September 2018 - mathfun@mailman.xmission.com

Re: [math-fun] Optimal polygonal pasture
by Dan Asimov 21 Sep '18

21 Sep '18

----- I'm actually not worried about units, but it's nice to know about squaring the perimeter. ----- I want to put in a plug for dimensionless quantities. The values of the non-dimensionless ones can very well depend on the choice of units, though admittedly varying by at most a positive factor at most. But also — whether or not the figures-of-merit I mentioned turn out to be useful — I think it's worth trying to find the, or a, figure-of-merit that gets as close to what your goal is as possible. To me this is a little like focusing a lens before peering through it. (This is actually a lesson I learned around 1998 from another math-fun member, and he knows who he is.) —Dan

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Re: [math-fun] Optimal polygonal pasture
by Dan Asimov 21 Sep '18

21 Sep '18

Many figures of merit have the same regular N-gon as the optimum. A common one when talking about the isoperimetric inequality, for any rectifiable simple closed curve C in the plane, is f_2(C) := length(C)^2 / area(C) and this figure of merit f_2(C) is dimensionless, so independent of units. Then the inequality is just f_2(C) <= 4 with equality only for the circle. ***>>> I think this is a fascinating area to explore. It may be that nothing too interesting happens in 2D. But in 3D, where a corresponding figure of merit for a rectifiable closed surface S in R^3 could be f_3(S) = area(S)^3 / volume(S)^2, and then the inequality is f_3(S) <= 36 again with equality only for the 2-sphere. Then for any given combinatorial type of polyhedron (the equivalence class of polyhedra having 1-1 correspondences between their faces of each dimension preserving incidence), we can ask for the one(s) that maximize f_3(S). A figure-of-merit with similar properties is the ratio of the radius of the inscribed sphere to that of the circumscribed sphere: R_inscr / R_circum <= 1 (again with equality only for the sphere; this also holds in n dimensions). Another figure-of-merit might be to try to make the vertices as far apart as possible, again among polyhedra of the same type. In 3D one way is to sum up all squared distances between points, with distances reckoned through 3-space. This kind of thing would not work if there existed no stationary point (for the figure of merit) within the given combinatorial type. But making the vertices as far apart as possible tends to make them equal, so this is OK. —Dan Allan Wechsler wrote: ----- For a fixed N, suppose we want to make a polygonal pasture with N sides, with the largest possible area given a fixed perimeter. Is the unique optimum always a regular N-gon? I am looking for a figure of merit that tells how regular a polygon is, and perimeter/area seems promising. -----

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[math-fun] Optimal polygonal pasture
by Allan Wechsler 21 Sep '18

21 Sep '18

For a fixed N, suppose we want to make a polygonal pasture with N sides, with the largest possible area given a fixed perimeter. Is the unique optimum always a regular N-gon? I am looking for a figure of merit that tells how regular a polygon is, and perimeter/area seems promising.

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Re: [math-fun] Chebonacci numbers
by Bill Gosper 21 Sep '18

21 Sep '18

Is this about theoretical complexity, or a programming competition? Fibonaccis are so easy to calculate (e.g. x^n mod x^2-x-1, or {{1,1},{1,0}}^n) that I never gave it much thought. My interest just now was how to define them for fractional n. The Binet and Cheby formulæ tend rapidly to agreement, producing interesting(?) algebraic near-identities. E.g., Binet(11/2) = 𝛕^(11/2)/√5 ~ 6.30880376931698 Cheb(11/2) = √(89/√5) ~ 6.30888341939335 —rwg On 2018-09-19 21:21, Andres Valloud wrote: > Bill, > > Some claim Dijkstra's recursion: > > http://www.cs.utexas.edu/users/EWD/ewd06xx/EWD654.PDF > > is very fast at calculating F_n, however it seems that using > > F_{a+b} = F_{a+1} F_b + F_a F_{b-1} > > recursively and splitting n = a+b such that a = 2^k leads to caching > just triplets around each 2^k pivot (k greatest such that 2^k < n). > This caching is harder to do as effectively with Dijkstra's algorithm, > so the splitting F_{a+b} method becomes competitive while using less memory. > > Do you have an opinion on the subject? > > Andres. > > On 9/19/18 20:11 , Bill Gosper wrote: >> World's smallest(?) fibonacci formula: F_n+1 = U_n(i/2)/i^n. >> where U:= Чебышёв polynomial, 2nd kind := >> sin((n+1)acos z)/√(1-z)/√(1+z). This provides an alternate (complex) >> interpolation for nonintegers. Slightly nonobvious: With this definition, >> Im(F_x+1)/Im(F_x) = -1/GoldenRatio (x real noninteger), which can >> be coaxed out of Mathematica. —rwg

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[math-fun] How many pieces? puzzle
by Dan Asimov 20 Sep '18

20 Sep '18

Let the n-dimensional torus T^n be defined as the n-cube Q_n = [-1, 1]^n with its opposite faces identified. (That is, any point x of Q_n with coordinate x_k = ±1 for some k is identified with the point having the same coordinates except with x_k changed to -x_k.) The vertices of Q_n are the 2^n points {-1, 1}^n. Now we are going to cut T^n along each of the 2^(n-1) hyperplanes defined as the perpendicular bisectors of each pair of antipodal vertices of Q_n. This requires extending each hyperplane throughout the n-torus. Puzzle: How many pieces do these planes cut the torus T^n into? E.g., for n = 2 there are two pieces. —Dan

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Re: [math-fun] How many pieces? puzzle
by Dan Asimov 20 Sep '18

20 Sep '18

What is the connection between the tessellation and the sphere-packing? —Dan Fred Lunnon wrote: ----- Duh --- light dawns! Unwinding Dan's n-torus into its universal covering tesselation of Euclidean space yields simply what seems to be known only as " n-space analogue of the face-centred cubic crystal tiling" or such (maybe Neil can correct this?): the unit-sphere packing where centres have evenly many odd integer Cartesian components. From the combinatorics of the hypercube it is now straightforward that the number of regions in the torus equals 2^n 3^(n-1) ; also the number of vertices equals 2^(n-1) . I might flesh this out with more detail on request. -----

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[math-fun] Titans of Mathematics Clash Over Epic Proof of ABC Conjecture
by Ray Tayek 20 Sep '18

20 Sep '18

https://science.slashdot.org/story/18/09/20/207227/titans-of-mathematics-cl… Two mathematicians have found what they say is a hole at the heart of a proof that has convulsed the mathematics community for nearly six years. Quanta Magazine:/In a report <http://www.kurims.kyoto-u.ac.jp/%7Emotizuki/SS2018-08.pdf> /[PDF]/ posted online Thursday, Peter Scholze of the University of Bonn and Jakob Stix of Goethe University Frankfurt describe what Stix calls a "serious, unfixable gap" within a mammoth series of papers by Shinichi Mochizuki <https://www.quantamagazine.org/titans-of-mathematics-clash-over-epic-proof-…>, a mathematician at Kyoto University who is renowned for his brilliance. Posted online in 2012, Mochizuki's papers supposedly prove the /abc/ conjecture <https://mathbabe.org/2015/12/15/notes-on-the-oxford-iut-workshop-by-brian-c…>, one of the most far-reaching problems in number theory. Despite multiple conferences dedicated to explicating Mochizuki's proof <https://www.quantamagazine.org/hope-rekindled-for-abc-proof-20151221/>, number theorists have struggled to come to grips with its underlying ideas. His series of papers, which total more than 500 pages, are written in an impenetrable style, and refer back to a further 500 pages or so of previous work by Mochizuki, creating what one mathematician, Brian Conrad of Stanford University, has called "a sense of infinite regress." Between 12 and 18 mathematicians who have studied the proof in depth believe it is correct, wrote Ivan Fesenko of the University of Nottingham in an email. But only mathematicians in "Mochizuki's orbit" have vouched for the proof's correctness, Conrad commented in a blog discussion last December. "There is nobody else out there who has been willing to say even off the record that they are confident the proof is complete." Nevertheless, wrote Frank Calegari of the University of Chicago in a December blog post, "mathematicians are very loath to claim that there is a problem with Mochizuki's argument because they can't point to any definitive error." That has now changed. In their report, Scholze and Stix argue that a line of reasoning near the end of the proof of "Corollary 3.12" in Mochizuki's third of four papers is fundamentally flawed. The corollary is central to Mochizuki's proposed abc proof. "I think the abc conjecture is still open," Scholze said. "Anybody has a chance of proving it."/ -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/

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[math-fun] Chebonacci numbers
by Bill Gosper 19 Sep '18

19 Sep '18

World's smallest(?) fibonacci formula: F_n+1 = U_n(i/2)/i^n. where U:= Чебышёв polynomial, 2nd kind := sin((n+1)acos z)/√(1-z)/√(1+z). This provides an alternate (complex) interpolation for nonintegers. Slightly nonobvious: With this definition, Im(F_x+1)/Im(F_x) = -1/GoldenRatio (x real noninteger), which can be coaxed out of Mathematica. —rwg

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[math-fun] Very old solid angle question I never solved
by Dan Asimov 19 Sep '18

19 Sep '18

This reminds me of a solid angle problem that's been bouncing around my visual cortex: Suppose we start with a cone (a single nappe) in 3-space. Remove the conepoint. What's left is topologically a cylinder: Cyl = S^1 x [0, oo). Now elongate the cone and wrap it around the unit circle in the xy-plane, in such a way that it avoids self-intersection. Now, make it wrap more and more around that same circle, always avoiding self- intersection, and finally limiting on the circle as the cone gets pointier and pointier (without the conepoint at all). Here's the question: We can assume this infinitely helical cone is *smooth* (away from the missing conepoint), and so has a well-defined outward unit normal vector at each point. These unit normal vectors define a map G : Cyl = S^1 x [0, oo) —> S^2 by translating each vector to the origin; this is the Gauss map G. Now consider the function f : [0, oo) —> R defined by f(t) = vol_+(G([t, oo)) where vol+ denotes signed area on the unit sphere S^2. Question: --------- What is the limit lim f(t) ??? t—>oo (I believe the answer is independent of choices and is determined by the conditions defining the cylinder in R^3.) —Dan Bill Gosper wrote: ----- I just computed the solid angles to be ... ... -----

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[math-fun] Julian's tympanohedron
by Bill Gosper 19 Sep '18

19 Sep '18

http://gosper.org/dodohedron.png (equilateral dodecahedron topologically equivalent to the regular one) has two kinds of vertices, ten convex and ten nonconvex. I just computed the solid angles to be ArcCos[27 - 42 √(2/5) - 62/(45 √2 + 29 √5)] ~ .514544425310092 π (slightly > than a cube corner), and 2 π + ArcCos[1/27 (-167 + 118 √2 - 70 √5 + 58 √10)] ~ 2.04206803084996 π (slightly > than a half-space). (Which you used to be able to type by holding down the space bar.) —rwg (Could I have neglected to propose this figure as a G4G12 decoration, or was it deemed too ugly?)

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