March 2019 - mathfun@mailman.xmission.com

Re: [math-fun] Statistics for a class of random solids, a drunkard's walkthrough
by Dan Asimov 14 Mar '19

14 Mar '19

I suspect X and Y here are random variables. pedantic interlude: ----- The word Gaussian applies to changes in both scale and location (standard deviation and mean) to the standard N(0,1) density d(x) = sqrt(1/2π) exp(1x^2/2), or in other words applying an affine linear function f(x) = ax + b to a standard Gaussian random variable X: Y = aX + b for any real numbers a ≠ 0 and b.) ----- So in fact an "orthogonal linear combination" of independent Gaussians like what Adam suggests will indeed result in a Gaussian. If the original ones are standard, then this careful linear combination will also be standard. (But it's also true that any affine linear combination of arbitrary Gaussian random variables, X_1 and X_2, like Y = a_1 X_1 + a_2 X_2 + b will result in Y's being Gaussian (unless it's degenerate). —Dan Adam Goucher écrit: ----- If I'm not mistaken, the Gaussian distribution is unique (up to scale) in that if X and Y are *independently* Gaussian, then every combination cos(theta) X + sin(theta) Y is also Gaussian. (Equivalently,, it's the only axisymmetric product distribution.) -----

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Re: [math-fun] Venn diagram
by Dan Asimov 14 Mar '19

14 Mar '19

(Apologies if someone already mentioned this but) a very nice result from that paper "The Search for Simple Symmetric Venn Diagrams" is this: ----- There exists a simple closed curve C in the plane such that the union of its N rotated images by multiples of 2π/N forms a Venn diagram for N items if and only if N is prime. ----- I suppose N >= 2 is required, but: Who'da thunk it? —Dan Steve Witham writes: ----- There is also the xkcd style in the Python library Matplotlib to make plots look like xkcd cartoons. --Steve > From: Stuart Anderson <stuart.errol.anderson(a)gmail.com> > Date: 3/11/19, 4:53 PM > > software for producing nice-looking Venn diagrams -----

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Re: [math-fun] Statistics for a class of random solids, a drunkard's walkthrough
by Steve Witham 13 Mar '19

13 Mar '19

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > 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. Allan Wechsler <acwacw(a)gmail.com> answered this on 3/10, but I didn't find his answer searching for "do you mean," and also there are some extra quirks in the following response: Ignoring constants, the normal distribution is like 1/exp(x^2). Independent distributions for the three axes combine by multiplying: 1 / (exp(x^2) exp(y^2) exp(z^2) )= 1/exp(x^2 + y^2 + z^2) Say r is the distance from the origin, r^2 = x^2 + y^2 + z^2. So the distribution is like 1/exp(r^2). Except for constants. And I think the converse has to be true too. Reminds me that a random walk of 2D grid steps gives a distribution whose contours approach circles. Two semi-related entries in HAKMEM: ITEM 26 (? via Salamin): (typos from scanning?) Let x be uniform on [0, 1], y be uniform on [0, 2ir], x and y independent. (2pi?) Calculate r = /-log x. (sqrt(-log x))?) Then r cos y and r sin y are two independent Gaussian distributed random numbers. ITEM 27 (Salamin): PROBLEM: Generate random unit vectors in N-space uniform on the unit sphere. SOLUTION: Generate N Gaussian random numbers and normalize to unit length. --Steve

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Re: [math-fun] Venn diagram
by Steve Witham 13 Mar '19

13 Mar '19

There is also the xkcd style in the Python library Matplotlib to make plots look like xkcd cartoons. --Steve > From: Stuart Anderson <stuart.errol.anderson(a)gmail.com> > Date: 3/11/19, 4:53 PM > > software for producing nice-looking Venn diagrams > > https://stackoverflow.com/questions/32440128/nice-looking-five-sets-venn-di… > >> > ------------------------------ >> > >> > Date: Mon, 11 Mar 2019 11:18:18 -0600 >> > From: rcs(a)xmission.com >> > >> > Today's XKCD has the best 5-variable Venn diagram >> > drawing that I've seen. >> > https://xkcd.com/2122/

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Re: [math-fun] 33 as sum of 3 integer cubes
by Dan Asimov 13 Mar '19

13 Mar '19

So I'm wondering if there's a way to capture the weirdness of these solutions (x, y, z) in Z^3 to x^3 + y^3 + z^3 = N, namely the extreme largeness of the "smallest" solutions for some integers N. I don't recall if anyone defined "smallest", probably because several notions of smallest were / are believed to coincide in these cases. But let's use (distance? No. Better is:) squared distance of (x, y, z) to the origin, D(x,y,z) = x^2 + y^2 + z^2. The goal: to measure the extent that (x,y,z) |—> x^3 + y^3 + z^3 takes large input to small output. How best to do this? The values of distance^2 (henceforth: dist2) from the origin for the input has one distribution while those values for the output has another. [Here we switch to using just x or y for an entire input point.] Each datum in a distribution has some fraction of all data less than it and some fraction greater than it. In this case these fractions will add up to 1. To find how much a function "disorders" its input over a finite measure domain, we can calculate the average value E(fr_input(x) - fr_output(x)) of the difference between input fraction (the measure of the set of points fr_input(x) = measure of {y in domain | dist2(y) <= dist2(x)}, and the output fraction fr_output(x) = measure of {y in domain | f(y) <= f(x)} where these two numbers are each between 0 and 1. Hence their difference and its expected or average value will also lie in the unit interval. (But I suspect the maximum amount you can disorder something to be less than !. Even if you turn something upside down it's only disordered on average by 1/2.) —Dan James Buddenhagen writes: ----- ... https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf Also, D.R. Heath-Brown has conjectured that if k is not +-4 mod 9 then there are infinitely many solutions in integers to x^3 + y^3 + z^3 = k. Apparently this conjecture is thought to be true by Booker and (some?) others familiar with the problem. -----

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[math-fun] Statistics for a class of random solids
by Allan Wechsler 12 Mar '19

12 Mar '19

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. 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 can think of many more questions of this sort, and am wondering if anybody here knows what's known.

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[math-fun] 33 as sum of 3 integer cubes
by James Buddenhagen 12 Mar '19

12 Mar '19

33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3 found by Timothy Browning communicated on Quora by Alon Amit

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Re: [math-fun] Statistics for a class of random solids
by Keith F. Lynch 12 Mar '19

12 Mar '19

Allan Wechsler <acwacw(a)gmail.com> wrote: > A remaining question: Keith pointed out early on that with > probability 1 all the faces are triangles. This implies that the > expected degree of a vertex is 6 - 12/V. but I don't have any > decent intuition for the actual distribution of these degrees. > Are they really almost all 6, or are there lots of 5s and 7s too? > How does this change with N? Are there big patches of degree-6 > triangular lattice like in a geodesic dome? I too look forward to an answer to that question. And to other questions, such as what the odds are that the origin is outside the polyhedron. Intuitively, that one ought to have a closed-form solution. Consider the case N=4. The origin will be outside the tetrahedron iff there's a plane going through the origin such that all four points are on the same side of it. My first thought was that since the X, Y, and Z coordinates are statistically independent, the answer must be 1 - (7/8)^3, which is exactly 0.330078125. But then I realized that the points can all be on the same side even if they're bipolar in each of X, Y, and Z. For instance if all the points were on the plane X+Y+Z = 1.

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Re: [math-fun] 33 as sum of 3 integer cubes
by Keith F. Lynch 12 Mar '19

12 Mar '19

There's an infortmative new video about the solution: https://www.youtube.com/watch?v=ASoz_NuIvP0 It includes video of the computer cluster that was used.

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Re: [math-fun] Why isn’t PentakisDodecahedron Johnson 93??
by Bill Gosper 12 Mar '19

12 Mar '19

GAAAAAAA!! In[12]:= PolyhedronData["PentakisDodecahedron", "EdgeLengths"] // DeleteDuplicates Out[12]= {1, 1/6 (9 - Sqrt[5])} I mistook pentakis to mean augmented! Apologies to everyone. On 2019-03-11 11:57, Tom Karzes wrote: > I see 6 triangles meeting at a vertex. If the triangles are > regular, then either they are coplanar or the polyhedron is > concave. > > Tom > > Bill Gosper writes: > > Johnson 93? > > < https://www.wolframcloud.com/objects/810fbe92-6ab3-480b-9f86-e4869990cf46> > > —rwg > > And why can't we tumble that figure? We probably can if we refresh the page once or twice. —rwg

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