August 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] 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] square root approximation?
by Stuart Anderson 07 Nov '18

07 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] FromContinuedFraction@{0, 1, 1, 2, 2, 3, 3, 4, 4, . . .}
by Bill Gosper 15 Sep '18

15 Sep '18

Finch, p424 (Euler-Gompertz Constant): "No one knows the exact outcome if we instead repeat each denominator in c1 [ := FromContinuedFraction@{0, 1, 2, 3, 4,. . .} = I_1(2)/I_0(2)] ~ .6977746579640] although numerically we find that c2 = .5851972651050... ". Using my mysteriously wrong-but-correctible matrix-to-CF (non-regular) converter, c2= 1/2 + 1/(12 + 2*ContinuedFractionK[1 - k^2, 1 + 3*k + 2*k^2 + k^3, {k, 2, ∞}]). Julian & I had a way to further convert these to log-derivatives of pFqs, if anyone is really interested. --rwg

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[math-fun] Oligomorphic polytopes
by Allan Wechsler 10 Sep '18

10 Sep '18

I've gotten interested in enumerating symmetric convex polytopes in various dimensions, intrigued by the fact that in dimensions as low as 5, we don't yet have a proven-complete census of the uniform polytopes. Every polytope in N dimensions has a single N-cell, and then a graded collection of cells of dimensions from N-1 down to 0 (the vertices). The symmetry group of a polytope divides this collection of cells into transitivity classes, and I propose to focus on polytopes with a small number of classes of cells. Since symmetries never change the dimensionality of a cell, the number of classes is always at least one more than the dimension of the polytope as a whole. I've chosen to apply the term "oligomorphic" ("few-forms") to a polytope that is interesting because its cells fall into a small number of symmetry classes. The only monomorphic polytope is the trivial 0-dimensional isolated vertex. The only bimorphic polytope is a line segment; its two vertices clearly form one symmetry class, while the body of the segment provides the other. We can say it has a "profile" of 1:1 because it has 1 kind of 1-cell and 1 kind of 0-cell. The regular polygons form an infinite call of trimorphic polytopes with profile 1:1:1 (1 kind of 2-cell, 1 kind of 1-cell, 1 kind of 0-cell). There are no other trimorphic polytopes. When we come to consider tetramorphic polytopes, there are the 5 classical Platonic solids in 3 dimensions, with profile 1:1:1:1, but there are also some interesting 2-dimensional examples. A non-square rectangle and a non-square rhombus are examples of trimorphic polygons with signatures 1:2:1 and 1:1:2, respectively. There are similar examples with any even number of edges. (1:1:2 means 1 kind of 2-cell, 1 kind of 1-cell, and 2 kinds of 0-cells.) Up until this point I am pretty sure that I have enumerated all the possibilities. But when we get to pentamorphic polytopes, I quickly start to lose confidence. Here is my list, but except in four dimensions I do not know if it is complete: In four dimensions there are the six "classical" regular polychora, profile 1:1:1:1:1. In three dimensions the only ones I have been able to think of are: the cuboctohedron and the icosidodecahedron, with profile 1:2:1:1 (because they have two kinds of faces); their duals, the rhombic dodecahedron and the rhombic triacontahedron, 1:1:1:2 (because they have two kinds of vertices); and the "isoceles" tetrahedron, 1:1:2:1 (with two kinds of edge). Note that the duality operation reflects the profile, except for the boilerplate leading 1. The dual of an isoceles tetrahedron is a different isoceles tetrahedron. (In combinatorial algebraic topology, they often add on a single "-1-cell" with which every vertex is presumed to be incident, just to complete this kind of duality symmetry.) In two dimensions there is an interesting little menagerie of pentamorphic polygons. The simplest is the isoceles triangle, with two kinds of edges and two kinds of vertices, profile 1:2:2. This can be generalized into a family of polygons with 3k sides, with a vertex-edge pattern (AaAbBb)^k. (I hope it is clear what my silly notation means here). There is another family, of which the (non-rhombic) parallelogram is the simplest exemplar. Here, bigger examples have 2k sides, with a vertex-edge pattern (AaBb)^k. The profiles 1:1:3 and 1:3:1 are clearly impossible, since a single class of edge cannot be incident with three classes of vertex, and vice versa. It would absolutely not surprise me if someone were to come up with pentamorphic examples in 2 and 3 dimensions that were not covered by my sketched catalog above. Regarding hexamorphic polytopes, I have no confidence whatsoever that I have managed to exhaust all the possibilities, except that we know there are only three hexamorphic 5-dimensional polytopes. The isoceles trapezoids and their dual kites are 2-dimensional examples; right prisms and their dual right bipyramids are 3-dimensional ones; I cannot produce a single 4-dimensional one. Can anyone help complete the enumeration of hexamorphic types? The snub cube is octomorphic, with profile 1:3:3:1.

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