September 2013 - mathfun@mailman.xmission.com

Re: [math-fun] Julian's dodecahedron
by Bill Gosper 02 Sep '13

02 Sep '13

="Bill Gosper" <billgosper(a)gmail.com> we still have the problem of naming this solid! Marc LeBrun>tympanon tympanohedron? Note that it's like a pentagonal antiprism (DoD NSA opponent?) but with M-shaped equilateral pentagonal flanks instead of equilateral triangles. We could recognize the (countable) family of "M" antiprisms and call this the pentagonal one. --rwg

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[math-fun] Ultra-high energy cosmic rays -- appears to be an energy cutoff
by Warren D Smith 02 Sep '13

02 Sep '13

The so-called "GZK cutoff" has been predicted since the 1960s -- cosmic rays of too high energy would be unable to propagate long distances due to various loss effects, mainly (for protons and higher nuclei) collisions with the microwave background radiation (which from the particle's point of view is like incoming gamma rays) blasting them apart. However, this cutoff never seemed to actually happen, leading to mysteries -- what was going on? Why were the ultra-high cosmic rays there despite claims they could not exist? E.g. http://en.wikipedia.org/wiki/Oh-My-God_particle http://en.wikipedia.org/wiki/Greisen–Zatsepin–Kuzmin_limit Now new observations at the South Pole "ice cube" detector indicate the cosmic ray spectrum really does seem cut off starting at 10^18 eV energies: http://icecube.wisc.edu/news/view/141 -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Bit puzzle
by quad 02 Sep '13

02 Sep '13

Here is a little puzzle. I encourage not spelunking around for the answer on the Internet. ;) Given a 32-bit unsigned integer x, and a positive integer n, let a = XOR(x, RIGHTSHIFT(x, n)) or in C syntax a = x ^ (x >> n) find, if one exists, a function f such that f(a) = x. In other words, invert the operation on x, or show otherwise. Cheers, Robert

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Re: [math-fun] Julian's dodecahedron
by Bill Gosper 02 Sep '13

02 Sep '13

DanA>Taking that idea one step backwards: "the drum" ? --Dan On 2013-09-01, at 11:27 AM, Marc LeBrun wrote: ="Bill Gosper" <billgosper(a)gmail.com> we still have the problem of naming this solid! tympanon Dodecahedrum. --rwg

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[math-fun] Julian's dodecahedron
by Bill Gosper 01 Sep '13

01 Sep '13

is made by replacing ten of the regular pentagons of a dodecahedron with concave equilateral pentagons, preserving the topology: http://gosper.org/dodohedron.png ( !#%&*@! was deleted again. Apologies to any of you who looked for it.) http://en.wikipedia.org/wiki/Dodecahedron glaringly omits it: [...] Topologically distinct dodecahedra include: - Pentagonal dodecahedra: - Regular dodecahedron, 12 pentagonal faces, Ih<http://en.wikipedia.org/wiki/Icosahedral_symmetry>symmetry, order 120 - Pentagonal truncated trapezohedron<http://en.wikipedia.org/wiki/Truncated_trapezohedron>- same topology as regular form, but D5d <http://en.wikipedia.org/wiki/Dihedral_symmetry> symmetry, order 20 - "Pyritohedron <http://en.wikipedia.org/wiki/Pyritohedron>" - same topology as regular form, but Th<http://en.wikipedia.org/wiki/Tetrahedral_symmetry>symmetry, order 12 - Uniform polyhedra: - Decagonal prism <http://en.wikipedia.org/wiki/Decagonal_prism> – 10 squares, 2 decagons, D10h<http://en.wikipedia.org/wiki/Dihedral_symmetry>symmetry, order 40 - Pentagonal antiprism<http://en.wikipedia.org/wiki/Pentagonal_antiprism>– 10 equilateral triangles, 2 pentagons, D5d <http://en.wikipedia.org/wiki/Dihedral_symmetry> symmetry, order 20 - Johnson solids <http://en.wikipedia.org/wiki/Johnson_solid> (regular faced): - Pentagonal cupola <http://en.wikipedia.org/wiki/Pentagonal_cupola>– 5 triangles, 5 squares, 1 pentagon, 1 decagon, C5v <http://en.wikipedia.org/wiki/Cyclic_symmetry> symmetry, order 10 [...] reinforcing the suspicion that Julian was not anticipated. (Perhaps a contributing factor to the news not reaching Wikipedia was its repeated disappearance from my site.) Anyway, each of the three pentagonal dodecahedra has its own article, as well as mention in the above list. Obviously there should be an article on Julian's. (Julian, do you still have the coordinates?) It would be amusing to create the article, and watch the Wikiwardens delete it as Original Research. Do we really need to create Propp's Electronic Journal of Unoriginal Research? Even if those <PEJURative> Wikiwardens relented, we still have the problem of naming this solid! --rwg

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[math-fun] 2D Einstein tile in hyperbolic plane by Penrose
by Warren D Smith 01 Sep '13

01 Sep '13

Roger Penrose devised a tile, which will tile the hyperbolic (nonEuclidean) plane but will only do so "aperiodically." To describe it, you need to know about the halfspace model of the hyperbolic plane H, http://en.wikipedia.org/wiki/Poincare_half-plane_model i.e. H is represented by the Euclidean upper halfplane y>0, using the distance function dist = arccosh( 1 + ((x1-x2)^2 + (y1-y2)^2) / (2*y1*y2) ). The x-axis plus point at infinity together are the "horocircle at infinity" and translation in x-direction, and rescaling, both are isometries. OK, the tile is described Euclideanly as a unit square with decorated edges. Left edge: small circular knob, fitting into small circular matching hole on Right edge. Top edge: Dilbert-shaped protrusion of size 2*epsilon at center of edge. Bottom edge: two Dilbert-shaped intrusions of size 1*epsilon at the 1/4 and 3/4 points along edge. Now to tile, 1. place a line of tiles with all lower edges along the Euclidean line y=1, tiling, basically, the strip 1<y<2. One of the tiles has left edge on y-axis. 2. scale that line of tiles by factors S that are powers of 2 (...1/8, 1/4, 1/2, 1, 2, 4, 8,...), that is apply the map (x,y) --> S*(x,y). The result is an "Escher" hierarchical tiling of the hyperbolic plane by pseudo-square tiles which seems obviously unique. I'm not sure whether we should count this tiling as "aperiodic." I mean, this tiling has an infinite symmetry group (the scalings by powers of 2). However, it is certainly less symmetric than, e.g, the tiling of the hyperbolic plane by regular 7-gons, with size chosen so that each angle is 120 degrees, which has a oriented-tile-transitive symmetry group. Penrose's tiling in contrast has an infinite number of different "tile types" i.e, there is no symmetry of the tiling which will map a tile in type A to one in type B.

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[math-fun] Okay, what's the name of this tiling?
by Adam P. Goucher 01 Sep '13

01 Sep '13

I think that this is the same basic structure as the tiling of Penrose's hyperbolic tiles decribed by Warren in his penultimate e-mail: http://cp4space.files.wordpress.com/2013/09/bs-tiling.png Does it actually have a name? I've always called it a `Baumslag-Solitar tiling' (c.f. B-S group), but that's certainly non-standard terminology. Warren described it as an `Escher hierarchical tiling', and upon searching it indeed features as the basic structure of Square Limit: http://www.tess-elation.co.uk/self-similar-tessellations It isn't equivalent to any nice tessellation of the Euclidean or hyperbolic plane, since it doesn't have an orbifold. Baumslag-Solitar tilings arise naturally in the complement of the Mandelbrot set, as shown below: http://cp4space.files.wordpress.com/2013/09/bs-mandelbrot.png Indeed, if we have a complex polynomial iteration and the polynomial has degree d, then we get a B-S tiling of order d. Sincerely, Adam P. Goucher http://cp4space.wordpress.com

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[math-fun] solving, denesting, and reading comprehension
by Bill Gosper 01 Sep '13

01 Sep '13

reg>http://www.math.harvard.edu/~elkies/trinomial.html <http://www.math.harvard.edu/%7Eelkies/trinomial.html> gives a (rational) continuum of irreducible, supposedly solvable sextics: (125 - u)*x^6 + 12*u*(u + 3)^2*x + u*(u - 5)*(u + 3)^2 but my sextic solver fails for u≠0. (≠-3,≠5) Can anybody propose a correction? CG>I have no correction, but I concur with your result. For example, with u = 4 I get 121x^6 + 2352x - 196 with Galois group PSL(2,5): polgalois(121*x^6 + 2352*x - 196) %1 = [60, 1, 1, "L(6) = PSL(2,5) = A_5(6)"] which is not solvable. Charles Greathouse Analyst/Programmer Case Western Reserve University Mea gufa. With more patience than I deserve, Elkies privately explained that his paper claims only that the Galois groups are atypically small, not necessarily small enough to solve. He also had no news re the possibility of a solvable, irreducible septic trinomial. Come to think of it, I don't even recall seeing a solvable, irreducible x^6+ax+b. --rwg

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