September 2013 - mathfun@mailman.xmission.com

Re: [math-fun] Probability that all molecules of a gas are in one half of the container
by Warren D Smith 19 Sep '13

19 Sep '13

So Veit E.'s idea (as he now describes) involves what has been called "spherical polars" or "duality." I'll now re-do his solution so I can understand it (you too, maybe). That is, each new point on the sphere, defines an excluded hemiball (centered at it) in which the empty hemiball's center (if any) cannot lie. I.e. the sphere is cut by a new centered hyperplane every time you add new point. After N points added, we've cut the sphere using N "polar" hyperplanes into 2^N regions (some of these "regions" may not actually exist, i.e. are empty sets). Only the ++++...+++ region (if any) is eligible for occupancy by an empty-hemiball center (if any). The probability an empty hemiball exists, is the probability that this region exists. Now then Veit continues by arguing that by considering replacing points by their antipodes, all regions are "equivalent" i.e. under some assignment of {points, antipodes} any region you like becomes the +++...++++ region. Hence, if there are Reg(N,D) genuine regions, Veit argues the probability a +++...++++ region exists, is just Reg(N,D)/2^N. And Reg(N,D) is a combinatorial quantity that can be found using recurrences. In particular Reg(N,2) = 2*N obviously, explaining Salamin's formula. And Reg(N,3) = 2*(N-1) + Reg(N-1, 3) more generally Reg(N,D) = Reg(N-1, D-1) + Reg(N-1, D) by considering adding 1 new cut-hyperplane and considering the D-1 dimensional situation within that hyperplane. Hence Reg(N,D) always is a degree=D-1 polynomial in N which you can determine by fitting to the data for N=0,1,2,...,D-1. So the problem is now solved in every dimension. Very nice. Elegant.

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[math-fun] math-fun no longer ASCII ??
by Henry Baker 18 Sep '13

18 Sep '13

Is it just me, or has anyone else noticed that math-fun is coming through in html rather than ascii ? This seems to be a relatively new development.

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[math-fun] Probability that all molecules of a gas are in one half of the container
by Eugene Salamin 18 Sep '13

18 Sep '13

A spherical vessel contains n molecules of gas. What is the probability that all the molecules can be found in one hemisphere? For a given hemisphere, it is 1/2^n. For 6 hemispheres arranged like the faces of a cube, inclusion-exclusion gives 6/2^n - 12/4^n + 8/8^n. But what is the probability if any hemisphere is allowed? I'm stuck on this problem. In two dimensions, the probability that all n molecules lie in some semicircle is 2n/2^n. A related question is: Given n vectors x[1], ... , x[n], how do we test if they all lie in some half-space, i.e. does there exist a vector u such that u.x[i] > 0 for all i? I'm stuck on this as well. -- Gene

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[math-fun] A Jewel at the Heart of Quantum Physics
by Ray Tayek 18 Sep '13

18 Sep '13

https://www.simonsfoundation.org/quanta/20130917-a-jewel-at-the-heart-of-qu… Physicists have discovered a jewel-like geometric object that dramatically simplifies calculations of particle interactions and challenges the notion that space and time are fundamental components of reality. "This is completely new and very much simpler than anything that has been done before," said Andrew Hodges, a mathematical physicist at Oxford University who has been following the work. --- co-chair http://ocjug.org/

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[math-fun] pretty cool
by Simon Plouffe 18 Sep '13

18 Sep '13

Hello, everyone, on this day, google came up with something quite amusing on the front page. https://www.google.fr/ or https://www.google.com/ it depends where you are located, (they know where you are you know ...) I would say it is the <swing of the day>. have fun, Simon Plouffe

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[math-fun] Salamin's empty hemisphere problem
by Warren D Smith 18 Sep '13

18 Sep '13

You can get upper and lower bounds on the answer. The problem is: N gas atoms in an 0-centers D-dimensional ball. (D>0.) What is chance there exists an empty hemi-ball? Lower bound: The left hemiball (x<0) is empty with chance 2^(-N). Better lower bound: Pempty >= 2^(1-N) by considering both the x<0 and x>0 hemiballs. Still better lower bound: Pempty >= N * 2^(1-N) by considering all hemiballs arising from semicircles in xy plane (assuming Salamin correct about the exact solution when D=2). And when D=4 this would give the lower bound (2*N)^2 * 2^(-N) by considering xy and zt planes. And when D=3 it would give (2*N+2)*2^(-N) by considering xy plane and the z<0 and z>0 hemiballs. Upper bound: If all 2^D quadrants are occupied, then there is no empty hemiball, because any hyperplane passing thru ball-center must contain at least one entire quadrant on one side. Better upper bound: divide D-space into D+1 equal regions by "coning off" the D+1 faces of a regular D-simplex centered at 0, to the origin. If every region is occupied, then there can be no empty hemiball. (Because any hyperplane passing thru ball-center must contain at least one entire region on one side.) The chance that all D+1 regions are occupied is lower bounded by D^N / (D+1)^N <= 1-Pempty. On-sphere lemma: (As somebody already noted) If all N points lie on the unit sphere, rather than in it, that makes no difference (Pempty unaffected). Special argument when D=2: The question is equivalent to this: is the longest side of the N-gon formed by N random points on the unit circle, on the "wrong" side of 0? (I.e. 0 lies on opposite side from all the other points.) There are N sides. Special argument when D=3: The question is equivalent to this: is the largest-circumradius-triangular-face of the polyhedron that is the convex hull of N random points on the unit sphere, on the "wrong" side of 0? (I.e. 0 lies on opposite side from all the other points.) There are 2*N-4 such faces (generically triangular, i.e. the probability a nontriangular face exists, equals 0). It seems to me that in both the D=2 and D=3 cases, the answer can be written in closed form as a definite integral using (a) the two special observations above, (b) the fact that it is impossible for more than 1 "special face" to exist, it automatically is unique one. So I think when D=2 you are going to get Pempty = N * int(0<u<1) (u/2)^(N-2) * du where u=theta/pi, if you see what I mean. I admit, this is off by a factor N-1 versus Salamin's claimed answer, so I'm probably missing something. Perhaps I was suppose to divide by int(0<u<2) (u/2)^(N-2) * du to normalize probability distribution or something. I think that yields Salamin's claimed answer. When D=3 you similarly (presumably also wrongly + missing something) would get Pempty = C * (2*N+4) * int(0<u<1) (u/2)^(N-3) * (1-(1-u)^2)^(3/2) * du using Archimedes in the direction orthogonal to the face, for some positive constant C I have not bothered to work out. If the same "normalize by dividing" fix worked then Pempty = (2*N+4) * int(0<u<1) (u/2)^(N-3) * (1-(1-u)^2)^(3/2) * du / int(0<u<2) (u/2)^(N-3) * (1-(1-u)^2)^(3/2) * du. So anyhow, I have not worked out the full details, but if this approach works you should be able using my hints, to get exact 2D and 3D answers; plus I have shown how to find upper & lower bounds in any dimension (which are undoubtably improvable).

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Re: [math-fun] math-fun Digest, Vol 127, Issue 19
by Warren D Smith 17 Sep '13

17 Sep '13

> Better upper bound: > divide D-space into D+1 equal regions by "coning off" the D+1 faces of > a regular D-simplex centered at 0, to the origin. > If every region is occupied, then there can be no empty hemiball. > (Because any hyperplane passing thru ball-center must contain at least > one entire region on one side.) > > The chance that all D+1 regions are occupied is lower bounded by > D^N / (D+1)^N <= 1-Pempty. --wrong. I should have said cone off the 2*D faces of an 0-centered hypercube to get 2*D regions. Need a centrosymmetric thing like hypercube; the regular D-simplex not being centrosymmetric, does not work. Then the corrected chance that all 2*D regions are occupied is lower bounded by (2*D-1)^N / (2*D)^N <= 1-Pempty. --still wrong (duh): the left hand side actually gives (a lower bound on) the probability that NOT all 2*D regions are occupied, so really Pempty <= (1 - 1/(2*D))^N = (2*D-1)^N / (2*D)^N is the upper bound I was seeking. Brilliant. Anyhow, the point is, the upper bound is exponentially shrinking with N, and so is the lower bound, though sadly their growth factors do not match so we get a goodly gap.

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Re: [math-fun] Probability that all molecules of a gas are in one half of the container
by Adam P. Goucher 17 Sep '13

17 Sep '13

And now for a related piece of trivia, which may or may not help with this problem: If you distribute points uniformly on the surface of a unit sphere, their x-coordinates are uniformly distributed in the interval [-1,1]. This is specific to the 2-sphere in 3-space, and was almost certainly first noted by Archimedes. Sincerely, Adam P. Goucher http://cp4space.wordpress.com > ----- Original Message ----- > From: Eugene Salamin > Sent: 09/17/13 08:43 PM > To: math-fun > Subject: Re: [math-fun] Probability that all molecules of a gas are in one half of the container > > And more generally, the particles can have any spherically symmetric distribution. If preparing to integrate over space, a useful distribution is the Gaussian, exp(-r^2) = exp(-x^2) exp(-y^2) exp(-z^2). > > -- Gene > > > > >________________________________ > > From: Gareth McCaughan <gareth.mccaughan(a)pobox.com> > >To: math-fun(a)mailman.xmission.com > >Sent: Tuesday, September 17, 2013 12:28 PM > >Subject: Re: [math-fun] Probability that all molecules of a gas are in one half of the container > > > > > >On 17/09/2013 20:16, Mike Stay wrote: > >> For each possible point on the sphere, the probability that all the > >> particles are in the hemisphere centered on that point is 2^{-n}. > >> Pull the constant out and integrate over the sphere to get 2^{-n} * 1. > >> Or am I missing something? > > > >I think so. The cases overlap, so the probabilities don't just add. > > > >I have the feeling I've seen this problem, or one awfully similar to it, > >before. If so, I don't remember the answer. > > > >One trivial remark: The answer is the same whether you choose > >particles uniformly *on the surface of the sphere* or *in the > >ball bounded by the sphere*. > > > >-- > >g > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Probability that all molecules of a gas are in one half of the container
by Adam P. Goucher 17 Sep '13

17 Sep '13

> On 9/17/2013 12:02 PM, Eugene Salamin wrote: > > A spherical vessel contains n molecules of gas. What is the probability that all the molecules can be found in one hemisphere? For a given hemisphere, it is 1/2^n. For 6 hemispheres arranged like the faces of a cube, inclusion-exclusion gives 6/2^n - 12/4^n + 8/8^n. But what is the probability if any hemisphere is allowed? I'm stuck on this problem. > > I don't understand the arrangement. Are the hemispheres stuck onto the faces of a cube? They overlap. > > In two dimensions, the probability that all n molecules lie in some semicircle is 2n/2^n. > > That doesn't look right. n=2 -> P=1 Yes, unless they're antipodal (with probability zero), we can take a line (say, the diameter orthogonal to their perpendicular bisector) partitioning the circle into two semicircles, one of which contains both points. Sincerely, Adam P. Goucher

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Re: [math-fun] The Bernoulli Manifesto
by Steve Witham 17 Sep '13

17 Sep '13

>From: Dan Asimov <dasimov(a)earthlink.net> > >Minuscule point: It seems to me that in Knuth's response to >Luschny, he writes cos(pi s) where he means to write cos(2pi s). He probably uses the New Convention privately and missed translating. Wouldn't a LaTeX macro handle all this? Write once, typeset in either Convention. And not just on paper: with online versions there would be a button to view either way. In fact you could typeset in any draft or forked convention, giving us all time for consideration. --Steve P.S. check out the Bernoulli disambiguation page: http://en.wikipedia.org/wiki/Bernoulli

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