get the other noise source, I'd rotate the plane around the origin by an angle, and use the new X coordinate of the same successive points as the second noise source. The correlation depended on the angle, 1 for an angle of 0, 0.5 for an angle of 45 degrees, 0 for an angle of 90 degrees, -1 for an angle of 180 degrees, etc. (Digression: It's interesting that people can easily distinguish between white noise sources with different correlations when they're played into headphones. I'm not sure if anyone has figured out how this is neurologically possible.) To generate the random points, I generated a random (well, pseudorandom) K-bit bit string (as distinct from a random K-bit number in that leading zeros are allowed), where K is large, and set the distance from the origin equal to the number of one-bits in the string minus K/2 all divided by some appropriate scaling factor. The angle from the origin was then randomly generated uniformly from 0 to 360 degrees. But how would I do the same in three dimensions, as would be necessary for your problem? Doing the same, but separately with longitude and latitude, would cause a strong bias toward the poles. I finally decided that I'd generate points by randomly generating X coordinates uniformly between -1 and +1, likewise with Y coordinates and with Z coordinates. I would then discard any points whose distance from the origin exceeded 1 to carve the cube into a sphere. I'd stop when I had accumulated N non-discarded points. These should give uniformly distributed random directions from the origin. I'd combine those directions with the distances generated by counting 1-bits in N random K-bit strings. To find the convex closure, I'd check every set of three of these N points. If all other of the N points were on the same side of the plane defined by those three, then the resulting triangle is one of the faces of the convex closure polyhedron. I'm not really happy with that, since it's O(N^4), which is problematic for large N. It would be better if I could start by identifying which of the N points were part of the convex closure polyhedron. I could do that by seeing, for each of the N points, whether all other points were on the same side of some plane passing through the selected point. But how can I efficiently do that? There are so many possible planes to choose from.

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Re: [math-fun] 33 as sum of 3 integer cubes
by Dan Asimov 09 Mar '19

09 Mar '19

Very interesting, Alon Amit! (Can I please have some anonymized network data?) Suppose f : Z^3 —> Z via f(x,y,z) = x^3 + y^3 + z^3. Then f can evidently take some very large triples to very small integers. Question: ----- Given an integer N in Z: What is the asymptotic behavior of the set S_3(N) = ((x,y,z) in Z^3 | x^3 + y^3 + z^3 = N} ??? Meaning above all: What is the asymptotic behavior of the monotone function #_3(N): Z+ —> Z>=0 where #_3(N)(t) denotes the number card({(x,y,z) in Z^3 | x^3 + y^3 + z^3 = N —and— x^2 + y^2 + z^2 <= t} And same for other polynomials. —Dan ----- James Buddenhagen writes: ----- 33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3 found by Timothy Browning communicated on Quora by Alon Amit -----

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[math-fun] Tilings with 2, φ, ψ, χ, ρ
by Ed Pegg Jr 07 Mar '19

07 Mar '19

I've written a blog about new substitution tilings and other properties based on the constants 2, φ, ψ, χ, ρ. https://blog.wolfram.com/2019/03/07/shattering-the-plane-with-twelve-new-su… If I missed something, I'd love to know about it. --Ed Pegg Jr

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[math-fun] entropy vs Lyapunov exponent
by Veit Elser 06 Mar '19

06 Mar '19

The distribution of the continued fraction partial quotients of a random number, say 2^(1/3), was worked out already by Gauss: p(a) = log_2(1+1/a) - log_2(1+1/(a+1)) If you compute the entropy of this distribution you get, numerically, H(A) = 2.3792… (A is my symbol for the random variable). On the other hand, the Lyapunov exponent of the continued fraction map x -> 1/x - floor(1/x) can be worked out analytically: L = zeta(2)/log(2) = 2.3731… Why are these numbers so close? Consider the joint probability of consecutive partial quotients (a, b): p(a, b) = log_2(1+1/(a+1/(b+1))) - log_2(1+1/(a+1/b)) I don’t know if Gauss worked this out too, but I’m sure someone (Knuth?) has. It should be obvious from the formula how it generalizes to triples, etc. In any case, from this you can compute, numerically, the mutual information between consecutive partial quotients: I(A,B) = 0.0056… That's most of the small difference — why? Consider the conditional entropy H(A|B) = H(A) - I(A,B) = 2.3736… This is closer to what we mean by the “entropy of partial quotients”, because partial quotient A (of a random number) always has a predecessor B, and the two are not independent. Surely someone — references welcome! — has proved L = H(A|BCD…) Trivia question: What real number, distinguished by its partial quotients, is also (like L) the ratio of two “special” functions at argument 2? -Veit

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[math-fun] 𝜗₃(𝕖^(-20π))
by Bill Gosper 06 Mar '19

06 Mar '19

https://www.open.wolframcloud.com/objects/92e4170d-590a-4f41-91af-595a443c6… gives a nondescript nested radical times 𝚪(1/4)/π^¾, and then rewrites the radical as a product of binomial surds. Expanding the 𝜗 gives a really fast series for 𝚪(1/4)/π^¾. I'm unsure whether 𝜗s of yet higher powers of 𝕖^-π can be put in the binomial surds form. (Even if they can, they might burgeon infelicitously.) Also note Mathematica's bizarre insistence on writing the radical as reciprocal(√(reciprocal)) instead of just √. (This annoyed the heck of of Corey while he was writing his denester.) —rwg (And I have no idea how long these Cloud bits stick around.)

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[math-fun] Primality-proving program for Windows?
by Robert Baillie 04 Mar '19

04 Mar '19

Hi Everyone, I am looking for a primality-proving program (presumably, based on ECPP) that runs on Windows, and which will work with numbers up to several thousand digits. For numbers that are not too large, Mathematica has the ability to produce primality certificates. I think it uses (recursively) partial factorizations of n + 1 and n - 1. However, on Linux, the primo program http://www.ellipsa.eu/public/primo/primo.html can prove primality of much larger numbers. Thanks for any info you might have! Bob Baillie

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Re: [math-fun] What is the ratio, Kenneth?
by Dan Asimov 02 Mar '19

02 Mar '19

My usual "Arggh!" I should not post that late at night! Let me restate the problem this way: Let the 2^n n-cube vertices be given by C(n) = {0, 1}^n with distances determined in R^n, and let the (n+1) n-simplex vertices be given by a orthonormal set S(n) = {e_j in R^(n+1) | <e_j, e_k> = delta(j,k)}. Then for n >= 2, define f(n) = the smallest number of isometric copies of S(n) that cover C(n). I.e., if ∆_1, ..., ∆_r are subsets of the n-cube C(n) each congruent to S(n) whose union is C(n): C(n) = ∆_1 u ∆_2 u ... u ∆_r then f(n) = the least such r. Questions: What is f(n) and what is it asymptotically as n —> oo ??? —Dan

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