September 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] Branko Grünbaum (1929-2018)
by James Buddenhagen 30 Sep '18

30 Sep '18

Perhaps already mentioned on this list, but I don't remember it. https://math.washington.edu/news/2018/09/18/branko-grunbaum-1929-2018

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Re: [math-fun] Curve-fitting methods ?
by Keith F. Lynch 28 Sep '18

28 Sep '18

Brent Meeker <meekerdb(a)verizon.net> wrote: > Keith F. Lynch wrote: >> But any three points that obey the triangle inequality are >> compatible with any radius of curvature whatsoever. (Except for a >> radius less than that which makes them coplanar with the center of >> the circle they're on.) > No they aren't. The spherical law of cosines is > cos a = cos b cos c + sin b sin c cos A > where lower case are edges and upper case are interior angles. I know. I should have said any three *sides* that obey the triangle inequality. For one thing, speaking of three *points* that obey the triangle inequality makes little sense, as no three points can fail to do so. Suppose all you have are the lengths of three sides of a triangle, and the knowledge that the three vertices of the triangle are equidistant from some "center" point in space, possibly a point at infinity, possibly a point in the same plane with the triangle. You don't know how curved the sides are, except that they all have the same radius of curvature, the curvature such that every point on all the sides is also equidistant from that same center point. Wouldn't you agree that with the information you have, you can't rule out any radius above a certain minimum? I'm not a visual thinker, so I won't guess at whether the locus of possible center points is a line or some other kind of curve. If the three sides of the triangle were equal, I assume it would have to be a line, by symmetry, unless it splits into three curves. > So you guess some radius R and solve for A and by permutation also B > and C. Then you see whether the law of sines is satisfied. > sin A/sin a = sin B/ sin b = sin C/sin c > In general it won't be but one side, say (sin a/sin b)? depends on > R while the other side (sin A/sin B) doesn't. So you adjust R and > iterate. > I guess I should try to this see if you really works. :-) > I tried it. Doesn't work in a practical sense as written...just not > enough resolution. Maybe I can fix it. Again, I don't think you have enough information. The law of sines should *always* be satisfied, unless you choose a radius less than that of the circle the three points are on. This is complicated by the fact that you can't assign any definite coordinates for the points or any straight-line distances between them until you know the radius. Any three points in space are on a unique plane and are on a unique circle in that plane. They're also on an infinite number of spheres, with all radii starting with that of the circle. Again, this is complicated by the fact that our points aren't fixed, but move around as the fixed-length sides change curvature, but I think the principle is the same. For extra credit, extend the scenario to negative curvature. See https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines

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