November 2019 - mathfun@mailman.xmission.com

[math-fun] Rolling polyhedra
by James Propp 26 Dec '20

26 Dec '20

I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees. For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is). Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)? Jim Propp

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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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[math-fun] One parameter quintic solution
by Bill Gosper 01 Nov '20

01 Nov '20

This is probably ho-hum, but while trying to simplify a Chowla-surdberg, I found that 0 = a + 80 x + 20 x^3 + x^5 has solutions {((-a - Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5), ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), (-a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^( 1/5), (-a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^(1/5)} Is it from a two-parameter family? C.f. https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics --rwg

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Re: [math-fun] FactorInteger
by Bill Gosper 20 Aug '20

20 Aug '20

Slight improvement: Mathematica's front end by default displays a centerdot for implicit multiplication. Strangely, this doesn't work in old the FI function below, which displays blanks. Blanks also (by default) separate digits in blocks of 3. So the display is sometimes ambiguous. But Wolfram CE Harry Calkins showed me this workaround: FI⨉[n : _Rational | _Integer] := Defer @@ {FactorInteger@n} /. {{a_Integer, 1} -> a, {a_Integer, b_} -> a^b} /. {a_} -> a /. List -> Times E.g. In[144]:= FI⨉@BernoulliB@50]@// tim During evaluation of In[144]:= 0.006522,1 (seconds) Out[144]= (5^2 417202699 47464429777438199)/(2 3 11) (The first Bernoulli number with a square factor!) Dirty trick. I used a tiny ⨉ symbol instead of x, so you need Front End completion to input FI⨉. You may want to edit the definition. Instead of little centerdots between the factors, there's a Preferences variable Appearances|Numbers|Multiplication|Automatically display multiplication symbol between multiplied numbers Use symbol: which you can bash to be a mini ⨉. Sorry, I don't know how to bash it under program control, as in an init file. Unfactor is way nicer than with FactorInteger: UnFI⨉@Defer[x_] := x In[153]:= UnFI⨉@%144 Out[153]= 495057205241079648212477525/66 —rwg On Wed, Jan 4, 2017 at 12:44 AM Bill Gosper <billgosper(a)gmail.com> wrote: > There are > > In[14]:= 52! > Out[14]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > ways to shuffle a deck. > That's just a "little" over a googol ... > > In[366]:= Log[10.^100, %] > Out[366]= 0.679066483922048 > > to the 2/3 power. The square of the cube root. And vice versa. So if > anybody asks how many > ways to shuffle a deck, "Oh, about the cube root of a googol squared." > But this is actually low > by a fairly substantial factor of > In[372]:= (10.^100)^(.679 - 2/3) > > Out[372]= 17.1132830416181 > > The actual number of shufflings is enormously composite: > > In[15]:= FactorInteger@%14 > Out[15]= > {{2,49},{3,23},{5,12},{7,8},{11,4},{13,4},{17,3},{19,2},{23,2},{29,1},{31,1},{37,1},{41,1},{43,1},{47,1}} > > But this is ugly and hard to read. The doc suggests > > In[16]:= CenterDot@@(Superscript@@@%) > Out[16]= > 2^49\[CenterDot]3^23\[CenterDot]5^12\[CenterDot]7^8\[CenterDot]11^4\[CenterDot]13^4\[CenterDot]17^3\[CenterDot]19^2\[CenterDot]23^2\[CenterDot]29^1\[CenterDot]31^1\[CenterDot]37^1\[CenterDot]41^1\[CenterDot]43^1\[CenterDot]47^1 > > This actually looks neat. Except for the ^1s. Try it. But it's unclear > (actually, a nice little Mathematica exercise) how to get your original > number back. > So I wrote something called FI > > In[364]:= FI[52!] > Out[364]= 2^49 3^23 5^12 7^8 11^4 13^4 17^3 19^2 23^2 29 31 37 41 43 47 > > In[365]:= ReleaseHold@% > Out[365]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Here is the definition: > FI[n : _Rational | _Integer] := > Times @@ ((HoldForm @@ {#1[[1]]})^#1[[2]] &) /@ FactorInteger[n] > > And here's a spoiler for how to invert FactorInteger: > > In[367]:= Times @@ (Power @@@ FactorInteger[73 137]) > Out[367]= 10001 > > which also works on that CenterDot expression: > > In[368]:= Times @@ (Power @@@ %16) > > Out[368]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Finally, you might enjoy finding a fairly nice formula for how many 0s are > on the end of n!, and, > more generally, what are exponents of the various prime factors. > --Bill >

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[math-fun] How does a dull spoon manage to cut a stringy piece of meat
by Bill Gosper 14 Jul '20

14 Jul '20

while sliding back an forth against the inside of a slippery bowl? --rwg

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[math-fun] Another pic of Hilbert's spacefill
by Bill Gosper 19 Apr '20

19 Apr '20

to accompany should've been "sampHilbert" <http://www.tweedledum.com/rwg/sampeano.htm> Hilbert's [spacefilling] function is dense with quadruple points, i.e. points in [0,*i*] with four distinct preimage points in [0,1]: gosper.org/hilbquad.png . E.g., the preimage of 1/2 + *i*/4 is {5/48, 7/48, 41/48, 43/48}. This picture connects, in the order they were swept, all the points in [0,*i*] with preimages having denominators ≤ 3⨉4096. (All quadruple point preimages have denominators 3⨉2ⁿ.) E.g., the lower-leftmost vertex is 1/32 + *i*/64, the image of 5/12288, 7/12288, 41/12288, and 43/12288. The white segments are retraced boundary, retroflexed in the middle, (making quadruple points). (Spacefilling functions map closed intervals to closed sets.) The Heighway Dragon is dense with mere triple points. Here's one: Hi res Heighway <http://gosper.org/dragtrip!.png> . —rwg

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[math-fun] Heighway triple points redux
by Bill Gosper 21 Mar '20

21 Mar '20

I thought OEIS A260482, A260747..A260750 completely described the (initial) Dragon triple points, all of which have preimages a(n)/15/2^k. But I just noticed an editor's remark that there is another infinite family of the form a(n)/7/2^k. These are much less "photogenic" than the old set (e.g. gosper.org/dragtrip!.png) <http://gosper.org/dragtrip!.png> because two preimages always(?) differ by 2/7/2^k, and ~75% of them have the 3rd preimage within 4/7/2^k. But they can get "arbitrarily" lopsided: In[508]:= undrag[(69 + 27 I)/160] Out[508]= {1539/3584, 2551/3584, 2553/3584} (Differences presumably limited to 40%.) Also, Julian has privately remarked that there are no irrational triple points. --rwg

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[math-fun] Mathematicians Catch a Pattern By Figuring Out How To Avoid It
by Ray Tayek 16 Jan '20

16 Jan '20

https://science.slashdot.org/story/19/11/28/180213/mathematicians-catch-a-p… We finally know how big a set of numbers can get before it has to contain a pattern known as a "polynomial progression." <https://www.quantamagazine.org/mathematicians-catch-a-pattern-by-figuring-o…> From a report:/A new proof <https://arxiv.org/abs/1909.00309> by Sarah Peluse of the University of Oxford establishes that one particularly important type of numerical sequence is, ultimately, unavoidable: It's guaranteed to show up in every single sufficiently large collection of numbers, regardless of how the numbers are chosen. "There's a sort of indestructibility to these patterns," said Terence Tao of the University of California, Los Angeles. Peluse's proof concerns sequences of numbers called "polynomial progressions." They are easy to generate -- you could create one yourself in short order -- and they touch on the interplay between addition and multiplication among the numbers. For several decades, mathematicians have known that when a collection, or set, of numbers is small (meaning it contains relatively few numbers), the set might not contain any polynomial progressions. They also knew that as a set grows it eventually crosses a threshold, after which it has so many numbers that one of these patterns has to be there, somewhere. It's like a bowl of alphabet soup -- the more letters you have, the more likely it is that the bowl will contain words. But prior to Peluse's work, mathematicians didn't know what that critical threshold was. Her proof provides an answer -- a precise formula for determining how big a set needs to be in order to guarantee that it contains certain polynomial progressions. Previously, mathematicians had only a vague understanding that polynomial progressions are embedded among the whole numbers (1, 2, 3 and so on). Now they know exactly how to find them./ -- 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] Terence Tao's Blog: The Missing Proof.
by Brad Klee 04 Dec '19

04 Dec '19

https://terrytao.wordpress.com/2019/08/13/eigenvectors-from-eigenvalues/ Here is the proof I wrote, in Mathematica: ZS[n_] := M[n, n] (M[1, 1] + M[2, 2] + M[3, 3]) - M[n, n] (L1 + L2 + L3) ZS[n_, Li_] := Li (-M[n, n] - t @@ Complement[Range[3], {n}] + L1 + L2 + L3) ZS0 = Plus[M[1, 1] M[2, 2] - M[1, 2] M[2, 1], M[1, 1] M[3, 3] - M[1, 3] M[3, 1], M[2, 2] M[3, 3] - M[2, 3] M[3, 2], -(L1 L2 + L1 L3 + L2 L3)]; ZSComp[n_, Li_] := Subtract[ZS[n], ZS[n, Li] + ZS0] evsq[Li_] := With[{Ann = Complement[{L1, L2, L3}, {Li}]}, Part[Dot @@ ((Array[M, {3, 3}] - # IdentityMatrix[3]) & /@ Ann), #, #] & /@ Range[3]] Expand[evsq[L1] - (ZSComp[#, L1] & /@ Range[3])] // MatrixForm Expand[evsq[L2] - (ZSComp[#, L2] & /@ Range[3])] // MatrixForm Expand[evsq[L3] - (ZSComp[#, L3] & /@ Range[3])] // MatrixForm The so-called "new identity" is just the projector formula modulo a few zero-sums. The zero-sums are well known in the theory of matrix invariants, see for example: https://en.wikipedia.org/wiki/Invariants_of_tensors#Principal_invariants --Brad

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[math-fun] Brain hemisphere spacefills likely with sextuple points.
by Bill Gosper 02 Dec '19

02 Dec '19

Julian & I (again mostly Julian) found exactly the two continuous functions schematized by this old Macsyma plot <http://gosper.org/hemis.gif> of a closed and unclosed curve. So now we can sample the functions at arbitrary rationals. E.g., triple frequency <http://gosper.org/semihexes3x.png>. But actually the median curve <http://gosper.org/semihexesmedian.png> most resembles the Macsyma plot. Since Julian recently found a sampling of Heighway's Dragon equal to its median curve, it's likely that there are samplings of these semihexes coincident with their median curves. This will be easy to test when we manage to invert the semihex functions. Which will also let us directly exhibit the sextuple points. Failing that, it should be possible to tediously guess and check a few. (Actually, since the Dragon truly spacefills, there is obviously some way to hit all the vertices on the median curve. But "luckily" not crazy and disordered.) Heretofore, the only spacefill dense with sextuple points I'd seen was Julian's huge Fourier triangle (@http://howwords.com/triangles/artcircle, which seems to be down), made from 25 copies of itself. Unfortunately, sextuple points are visited very furtively, so the Fourier plot is unconvincing, never (in six tries) coming very close to the actual sextuple points. For the exact definition, see my 21 Aug 2016 math-fun, which gives the six preimages of In[381]:= untrifil25[1/2 + I Sqrt[3]/10] Out[381]= {8/75, 11/75, 14/75, 17/75, 4/15, 23/75} (In arithmetic progression. Always?) —rwg

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