August 2013 - mathfun@mailman.xmission.com

[math-fun] Fwd: Calculus of variations problem: the one-wire cage.
by Allan Wechsler 30 Aug '13

30 Aug '13

Erich Friedman complains that he can't post to math-fun, and asks me to forward this. This seems to almost put the problem to rest. Rich, can you troubleshoot to see what Erich's posting trouble is? He's a valuable funster. ---------- Forwarded message ---------- From: Erich Friedman <efriedma(a)stetson.edu> Date: Fri, Aug 30, 2013 at 1:01 PM Subject: Re: [math-fun] Calculus of variations problem: the one-wire cage. To: Allan Wechsler <acwacw(a)gmail.com> in 1946, p.a.p. moran showed that as long as the curve contains no 4 coplanar points, a reasonable assumption, that the maximum volume is achieved by one turn of a circular helix of height 1/sqrt(3) and radius 1/pi sqrt(6). unfortunately, i can't post to math-fun, so please do so for me. erich friedman On 8/30/13 12:02 PM, "Allan Wechsler" <acwacw(a)gmail.com> wrote: >Which three-dimensional continuous differentiable curve of unit arc-length >encloses the greatest volume in its convex hull? >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >

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Re: [math-fun] Looking for Javascript Life engine
by Phil Carmody 30 Aug '13

30 Aug '13

From: "scott(a)scottkim.com" <scott(a)scottkim.com> > I'm proposing to Google an interactive Google Doodle to honor Martin > Gardner, which would contain 6 minigames (one for each letter of > GOOGLE), highlighting six things Gardner wrote about. One of those > games of course should be a version of Life. Does anyone know of > or have a Life engine that could be modified for this use? I'd prefer > to open-source the code, but that is not essential. I'm guessing it > should be in Javascript to be browser friendly (the Turing Google > Doodle was open-source Javascript). A noble idea. Here's a tiny modern JS-based life engine, only 376 bytes: http://js1k.com/2013-spring/demo/1478 Phil -- () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani]

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Re: [math-fun] Looking for Javascript Life engine
by Henry Baker 30 Aug '13

30 Aug '13

For extra credit, use those shiny new HTML5 graphics capabilities & take advantage of the GPU's to make it really fast! I could imagine getting someone from AMD/ATI or nVidia to do this over this long weekend. At 05:18 PM 8/29/2013, Tom Rokicki wrote: >I suspect you'll do best putting what simple key functionality you want >in a few dozen lines of your own JavaScript. > >For the grid size you probably want, a standard two-array, nested-loop >approach will assuredly be plenty fast. > >It will take you much less time to write it yourself than to adopt all these >other solutions. > >Even adding RLE parsing support is pretty trivial. > >For interest you might consider using colored life rules (like three-color >voting); most of them subsume the standard two-state rules but add a >bit of shiny. > >-tom > >On Thu, Aug 29, 2013 at 5:13 PM, Mike Stay <metaweta(a)gmail.com> wrote: >> Javascript Golly in under 10K, MIT license: >> https://github.com/sbarski/jolly >> >> On Thu, Aug 29, 2013 at 6:05 PM, Mike Speciner <ms(a)alum.mit.edu> wrote: >>> A quick google search revealed >>> http://pmav.eu/stuff/javascript-game-of-life-v3.1.1/ among many others. This >>> one seems to have an MIT license. >>> >>> On 2013-08-29 19:43, scott(a)scottkim.com wrote: >>>> >>>> I'm proposing to Google an interactive Google Doodle to honor Martin >>>> Gardner, which would contain 6 >>>> minigames (one for each letter of GOOGLE), highlighting six things Gardner >>>> wrote about. One of those >>>> games of course should be a version of Life. Does anyone know of or have a >>>> Life engine that could be >>>> modified for this use? I'd prefer to open-source the code, but that is not >>>> essential. I'm guessing it should >>>> be in Javascript to be browser friendly (the Turing Google Doodle was >>>> open-source Javascript). >>>> -- Scott Kim

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[math-fun] Looking for Javascript Life engine
by scott＠scottkim.com 30 Aug '13

30 Aug '13

I'm proposing to Google an interactive Google Doodle to honor Martin Gardner, which would contain 6 minigames (one for each letter of GOOGLE), highlighting six things Gardner wrote about. One of those games of course should be a version of Life. Does anyone know of or have a Life engine that could be modified for this use? I'd prefer to open-source the code, but that is not essential. I'm guessing it should be in Javascript to be browser friendly (the Turing Google Doodle was open-source Javascript). -- Scott Kim -------------------------------------------------------------------- mail2web.com - Microsoft® Exchange solutions from a leading provider - http://link.mail2web.com/Business/Exchange

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[math-fun] Packing puzzle
by David Wilson 30 Aug '13

30 Aug '13

Consider a 3-D tile made of 5 cubies arranged like a P-pentomino, 1 cubie thick. You pack these into a 3x3 cube (together they fill all but 2 cubies). Suppose one of the unpacked cubies is the center cubie. Can the remaining unpacked cubie be at a vertex of the 3x3? An edge? A face center?

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[math-fun] MathWorld x^5+ax+b solver
by Bill Gosper 29 Aug '13

29 Aug '13

Looking for solvable irreducible sextics to test my (& Julian's) solver, I found a paper by Boswell & Glasser which, for no apparent reason, states in the introduction an amazingly succinct quintic solver, (apparently lifted straight out of MathWorld). NeilB and I further distilled it to BJQuintic[a_, b_] := ϵ*e*E^(2*π*I*j*Range[4]/5). (Sign[#]*Abs[#]^(1/5) & /@ ({v1^2*v3, v3^2*v4, v2^2*v1, v4^2*v2}/d^2)) /. {v1 -> √d + √(d - ϵ*√d), v2 -> -√d - √(d + ϵ*√d), v3 -> -√d + √(d + ϵ*√d), v4 -> √d - √(d - ϵ*√d)} /. {d -> c^2 + 1, ϵ -> Sign[c]} /. Solve[{5*e^4*(3 - 4*c)/(c^2 + 1) == a, b == -4*e^5*(11 + 2*c)/(c^2 + 1)}, {e, c}, Rationals][[1]] E.g., In[134]:= BJQuintic[20,32] Out[134]= -((2^(4/5) (Sqrt[5]/2-Sqrt[5/4-Sqrt[5]/2])^(1/5) (Sqrt[5]/2+Sqrt[5/4+Sqrt[5]/2])^(2/5) E^((2 I j \[Pi])/5))/5^(2/5))-(2^(4/5) (Sqrt[5]/2-Sqrt[5/4-Sqrt[5]/2])^(2/5) (-(Sqrt[5]/2)+Sqrt[5/4+Sqrt[5]/2])^(1/5) E^((4 I j \[Pi])/5))/5^(2/5)+(2^(4/5) (Sqrt[5]/2+Sqrt[5/4-Sqrt[5]/2])^(2/5) (Sqrt[5]/2+Sqrt[5/4+Sqrt[5]/2])^(1/5) E^((6 I j \[Pi])/5))/5^(2/5)-(2^(4/5) (Sqrt[5]/2+Sqrt[5/4-Sqrt[5]/2])^(1/5) (-(Sqrt[5]/2)+Sqrt[5/4+Sqrt[5]/2])^(2/5) E^((8 I j \[Pi])/5))/5^(2/5) In[136]:= Chop[#^5+20*#&/@N[Table[%%,{j,5}],22]] Out[136]= {-32.0000000000000000000,-32.0000000000000000000,-32.0000000000000000000,-32.0000000000000000000,-32.00000000000000000000} (More convincing would be to get exact integers from RootReduce, but it is broken.) BJQuintic dodges the issue of putting the quintic into "Bring-Jerrard form", and relies on Mathematica to magically find rational solutions to the a and b equations. But the latter isn't hard. Just look for linear factors after eliminating e or c with a Resultant. The rest of the B&G paper breaks x^6+x^2+c*x+d sextics into two Galois cases, the first of which seems too badly misprinted to merit salvaging, despite Neil's clever reconstruction of the Mathematica from the PostScript. Why don't people computer- check their formulas? Why don't they include TeX sources including the input expressions? Even Macsyma's TeX command does that. --rwg

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[math-fun] Solving polynomial diophantine equations
by Warren D Smith 29 Aug '13

29 Aug '13

This question is called "Hasse / local-global principle" and there is a large literature on its truth and falsity in various situations. For example, re Goucher's 4 squares example, relevant is "Hasse's theorem": Consider F(x1,x2,...,xm)=0 with F a homogeneous quadratic polynomial. This has a solution over Z^m iff it has a solution over Q^m, iff it both has a solution mod N for every N>1 and has a real-number solution, iff it both has a solution mod P^K for every K>0 and prime P and has a real-number solution. However, the Hasse principle fails for homogeneous cubics. Example by Selmer 3*X^3 + 4*Y^3 + 5*Z^3 = 0 has no nontrivial integer solutions despite having both real solutions and solutions modulo N for every N>1. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Older & younger versions of our sun
by Warren D Smith 28 Aug '13

28 Aug '13

http://www.eso.org/public/news/eso1337/ astronomers tried to search for stars as close as possible to being our sun (same mass & composition), but older & younger. They claim that the sun for some reason is destroying lithium, which might be a resolution of the "lithium problem" which is that there seems to be way less lithium in the universe than big bang models say there ought to be. But surely to make that case one would need evidence that non-stellar material has the expected high lithium content, and that very young stars do too? Where is that evidence? If it cannot be found, the "lithium problem" remains a problem.

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Re: [math-fun] Solving polynomial diophantine equations
by Adam P. Goucher 27 Aug '13

27 Aug '13

Dan, I can immediately answer two of your questions: > 2) Is (*) equivalent to saying P(k) == 0 mod q has an integer solution k(q) for all prime powers q ? Yes, by Chinese Remainder theorem. Or, more truthfully, there is a natural ring isomorphism between Z_n and the direct sum of a bunch of Z_queues. This generalises to multivariate polynomials. > 3) What is known about the questions analogous to 1) and 2) for integer polynomials in 2 integer variables??? More than 2 integer variables? In four integer variables, take the equation: P(w,x,y,z) = w^2 + x^2 + y^2 + z^2 + 1 = 0 solvable modulo every N by Lagrange's four-square theorem, but trivially not solvable in Z. I think that this trick also works for three squares (in fact it definitely does), but it miserably fails for two squares. Sincerely, Adam P. Goucher

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Re: [math-fun] solving and denesting
by Bill Gosper 27 Aug '13

27 Aug '13

http://www.math.harvard.edu/~elkies/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. Can anybody propose a correction? 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 Thank you Charles! May I ask the nature of your Galwogger? I assume you observed no reducible cases for other (non0 u). If you find time, could you try one or two u and v in Matzat's allegedly solvable sextics? Not that I understand a lick of it, but this single measly bit of solvability information is the only contribution I have seen from Galois theory worth a damn to computer-algebraic denesting and solving. Has anybody yet implemented Landau's algorithm? For a ridiculous example of erudite inutility applied to solving sextics, www.sciencedirect.com/science/article/pii/S002186930098428X , which handles every possible Galois group separately, requiring 15th degree resolvents and formulæ that run on for pages on end. Excerpt: "Problem Ža. is straightforward, though with of high degree and G G48 , there are some concerns regarding memory and speed. For example, the coefficient c10 of f10Ž x. requires on the order of 9^10 arithmetic operations to be expressed. All calculations for this paper were done using Mathematica on a Sun 5.6 and this step took at most one day for any of the polynomials calculated. Solving Problem Žb. was more difficult. ..." But since the author used a computer, his worked example was printed correctly: x^6 + x^4 - x^3 - 2*x^2 + 3*x - 1 But not worked to completion, due to the godawful machinery. Anyway Ssexy got it immediately, a typical root being (1/2 (-9+Sqrt[93]))^(1/3)/(2 3^(2/3))-(1/2 (9+Sqrt[93]))^(1/3)/(2 3^(2/3))-1/(2 Sqrt[6/(-4-22 (2/(1937+201 Sqrt[93]))^(1/3)+2^(2/3) (1937+201 Sqrt[93])^(1/3))]) , an atypically simple expression. But I want simpler. The quintics that led to the denestings had as roots the sums of 5th roots of monomials. How do I find sextics and septics like that? On Mon, Aug 26, 2013 at 7:10 AM, Bill Gosper <billgosper(a)gmail.com> wrote: According to http://www.math.harvard.edu/~elkies/trinomial.html, modulo canonicalization, there are only five solvable, irreducible quintics with quadratic middle terms. Only one, x^5 - 100 x^2 - 1000, has roots that are the sums of fifth roots of monomials. Its real root is 2 2^(1/5) + 2 2^(2/5) - 2^(3/5) + 2^(4/5). Raising this to consecutive powers as before, In[199]:= Table[FullSimplify[Expand[%^k]], {k, 12}] Out[199]= {Root[-1000 - 100 #1^2 + #1^5 &, 1], 10 (2^(1/5) + 2^(3/5)), 10 (6 - 2 2^(1/5) + 4 2^(2/5) + 2 2^(3/5) + 2^(4/5)), 100 2^(1/5) (2 + 2^(1/5) + 2 2^(3/5)), 1000 (1 + 2^(1/5) + 2^(3/5)),[...]} The square gives this familiar-looking denesting that I can't find in my notes: In[193]:= DenestRadicals3[Sqrt[1 + 4^(1/5)]] [...] Out[193]= (-1 + 2^(1/5) + 8^(1/5) + 16^(1/5))/Sqrt[5] (Heuristic denester by Corey Ziegler Hunts.) Does anybody recognize this? I can't imagine missing it when I did sqrt(2-2^(1/7)). The fifth power gives a trinomial^(1/5) -> quadrinomial denesting In[206]:= DenestRadicals3[(1 + 2^(1/5) + 8^(1/5))^(1/5)] [...] Out[206]= (-1 + 2^(1/5) + 8^(1/5) + 16^(1/5))/125^(1/5) This I would nominate to replace the "more complicated" yet meaningless, undenestable trinomial^(1/3) inhttp://en.wikipedia.org/wiki/Nested_radical . Well, not if the article is really more about nesting than denesting. Better would be some regular polygon area formulas. But that would smack of the ever contemptible Original Research. Also missing from that article is a remark to the effect that denestings with outer radicand of the form a+b^(k/n) are quite rare when n>2. Citation needed. --rwg Actually the reason is simple: Raising a polynomial to a positive integer power hardly ever shortens it. --rwg http://www.math.harvard.edu/~elkies/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. Can anybody propose a correction? I found a djvu of B.H. Matzat: *Konstruktive Galoistheorie*, LNM 1284 (1987), where Elkies claims he got the sextic, but instead Matzat has the two families x^6 + (200000 (5 + u) (8 + u)^2 (-5 + 6 x))/((-120 + u) u^5) and x^6 - (84375 (-16 + v) v^3 (-5 + 6 x))/(4 - 14 v + v^2) Maybe one of these is Elkies's under some tricky change of variable? But woe, my sextic solver fails on these as well! Perhaps it's just broken. But it works on the cases I can find which are substantiated with actual solutions, leaving me to wonder if *both* Elkies and Matzat failed to computer-check their drafts. --rwg

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