November 2017 - 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] 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] Unlucky thirteen
by James Propp 24 Mar '18

24 Mar '18

#1. Puzzle: Show that for every positive integer n less than thirteen, there exists a Laurent polynomial p(x) with coefficients in {0,1,2,3,4} satisfying p(1/2) = p(1/3) = n. (Hint: Use an iterative construction n -> n+1.) #2. Problem: Show that this NOT the case for thirteen or any larger positive integer. (I don't know how to prove this.) If #1 seems tricky, here's a simpler version: Show that for every positive integer n there exists a Laurent polynomial p(x) with coefficients in {0,1,2} satisfying p(1) = p(2) = n. Equivalently, show that every positive integer n can be written as the sum of n powers of 2, each of which may appear at most twice (where "power of 2" means "2^k for some k in Z"). Jim Propp

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[math-fun] Unconditionally stable nesting coffee cups?
by Henry Baker 22 Feb '18

22 Feb '18

I don't know the answer to the following question, but it would be fun to come up with one. "Mathematicians turn coffee into theorems." -- unknown ?? I drink a lot of coffee, but when I shuffle papers on my desk, the coffee cup turns over & spills coffee all over the papers. Therefore, I'd love a cup design that is strongly resistant to turning over. This implies that the base should be larger than the top, and that the center of gravity -- even when full of coffee -- is very low. I also don't have a lot of room for storing coffee cups, so I'd still like them to _stack_ in such a way that they _nest_ & therefore don't occupy as much volume when stored. Finally, being a human, I'd still like to have a handle through which a finger could fit. The handle needn't touch in two places, so topologically it could be simply-connected. I envision some sort of "star-like" shape which will nest when the second cup is rotated 2pi/N. Like I said, I don't have a candidate shape.

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Re: [math-fun] RMS paradox
by Keith F. Lynch 12 Dec '17

12 Dec '17

Eugene Salamin <gene_salamin(a)yahoo.com> wrote: > Actually, the conjecture for sums of multiple sine waves is false. Oops, you're right. When it comes to the sum of sine waves, I'm more of a radio and electronics guy than a mathematician. As was perhaps evident from my using the term "RMS" rather than "quadratic mean." > What is true is that the mean square MS = RMS^2 of the sum equals > the sum of the mean squares of the component sine waves, with the > proviso that the components are orthogonal. Yes, that's what I was probably thinking of. The first time I learned that power was proportional to the square of the amplitude, that struck me as very strange. Two identical radio transmitters emit four times the power of either of them alone? Eventually, I realized that a similar principle applies in lots of contexts. For instance: * In quantum mechanics, the probability of finding a particle at a given point is proportional to the square of the absolute magnitude of the complex amplitude there. (Or, equivalently, the product of the amplitude with its complex conjugate.) * The power in a sound wave is proportional to the square of the RMS overpressure. * The power in a water wave is proportional to the square of the RMS height of the wave. * The power of a gravitational wave is proportional to the square of the RMS strain (i.e. of the degree to which space is stretched or squished). * The energy in an electric or magnetic field is proportional to the square of the field strength. * Bosons tend to be in the same state as their neighbors because we're adding amplitudes, then measuring the squares of those amplitudes. * The work it takes to shovel snow is proportional to the square of the snow depth. (X times the depth of snow means X times as much snow per unit area, and also means you have to lift it X times as high to get it above the surrounding snow.) * The work it takes to dig a hole is proportional to the square of the depth of the hole. My intended solution to the RMS paradox was that as you add more and more sine waves to make a square wave, the corners continue to have spikes that rise above the flat part. Adding more terms will make those spikes narrower, but never make them shorter. At the limit, they're infinitely narrow but still no less tall than they ever were. How about the radio paradox I alluded to above? How can two identical transmitters radiating the same frequency and phase radiate four times the power as either one alone? One answer is that the antennas are generally in different places, so the waves are out of phase in some places and in phase in others. But what if the antennas are an infinitesimal distance apart, e.g. two identical parallel dipoles separated by much less than one wavelength? Then the waves would be in phase everywhere. Brent Meeker <meekerdb(a)verizon.net> wrote: > If I add two sine waves of equal amplitude but with 180deg > phase difference, I get zero - clearly not an RMS of the > amplitude/sqrt(2). Actually, my mistaken claim does work there, since the peak amplitude of the sum of those two waves is zero, and if you divide that by the square root of 2, you get zero, which is indeed the RMS.

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