June 2019 - mathfun@mailman.xmission.com

[math-fun] Is it me, or is math.stackexchange.com controlled by morons?
by Bill Gosper 24 Jun '19

24 Jun '19

https://math.stackexchange.com/questions/3268714/what-are-the-semiaxes-a-b-… —Bill

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Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
by Dan Asimov 24 Jun '19

24 Jun '19

As long as we're advising on possessives, it's a lot more acceptable to pluralize proper names with apostrophe + S (in particular, even when the name ends with an S). So it would be "Archimedes's" imho. Exceptions are only a few traditional phrases like Achilles' heel and Jesus' sake ..., but alas! not including poor old Archimedes. —Dan ----- No worries. Your latest statement sounds equivalent to the one from "Mathematics Magazine". So we've got it all straightened out except that the spelling "Archimede's" is really not an acceptable approximation to "Archimedes' ". Ouch! I guess I need to watch those apostrophes more closely -----

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Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
by Dan Asimov 24 Jun '19

24 Jun '19

Come to think of it, how can we prove that the mapping (0,oo)^3 —> (0,oo)^3 of positive octants via (a,b,c) |—> (girth_ab, girth_bc, girth_ca) is one-to-one? Or is it? —Dan

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Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
by Dan Asimov 24 Jun '19

24 Jun '19

PS: What about the surface area S(a,b,c) of the ellipsoid E = E(a,b,c) having semiaxes a,b,c ??? If g(a,b), g(b,c), g(c,a) are the respective girths, aka circumcrferences of ellipses e(a,b), e(b,c), e(c,a) ... ... can we express S(a,b,c) in closed form in terms of the g( , )s ??? Nicest of all'd be if S(a,b,c)^n is a polynomial in g(a,b), g(b,c), g(c,a) for some n. Hmm, so for the sphere we'd have (4π)^3 = 2π * 2π * 2π / (π^2 / 8), so conceivably, S(a,b,c)^3 = (8/π^2) g(a,b) * g(b,c) * g(c,a) Wouldn't that be nice. —Dan I wrote: ----- Let g(a,b) denote the circumference of the ellipse (x/a)^2 + (y/b)^2 = 1. Is there an closed expression for ∂g/∂a or ∂^2g/∂a∂b ? -----

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Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
by Dan Asimov 24 Jun '19

24 Jun '19

To be clear, I didn't object at all to discussing the problem RWG encountered with Math StackExchange (MSE) — they could potentially affect any of us. (I just missed having more of a discussion about the math Bill posted.) There are two reasons occurring to me that these problems happen on MSE. One is that it's so big and anonymous that the people on it often don't know each other or even e-know e-chother. The other is that some people see their role as one of enforcing rules, and some such people don't bother to make sure their enforcing is reasonable. (Sometimes called "authoritarian personalities".) Let g(a,b) denote the circumference of the ellipse (x/a)^2 + (y/b)^2 = 1. Is there an closed expression for ∂g/∂a or ∂^2g/∂a∂b ? —Dan Fred Lunnon wrote: ----- << The mathematics is more interesting than the politics of math.stackexchange. >> Agreed; but politics has an unpleasant habit of intruding on our interests uninvited. RWG ain't the first to fall victim to this particular unforeseen consequence, and sure as eggs is eggs, he won't be the last. I thought WGD's analysis informative and convincing, as far as it goes --- and a salutary reminder of how fortunate we are to have math-fun run as smoothly as it does! No doubt most people on this list will sympathise heartily with a Q&A site attempting to avoid transformation into a free online plagiarism factory. So why do such efforts regularly and spectacularly backfire in this fashion? One factor --- difficult to combat --- is exemplified by the experience of a chess IM acquaintance who once found himself playing, for the team representing a very small country, on top board against Anatoly Karpov. He reported that the world champion just sat there, (apparently) aimlessly shunting rooks to and fro along the back rank, until it suddenly dawned on yer man that (as he put it) "his options had run out". For chess, the score line is eventually on hand to distinguish between naïve incompetence and incomprehensible superiority; however in mathematics, as in many real-life situations, a convenient discrimination algorithm is often unavailable. << As for the five who jumped me: "Off topic" my asymptote. >> Another recurrent feature --- on the face of it, much easier to fix --- is implicit segregation. When the club doorman refuses you entry on the grounds that you have the wrong sex, colour, religion, age, etc. you might well consider such discrimination unacceptable, but at least you know where you stand and can consider constructively how to resolve the situation. If he attempts to avoid confrontation by concocting some obviously irrelevant excuse, you are merely left bewildered and possibly angry. In a more straightforward world, SE would simply call a spade a spade: apologetically admit that an enquiry appeared potentially plagiaristic, before requesting supporting evidence that it was genuine (background, citations, etc). By attempting to avoid offending a small army of cheats, they manage instead to upset the very people they (presumably) want onboard. ... -----

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Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
by Dan Asimov 24 Jun '19

24 Jun '19

The average projected length of a unit segment in the plane over orthogonal projections onto lines in all directions is (1/π)(Integral_{0 to π} cos(t) dt = 2/pi. If you approximate a convex closed curve S in the plane by a polygon P, you can sum the above equation over all segments of P to see that the average projected length of S is length(S) / π. This generalizes to higher dimensions. —Dan I wrote: ----- I think Cauchy proved the 2-dimensional version of this by averaging orthogonal projections over all different directions. So there may be a proof like that in higher dimensions. Convex surfaces are almost everywhere C^2, which may help. -----

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Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
by Dan Asimov 24 Jun '19

24 Jun '19

I think Cauchy proved the 2-dimensional version of this by averaging orthogonal projections over all different directions. So there may be a proof like that in higher dimensions. Convex surfaces are almost everywhere C^2, which may help. —Dan Jim Propp wrote ----- I believe the two-dimensional version of this assertion was made by Archimedes, and was intended to serve as a way of characterizing arc-length. On Sun, Jun 23, 2019 at 8:09 PM Adam P. Goucher <apgoucher(a)gmx.com> wrote: ----- Excellent proof! By the way, for the 'obvious' lemma that the girth is an increasing function of the two semi-axes, it follows from: Theorem: If L is a compact subset of R^n and K is a convex compact subset of L, then the surface area of K is bounded above by the surface area of L, with equality if and only if K = L. ... ----- -----

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Re: [math-fun] optimal advance tournament scheduling
by Keith F. Lynch 23 Jun '19

23 Jun '19

Tomas Rokicki <rokicki(a)gmail.com> wrote: > Nauty comes with a pretty good description of the algorithm it uses > to solve the graph isomorphism problem and enumerate the graphs. Thanks. Can you be more specific? Is it in this 103-page manual: http://users.cecs.anu.edu.au/~bdm/nauty/nug26.pdf and if so, at what page number? > The cube with long diagonals is just K4,4 (a bipartite graph with > four nodes on each side and degree four). > I'm not sure what you're saying about permutating that matrix; I can > permute it (rows and columns identically) into > 0 0 0 0 1 1 1 1 > 0 0 0 0 1 1 1 1 > 0 0 0 0 1 1 1 1 > 0 0 0 0 1 1 1 1 > 1 1 1 1 0 0 0 0 > 1 1 1 1 0 0 0 0 > 1 1 1 1 0 0 0 0 > 1 1 1 1 0 0 0 0 I screwed up. I apologize. I'm not a visual thinker. And I was perhaps misled by the fact that all vertices in that graph are isomorphic to each other (if that's the right term). I hadn't fully realized that there's only one bipartite graph with 8 vertices and 16 edges. Am I correct that it could also be represented by a 1 on every black square of the checkerboard and a 0 on every red square? Indeed, before laying them out on an actual checkerboard, I thought that was the pattern I'd get. But wait, I had convinced myself that the number of 1-bits on black squares was unchanged with graph automorphisms, since moving a bit by N rows and also by N columns keeps it on the same color. (And indeed it was unchanged in your matrix.) So now I'm even more confused. The all-black matrix represents each even-numbered player playing all odd-numbered players and no even-numbered players, and vice versa. And isn't that just K4,4 again? (The (Thue-Morse) labeling I used was based on treating the coordinates of the vertices on the cube as binary numbers, so 0,0,0 became player 0, 0,0,1 became player 1, 0,1,0 became player 2, etc.) Mike Beeler <mikebeeler2(a)gmail.com> wrote: > I think what you call ?average bits? is what Keith calls ?score?, > the number of player pairs whose ordering is found by the tournament > results. Here I think I'm on slightly firmer footing. Right. The "number of bits of information" is somewhat ambiguous, as with asking how many bits are in the first thousand decimal digits of pi. Or asking how many bits are in Borges's Library of Babel, which contains all possible books. If all 8 players play each other, and any result is possible, including ones that violate consistency and transitivity, that conveys 28 bits. With consistency and transitivity, most patterns of 28 bits are impossible. A complete ordering conveys log2(8!) bits, about 15.3 bits. Either way, with just 16 games we can't get more than 16 bits. Since 16 > 15.3 it sounds like it might be possible to get a complete ordering. But apparently not, at least not without deciding on who plays whom before knowing the results of any games. However, if, even before I thought about the problem at all, someone had suggested that 15 games would be enough, I'd know they were wrong, since 15 < 15.3. Saying the result of a tournament conveys 15.3 bits implicitly assumes that the result of a tournament is always a complete ordering. If, as seems to be the case, the result may be a partial ordering, we have to count how many of those are possible too. For instance if the result is always either a complete ordering or an ordering with just 1 of the 28 possible games having an unknown result, i.e. with two players "tied" in the sense that we don't know which one is better, it would require log2(8! + 7*7!), about 16.2 bits, to encode. Assuming any amount of ignorance is possible, including having no results at all, A000670 says there are 545835 possibilities for 8 players. That would require 19.1 bits to encode. Given that even the worst graph will return *some* results, the floor is somewhat above total ignorance. So somewhere between 16.2 and 19.1 bits in that sense. Confusingly, "partial ordering" doesn't seem to mean what I thought it meant. A000670 differs from A001035.

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Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
by Dan Asimov 23 Jun '19

23 Jun '19

Same here. And I'm also interested in the related problem I (mis-)quoted (Thanks, Gene for your earlier reply to it): In terms of the girths, find the volume (4π/3)abc ... which may be easier than finding the semiaxes. —Dan Gene Salamin wrote: ----- Has anyone solved the problem of determining the axes of an ellipsoid given the three girths? It has me intrigued. The mathematics is more interesting than the politics of math.stackexchange. -----

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[math-fun] more ries use cases
by Brad Klee 23 Jun '19

23 Jun '19

A success case: Input: ries 0.627741162589377 Output: (3/2)*Log[2/3^(1/4)] A related fail case: input: ries 0.0485338913375231 output: no match but I have now that: 8*(Log[3]/2 - (12*ArcSin[1/3]*Log[2/3^(1/4)])/Pi) These numbers fall out when constructing a functional form for the complex, nearest-neighbor tunnelling action between C_4 symmetric regions of the unique, degree-four octahedral R.E.S. ( more on this soon ). Fun program! Thanks for the help! --Brad

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