November 2006 - mathfun@mailman.xmission.com

Re: [math-fun] Ambiguously placed cities
by Michael Reid 30 Nov '06

30 Nov '06

> Before people go too far looking for such charts, I have a question: is > it possible to have 4 points in a plane, no 3 colinear, so that all 6 > distances are integers? And with the additional constraint that no 2 > distances are equal? it is possible to have arbitrarily many points in the plane, no three collinear, such that all distances between them are integers, and no two such distances are equal. mike

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[math-fun] integrate(x^p floor(x)^q,x)
by R. William Gosper 29 Nov '06

29 Nov '06

Maple and Macsyma incorrectly give x floor x for integrate(floor(x),x). (Integrals are continuous.) Mma abstains. The answer is ceiling(x) floor(x) + 1 (x - ----------) (ceiling(x) - 1) = floor(x) (x - ------------), 2 2 (products of discontinuous functions). More generally, one can find (e.g., by undetermined coefficients) a polynomial(floor x) of degree p+q which, when added to x^p floor(x)^q, renders the sum continuous. E.g., / 4 4 8 7 [ 3 4 x floor (x) floor (x) 2 floor (x) I x floor (x) dx = ------------ - --------- - ----------- ] 4 8 7 / 5 3 3 floor (x) floor (x) 23 floor(x) + ----------- - --------- + -----------. 10 6 840 How does one seek in EIS a two-parameter family of polynomials with rational coefficents (presumably named Smarandache)? --rwg UNDERAGES UNGREASED DUNGAREES

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Re: [math-fun] integrate(x^p floor(x)^q,x)
by Daniel Asimov 29 Nov '06

29 Nov '06

Apologies to whoever posted this, whose identity I've forgotten: << When I participated in Maple beta program, I submitted this bug. It was a heated discussion there about it, and I was told by Maple developers that in Maple indefinite integrals are defined up to "a piecewise constant", i.e. integrals of continuous functions can be discontinuous. The bug was classified as "works as designed". >> I find this philosophy rather ridiculous for antiderivatives of continuous functions. But for antiderivatives of functions continuous but on natural domains that are not connected, like the ubiquitous f(x) = 1/x, calculus books should give its antiderivative as ln|x| + C_1, x > 0 F(x) = { ln|x| + C_2, x < 0 for arbitrary choice of constants C_1, C_2, and likewise for 1/x^p, p > 0, p <> 1 (and mutatis mutandum for sec(x), csc(x), etc.). --Dan

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Re: [math-fun] Ambiguously placed cities
by Fred lunnon 27 Nov '06

27 Nov '06

Ed has pointed out what (in the cold light of day) should have been glaringly obvious: 5+8=13, and my example is a bit of a damp squib. I could modify the program to eliminate these; but a 5000 sec search up to sum of edges = 100 yielded only one other example, which suffers the same flaw. A triplet of tetrahedra with equal edges and the same volume turns out equally dubious. Heigh-ho --- back to the drawing-board! WFL On 11/27/06, Ed Pegg Jr <ed(a)mathpuzzle.com> wrote: > Very nice. This is the sort of configuration set I was looking for. > > Unfortunately, in the first set, AB+BC=AC, so that part of it is a straight > line, > which is a slight flaw. > > --Ed > > Fred lunnon <fred.lunnon(a)gmail.com> wrote: > Let the road lengths be arranged [AB, BC, CA, BD, AD, CD], > where A,B,C,D denote cities. Then the follwoing two charts are > distinct and planar, and both permute the same set of integers: > > [5, 8, 13, 11, 9, 17] > [5, 17, 13, 11, 9, 8] > > [Given my capacity for wishful thinking, not to mention Maple's > for wishful computation, I'd appreciate somebody verifying this > independently!] > > Fred Lunnon > > >

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Correction Re: [math-fun] Ambiguously placed cities
by Daniel Asimov 26 Nov '06

26 Nov '06

Never mind the part about integral distances in the 4-town configurations.

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RE: [math-fun] Embedding simplicial complexes in Euclidean space
by Daniel Asimov 25 Nov '06

25 Nov '06

Andy wrote: << More simply, take two equilateral triangles, identify all three corners of one with all three corners of the other. This does not have an affine embedding in R^n for all n. This (along with identifying some of the edges of the triangles) is exactly what you've done with the two center triangles of your strip, in your more complex example. Even simpler, in one dimension less, take two line segments, and identify their endpoints. This does not embed in R^n affinely for any n. . . . >> The problem is, in a simplicial complex any two simplices that intersect must do so in one common face (of any dimension). So two 2-simplices that intersect in all three 0-faces doesn't qualify. Ditto, of course, for two 1-simplices that intersect in both their endpoints. --Dan

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Re: [math-fun] Embedding simplicial complexes in Euclidean space
by Daniel Asimov 25 Nov '06

25 Nov '06

Thanks to Fred, Bill, and Andy for pointing out my blunder with the allegedly 4-triangle Moebius band. --------------------------------------------------------- A natural extension of the question of what dimension a simplicial complex embeds in, affinely on simplices (AOS), is this: QUESTION: Given a simplicial complex K and its abstract symmetry group, what is the smallest N such that K embeds AOS in R^N, such that each original symmetry of the embedded complex is realized by a rotation of R^N ??? 1. E.g., the projective plane can be constructed by identifying the faces of the icosahedron by the antipodal map, giving a simplicial complex K of 10 triangles, having a symmetry group of order 60 (= the smallest non-abelian simple group?). So . . . what is the lowest-dimensional Euclidean space into which this embeds AOS, such that all 60 symmetries are realized by rotations? 2. Ditto for the lovely torus someone mentioned -- given by a simplicial complex of 14 triangles (6 per vertex). Its symmetry group has order 84. (This complex is the dual of the amazing 7-hexagon tiling of the torus.) 3. Likewise for the beautiful rendition of the 3-holed torus as 56 triangles (7 per vertex). Its symmetry group has order 336. (This complex is the dual of the gorgeous 24-heptagon tiling of the surface of genus 3.) (Or, for 2. and 3. one could just ask to realize as rotations the *orientation- preserving* symmetry groups -- of orders 42, and 168 (= the second-smallest non-abelian simple group), respectively.) --Dan

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Re: [math-fun] Embedding simplicial complexes in Euclidean space
by Daniel Asimov 25 Nov '06

25 Nov '06

<< Finally, turning a previous comment on its head, this result seems to give a very cheap proof of a weaker version of the manifold embedding theorem: just triangulate a d-manifold arbitrarily densely, then continuity ensures it can be embedded in (2d+1)-space! [Have I overlooked anything here?] Now then, how might the extra dimension be unloaded, I wonder ... >> Only that with a little more work, it can be seen that d-manifolds embed in 2d-dimensional space. But I suspct there's a simplicial complex whose highest-dimensional simplex is d, but which doesn't embed in 2d-space. Quite possibly the "Hilbert heptahedron" consisting of every other face of an octahedron, plus three squares (topologically a projective plane) -- after each square is subdivided into two triangles to get a simplical complex -- does not embed (affinely on simplices) in 2*2 = 4-space. (But -- as I now realize -- it must embed AOS in 5-space since it has only 6 vertices and is a subcomplex of the 5-simplex.) --Dan

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Re: [math-fun] Embedding simplicial complexes in Euclidean space
by Daniel Asimov 24 Nov '06

24 Nov '06

Make a band of four equilateral triangles in the plane, forming a parallelogram. Now identify the left and right edges of the parallelogram, creating a simplicial complex K, containing four 2-simplices, that's topologcially a Moebius band. For all n, there is no topological embedding h: K -> R^n such that h is affine on simplices. (Exercise: prove this.) That settles one question. --Dan

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RE: [math-fun] Embedding simplicial complexes in Euclidean space
by Andy Latto 23 Nov '06

23 Nov '06

> -----Original Message----- > From: > math-fun-bounces+andy.latto=pobox.com(a)mailman.xmission.com > [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission > .com] On Behalf Of Daniel Asimov > Sent: Thursday, November 23, 2006 4:54 AM > To: math-fun > Subject: Re: [math-fun] Embedding simplicial complexes in > Euclidean space > > > Make a band of four equilateral triangles in the plane, > forming a parallelogram. > > Now identify the left and right edges of the parallelogram, creating > a simplicial complex K, containing four 2-simplices, that's > topologcially > a Moebius band. > > For all n, there is no topological embedding h: K -> R^n > such that h is affine on simplices. More simply, take two equilateral triangles, identify all three corners of one with all three corners of the other. This does not have an affine embedding in R^n for all n. This (along with identifying some of the edges of the triangles) is exactly what you've done with the two center triangles of your strip, in your more complex example. Even simpler, in one dimension less, take two line segments, and identify their endpoints. This does not embed in R^n affinely for any n. If a complex is to be embedded affinely in R^N, it must not contain two distinct simplices with the same vertex set. Is this necessary condition also sufficient? Andy Latto andy.latto(a)pobox.com

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