January 2016 - 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] 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] 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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[math-fun] muffin problem -- clusters and bounds
by Huddleston, Scott 01 Dec '17

01 Dec '17

I enclose examples of some of the bounds in my 28 Feb. post, and a derivation of the most involved example. S(12/7) = 3/7 = r/4 = (3-r)/3 Muffins: 1/7 * { [3,4]*12 } Persons: 1/7 * { [4*3]*4, [3*4]*3 } S(7/4) = 5/12 = (3-r)/3 = (2r-1)/6 Muffins: 1/12 * { [5,7]*6, [6*2] } Persons: 1/12 * { [5*3,6]*2, [7*3]*2 } S(40/23) = 19/46 = (2r-1)/6 = (12-5r)/8 Muffins: 1/46 * { [19,27]*24, [20,26]*12, [23*2]*4 } Persons: 1/46 * { [19*3,23]*8, [20*4]*3, [27*2,26]*12 } S(228/131) = 54/131 = (12-5r)/8 = (7r-6)/15 Muffins: 1/131 * { [54,77]*120, [55,76]*24, [57,74]*60, [65,66]*24 } Persons: 1/131 * { [54*3,66]*24, [54*2,55,65]*24, [57*4]*15, [77*2,74]*60, [76*3]*8 } S(449/258) = 319/774 = (7r-6)/15 = (73-32r)/42 Muffins: 1/774 * { [319,455]*210, [320,454]*30, [325,449]*42, [336,438]*30, [337,437]*90, [384,398]*42, [387*2]*5 } Persons: 1/774 * { [319*3,390]*42, [319*2,325,384]*42, [320*3,387]*10, [336,337*3]*30, [455*2,437]*90, [455,454,438]*30, [449*3]*14 } Here's a derivation of 449/258 example. 449 muffins -> 898 pieces, used in 124 4-parts with 496 pieces and 134 3-parts with 402 pieces. We split muffins between a 3-part and a 4-part when possible, but 4-parts contain 94 more pieces than 3-parts. We divide these 94 pieces (47 muffins) as evenly as possible among the 124 4-parts. 47 pairs of 4-parts contain both halves of a muffin, which I'll call an A^1 cluster and write A^1 = [a1*3,m]*2 where m denotes the two muffin halves and the other pieces are denoted a1, indicating a set of pieces of possibly different sizes that occure in A^1 clusters. 34 4-parts contain only muffin pieces whose mates appear in 3-parts. These I call A^0 clusters and write A^0 = [a0*4] The 449/258 example uses 30 A^0 clusters ([a0*4])*30 (120 a0 pieces, 30 4-parts) 47 A^1 clusters ([a1*3,m]*2)*47 (282 a1 pieces, 94 4-parts) Note that the A^1 cluster by itself sets an upper bound on S -- the cluster is 2 persons (size 2r), it cntains 1 complete muffin (size 1, pieces m*2), and it has 6 unmatched muffin pieces (a1*6), giving S(449/258) <= avg(a1) = (2r-1)/6. A^0 clusters also bound S by S(449/258) <= avg(a0) = r/4, but the A^1 cluster bound is tighter. Note that this (2r-1)/6 bound holds for examples 7/4 and 40/23. The muffin mates of a0 pieces (resp. a1 pieces) are z0 pieces (resp. z1 pieces). We distribute z0 and z1 pieces as evenly as possible across the 134 3-parts (402 pieces). With 282 z1 pieces, at least 14 3-parts must consist of only z1 pieces. These are Z^0 clusters, Z^0 = [z1*3] = [z10*3] where I refine piece labels so that z1 pieces appearing in Z^0 clusters are labelled z10. I create general Z^k clusters by finding minimal sets of persons that contain complete a0/z0 muffins. Maximally even distribution of the 120 z0 pieces places one per 3-part. Then a Z^1 cluster, closed under a0/z0 muffins, is Z^1 = [a0*4],[z0,z1*2]*4 = A^0,[z0,z11*2]*4 where z1 pieces appearing in Z^1 clusters are relabelled z11. Z^0 clusters bound S by S(449/258) <= avg(a10) = 1-avg(z10) = 1-r/3 = (3-r)/3 and Z^1 clusters provide an even tighter bound S(449/258) <= avg(a11) = 1-avg(z11) = 1-(5r-4)/8 = (12-5r)/8. The 449/258 example uses 14 Z^0 clusters = ([z10*3])*14 ( 42 z10 pieces, 14 3-parts) 30 Z^1 clusters = (A^0,[z0,z11*2]*4)*30 (240 z11 pieces, 120 3-parts) As before, we refine the labelling of a1 pieces into a10 and a11 pieces, according to z10 or z11 complements. The next level of clustering is constructed by taking closure under a10/z10 muffins. The 282 a1 pieces in the 47 A^1 clusters are divided into 240 a11 pieces and 42 a10 pieces. Thus at least 5 A^1 clusters must contain only a11 pieces, and with maximally even distribution we have 42 A^1 clusters with 1 a10 piece and 5 a11 pieces. Continuing to refine pieces according to their appearance in A^10 or A^11 clusters, we have A^10 = A^1 = {a110*6, ...} (6 a110 pieces and some other pieces) A^11 = Z^0, (A^1)*3 = (a111*15, ...) with 449/258 consisting of 5 A^10 clusters ( 5 A^1 clusters, 30 a110 pieces) 14 A^11 clusters (42 A^1 clusters, 252 a111 pieces) A^11 clusters provide bound S(449/258) <= avg(a111) = (7r-6)/15 The 449/258 example lies on the intersection of this bound and the next level cluster bound. The next cluster level takes a110/z110 muffin closure and induces clusters Z^10 = Z^1 Z^11 = A^10, (Z^1)*6 with 449/258 consisting of 0 Z^10 clusters 5 Z^11 clusters (i.e., only Z^11 clusters) and bound S(449/258) <= avg(a1111) = 1-avg(z1111) = (73-32r)/42. I conclude with some computer generated additional information on these bounds. If anyone sort of understands this post and would like to use my program as is (i.e., no manual and documentation is mostly my posts), drop me a note and I'll send the C++ source (300 lines). B<4>(r) = (1r-0)/4; S'(r) <= B<4>(r) on (3/2,2/1] tight bound on (3/2,8/5] and at r = (4+8k)/(2+5k) (3/2,8/5] contains mediant(3/2,8/5) = 11/7 S'( (3/2,8/5] ) = (3/8,2/5]; S'(11/7) = 11/28 B<>(r) = (1-0r)/2 B<4>(r) = B<>(r) at r = 2/1 = 2.0000000 S'(2/1) = 1/2 = 0.50000000 B<4,1>(r) = (3-1r)/3; S'(r) <= B<4,1>(r) on [12/7,2/1) tight bound on [9/5,2/1) and at r = (12+9k)/(7+5k) [9/5,2/1) contains mediant(9/5,2/1) = 11/6 S'( [9/5,2/1) ) = [2/5,1/3); S'(11/6) = 7/18 B<4>(r) = (1r-0)/4 B<4,1>(r) = B<4>(r) at r = 12/7 = 1.7142857 S'(12/7) = 3/7 = 0.42857143 B<4,1,1>(r) = (2r-1)/6; S'(r) <= B<4,1,1>(r) on (12/7,7/4] tight bound on (12/7,59/34] and at r = (21+59k)/(12+34k) (12/7,59/34] contains mediant(12/7,7/4) = 19/11 S'( (12/7,59/34] ) = (17/42,7/17]; S'(19/11) = 9/22 B<4,1>(r) = (3-1r)/3 B<4,1,1>(r) = B<4,1>(r) at r = 7/4 = 1.7500000 S'(7/4) = 5/12 = 0.41666667 B<4,1,1,1>(r) = (12-5r)/8; S'(r) <= B<4,1,1,1>(r) on [40/23,7/4) tight bound on [148/85,7/4) and at r = (80+148k)/(46+85k) [148/85,7/4) contains mediant(47/27,7/4) = 54/31 S'( [148/85,7/4) ) = [7/17,13/32); S'(54/31) = 51/124 B<4,1,1>(r) = (2r-1)/6 B<4,1,1,1>(r) = B<4,1,1>(r) at r = 40/23 = 1.7391304 S'(40/23) = 19/46 = 0.41304348 B<4,1,1,1,1>(r) = (7r-6)/15; S'(r) <= B<4,1,1,1,1>(r) on (40/23,228/131] tight bound on (40/23,1119/643] and at r = (228+1119k)/(131+643k) (40/23,1119/643] contains mediant(40/23,47/27) = 87/50 S'( (40/23,1119/643] ) = (142/345,265/643]; S'(87/50) = 103/250 B<4,1,1,1>(r) = (12-5r)/8 B<4,1,1,1,1>(r) = B<4,1,1,1>(r) at r = 228/131 = 1.7404580 S'(228/131) = 54/131 = 0.41221374 B<4,1,1,1,1,1>(r) = (73-32r)/42; S'(r) <= B<4,1,1,1,1,1>(r) on [449/258,228/131) tight bound on [9349/5372,228/131) and at r = (1347+9349k)/(774+5372k) [9349/5372,228/131) contains mediant(134/77,47/27) = 181/104 S'( [9349/5372,228/131) ) = [1107/2686,2267/5502); S'(181/104) = 75/182 B<4,1,1,1,1>(r) = (7r-6)/15 B<4,1,1,1,1,1>(r) = B<4,1,1,1,1>(r) at r = 449/258 = 1.7403101 S'(449/258) = 319/774 = 0.41214470 B<4,1,1,1,1,1,1>(r) = (61r-60)/112; S'(r) <= B<4,1,1,1,1,1,1>(r) on (449/258,764/439] tight bound on (449/258,149516/85913] and at r = (10696+149516k)/(6146+85913k) (449/258,149516/85913] contains mediant(449/258,764/439) = 1213/697 S'( (449/258,149516/85913] ) = (11909/28896,35408/85913]; S'(1213/697) = 32173/78064 B<4,1,1,1,1,1>(r) = (73-32r)/42 B<4,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1>(r) at r = 764/439 = 1.7403189 S'(764/439) = 2533/6146 = 0.41213798 B<4,1,1,1,1,1,1,1>(r) = (1101-487r)/615; S'(r) <= B<4,1,1,1,1,1,1,1>(r) on [160212/92059,764/439) tight bound on [1313469/754729,764/439) and at r = (160212+6567345k)/(92059+3773645k) [1313469/754729,764/439) contains mediant(19549/11233,764/439) = 20313/11672 S'( [1313469/754729,764/439) ) = [1555262/3773645,111271/269985); S'(20313/11672) = 986147/2392760 B<4,1,1,1,1,1,1>(r) = (61r-60)/112 B<4,1,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1,1>(r) at r = 160212/92059 = 1.7403187 S'(160212/92059) = 37941/92059 = 0.41213787

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[math-fun] Big Splat theory
by James Propp 04 Sep '17

04 Sep '17

Has anyone ever proposed in the Annals of Improbable Research or a similar venue that, in addition to undergoing periodic reversals of its magnetic field, Earth is also subject to occasional short-lasting reversals of its gravitational field? Such a phenomenon would handily explain selective extinction of megafauna: a mouse could survive an upward fall for five seconds followed by a downward fall for five seconds followed by an extreme deceleration, but a mastodon would not fare as well. :-) Jim Propp

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[math-fun] Newman = -1/Oldman
by Bill Gosper 30 Mar '17

30 Mar '17

This must have been found by Newman himself but I only just noticed. Enumerate the nonnegative rationals by iterating "Newman's function": In[96]:= NestList[1/(2*Floor[#] + 1 - #) &, 0, 9] Out[96]= {0, 1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3} "Oldman's function" runs the sequence back to 0, counting the rationals: In[97]:= NestList[2*Ceiling[1/#] - 1 - 1/# &, 4/3, 9] Out[97]= {4/3, 1/4, 3, 2/3, 3/2, 1/3, 2, 1/2, 1, 0} E.g., 0 is the zeroth rational, 4/3 is the ninth. Oldman is completely unnecessary! Just negate and reciprocate the last value, and Newman runs backwards: In[100]:= NestList[1/(2*Floor[#] + 1 - #) &, -3/4, 9] Out[100]= {-3/4, -4, -1/3, -3/2, -2/3, -3, -1/2, -2, -1, ComplexInfinity} (with a much louder completion announcement). Or if you prefer, Oldman runs the negative rationals forwards: In[99]:= NestList[2*Ceiling[1/#] - 1 - 1/# &, -1, 9] Out[99]= {-1, -2, -1/2, -3, -2/3, -3/2, -1/3, -4, -3/4, -5/3} obviating Newman. --rwg

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[math-fun] Panel puzzle
by James Propp 18 Mar '17

18 Mar '17

You can join four isosceles right triangles along hinges (I use tape) to form a square, and then join four such square panels (again with tape) to form a long rectangle, and lastly join the first and last panels together with tape to form a closed band. If you color the two sides of the band with different colors, it's a fun puzzle to turn the band inside out, exchanging the colors. I learned about this in the 80s, and amused myself by learning how to do the moves behind my back. Does anyone know the name and/or provenance of this puzzle? Jim

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[math-fun] A problem about finite sets
by Victor Miller 10 Sep '16

10 Sep '16

Say that a finite set of integers, A, has *property P* is there are two non-empty disjoint subsets S and T of A, such that sum(S) = sum(T) (where sum(X) is the sum of the elements of X). 1) Show that every 10 element subset of {1,...,100} has property P (this is fairly easy). 2) Is it true that every 9 element subset of {1,...,100} has property P? Victor

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