September 2012 - mathfun@mailman.xmission.com

[math-fun] inelastic vs other probability models of dice
by Warren Smith 22 Sep '12

22 Sep '12

I had my own debatable models in mind. One was the "ergodic plus slow energy loss" model. In this model, all orientation and rotation-axis (uniform on sphere) states are equally likely at any time, and energy (rotational kinetic plus gravitational potential) is assumed to be lost gradually and very slowly. At some moment it will happen that the dice becomes trapped in a local energy basin which it can thereafter never escape from because the energy keeps decreasing after the moment energy falls below the barrier height. Then only orientations within the current basin are permitted (uniform within) -- well more generally we only permit orientations below the current energy. So then there is a "landscape" of gravitational energy as function of orientation, and the basin-measures correspond to the probabilities we are seeking. It actually is not as nasty as all that may sound because for a convex-polyhedron-die, the barrier heights are easy to work out (they arise from polyhedron edges...) and the basins correspond to certain polygonal regions drawn on the surface of the sphere of orientations using circular arcs as polygon edges. In full generality life could be complicated, but for a symmetrical pyramid die, which was what I originally had in mind, it appears it really is within reach to work it out in full in this model.

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Re: [math-fun] freely spinning body
by Bill Gosper 22 Sep '12

22 Sep '12

DanA> I was not able to access Gene's writeup since I got a 404 error from the URL below, so can't compare what he wrote to the above. Gene? --Dan GMail ior your mail reader got faked out by the blanks in the URL. If the following fails, try "physically" pasting into the top browser pane. http://gosper.org/Rigid <http://gosper.org/Rigid%20Body%20(1).pdf> http://gosper.org/Rigid Body (1).pdf --rwg

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[math-fun] freely spinning body
by Bill Gosper 22 Sep '12

22 Sep '12

DanA> Yeah, I forgot. There's the Jacbobi sn, cn, dn, giving the solutions to the ideal rotations on the surface of an ellipsoid. (Showing why you can't flip a book in midair, halfway around its middle axis.) Hmm, but how would you assign probabilities to the initial conditions? --Dan On 2012-09-21, at 6:13 AM, Veit Elser wrote: > On Sep 21, 2012, at 12:32 AM, meekerdb <meekerdb(a)verizon.net <http://gosper.org/webmail/src/compose.php?send_to=meekerdb%40verizon.net>>> wrote:> >> On 9/20/2012 9:02 PM, Dan Asimov wrote:>>> Back in H.S. I wondered about this for any polyhedron, and guessed a face's probability of landing down is proportional to the solid angle it subtends from the center of gravity.>> >> And whether the projection of the CG onto the plane of the face falls within the face. :-)>> >> Brent> That's close to my physical model. It came up in connection with estimating the probability of a coin landing on its side. I had my students> design a cylindrical dice that has equal odds for heads, tails and "side". But because this was a physics course we did not make the> ad hoc assumption that the rotation group is sampled uniformly in a toss. Instead we used the coin tossing model where the cylinder axis rotates rapidly about a horizontal axis. Non-uniform sampling is especially valid for symmetric objects, where the three moments of inertia are equal and there is no free precession. So I don't find Michael Kleber's "fairness by symmetry" very compelling. I bet if one "rolled" a cubic dice more like the standard coin toss one could strongly bias the outcome (against the faces intersected by the rotation axis). I've always wanted an ISS astronaut to demonstrate this with some sort of brick with a little dimple in the center of each face, held between pencil points, then spun (with breath?) and released. Unfortunately, the "breath" part reminds us that the experiment is tainted by aerodynamics. Dan, it sounds like you have a fairly simple geometric argument. Several years ago, Gene wrote up a fairly complicated physics argument <http://gosper.org/Rigid Body (1).pdf>. Does this mean you can simplify it? --rwg

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[math-fun] Dice problem continued
by Warren Smith 22 Sep '12

22 Sep '12

I like Wechsler's solution better than my planned one: A long regular N-sided right prism topped with short pyramids; only the rectangle faces of the prism, not the triangle faces on the pyramids, are stable resting positions. Nice and elegant. [I was planning on an (N-1)-sided pyramid with the exact right value for the height/diam ratio to make the land-on-base probability come out equal to the land-on-each-other-face probability...] re Adam Goucher: rhombic hexecontahedron is a polyhedron such that the projection of the centroid to any face lies outside that face. that's a nonconvex polyhedron. For a convex one, any face whose plane has minimum distance to the center of mass, is always a stable resting face, and it is never stable to rest on an edge or corner not a face. This all also is true for polyhedra of nonuniform internal composition.

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[math-fun] polyhedral dice problem
by Warren Smith 21 Sep '12

21 Sep '12

Well first of all, if the number N>4 of faces is even, you can use a doubled-version of an (N/2)-sided pyramid [glue two of them together at the common (N/2)-gon base face] to get a dice which clearly has exact uniform probabilities because each face is the same under a symmetry as each other. Second, re Veit's 2-sided coin problem, the easier and probably better solution for it is to make a non-polyhedral smooth-surfaced convex "coin" got by, e.g. rotating about the y axis, the curve got from a circle of radius=1 centered at (9,0) and horizontal line segments at |y|=1 with |x|<9. The resulting object will have exactly two stable positions on a tabletop. However, when N is odd such as a 5-sided dice, it is less obvious how to proceed. I have solutions in mind, but I'll keep them secret for a while to get you suitably frustrated, plus their validity is debatable anyhow.

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[math-fun] Huh?
by Emma Cohen 21 Sep '12

21 Sep '12

Can someone explain the results here? http://www.wolframalpha.com/input/?i=n%5E4+%2B+n%281%2F3%29 I can come up with a hypothesis which explains some of the results, but can't think of anything at all which would explain the "alternate form." Any ideas? --Emma Cohen

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[math-fun] pizza theorems
by James Propp 21 Sep '12

21 Sep '12

In the past decade there's been a fair bit of attention paid to the "pizza theorem" and its generalizations. The basic pizza theorem says that if you choose an arbitrary point P in the interior of a disk, draw 4 lines through P that meet at 45 degree angles (by which I mean "45 degrees and multiples thereof", for those of a pedantic turn of mind), and color the 8 resulting wedges alternately black and white, the total area of the black wedges equals the total area of the white edges. I mentioned the result to Pete Winkler, and he quickly came up with an approach that allows one to give a purely mental proof of the result. I'm wondering if the proof is new. I'll give people a chance to come up with it on their own before giving hints or giving details. Of course, reasonable people may differ about what counts as an argument one can construct "in one's head". In both cases, there was an algebraic calculation that needed to be done, which I was able to do in my head with some slight effort. In the course of finding the proof, I inadvertently proved a slightly different result as well, in which the pizza is divided into 8 pieces in a different fashion, namely, by joining the (arbitrary internal) point P to 8 equally-spaced points on the boundary of the disk. In both cases, the method of proof automatically establishes a stronger result, in which you divide the pizza into 4n pieces and divide the pieces among n people in cyclic fashion, so that each of the n people gets 4 slices whose areas turn out to sum to 1/n times the area of the pizza. I'm hoping that people who know the pizza-problem literature will chime in at some point, after the non-initiates have had a chance to mull over the problem on their own over the next couple of days. Jim Propp

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[math-fun] News for Dice Men
by Alex Bellos 20 Sep '12

20 Sep '12

Anyone able to work out how many unique solutions there are to the Go First dice problem? http://www.guardian.co.uk/science/alexs-adventures-in-numberland/2012/sep/1…

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[math-fun] π iteration, again(?)
by Bill Gosper 19 Sep '12

19 Sep '12

I'm pretty sure I sent this before, but since it took me days to dig it back up, here it is again, that quadratically convergent π iteration using "Newton's method for √1". (= Coth double angle formula.) There's also a √ in the loop, but you need only four of them for 34 digits. You also need ln 2 and ten assorted powers of 2: In[323]:= sum[k_, m_] := Sum[HoldForm @@ {1/2^(k*n*(n - 1)/2)}, {n, m}]^2*k/2^(k/4) Note these sums represent trivial effort--just sprinkling m 1s into a string of 0s. In[326]:= sum[4, 7]/sum[8, 5] Out[326]= 1 1 1 1 Power[-------------------------- + ------------------- + ------------- + -------- + 19342813113834066795298816 1152921504606846976 1099511627776 16777216 1 1 ---- + -- + 1, 2] / 4096 16 1 1 1 1 2 (------------------------- + --------------- + -------- + --- + 1) 1208925819614629174706176 281474976710656 16777216 256 IIn[363]:= N[ReleaseHold[%326], 34] Out[363]= 1.120652900418845687263888977941327 In[364]:= Sqrt[(% + 1/%)/2] Out[364]= 1.0032422086165441482848802264099015 In[365]:= Sqrt[(% + 1/%)/2] Out[365]= 1.0000026194828265385324048741581395 In[366]:= Sqrt[(% + 1/%)/2] Out[366]= 1.0000000000017154180761229126907979 In[367]:= Sqrt[(% + 1/%)/2] Out[367]= 1.00000000000000000000000073566479397 In[368]:= %*%%*%%%*%%%%*ReleaseHold[Numerator[%326]]*Log[16] Out[368]= 3.141592653589793238462643383279502 With three more powers of 2, but no more √ s, we get ~49 digits: In[381]:= sum[4, 9]/sum[8, 6] In[382]:= N[ReleaseHold[%], 50] In[383;;386]:= Sqrt[(% + 1/%)/2] In[387]:= π-Times @@ (Out /@ Range[383, 386])*ReleaseHold[Numerator[%381]]*Log[16] Out[387]= 4.*10^-49 But preoccupation with minimizing √ s is anachronistic. They should be no costlier than division. So keeping the sum lengths 7 and 5, but doubling k, we get ~67 digits with six √ s: In[370]:= sum[8, 7]/sum[16, 5] In[371]:= N[ReleaseHold[%], 69] Out[371]= \2.01559424540388920698155253866341100838034807969448027057273589434627 In[372;;377]:= Sqrt[(% + 1/%)/2] Out[377]= 1.0000000000000000000000000000000000000000000000001353006722721156829769 In[379]:= Times @@ (Out /@ Range[372, 377])*ReleaseHold[Numerator[%370]]*Log[4] Out[379]= 3.14159265358979323846264338327950288419716939937510582097494459230751 In[380]:= % - π Out[380]= -3.1*10^-67 --rwg

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[math-fun] proof of ABC conjecture?
by Warren Smith 19 Sep '12

19 Sep '12

If this is true, then definitely Mochizuki deserves the Fields Medal, the works. I have no evidence it is true, though. His "inter-universal" work is far outside MY universe, I pretty much can't understand a word. And I doubt many number theorists can either. It doesn't look like number theory normally looks. So I actually had sent a little tip to the NY Times science section saying "possible big story re ABC." Then I actually got a reply, basically saying "Nah, we at the NY Times will be highly interested in ABC, so thanks for the tip, *but* only when proof is confirmed, we don't report on maybe-proofs, we're not that kind of low-class paper." But, lo and behold, in today's science section there is a story on ABC by K.Chang. -- Warren D. Smith

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