October 2008 - mathfun@mailman.xmission.com

Re: [math-fun] tetraroller volume
by Dan Asimov 21 Oct '08

21 Oct '08

I wrote: << Imagine the intersection of N with the unit sphere S^2 about the origin. That will be a number of great circles of S^2. >> This isn't right; rather it's just each 2D cylinder that intersects the unit sphere in a great circle. Now look at how each of these great circles on S^2 is cut by the others; each resulting arc corresponds to one "face" of the boundary of the intersection solid. I hope that was clearer and truer. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] tetraroller volume
by Dan Asimov 21 Oct '08

21 Oct '08

I wrote: << was able to figure out the face structure of the four equal cylinders (which could be said to intersect as do the great diagonals of a cube). Same method should apply to the higherhedra. >> Here's the method (but I've patented it, and you may need to pay me royalties): Suppose n solid infinite cylinders of unit radius are arranged in 3-space so their axes each pass through the origin. Then the solid K where they all intersect has boundary surface N, where N has "faces" that are maximal portions of 2D unit cylinders. Imagine the intersection of N with the unit sphere S^2 about the origin. That will be a number of great circles of S^2. These circles cut each other up into (uncut) great circle arcs. Each of these occurs in just one face of N. So # faces of N = # largest uncut great circle arcs on S^2. The faces will resemble Voronoi regions of the great circle arcs. In a symmetrical situation, e.g., where the axes are the diagonals of a higherhedron, this will be exact (if the faces of N are radially projected onto S^2). E.g., for 4 main diagonals of a cube, we can pretend the cube is a unit sphere, and the great circles will be equators w.r.t. opposite pairs of cube vertices -- giving us 4 regular hexagons inscribed on the cube. Each hexagon is cut in sixths by the other hexagons, giving us 24 "great circle" arcs, hence 24 faces of the intersection solid K. The Voronoi regions on the sphere are each a quadrilateral with one axis of symmetry. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] tetraroller volume
by Dan Asimov 20 Oct '08

20 Oct '08

Seb. Perez sent links to some lovely polyhedra nets. But where he wrote: << On Mon, Oct 20, 2008 at 19:18, Dan Asimov <dasimov(a)earthlink.net> wrote: > << > PS, did anybody look at intersecting four cylinders instead of cones? > I'm not even sure how many faces this makes (http://gosper.org/face.jpg) > Likewise the intersections of 6 and 10 cylinders corresponding to the > higherhedra. >>> >> in fact the PS... words are those of Mr. Gosper, which in my e-mail I was merely quoting. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] tetrahedron volume
by Michael Reid 20 Oct '08

20 Oct '08

> True or False Quickie: The volume of a tetrahedron is determined by the > areas of its faces. > _______________________________________________ > > False. There exists a face area preserving continuous deformation that alters the volume. Start with a regular tetrahedron. Perform a dilatation along the mutual perpendicular of two opposite edges, while simultaneously performing a dilatation in the transverse plane so as to preserve the face areas. This is possible because all four faces are inclined with respect to that mutual perpendicular by the same angle. The volume is not preserved, in particular the volume goes to zero as the tetrahedron is squeezed flat. an easy way to see this is to take four alternating vertices of a rectangular parallelepiped; then all four faces are congruent. i've heard this figure called an "isosceles tetrahedron". if one dimension of the rectangular parallelepiped is very small, then the volume of the tetrahedron is very small, but the area of the faces is not necessarily small. specifically, if the rectangular parallelepiped has dimensions a x b x c , then the volume of the tetrahedron is abc / 3 , and the area of the congruent faces is sqrt(a^2 b^2 + b^2 c^2 + c^2 a^2) / 2 . mike

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Re: [math-fun] tetraroller volume
by Dan Asimov 20 Oct '08

20 Oct '08

<< PS, did anybody look at intersecting four cylinders instead of cones? I'm not even sure how many faces this makes (http://gosper.org/face.jpg) Likewise the intersections of 6 and 10 cylinders corresponding to the higherhedra. >> Don't think I ever calculated the volumes, but was able to figure out the face structure of the four equal cylinders (which could be said to intersect as do the great diagonals of a cube). Same method should apply to the higherhedra. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] P.E.N.T.A.G.O.N.A.L
by Eric Angelini 20 Oct '08

20 Oct '08

Hello Math-Fun, according to this page, http://www.primepuzzles.net/puzzles/puzz_457.htm "the smallest prime number whose name written in English contains the letters "PRIME" in order embedded in it" is ("For American and Modern British English"): 1000000000000000035000061 one sePtillion thiRty-fIve Million sixty-onE - - - - - Similarily, what would be the smallest PENTAGONAL? Best, É.

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Re: [math-fun] Theta_3(0,q) near the unit circle
by Eugene Salamin 20 Oct '08

20 Oct '08

----- Original Message ---- From: "rwg(a)sdf.lonestar.org" <rwg(a)sdf.lonestar.org> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Tuesday, October 7, 2008 4:02:59 AM Subject: [math-fun] Theta_3(0,q) near the unit circle ... PS, how can a natural boundary grow only like 1/sqrt(-log q)? The other thetas are *much* wilder. _______________________________________________ Einar Hille gives the following example (Analytic Function Theory, vol. 1, p.133). sum( z^(2^n) / n!, n=0..infinity) This function and its derivatives of all orders are continuous and bounded in the closed unit disk. Yet the unit circle is a natural boundary. Gene

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[math-fun] Plouffe's inverter down?
by rcs＠xmission.com 18 Oct '08

18 Oct '08

I'm getting "no such website" when I try to use the Plouffe inverter at http://pi.lacim.uqam.ca/eng/ Rich

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Re: [math-fun] Old cute math problem
by Dan Asimov 17 Oct '08

17 Oct '08

Steve W. wrote: << [I wrote] << Let x_n be a dense sequence in [0,1]. x_1 divides [0,1] into a left piece of length a_1 and a right piece b_1. Inductively, x_(n+1) divides one of the subintervals that {x_1,...,x_n} divide [0,1] into a left piece of length a_n and a right piece of length b_n. >> Shouldn't x_(n+1) divide into a_(n+1) and b_(n+1) ? >> Yes, that's what I should have written. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] what's the best way of finding n unknown integers from their sums taken k at a time
by Michael Reid 17 Oct '08

17 Oct '08

> The corresponding problem where sums of {\em triples} > of elements of a set are given has been settled by > Boman \& Linusson. The exceptions are precisely > 3, 6, 27, 486. For $n=27$ they give five examples of > which the simplest is $\{-4,-1^{10},2^{16}\}$ and > its negative, where exponents denote repetitions. i think this is not right, but if i switch the multiplicities of -1 and 2 , then it seems to work: for the multiset { -4, -1^16, 2^10 } , its 3-at-a-time-sums multiset is { -6^120, -3^720, 0^1245, 3^720, 6^120 } , which is symmetric about 0 . > For $n=486$ they give > $\{-7,-4^{56},-1^{231},2^{176},5^{22}\}$ and its > negative. this one seems to be fine, with 3-at-a-time-sums multiset { -15^1540, -12^40656, -9^392161, -6^1800568, -3^4358893, 0^5826304, 3^4358893, 6^1800568, 9^392161, 12^40656, 15^1540} , which is symmetric about 0 . mike

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