May 2019 - 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] Another pic of Hilbert's spacefill
by Bill Gosper 19 Apr '20

19 Apr '20

to accompany should've been "sampHilbert" <http://www.tweedledum.com/rwg/sampeano.htm> Hilbert's [spacefilling] function is dense with quadruple points, i.e. points in [0,*i*] with four distinct preimage points in [0,1]: gosper.org/hilbquad.png . E.g., the preimage of 1/2 + *i*/4 is {5/48, 7/48, 41/48, 43/48}. This picture connects, in the order they were swept, all the points in [0,*i*] with preimages having denominators ≤ 3⨉4096. (All quadruple point preimages have denominators 3⨉2ⁿ.) E.g., the lower-leftmost vertex is 1/32 + *i*/64, the image of 5/12288, 7/12288, 41/12288, and 43/12288. The white segments are retraced boundary, retroflexed in the middle, (making quadruple points). (Spacefilling functions map closed intervals to closed sets.) The Heighway Dragon is dense with mere triple points. Here's one: Hi res Heighway <http://gosper.org/dragtrip!.png> . —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] Hot tea sucks.
by Bill Gosper 25 Nov '19

25 Nov '19

I brew it in a 46oz pickle jar that easily tips over, and leaks if the lid isn't snug. But if I forget to momentarily loosen the lid within a few minutes, a powerful vacuum develops that makes it virtually impossible to unscrew. What is the trivial solution? —Bill

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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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Re: [math-fun] Sugar + water paradox:
by Bill Gosper 01 Sep '19

01 Sep '19

On Wed, Mar 20, 2019 at 11:37 AM <Shy respondent> wrote: > [just us] > > I think you may just need to wait longer. > Yes. > If the water isn't stirred (either manually or by boiling) > then the liquid near the sugar pile will be highly saturated > and you'll have to wait for random thermal stirring to transport it away > But for several minutes, at least, the amount of dissolved sugar actually declines. On 2019-03-20 12:17, Allan Wechsler wrote: > The obvious question, for me, is how are you measuring this process? I > imagine that you could extract a cc of solution each minute, then evaporate > them and weigh the amount of sugar in each. But I think you're probably not > doing that, but rather relying on some kind of visual measure. What is the > observed dynamic? Does it dissolve for a while until the concentration hits > a plateau, and then sugar starts to precipitate out? Exactly! There's way more sugar in the bottom after it cools for a while. Huge density gradient from lack of stirring. You should have no trouble observing this. Unless I decontrolled it with a teabag of Stash Double Bergamot. —rwg > > On Wed, Mar 20, 2019 at 1:57 PM Bill Gosper <billgosper(a)gmail.com> wrote: > >> Dump a couple of Tbsp of sugar into a quart or two jar of tepid water. >> Don't stir. >> 𝜇wave a couple minutes, hot but not boiling. Let it sit. Experimental >> claim: >> The total dissolved sugar is not an increasing function of time, even >> though a >> couple of Tbsp is << enough to saturate even cold water. >> —rwg > > > > On Mar 20, 2019, at 10:57 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > > > > Dump a couple of Tbsp of sugar into a quart or two jar of tepid water. > > Don't stir. > > 𝜇wave a couple minutes, hot but not boiling. Let it sit. Experimental > > claim: > > The total dissolved sugar is not an increasing function of time, even > > though a > > couple of Tbsp is << enough to saturate even cold water. > > —rwg > > _______________________________________________ > >

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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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