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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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[math-fun] Every N = p1+p2+p3; pi a palindrome
by Henry Baker 19 Sep '18

19 Sep '18

FYI -- This result may have been discussed here already... https://arxiv.org/abs/1602.06208v2 Every positive integer is a sum of three palindromes Javier Cilleruelo, Florian Luca, Lewis Baxter (Submitted on 19 Feb 2016 (v1), last revised 17 Jun 2017 (this version, v2)) For integer $g\ge 5$, we prove that any positive integer can be written as a sum of three palindromes in base $g$.

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Re: [math-fun] Dense unit distance graphs.
by Dan Asimov 17 Sep '18

17 Sep '18

Very interesting question, whether or not I may have heard it before. And Ed Pegg's graphs are fantastic. I suppose the same question can be asked about a metric surface, especially one with a lot of symmetry like the sphere or one of the tori (here C = complexes, and ζ_n denotes the primitive nth root of unit with smallest positive angle). T_g = C / Z[ζ_4] (the Gaussian torus) and T_e = C / Z[ζ_6] (the Eisenstein torus). (Here Z[α] denotes the subring of the complex numbers generated by the complex number α, i.e. the set of values {P(α)} where P runs through all polynomials of any degree with integer coefficients. The expression C / Z[α] means the quotient of C by the additive group of Z[α], viewed as translations of the complex plane.) The choice of distance to try to repeat is not scale-free, as it is on the plane. Still, the distance of 1 has special significance on T_g and T_e since it is a generator of the additive groups of Z[ζ_4] and Z[ζ_6]. T_e has a *special* relationship with this problem because of all the equilateral triangles among the points of Z[ζ_6]. In any case see the set of vertices of many toral tilings here: [http://www.weddslist.com/groups/genus/1/] —Dan Allan Wechsler wrote: ----- For a given set of k points in the plane, some number n of the mutual distances are 1. How big can we make n for a given k? A186705 at OEIS given the answers we know, but only up to 14 vertices. Where can I see pictures of example maximal unit distance graphs? The first "surprise" is a(9)=18; I would love to see that one. -----

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[math-fun] music and math
by Cris Moore 17 Sep '18

17 Sep '18

Enjoy! https://www.pbs.org/video/majesty-of-music-and-math-egpyyf/ <https://www.pbs.org/video/majesty-of-music-and-math-egpyyf/> - Cris

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