October 2020 - mathfun@mailman.xmission.com

Re: [math-fun] Relative convexity
by Dan Asimov 24 Oct '20

24 Oct '20

Yes. For any smooth metric surface (or n-dimensional manifold) M, a subset X of M is "geodesically convex" if for any points p, q of X there is a unique shortest geodesic curve C in M connecting p and q, and C lies entirely in X. —Dan Jim Propp wrote: ----- Is there a notion of "relative convexity" that would make an ordinary torus consisting of points at small fixed distance from a large circle "convex relative to the circle"? Thinking about different sorts of polyhedral tori people have come up with, I realize that part of what I want esthetically is some kind of relative convexity, but I don't know what it should mean! -----

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[math-fun] Relative convexity
by James Propp 24 Oct '20

24 Oct '20

Is there a notion of "relative convexity" that would make an ordinary torus consisting of points at small fixed distance from a large circle "convex relative to the circle"? Thinking about different sorts of polyhedral tori people have come up with, I realize that part of what I want esthetically is some kind of relative convexity, but I don't know what it should mean! Jim Propp

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Re: [math-fun] Help with Mathematica desired
by Dan Asimov 24 Oct '20

24 Oct '20

Kris Katterjohn wrote: [in essence] ----- Solve[Exp[I*(3*u+v)] + Exp[I*(3*v+w)] + Exp[I*(3*w+u)] == 0, {u,v,w}] ----- worked. Thanks, Kris!!! That "worked" for me, but I know from having solved the exact same equation in the past by hand that Wolfram Alpha's solution was VASTLY more complicated than necessary. So I tried FullSimplify, and then just Simplify, on it, and got no results at all either way!!! So I think the problem is that it times out, because the problem is just too hard for it to solve gracefully within its time/computation limit. —Dan

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[math-fun] Help with Mathematica desired
by Dan Asimov 24 Oct '20

24 Oct '20

I tried to submit this to Wolfram Alpha in correct syntax, but it never gave any indication that it had the vaguest idea that this was an equation to solve, no less the correct equation to solve for real numbers u, v, w. Solve[Exp[I(3u+v)] + Exp[I(3v+w)] + Exp[I(3w+u)] == 0, {u,v,w}, Reals] What am I doing wrong? Thanks, Dan

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Re: [math-fun] Help with Mathematica desired
by Dan Asimov 24 Oct '20

24 Oct '20

Jim Propp wrote << Did you try putting the equation in curly braces? I didn't, because I couldn't see how that would make a difference. (But I just tried it now, and it didn't.) —Dan

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Re: [math-fun] Help with Mathematica desired
by Dan Asimov 24 Oct '20

24 Oct '20

This time I tried a simplified, all-real command: Solve[Cos[3 x+y] + Cos[3 y+z] + Cos[3 z+x] == 0, {x,y,z}] (verbatim). Wolfram Alpha had absolutely no idea that it was even being asked to solve an equation. —Dan

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Re: [math-fun] Help with Mathematica desired
by Dan Asimov 23 Oct '20

23 Oct '20

In fact I already tried both 1. and 2. right after posting to math-fun, and there was no improvement! —Dan Gareth McCaughan wrote: ----- On 23/10/2020 22:36, Dan Asimov wrote: > I tried to submit this to Wolfram Alpha in correct syntax, > but it never gave any indication that it had the vaguest idea > that this was an equation to solve, no less the correct equation > to solve for real numbers u, v, w. > > Solve[Exp[I(3u+v)] + Exp[I(3v+w)] + Exp[I(3w+u)] == 0, {u,v,w}, Reals] > > What am I doing wrong? I don't know how helpful the following is, but: 1. If I put exactly that into Mathematica, it complains about the instances of I, on the grounds that if you're trying to solve an equation over the reals then all the constants involved must be real. 2. If I replace it with Cos[...]+...==0 && Sin[...]+...==0, then instead I get "The system cannot be solved with the methods available to Solve." -----

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Steve Witham 23 Oct '20

23 Oct '20

As the Sesame Street Martians say, Yyyyyyyep. Yyyyep. Yep-yep-yep-yep-yep-yep. GCD is the next best thing to factoring. GCD of Gaussian integers looks slightly trickier than the ordinary version. My brain says "I don't want to learn a whole new skill!" and I have to tell it, "You put your shoes on, stop hiding in your comfort zone, go outside and have some math-fun young man." --Steve > From: Allan Wechsler <acwacw(a)gmail.com> > Date: 10/22/20, 11:32 AM > > Yes. Then take the GCD of all the remaining points (considered as Gaussian > integers), and divide out by the GCD to obtain the solution. > > On Thu, Oct 22, 2020 at 1:19 AM Steve Witham <sw(a)tiac.net> wrote: > >> Not only does reflection not matter to the problem, >> but translate the problem so that one original point >> is (0, 0) and let the same point in the solution be (0, 0). >> >>> From: Henry Baker <hbaker1(a)pipeline.com> >>> Date: 10/21/20, 10:27 AM >>> >>> I'm guessing factorization of Gaussian integers is involved... >>> >>> At 12:51 AM 10/21/2020, Tom Karzes wrote: >>>> Here's a puzzle some of you might like: (fin)

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 23 Oct '20

23 Oct '20

Not at all. My answer was the result of very deep thought! —Dan Jim Propp wrote: ----- I was just overlooking the obvious! -----

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 23 Oct '20

23 Oct '20

For some reason this is reminding me of a very pretty polyhedral surface of genus 3 that does not exist as a subset of R^3 but instead as a subset of the torus R^3/Z^3. Truncate the corners of a unit cube so that what remains is a truncated octahedron (the corners become 8 hexagons and the faces become 6 smaller squares at 45º angles). Now just considering the 14 faces, delete the 6 squares and identify the opposite faces of the original cube so that it becomes the torus T^3 = R^3/Z^3. The truncated octahedron is now a polyhedral surface M_3 of genus 3 inside the torus T^3, tiled by 8 hexagons, 4 per vertex. There is an isometry of the whole M_3 to itself taking a given hexagon to any hexagon in any of 12 dihedral ways, for a total of 96 isometries altogether. —Dan Tomas Rokicki wrote: ----- Or take the eight cube solution and skew two opposite cubes so the visible parts are not rectangles. This introduces the necessary separating edges. -----

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