June 2018 - mathfun@mailman.xmission.com

[math-fun] Rolling polyhedra
by James Propp 25 Dec '20

25 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] One parameter quintic solution
by Bill Gosper 31 Oct '20

31 Oct '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 21 Oct '19

21 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] square root approximation?
by Stuart Anderson 06 Nov '18

06 Nov '18

Is this a known method? Not particularly accurate though https://mystudentvoices.com/how-an-old-mans-ravings-awakened-the-mathematic…

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[math-fun] When is base 2 not binary?
by Bill Gosper 01 Aug '18

01 Aug '18

When the digits are 0 and 1 and the other two cuberoots of 1. See gosper.org/gaskettalk.pdf pp 11 and 13. E.g., i/√3 = red.cyangreencyangreencyangreen... = cyan.greenmagentagreenmagentagreen... . Presumably we can expect a lot of these multiple representations when the unit patch has a fractal boundary. PUZZLE: Here of dimension what? We need |base|² digits to cover a positive area in the complex plane. The "France fractal" bounds the unit patch of the base 2+i^(2/3) with digits zero and the 6th roots of 1. |2+i^(2/3)| = √7. Base 2+i^(2/3) with three digits moved outward gosper.org/base2+w.bmp has a unit patch with a boundary of large dimension. (PUZZLE: D=?) --rwg

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[math-fun] Redundant trinomials --- are they really necessary?
by Fred Lunnon 08 Jul '18

08 Jul '18

I need an existence proof --- it's the missing link in a theorem establishing full-plane integer zero-free number walls over finite characteristic ( |F_2 excepted). CONJECTURE: Given any r > 1 , there exists over |F_2 a trinomial f(X) = X^s + X^t + 1 = g(X) h(X) , where g(X) is irreducible of degree r . This topic turns out to have been studied previously, since it is of interest in elliptic curve public-key cryptography. Such "redundant trinomials" are used to do arithmetic in binary finite fields |F^(2^r) lacking a trinomial generating polynomial: the idea is to calculate instead in a larger ring using trinomial f(X) with factor some field generator g(X) , afterwards reducing the result to the field. Of course, engineers don't care greatly about proving that such a trinomial exists for every field degree: they're just content that some have been found for every degree up to 10,000, comfortably exceeding anything currently required. I haven't uncovered a great deal about this in the literature: Christophe Doche (2005) <doche(a)ics.mq.edu.au> "Redundant Trinomials for Finite Fields of Characteristic 2 " http://web.science.mq.edu.au/~doche/redundant.pdf is cited by most of the rest; Ryul Kim, Wolfram Koepf <koepf(a)mathematik.uni-kassel.de> "Divisibility of Trinomials by Irreducible Polynomials over |F_2 " https://arxiv.org/pdf/1311.1366.pdf includes results about cofactor h(X) , along with baffling typos. If required I can post lists of sample trinomials and cofactors, currently for prime r < 4096 (in 3 hrs), also all r < 1024 (in 12 mins). Fred Lunnon

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