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[math-fun] Random nonempty intersection of balls
by James Propp 19 Dec '17

19 Dec '17

Let P be a point in Euclidean n-space and P' be a point chosen uniformly at random from the ball of radius 2 centered at P, so that the unit balls centered at P and P' have nonempty intersection. Show that the expected hypersurface area of the intersection of the balls (I guess the shape would be called a lunoid) is exactly equal to the hypersurface area of an (n-1)-sphere of radius 1/2. Is this observation new? What about the case n=2 where the lunoid is a lune? I know a proof, but it uses combinatorics in a way that I suspect is unnecessary. Jim Propp

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[math-fun] Problems suggested by "Random nonempty intersection of balls"
by Dan Asimov 19 Dec '17

19 Dec '17

Interesting question! Which also suggests some related questions: I) -- What about the expected intersection of two hemi-n-spheres in the n-sphere S^n ? In S^0 the average intersection has 0-area equal to 1/4 that of S^0. In S^1 the average intersection has 1-area equal to 1/4 that of S^1. In S^2 the average intersection has 2-area equal to ... ...Is it again 1/4 that of S^2 ??? (Dubious argument: Two random hemispheres are determined by two random *oriented* 2-planes through the origin of R^3 (or equivalently, two random unit vectors. The (least) angle between the planes is the same as that between the vectors, so is anywhere from 0 to 180 degrees, averaging 90 degrees. Two hemispheres at 90 degrees intersect in half a hemisphere, again giving 1/4 the n-area of S^n, this time for n = 2. Hey, I said it was dubious.) II) --- One could ask the analogous question to Jim's: What is the expected (n-1)-area of the intersection of two random hemispheres of S^n ??? III) ---- Another nice space to ask both the I) and II) (Jim's) types of questions about is the torus. The simplest torus in all dimensions is the cubical one, T^n = R^n / Z^n obtained from a unit cube by identifying corresponding points on opposite faces. The only "drawback" on T^n is that perhaps there is no obvious radius of a ball or sphere to ask about. So maybe here it's appropriate to ask, for any two given radii r > 0 and s > 0, about the expected intersection n-area (or boundary (n-1)-area) of two random balls with those radii. III) ---- Something I worked on a long time ago is the n-dimensional version(s) of this question*: Given two geometric circles (each the intersection of a 2-sphere and a 2-plane through its center) in the 3-torus T^3 of radii r and s, what is the probability that they *link* each other? This turns out to be pretty interesting. —Dan —————————————————————————————————————————————————————————————————— * which requires spheres S^k of radius r and S^m of radius s, with k + m = n - 1 so that linking can take place. Jim Propp wrote: ----- Let P be a point in Euclidean n-space and P' be a point chosen uniformly at random from the ball of radius 2 centered at P, so that the unit balls centered at P and P' have nonempty intersection. Show that the expected hypersurface area of the intersection of the balls (I guess the shape would be called a lunoid) is exactly equal to the hypersurface area of an (n-1)-sphere of radius 1/2. Is this observation new? What about the case n=2 where the lunoid is a lune? I know a proof, but it uses combinatorics in a way that I suspect is unnecessary. -----

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[math-fun] Can you solve it?
by Fred Lunnon 19 Dec '17

19 Dec '17

I enjoyed Alex Bellos' teasers in the Guardian this week: https://www.theguardian.com/science/2017/dec/18/can-you-solve-it-feast-on-t… WFL

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[math-fun] Where does this series converge?
by Dan Asimov 19 Dec '17

19 Dec '17

I recently tried to make a plot of (*) y = Sum_{1 <= n < oo} sin(nx) for some small values of infinity, and it distinctly hovered around the apparent value Sum_{1 <= n < oo} Im( exp(inx)) = Im(Sum_{1 <= n < oo} exp(inx)) =(???) Im( exp(ix) / (1 - exp(ix)) ) = sin(x) / (2 - 2*cos(x)). Clearly the original series converges only for x = n*pi for some integer n. Can this function f(x) = sin(x) / (2 - 2*cos(x)) be the *Cesaro sum* of the original series (*) ??? Or at least for certain values of x ??? —Dan

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Re: [math-fun] Where does this series converge?
by Dan Asimov 18 Dec '17

18 Dec '17

----- why do you say that clearly the original series converges only for x = n*pi? ----- If x <> 0 mod pi is a rational multiple of 2pi, then {sin(nx)} just repeats the same > 1 distinct values as n —> oo. If x is an irrational multiple of 2pi, then {nx mod 2pi} is dense in the interval [0, 2pi), so the terms sin(nx) do not approach 0 as n —> oo. —Dan

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[math-fun] Limit question
by Dan Asimov 17 Dec '17

17 Dec '17

What is lim ((1/n)^n + (2/n)^n + ... + (n/n)^n) n—>oo (exactly, if possible) ??? My bag of limit tricks doesn't readily seem to include how to solve this one. —Dan

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[math-fun] 4-orchard problem
by Ed Pegg Jr 17 Dec '17

17 Dec '17

I recently tackled the 3-orchard problem. Nice pictures are at http://oeis.org/A003035 . The 4-orchard problem is proving harder because the solutions don't seem as strong. In particular, the solutions for 22-25 points seem easily beatable. I've posted values and a picture at https://math.stackexchange.com/questions/2554709/the-4-orchard-problem If anyone can improve these solutions, please let me know. --Ed Pegg Jr

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[math-fun] Recent Grumbling about the ABC-Conjecture Proof
by Hans Havermann 17 Dec '17

17 Dec '17

It's been reported that "Mochizuki's (claimed) proof of the abc-conjecture had been vetted and considered fit to be published in a respectable journal" [John Baez, today on Google+]. Peter Woit has summarized why we should continue to maintain some skepticism: http://www.math.columbia.edu/~woit/wordpress/?p=9871

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[math-fun] endopentagram
by Bill Gosper 17 Dec '17

17 Dec '17

Each face of an endododecahedron is, as you might expect, two legs of an ordinary pentagram. Including pants. http://gosper.org/endogram.png --rwg

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[math-fun] Intermediate expression swell
by Bill Gosper 15 Dec '17

15 Dec '17

Five ways to compute the solid angle subtended by a face of a regular dodecahedron w.r.t. the centroid led to {2 (ArcCot[√15] + 2 ArcCot[√(3(9 + 4 √5))]), ArcCos[7/8] + 2 ArcCos[1/8 + (3 √5)/8], π/3, π/3, 4 (ArcTan[(-√5 + √6) Cot[1/4 (π + ArcSec[-3])]] + ArcTan[Root[1 - 220 #1^2 + 582 #1^4 - 220 #1^6 + #1^8 &, 5]] + ArcTan[Root[1 - 6 #1 + 2 #1^2 + 6 #1^3 + #1^4 &, 4] Tan[1/4 ArcSec[3/√5]]])} (Recall that) these should all be 4π/12 = π/3. Can FullSimplify manage? FullSimplify[ArcTan[Print[ByteCount /@ #]; #] &@TrigExpand@Tan@%] {33224,10912,128,128,81747368} I.e., the intermediate byte counts are 33K, 10K, 128, 128, and 81M! The 33K takes several minutes the first time. I don't expect to see it finish the 81M. I actually don't know a way to symbolically prove case 5. MinimalPolynomial[Tan[...]] seems to have the best shot to answer within a human lifespan, but not a span of hours. --rwg Apropos the timezone map, everybody knows about the 3½ hr discontinuity, right?

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