October 2020 - mathfun@mailman.xmission.com

Re: [math-fun] Torus tiling puzzle
by Dan Asimov 20 Oct '20

20 Oct '20

I asked this question about a month ago. I found this interesting because the connection graph of such a tiling (since any two tiles share an edge) is the complete graph K(N). I'll post the answer in a couple of days. —Dan ----- It's known that a certain torus T^2 (namely, the hexagonal torus obtained by identifying opposite edges of a regular hexagon) can be tiled by regular hexagons so that each hexagon shares an edge with each other one. Puzzle: For which N ≥ 1 does there exist a tiling of some torus (either the square or hexagonal one) by N regular polygons, any two of which share an edge? -----

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[math-fun] Tom Lehrer puts lyrics and music into public domain
by ed pegg 20 Oct '20

20 Oct '20

https://tomlehrersongs.com/ Meanwhile, Epsilon is getting very large.

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Re: [math-fun] 247 zeptosecond time interval measured
by Dan Asimov 20 Oct '20

20 Oct '20

----- Zepto = 10^-21. A single photon ejected the two electrons from a hydrogen molecule. The time interval between the ejections was 247 zs, more or less depending on the orientation of the molecule. https://news.yahoo.com/zeptosecond-femtosecond-163929049.html ----- I may be impressed when I know an upper bound for the maximum possible error. (It took me 246 zeptoseconds plus around a minute to write this post.) —Dan

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[math-fun] 247 zeptosecond time interval measured
by Eugene Salamin 20 Oct '20

20 Oct '20

Zepto = 10^-21. A single photon ejected the two electrons from a hydrogen molecule. The time interval between the ejections was 247 zs, more or less depending on the orientation of the molecule. https://news.yahoo.com/zeptosecond-femtosecond-163929049.html -- Gene

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[math-fun] Folding flat figures with holes
by James Propp 19 Oct '20

19 Oct '20

How many folds are needed to turn a square piece of paper into a flat figure with k holes? As an example, you can fold a square up into a long skinny rectangle like the wrapper a drinking straw comes in, and then bend that long skinny rectangle in lots of places to create lots of crossings (and lots of holes between the crossings). I can show that asymptotically you can get k holes with something like constant times sqrt(k) folds, but I don't know if this is best possible. Jim Propp

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Re: [math-fun] Factoring/normalizing quaternions
by Keith F. Lynch 19 Oct '20

19 Oct '20

Henry Baker <hbaker1(a)pipeline.com> wrote: > There's a *huge* difference between 'i' and 'j', so the eye > might not pick up on this subtlety. In the quaternions, 'i' > is perpendicular to 'j', and 'i' doesn't commute with 'j'. I thought there was perfect symmetry, i.e. that in any quaternion equation you can replace i with j if you also simultaneously replace j with k and k with i. And that you can also swap i with j if you simultaneously stick a minus sign on one (but not both) of them. This is analogous to how in complex numbers you can swap i with -i. And also analogous with the symmetry between electricity and magnetism if you make the Maxwell equations symmetric by removing the bald assertion that there are no magnetic monopoles. Sure, they haven't found any *yet*, but they haven't looked everywhere. As far as I know, nobody has checked under my bed. You can also swap positive for negative electric charges, or north for south magnetic charges.

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Re: [math-fun] Factoring/normalizing quaternions
by Henry Baker 19 Oct '20

19 Oct '20

Sorry, I was unable to follow much of your post due to the fact that I'm still living in an ASCII world. --- Just think of me as a 'steampunk' mathematician channeling Hamilton/Tait/Maxwell/Sylvester, except that I have access to a computer running a symbolic algebra system (Maxima) which enables me to handle large equation sets impossible for a 19th C. denizen. I'm trying to imagine what these folks would have accomplished if Babbage had been successful and had produced Maxima. I.e., what if I had slept through the entire 20th C. of Bourbaki mathematics ? At 06:39 PM 10/13/2020, M F Hasler wrote: >Le mar. 13 oct. 2020 Ã 17:17, Dan Asimov <dasimov(a)earthlink.net> a ÃÂ©crit : >> I skipped over sections of this post by Henry Baker to try to find the >> points that it included. >> But a general quaternion is *not* a product of quaternions >> that have zero j components and zero k components. > >There seem to be many more or less original compilations of formulas about calculations in quaternions around on the web and on this list. > >But there is the century old formalism of "standard" decompositions of complex 2x2 matrices into trace and traceless (aka "vectorial") part, based on standard representation theory. > >(Clebsch-Gordan, Young tableaux, ...) > >Those who have worked with Lie algebras and/or relativity willl probably know well the Clifford algebra of sigma or Pauli matrices: > >any 2x2 matrix can be written as A = aÃ¢ÂÂ° sÃ¢ÂÂ° + *a* Â *ÃÂ* *a* ÂÃ¢ÂÂ¢ *ÃÂ* = aÃÂ¹ ÃÂÃÂ¹ + aÃÂ² ÃÂÃÂ² + aÃÂ³ ÃÂÃÂ³ where ÃÂÃ¢ÂÂ° = *I* is the identity matrix and *ÃÂ* =* ÃÂ *= ( ÃÂÃÂ¹ , i ÃÂ , ÃÂÃÂ³ ) are the three hermitian traceless Pauli matrices verifying ÃÂÃÂ¹ ÃÂÃÂ² = i ÃÂµÃÂ¹ÃÂ²ÃÂ³ ÃÂÃÂ³ where indices 123 can be replaced by any indices *i,j,k* and ÃÂµ is the completely antisymmetric tensor with ÃÂµÃÂ¹ÃÂ²ÃÂ³ = 1. > >(I use boldface for vectors since I can't put an arrow over them, I hope this comes through the mailing list... otherwise, as a rule of thumb, any lowercase letter without an index will represent a 3-vector (possibly of matrices, for Ï & Ñ) except for i and Ñ .) > >Then: A is hermitian iff (aÃ¢ÂÂ°, aÃÂ¹, aÃÂ², aÃÂ³) are all real ; det A = (aÃ¢ÂÂ°)ÃÂ² - *a*ÃÂ² is the Minkowski "norm" of the 4-vector (aÃ¢ÂÂ°,*a*) ; and a product is given by A B = (aÃ¢ÂÂ° bÃ¢ÂÂ° +* a *Â *b*) ÃÂÂÃ¢ÂÂ° + (aÃ¢ÂÂ°* b* + bÃ¢ÂÂ°* a* + i* a* ÃÂ *b*) Â *ÃÂ* wheree the last part involving the cross product, i (*a* ÃÂ *b*) Â *ÃÂ *is also the normalised commutator [A,B] = ÃÂ½Â½(AB Â BA). > >Many more very nice formulas are easily disccovered, e.g. to extract components using the normalized trace: aÂ¹ = Â½ tr( A ÏÂ¹ ) and the same for all other indices 0 .. 3. > >These formulas allow very easy calculations for quaternions, too, possibly using rather antihermitian matrices Â *i* *ÃÂ**, * to "absorb" the imaginary unit in the structure constants (product & commutation relations), for the more popular (standard(?)) quaternion basis *i, j, k.* > >(Where this boldface *i* obviously is not the imaginary unit!) - Maximilian > >PS: Apologies to GR professionals who will prefer 4-vector index notation, use also 2-index (symmetric/traceless) sigma matrices, raise & lower 4-indices with eta = (-1,1,1,1) (or opposite according to religion) and matrix indices with the 2x2 epsilon tensor and its inverse, converting traceless <-> symmetric and scalar <-> antisymmetric.

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Re: [math-fun] Jaywalker geometry
by Keith F. Lynch 19 Oct '20

19 Oct '20

Marc LeBrun <mlb(a)well.com> wrote: > It sounds fascinating, but I'm really having trouble picturing this. > Are the grid points at the intersections of these narrow-most roads? Yes. > Are the jays walking "diagonally" along them or across them or what? Diagonally along them. Each straight section of road is a long narrow rectangle, and the jaywalker takes the diagonal of that rectangle if he's turning left at one end of it and right at the other. (Okay, it's ambiguous which side of the road he starts and ends his trip on. Might as well compromise on the middle as Gareth implicitly does.) > You imply the shortest path from 0,0 to 2,2 goes thru 1,1. Right. > But why? Given some arbitrary path from A to B, how do we compute > its length? Can you help by providing some worked examples or the > like? Thanks! Gareth McCaughan <gareth.mccaughan(a)pobox.com> wrote: > The total distance is some unpleasant thing involving square roots > of quadratics in h, but e.g. with h=0.001 it's about 3.994 whereas > the path that goes (0,0) - (2-h,h) - (2,2) has length about 3.998. It's really quite straightforward. It the road has length 1 and infinitesemal width w, the diagonal is sqrt(1^2 + w^2), which equals 1 + w^2/2. Proof by the FOIL method: (1 + w^2/2)^2 = 1 + w^2/2 + w^2/2 + w^4/4) = 1 + w^2. (The fourth power of an infinitesimal is vanishing in comparison with the square of the infinitesimal.) It the road has length L and infinitesemal width w, the diagonal is sqrt(L^2 + w^2), which equals L + w^2/2L. Proof: (L + w^2/2L)^2 = L^2 + Lw^2/2L + Lw^2/2L + w^4/(4L^2)) = L^2 + w^2. Since we're starting and ending in the middle of the width, it makes more sense to use Gareth's half-width h than my full-width w. You can use the same equations with w replaced by h, since you just have narrower rectangles. It's past my bedtime. I'll finishing working it out in detail tomorrow or Tuesday if someone doesn't beat me to it.

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Re: [math-fun] Jaywalker geometry (was Re: Disks in Euclidean approximation metrics ...)
by Dan Asimov 19 Oct '20

19 Oct '20

I'd like to politely complain that I'm having trouble understanding exactly what is intended here. Regardless of how a math problem may be couched in real-world terms (like jaywalkers and infinitesimally wide roads), would it be okay to also please express the problem in unambiguous mathematical terms so that the meaning of the question doesn't have to be questioned? —Dan ----- A variant of taxicab geometry I came up with is jaywalker geometry. Envision the roads having, not zero width, but infinitesimal width. Distance is defined, not as how far a taxicab drives, but as how far a jaywalker walks. If a right turn is to be followed by a left turn or vice versa, to minimize his walking distance he crosses the street diagonally, over the whole length of it that he traverses, much to the annoyance of the taxicabs. In taxicab geometry there are in general lots of shortest paths between any two grid points, but in jaywalker geometry, of all those paths, the paths with the most turns are shortest. I think those paths are the ones that approximate Euclidean paths most closely. For instance instead of walking ten blocks north then ten blocks east, the jaywalker would alternate between walking one block north and one block east. -----

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[math-fun] Jaywalker geometry (was Re: Disks in Euclidean approximation metrics ...)
by Keith F. Lynch 19 Oct '20

19 Oct '20

Brad Klee <bradklee(a)gmail.com> wrote: > I really love lattice walk problems, think that they are entirely > practical, and with many branch points into other areas of > mathematics and physics. A variant of taxicab geometry I came up with is jaywalker geometry. Envision the roads having, not zero width, but infinitesimal width. Distance is defined, not as how far a taxicab drives, but as how far a jaywalker walks. If a right turn is to be followed by a left turn or vice versa, to minimize his walking distance he crosses the street diagonally, over the whole length of it that he traverses, much to the annoyance of the taxicabs. In taxicab geometry there are in general lots of shortest paths between any two grid points, but in jaywalker geometry, of all those paths, the paths with the most turns are shortest. I think those paths are the ones that approximate Euclidean paths most closely. For instance instead of walking ten blocks north then ten blocks east, the jaywalker would alternate between walking one block north and one block east. Has anyone else played with this?

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