December 2017 - mathfun@mailman.xmission.com

[math-fun] Question suggest by solid angle issues
by Dan Asimov 21 Dec '17

21 Dec '17

Let S be the unit sphere in 3-space centered at (1, 0, 0). Now transform the x- and y-coordinates like the squaring operation (x + yi) |——> (x^2-y^2) + 2xyi, i.e., define a function f: R^3 —> R^3 via f(x, y, z) = (x^2-y^2, 2xy, z) and apply this to the surface S to get a new surface: X = {(x^2-y^2, 2xy, z) | (x-1)^2 + y^2 + z^2 = 1}. The origin 0 belongs to X. What is the solid angle SA(X; 0) subtended by the surface X at the point 0 ??? ((( Meaning: Let U be any open neighborhood of 0 in X. Let A(U) denote the area on the unit sphere of the set of all outward normal unit vectors to X at a point of U. Then we define the solid angle SA(X; 0) = inf {A(U) | U is an open neighborhood of 0 in X}. ))) —Dan

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[math-fun] Cutting a pie ...
by Guy Haworth 20 Dec '17

20 Dec '17

The problem is usually expressed as one of dividing a cake (or pie) and distributing it so that no-one is in a position to complain about the size of their slice ... There's this ... https://www.scientificamerican.com/article/the-mathematics-of-cake-cutting/ ... and then there's this by Ian Stewart https://www.amazon.co.uk/How-Cut-Cake-Mathematical-Conundrums/dp/0199205906 However, the best solution I know of (which Ian Stewart confirmed he had not heard of) came from the 13-year-old son of a colleague of mine .... Everyone gets a knife and makes a cut in a circular cake which is standing on a turntable ... with the aim of making equal-size slices. The cake is then spun, and everyone gets the slice opposite them. [ In other words, the randomness takes the decision about who gets what away from the hungry mathematicians ] Guy

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[math-fun] Fwd: Cutting a pie ...
by James Propp 20 Dec '17

20 Dec '17

More from Warren: ---------- Forwarded message ---------- From: Warren D Smith <warren.wds(a)gmail.com> Date: Wed, Dec 20, 2017 at 12:44 AM Subject: Re: [math-fun] Cutting a pie ... To: James Propp <jamespropp(a)gmail.com> I can't bring myself to believe F(n) obeys any linear upper bound; I think it must have a superlinear lower bound. Indeed F(2*n) >= 2*F(n)+n-1 holds for n=1,2,3,4 and perhaps 5, and it seems plausible F(2*n)>=2*F(n)+K*n holds for K>=0, with indeed K lower bounded by some positive constant on average... which would imply a n*log(n) style lower bound. So my conjecture is F(n) asymptotically is both superlinear and subquadratic, but I have no idea where it lies within that range. Very annoying! -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Random nonempty intersection of balls
by James Propp 20 Dec '17

20 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 19 Dec '17

19 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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