July 2015 - mathfun@mailman.xmission.com

[math-fun] Computational effort
by James Propp 01 Mar '16

01 Mar '16

I don't know what the relevant units of computation are --- floating-point operations are the ones I've heard about most, but what if you're not using floating point? --- but I'm wondering if anyone has tried to quantify "how much computing" took place in (a) code-breaking at Bletchley Park, (b) the Manhattan Project, and (c) the Apollo space program. It seems likely to me that my laptop, running some silly number-crunching problem in Mathematica and returning an answer after six hours that I could probably have figured out in my head if I'd thought a little bit more before jumping in to write some code, has them all beat. What do you think? (And while we're on the subject: What's the canonical example of wasted computational effort? Isn't there some well-known old story of someone who had an early computer calculate the determinant of a singular matrix?) Jim Propp

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[math-fun] San Jose Math Decagon
by Bill Gosper 17 Feb '16

17 Feb '16

A couple of years ago Julian and I showed some animations to the San Jose Math Circle. gosper.org/decazoom.gif suggests why the area of a circle is the radius times half the circumference, and displays the continued fraction of a very lame approximation to π. --rwg

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[math-fun] xkcd
by rcs＠xmission.com 17 Dec '15

17 Dec '15

Today's XKCD cartoon has a nice collection of approximations. http://xkcd.com/1047/ There's the ancient expression 6 pi^5 for the mass ratio of the proton to the electron, and the more accurate (e^8 - 10)/phi. Robert Munafo & REIS are at least partially responsible. Rich

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[math-fun] Julian pimped my pump
by Bill Gosper 17 Dec '15

17 Dec '15

To run it <http://gosper.org/pumper_2.mp4> continuously, download with Miro, QuickTime-View-Loop. The pumped shapes split lengthwise resemble boats, so a grand entrance to a math museum would be a chain of these boats in an annular canal, passing through four rotors acting as an (approximate) air|water lock, with another four for the exit. Oh, the insurance. --rwg No manual entry for overboard

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[math-fun] A gallery of math art from Bridges 2015
by Alex Bellos 16 Nov '15

16 Nov '15

http://www.theguardian.com/science/alexs-adventures-in-numberland/gallery/2… <http://www.theguardian.com/science/alexs-adventures-in-numberland/gallery/2…>

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[math-fun] Terminology: "Characteristic Generating Function"?
by Bill Gosper 23 Oct '15

23 Oct '15

"Indicator Generating Function"? "Predicate Generating Function"? What is the name for the G.F. that encodes an integer sequence in the exponents with coefficients in {0,1} ? E.g., the OGF for the squares is Sum[n^2*x^n, {n, ∞}] == -((x*(1 + x))/(-1 + x)^3), but the thing I'm looking for is Sum[x^n^2,{n,0,∞}]==(1/2)*(1 + EllipticTheta[3, 0, x]). (Years ago I sent some of you the GF for expressions for k as the sum of n (optionally distinct) members of such a set, in terms of the GF predicating membership in the set. It turned out to be a superspecial case of a formula in Generatingfunctionology that was so general I didn't even recognize it as the same problem.) The reason I ask is that Neil just came up with 1/2 (-1+EllipticTheta[2,0,x]/x^(1/4)+EllipticTheta[3,0,x]) predicating the quartersquares<http://en.wikipedia.org/wiki/Multiplication_algorithm#Quarter_square_multip…> A002620 <https://oeis.org/A002620>, from which he got (1/(2 (-1 + x)^2))((-1 + x) x^(3/4) EllipticTheta[2, 0, x] + x (3 - x + (-1 + x) EllipticTheta[3, 0, x])) generating what we think is the number of volumes you can pump with n of the recently discussed rotors, arranged optimally. --rwg

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Re: [math-fun] Another(?) stick length illusion
by Dan Asimov 25 Aug '15

25 Aug '15

Huh? At least to the nearest pixel, both lines are the same length. They are corresponding diagonals of congruent regular pentagons. I find this to be an impressive illusion. --Dan << << To me the stick that is in fact longer looks like it is in fact longer. If they're supposed to be the same length I think your image needs a bit of editing. >> On Thu, Aug 19, 2010 at 4:31 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > While hacking the pentagon fill, tutor Julian found > www.tweedledum.com/rwg/6pents.png, which seems to work backwards: > The stick larger than its context appears smaller than the > stick smaller than its context. >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] New(?) elliptic K transformation
by Bill Gosper 20 Aug '15

20 Aug '15

On Mon, Nov 5, 2012 at 11:44 PM, Bill Gosper <billgosper(a)gmail.com> wrote: Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r] E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4] For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form. [Added 2012-11-16 07:03 Namely, EllipticK[1/2 + I/2] == Sqrt[-1 + Sqrt[2]] E^(I π/4) EllipticK[4 - 2 Sqrt[2]] And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I] ] None of the eight sign guesses works for Re(r)>1/2. If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg No one seemed to recognize it, so, grotesque though it is, NeilB and I decided it's time to write this up. So Neil bashed on it with Mathematica, and it got a simpler. Then I bashed on it so hard that it got unnervingly ungrotesque: (*) EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t] How could anything this simple be new? It looks like a routine 2F1 quadratic transformation. But which? As before, it makes the strange little results EllipticK[1/4] == E^(-((I \[Pi])/6)) EllipticK[Sqrt[3] E^((I \[Pi])/6)] and EllipticK[I] == (E^((I \[Pi])/16) EllipticK[1/2 - 1/4 Sqrt[7/2 - I/2]])/2^(1/8) Are these familiar? And somehow, (*) has miraculously lost its Re[t]<1/2 restriction:. E.g. for t=2, EllipticK[1/2] == (-1)^(1/4) EllipticK[2], as one might hope. For t=(-1)^(1/6), EllipticK[1/4 (2 + (-2 + (-1)^(1/6))/Sqrt[1 - (-1)^(1/6)])] == (2 - Sqrt[3])^(1/8) E^(-((5 I \[Pi])/48)) EllipticK[E^((I \[Pi])/6)] Angling for a K'/K identity, solve 1/4 (2 + (-2 + t)/Sqrt[1 - t]) == 1 - t : t -> 1/32 (31 + 3 I Sqrt[7]) So, Log[EllipticNomeQ[1/32 (31 - 3 I Sqrt[7])]] == -(1/2) Sqrt[1/2 (-3 I + Sqrt[7])] E^((I \[Pi])/4) \[Pi] along with its formal conjugate. Stop me if you've heard this one. --rwg (Is there a (dlmf?) page of special values of nome? Or just garden varieties:-)

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[math-fun] two "trivia"
by Bill Gosper 02 Aug '15

02 Aug '15

A miscellaneous nontriangular matrix product: [ sqrt(%pi) ] [ 0 0 --------- ] [ 1 1 5 1 ] [ 2 3 ] oo [ 0 -- - ----- - - --- ] [ gamma (-) ] /===\ [ 16 2 4 8 j ] [ 4 ] | | [ 64 j ] [ ] | | [ ] = [ 2 3 ] | | [ 1 1 2 ] [ 4 gamma (-) ] j = 1 [ (- - ---) 0 1 ] [ 4 ] [ 2 4 j ] [ 0 0 ----------- ] [ ] [ 3/2 ] [ 0 0 1 ] [ %pi ] [ ] [ 0 0 1 ] (Does anybody know how to make these archaic ASCII displays in Mathematica, for our archaic math-fun server?) A260747, Consolidated Dragon Curve triple points. If D:[0,1] is a Dragon curve, then besides n, there are two other integers p and q with D(A(n)/(15*2^k)) = D(A(p)/(15*2^k)) = D(A(q)/(15*2^k)), where k is any integer > lg(max(A(n),A(p),A(q))/15) 13, 21, 23, 26, 37, 39, 42, 46, 47, 52, 73, 74, 78, 81, 83, 84, 92, 94, 97, 99, Note the gap of 73 - 52 = 21 at position 10. There are apparently no larger gaps, and the next gap of 21 is at position 17651. --rwg

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[math-fun] Formulation of Somsky Gears as Diophantine Problem
by Warren D Smith 31 Jul '15

31 Jul '15

Formulation of Somsky Gears as Diophantine Problem ========Warren D Smith====July 2015=============== Suppose the sun gear has S teeth, the outer ring gear has R teeth, the K planets have P1, P2, ..., PK teeth, and the distance between the sun-center and ring-center is equivalent to the radius of a gear with D teeth. Those involve K+2 positive-integer-valued variables S, R, P1, P2, ..., PK. Here D>0 need not be integer, it might merely be real. For each planet whose center is not collinear with SunCenter and RingCenter, it is possible to adjoin a "twin" planet reflected about the SunCenter-RingCenter line. I will ignore those twin planets, focusing here only on primal planets, all of which lie on same side of that line. Then by the law of cosines: SunCenter-RingCenter angle viewed from planetJ = arccos [D^2 - (PJ+S)^2 - (R-PJ)^2]/[2*(PJ+S)*(R-PJ)] RingCenter-PlanetJ angle as viewed from SunCenter = arccos [(R-PJ)^2 - (PJ+S)^2 - D^2]/[2*(PJ+S)*D] SunCenter-PlanetJ angle as viewed from RingCenter = arccos [(PJ+S)^2 - (R-PJ)^2 - D^2]/[2*(R-PJ)*D] Now the variously named "belt condition" or "alternating sum condition" is that the following quantity (regardless of the indices A and B with 0<A<B<=K) must be a rational multiple of pi. These arise from considering closed "belts" always contacting either sun, planetA, ring, or planetB, and traversing them in the turning direction. [The circumference of any such belt is demanded to be an integer number of teeth.] R*arccos [(PA+S)^2 - (R-PA)^2 - D^2]/[2*(R-PA)*D] - R*arccos [(PB+S)^2 - (R-PB)^2 - D^2]/[2*(R-PB)*D] + S*arccos [(R-PA)^2 - (PA+S)^2 - D^2]/[2*(PA+S)*D] - S*arccos [(R-PB)^2 - (PB+S)^2 - D^2]/[2*(PB+S)*D] - PA*arccos [D^2 - (PA+S)^2 - (R-PA)^2]/[2*(PA+S)*(R-PA)] - PB*arccos [D^2 - (PB+S)^2 - (R-PB)^2]/[2*(PB+S)*(R-PB)] If these rationality conditions are always satisfied (and if I did not screw up the signs in the first column; correct those if necessary) then the given S,R,P1,P2,...,PK work (at least if scaled up by some suitable integer multiple) as Somsky gear-tooth counts. It should suffice if these rationality conditions merely hold when A=1; the rationality of the others would then follow automatically. If so there are K-1 quantities which need simultaneously to be rational. To assure this we need to adjust K+2 integer quantities, plus one continuously variable quantity D. It is possible to rephrase the above in a way that avoids use of arctrig and trig functions, instead employing integer powers of complex numbers, and square root function, only. That makes it clear that D is always going to be an ALGEBRAIC number.

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