November 2018 - 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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[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] 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] Dean Hickerson
by James Propp 22 Oct '19

22 Oct '19

Anyone have a current email address for Dean Hickerson? "dean at math daht ucdavis daht edu" and "dean at ucdmath daht ucdavis daht edu" aren't working anymore. Thanks, Jim Propp

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[math-fun] ellipsoid from girths: multivariate series reversion
by Bill Gosper 25 Jun '19

25 Jun '19

Corey notes that ellipsoid girths can't be {large, tiny, tiny} and challenges us to find the "ellipsoid girth inequality" analogous to the triangle inequality. What are the dimensions (semiaxes a,b, and c) of say, the ellipsoidoidal rock by the road near the Foothill College fire station, given the girths gab, gac, gbc; a>b>c? (Assuming you can get the damned thing off your tape measure once you've rolled it on.) In terms of a, b, c, the girths are {gab -> 4 a EllipticE[1 - b^2/a^2], gac -> 4 a EllipticE[1 - c^2/a^2], gbc -> 4 b EllipticE[1 - c^2/b^2]} This is a routine problem for FindRoot, but it would be nice to see how a, b, c depend on the girths in the low eccentricity case, say, by reverting the series expansions of the girths. Julian, Corey, and I were surprised by the absence of multivariate reversion from Macsyma and Mathematica, and set about building our own. After several hours of cruel hoodwinking by Mma's Series facility, we emerged having only solved the ellipsoid case. Surprisingly(?), it has at least eight (2^3) solutions, which may help to explain the shortage of multivariate reverters. To fourth order, {a -> (gab/(2 Pi))(1 - eac + eac^2 + ebc + (eac ebc)/2 - (3 ebc^2)/2 + 1/8 (-8 eac^3 - eac^2 ebc - 9 eac ebc^2 + 18 ebc^3) + 1/16 (16 eac^4 + 2 eac^3 ebc - 3 eac^2 ebc^2 + 42 eac ebc^3 - 57 ebc^4)), a -> (gac/(2 Pi))(1 + ebc + 3/2 (eac ebc - ebc^2) + 3/8 (eac^2 ebc - 7 eac ebc^2 + 6 ebc^3) - 3/16 (7 eac^2 ebc^2 - 26 eac ebc^3 + 19 ebc^4))} {b -> (gab/(2 Pi))(1 + eac - (3 eac^2)/2 - ebc + (eac ebc)/2 + ebc^2 + 1/8 (18 eac^3 - 9 eac^2 ebc - eac ebc^2 - 8 ebc^3) + 1/16 (-57 eac^4 + 42 eac^3 ebc - 3 eac^2 ebc^2 + 2 eac ebc^3 + 16 ebc^4)), b -> (gac/(2 Pi))(1 + 2 eac - eac^2/2 - ebc - (eac ebc)/2 + ebc^2 + 1/8 (6 eac^3 - 5 eac^2 ebc + 7 eac ebc^2 - 8 ebc^3) + 1/16 (-21 eac^4 + 24 eac^3 ebc - 5 eac^2 ebc^2 - 14 eac ebc^3 + 16 ebc^4))} {c -> (gab/(2 Pi))(1 - eac + eac^2 + eac^4 - ebc - (eac ebc)/2 - (eac^3 ebc)/ 8 + ebc^2 - (5 eac^2 ebc^2)/16 - (eac ebc^3)/8 + ebc^4 + 1/8 (-8 eac^3 + eac^2 ebc + eac ebc^2 - 8 ebc^3)), c -> (gac/(2 Pi))(1 - ebc - (3 eac ebc)/2 - (3 eac^2 ebc)/8 + ebc^2 + (9 eac ebc^2)/8 - (3 eac^2 ebc^2)/16 - ebc^3 - ( 9 eac ebc^3)/8 + ebc^4)} where {eac -> gab/gac - 1, ebc -> gab/gbc - 1}, which are small for low eccentricity. E.g., for {a -> 5., b -> 4., c -> 3.}, the girth formulæ give {gab -> 28.3617, gac -> 25.527, gbc -> 22.1035}. Plugging these back into the reversions: {{a -> 4.97828, a -> 4.98561}, {b -> 4.00689, b -> 4.00371}, {c -> 3.0062, c -> 3.00353}}. --rwg

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[math-fun] Neutron stars as atoms
by Warren D Smith 23 Apr '19

23 Apr '19

Whitfield Diffie I recall reading the intreaguing observation that stable neutron structures run up to around atomic wieght 300 and then their is a long gap before you get to the neutron stars with atomic weights around 10^50. Perhaps we should regard the neutron stars as isotopes of very heavy elements. The atomic numbers would be miniscule by comparison with the atomic weights but a mildely positively charged neutron star doesn't seem impossible: would adding a billion protons make a difference? --WDS: A better way to view this is: replace the word "stable" with "stable under pressure." See, I would claim that it is an illusion that there is no stable nucleus with atomic weight 1000. Actually, if you put it under enough pressure, then it would presumably become stable. The only problem is humans have no hope of sustaining pressures that large and the only known way to reach them is inside a neutron star. Anyhow, with suitable pressure ramp up, probably there is no "gap" at all, and practically the full range of atomic weights become reachable with stability. Another thing is: the usual nuclei generally are in the ground state. Excited states that last long, are rare for nuclei. But for a neutron star, it presumably never reaches ground, there are too many excited states it gets trapped in for long periods. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] The Case Against Quantum Computing
by Hans Havermann 10 Jan '19

10 Jan '19

In the IEEE Spectrum last week, Mikhail Dyakonov presented his overview of the field: https://spectrum.ieee.org/computing/hardware/the-case-against-quantum-compu…

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[math-fun] Messages in pi
by James Propp 14 Dec '18

14 Dec '18

Keith Lynch writes: “Carl Sagan’s 1985 novel _Contact_ is a classic, though not very original, SETI novel. Unoriginal except for a subplot about the aliens having discovered messages hidden in the base-eleven digits of pi, which was unfortunately left out of the 1997 film of the same name. I find it an interesting question whether evidence for such a message in pi can ever overcome the possibilities that it's either an astonishing coincidence or someone hacking the computer that's doing the calculation. Presumably if there really was such a message, it would have to be from God, rather than from advanced aliens.” I’m not sure if we’ve discussed this before on math-fun, but I’ve never found this aspect of Sagan’s novel compelling or even coherent. I can imagine a God who intervenes in ways that locally violate the laws of physics, or a God who chooses the laws of physics, but not a God who chooses the laws of mathematics. Is there anything that this might mean? My guess is that, even though Sagan acknowledges in the novel that pi is a mathematical constant rather than a physical one, he didn’t really understand what this entails. Hiding a message in the digits of the base ten expansion of pi is no different from hiding a message in the digits of the base ten expansion of seventeen. (“Mathematicians will tell you that it’s all zeros after the decimal point, but how far out how they really checked?”) I know that the mathematician and science-fiction novelist Greg Egan plays thought experiments with the mutability of math, but I always get the sense with him that at least a fraction of his tongue is in his cheek when he does this, whereas I think Sagan was invoking mathematical ideas in an attempt to convey a sense of transcendence, much as Arthur C. Clarke had done in _2001: A Space Odyssey_ (when the newly transhuman Bowman chides his recently human self for failing to imagine that the numerical pattern governing the proportions of the monolith were limited to three dimensions). Trying to convey something beyond human comprehension is a tricky business; it’s akin to designing good technobabble, but harder. Jim Propp

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