November 2020 - mathfun@mailman.xmission.com

[math-fun] An identity involving the floor function
by Neil Sloane 22 Nov '20

22 Nov '20

Dear Math Fun, The OEIS is full of assertions whose status is unclear - are they theorems or conjectures? This one is stated unconditionally but without a proof. If we had a proof it would help with quite a lot of other sequences. It is surely true, and maybe not difficult to prove. Can someone help? T(k) = k*(k+1)/2, k >= 0, is a triangular number. Claim: For n >= 1, we have Sum_{ k >= 1, stop when T(k-1) >= n } (-1)^(k+1) * ( floor ( (n - T(k-1))/k ) )^2 = Sum_{k=1..n} sigma(k). The RHS is the sum of the divisors of all numbers from 1 to n, A024916(n), which can also be written as Sum_{d=1..n} d*floor(n/d) = n^2 - Sum_{d=1..n} n%d. The summands on the LHS look like [ (n-T(k-1))/k ]^2, and the sum stops when the quantity n - T(k-1) becomes zero or goes negative. For example, when n=8, the assertion is that 8^2 - [ (8-1)/2 ]^2 + [ (8-3)/3 ]^2 = 8^2 - 3^2 + 1^2 = 56 = 1+3+4+7+6+12+8+15 . Neil

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[math-fun] dvipdf
by James Propp 21 Nov '20

21 Nov '20

Has anyone else been having problems lately with dvipdf? My install of dvipdf now converts a multi-page dvi file with actual content into a one-page file containing nothing but white space. I tried it on a file that compiled properly back in May, and it's not working now. I reinstalled TexShop and restarted my laptop, but that didn't fix the issue. (I am trying to do the final upload for an article I've been working on for seven years. So of course something like this would happen...) Jim Propp

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[math-fun] The Walrus Binary Entropy Coder
by Steve Witham 21 Nov '20

21 Nov '20

"ABSTRACT: The Walrus coder does the same job as a binary arithmetic coder-- "binary" meaning having an input alphabet of just two symbols. The Walrus has the same API as arithmetic coders, and similar coding efficiency and latency, but the Walrus maps symbols to bits differently: at each step it designates one of the two symbols "walrus" and the other "eggman." The walrus is assigned an available bit string, which is then either output (if walrus is sent) or made unavailable (if eggman is sent). ... This file is "literate executable pseudocode" in Python... An appendix contains test code... A separate Jupyter notebook [N] does some analysis [and graphs] of the Walrus's efficiency both empirically and theoretically." http://www.mac-guyver.com/switham/2020/11/Walrus Â --Steve

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Re: [math-fun] dvipdf
by Henry Baker 20 Nov '20

20 Nov '20

MacOS Catalina & Big Sur break a lot of stuff, due to the new permission mechanisms, which work in tandem with Linux file permissions, but go beyond them. (If you know anything about Android, they have a similar problem with their 'SELinux' permission structure; many people give up and turn off SELinux completely.) Sometimes going to the security & privacy section and giving the failing app permission to read/write files may work. (The app still can't go around the normal Linux file permissions.) For example, Big Sur broke Emacs, so I had to Google around until I found a fix which involved giving Emacs more permissions. At 08:03 AM 11/20/2020, James Propp wrote: >The problem affects dvipdf on my new MacBook running Catalina 10.15.7 but >not my old MacBook running High Sierra 10.13.6. > >Jim Propp > >On Fri, Nov 20, 2020 at 10:49 AM James Propp <jamespropp(a)gmail.com> wrote: >> Has anyone else been having problems lately with dvipdf? >> >> My install of dvipdf now converts a multi-page dvi file with actual >> content into a one-page file containing nothing but white space. >> >> I tried it on a file that compiled properly back in May, and it's not >> working now. >> >> I reinstalled TexShop and restarted my laptop, but that didn't fix the >> issue. >> >> (I am trying to do the final upload for an article I've been working on >> for seven years. So of course something like this would happen...) >> >> Jim Propp

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[math-fun] That InverseFourier hack seems also to solve cubics and septics.
by Bill Gosper 17 Nov '20

17 Nov '20

(But the septic solver requires a sextic solver.) Here, with crucial help from Julian (despite his disdain for preoccupation with radicals), is the cubic (with rational coefficients) solver: In[107]:= scub[x^3-8x-10] Out[107]= {(45+Sqrt[489])^(1/3)/(3^(2/3) I^(4/3))+((45-Sqrt[489])^(1/3) I^(4/3))/3^(2/3), (45-Sqrt[489])^(1/3)/(3^(2/3) I^(4/3))+((45+Sqrt[489])^(1/3) I^(4/3))/3^(2/3), (45-Sqrt[489])^(1/3)/3^(2/3)+(45+Sqrt[489])^(1/3)/3^(2/3)} In[109]:= ContinuedFraction[%%[[3]],69] Out[109]= {3,3,7,4,2,30,1,8,3,1,1,1,9,2,2,1,3,22986, 2,1,32,8,2,1,8,55,1, 5,2,28,1,5,1,1501790, 1,2,1,7,6,1,1,5,2,1,6,2,2,1,2,1,1,3,1,3,1,2,4,3,1,35657, 1,17,2,15,1,1,2,1,1} scub[q_]:= ToRadicals[Block[{rts = (#1[[1,2]] & ) /@ NSolve[q == 0,WorkingPrecision->33],foos}, foos = (rts.{{1,1,1},{1,1/I^(4/3),I^(4/3)},{1,I^(4/3),1/I^(4/3)}})^3/27; foos = (Together[#1[[1,2]]]^(1/3) & ) /@ Solve[0 == Rationalize[Expand[Times @@ (#1 - Rest@foos)]]]; Rationalize@Mean@ rts + (foos . #1 & ) /@ {{I^(4/3),1/I^(4/3)},{1/I^(4/3),I^(4/3)},{1,1}}]] That's it! For triples, the InverseFourier open codes to this little matrix multiplication foos = (rts.{{1,1,1},{1,1/I^(4/3),I^(4/3)},{1,I^(4/3),1/I^(4/3)}})^3/27; and the searches over root permutations and roots of unity degenerate. Whoa, wait a minute, it calls Solve! But only on quadratics. Which we could open-code. In[111]:= scub[Echo@MinimalPolynomial[Cos[2\[Pi]/7],x]] -1-4 x+4 x^2+8 x^3 Out[111]= {-(1/6)+(7/2 (1+3 I Sqrt[3]))^(1/3)/(6 I^(4/3))+1/6 (7/2 (1-3 I Sqrt[3]))^(1/3) I^(4/3), -(1/6)+(7/2 (1-3 I Sqrt[3]))^(1/3)/(6 I^(4/3))+1/6 (7/2 (1+3 I Sqrt[3]))^(1/3) I^(4/3), -(1/6)+1/6 (7/2 (1-3 I Sqrt[3]))^(1/3)+1/6 (7/2 (1+3 I Sqrt[3]))^(1/3)} In[114]:= FullSimplify@ArcCos@%%% Out[114]= {ArcCos[-(1/6)+(7/2 (1+3 I Sqrt[3]))^(1/3)/(6 I^(4/3))+1/6 I^(4/3) 2.37\[Ellipsis]-1.18\[Ellipsis] I], ArcCos[-(1/6)+1/6 (7/2 (1+3 I Sqrt[3]))^(1/3) I^(4/3)+2.37\[Ellipsis]-1.18\[Ellipsis] I/(6 I^(4/3))], 2π/7}. Disclaimer: For sufficiently exotic coefficients, WorkingPrecision->33 would need to increase, But I'm surprised that such a concise formula isn't more popular. —rwg

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 16 Nov '20

16 Nov '20

Thanks, George! I look forward to watching the video, which unfortunately I don't have time to do this morning. But I'm wondering a couple of things from this post: 1. How do you define the product of two polygons as a polyhedron in 3-space? Abstractly, it makes perfect sense: The product of two vertices is a vertex; of a vertex and an edge (in either order) is an edge; and of two edges is a quadrilateral. So far this is just a combinatorial object, but it could even be an abstract geometrical object, as if each polygon were in a distinct plane and the product becomes a specific subset of 2+2 = 4-space. But that seems to allow a lot of possibilities for it to end up in 3-space. How do you see that any way of doing this results in a polyhedron that's self-intersecting? 2. What does "homogeneous" mean in this situation? It could refer just to the result being combinatorially homogeneous, in that the result has maps taking vertices to vertices, edges to edges, and faces to faces while preserving incidence, that will carry any vertex to any other (transitivity on vertices). Is that it? Thanks, Dan ----- The talk Jim had referred to is now online here in a more completed form. Polyhedra fans might enjoy it: https://www.youtube.com/watch?v=_PSdVX02Vbs The question which I thought might have been intended (during the CoM discussion, or initially by Jim) was whether one could make a *homogeneous* zonohedral surface in 3D which is topologically toroidal without self-intersection. By homogeneous, I mean following the algorithm described in the video from a given initial state. (One with all vertices of degree four would be homogeneous, but that isn't a necessary condition; homogeneous examples might have vertices of any order, e.g., polar zonohedra.) One can take a product of two polygons to create a toroidal surface composed of parallelograms and having all vertices of degree four, but it will be self-intersecting. And one can approximate any continuous surface by joining homogeneous patches, e.g., gluing together small cubes. (The video shows how to join larger homogeneous patches seamlessly.) But I would conjecture that a homogeneous zonohedral surface can only be toroidal if it is self-intersecting. -----

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 16 Nov '20

16 Nov '20

Yes (I presume this means in R^3). To get on with all faces rectangles, drill a hole straight through a rectangular solid and look at its boundary. Now apply a random linear mapping and all the rectangles will probably turn into parallelograms. —Dan Jim Propp wrote: ----- George Hart’s Celebration of Mind talk (just concluded) makes me wonder: is there a non-self-intersecting torus surface made of parallelograms? -----

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[math-fun] Several years ago I wrote a terse quintic solver-in-radicals-where-possible.
by Bill Gosper 15 Nov '20

15 Nov '20

Unfortunately, I no longer have enough brain cells to remember or figure out how it works. It somehow gets exact expressions by taking the inverse Fourier transform of numerically approximated roots, without all the Galois group and resolvent rigamarole. I.e., it claims to find the radicals when they exist, subject to adjustable limits on coefficient size. Two examples (tim is just a timing function): In[451]:= squint[x^5 - 5 x x - 3] // tim During evaluation of In[451]:= 0.319225 seconds, 5 roots Out[451]= {-((-1)^(3/5)/(2^(2/5) (5/(5 - Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5))) + ((-1)^(2/5) (1/5 (5 - Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^(1/5))/2^( 2/5) - (-1 - 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5)/2^( 2/5) - ((-1)^(3/5) (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^(2/5), 1/(2^(2/5) (5/(5 - Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5)) + (1/5 (5 - Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^(1/5)/2^( 2/5) + ((-1)^(4/5) (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^( 2/5) + (1 + 1/Sqrt[5] + 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5)/2^(2/5), ((-1)^(4/5) (1/5 (5 - Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5))/2^(2/5) - (1/5 (-5 + Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5)/2^(2/5) + ((-1)^(2/5) (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^( 2/5) + ((-1)^(2/5) (1 + 1/Sqrt[5] + 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^(2/5), (-1)^(4/5)/(2^(2/5) (5/(5 - Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5)) - (1/5 (-5 + Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^(1/5)/2^( 2/5) - ((-1)^(1/5) (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^( 2/5) - ((-1)^(3/5) (1 + 1/Sqrt[5] + 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^(2/5), (-1)^(2/5)/( 2^(2/5) (5/(5 - Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5)) - ((-1)^(3/5) (1/5 (5 - Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^(1/5))/2^( 2/5) + (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5)/2^( 2/5) + ((-1)^(4/5) (1 + 1/Sqrt[5] + 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^(2/5)} In[445]:= squint[-1 + 13755 x + 3740 x^2 + 1260 x^3 - 5 x^4 + x^5] Out[445]= {1 + 5 2^(1/5) - 10 2^(2/5) + 10 2^(3/5) - 5 2^(4/5), 1 + 5 (-1)^(4/5) 2^(1/5) + 10 (-1)^(3/5) 2^(2/5) + 10 (-1)^(2/5) 2^(3/5) + 5 (-1)^(1/5) 2^(4/5), 1 - 5 (-1)^(3/5) 2^(1/5) + 10 (-1)^(1/5) 2^(2/5) + 10 (-1)^(4/5) 2^(3/5) - 5 (-1)^(2/5) 2^(4/5), 1 + 5 (-1)^(2/5) 2^(1/5) - 10 (-1)^(4/5) 2^(2/5) - 10 (-1)^(1/5) 2^(3/5) + 5 (-1)^(3/5) 2^(4/5), 1 - 5 (-2)^(1/5) - 10 (-2)^(3/5) - 10 (-1)^(2/5) 2^(2/5) - 5 (-1)^(4/5) 2^(4/5)} (It can take Mathematica several minutes to check one of these symbolically.) Here is the definition: squint[q_] := ToRadicals[ Block[{rts = (#1[[1, 2]] & ) /@ NSolve[q == 0, WorkingPrecision -> 99], boa, foos}, foos = (InverseFourier[rts[[#1]]/Sqrt[5]]^5 & ) /@ Permutations[Range[5]]; foos = (Together[#1[[1, 2]]]^(1/5) & ) /@ Solve[0 == Rationalize[Expand[Times @@ (#1 - Rest[MinimalBy[foos, Denominator@Rationalize[Plus @@ #, 9.^-69] &,1][[1]]])]]]; boa = Rationalize@Mean@rts; boa + Union[(foos . #1 & ) /@ Select[Tuples[I^(4 Range@5/5), 4], MemberQ[Chop[#1 . foos + boa - rts], 0] & ]]]] One thing I remember is boa is "b over a", as in a x^5 + b x^4 + . . . note boa = Rationalize@Mean@rts; Something I don't remember is feeling any surprise that this works. It would be funny if it comes as a surprise to algebraists. Actually, it wouldn't be funny because then I'd have to publish something. That I don't remember. Replacing 5 by 7, 4 by 6, and Solve[0==. . .] with a sextic solver, Julian and I found the analogous septic solver. He expressed mistrust, but it has yet to fail. —rwg

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[math-fun] Chess variant
by James Propp 14 Nov '20

14 Nov '20

Ordinary chess forbids moving into check (and hence from moving into checkmate), which means you automatically lose if there's no way to avoid moving into check. Is there a variant of chess in which you're not allowed to make a move that would cause you to lose 2 moves in the future? (Replacing 2 by larger integers, we get a game in which the legality of a move can become rather murky!) Jim

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[math-fun] Chess variant
by David Wilson 13 Nov '20

13 Nov '20

In Paul Morphy's "Opera House" chess game, Morphy never moves a piece backward. I propose Opera House Chess, with initial position and moves as in chess, except pieces cannot move backward. Is it playable? With victory conditions modified?

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