May 2015 - mathfun@mailman.xmission.com

[math-fun] Nash's letter to the NSA
by Stuart Anderson 26 May '15

26 May '15

Nash wrote to the NSA in 1955 with proposals for an encryption/decryption machine. https://www.nsa.gov/public_info/_files/nash_letters/nash_letters1.pdf Nothing new in these proposals now but much of this must have been novel at the time. Sent from my iPhone

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Re: [math-fun] 4 solid angles
by Warren D Smith 25 May '15

25 May '15

If the sum S of the 4 corner solid interior angles are used, then Propp is wrong. A flattened sliver tetrahedron has S=0, a flattened regular tet has S=2*pi, etc. there is no constant value. Propp presumably meant, the 4 tet faces viewed from a point inside; then obviously the "whole sky" has total solid angle 4*pi, the end.

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[math-fun] Gosper's EllipticK formula
by Warren D Smith 25 May '15

25 May '15

> Bill Gosper: > 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? > > WDS: > Let s=Sqrt[1-t]. Then (*) is rewritten as > EllipticK[ 1/2 - (1+s^2)/(4s) ] == EllipticK[ -(1-s)^2/(4s) ] == > s^(1/2) EllipticK[1-s^2] > Now let s=1-m. Then this is > EllipticK[ -m^2/(4(1-m)) ] == (1-m)^(1/2) EllipticK[(2-m)m]. > E.Salamin: > Complains WDS formula is wrong. --WDS: arg. Well, I just started with Gosper's "ungrotesque" formula taking its correctness on faith, then worked by hand to produce mine. Doing same thing over again now: Gosper: EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t] Let t=1-s^2: EllipticK[1/4 (2 - (1+s^2)/s)] == (s)^(1/2) EllipticK[1-s^2] EllipticK[(2s-1-s^2)/(4s)] == (s)^(1/2) EllipticK[1-s^2] EllipticK[-(1-s)^2/(4s)] == (s)^(1/2) EllipticK[1-s^2] which is same as I got before. Now let s=1-m: EllipticK[-m^2/(4(1-m))] == (1-m)^(1/2) EllipticK[1-(1-m)^2] EllipticK[-m^2/(4(1-m))] == (1-m)^(1/2) EllipticK[2m-m^2] EllipticK[-m^2/(4(1-m))] == (1-m)^(1/2) EllipticK[(2-m)m] which is same as I got before, OK? So either I made same error over again because my brain is truly decaying, or Gosper was wrong, or Salamin was wrong, you figure out which. I was just trying to produce a nicer equivalent version. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] gcF??
by Bill Gosper 25 May '15

25 May '15

Perhaps because they're taught factorization rather than Euclid's algorithm, gradeschoolers are now being forced to say gcf ("greatest common factor") instead of gcd. Or maybe just to make it sound easier. So I asked a kid: 12 contains how many factors of 3? And how many factors of 2? And how many factors of 1? "Tell your teacher 1 can't be a factor of anything. 2 and 3 have no greatest common factor." "I'll get in trouble." A lot of stuff we're taught is oversimplified. We have to unlearn it and relearn it later. But this is ridiculous. Is there some way we funsters can gang up against this idiocy? --rwg They're also taught that improper fractions are not in lowest terms.

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[math-fun] What is elliptic K and E closed form?
by Bill Gosper 25 May '15

25 May '15

I think the answer has changed in the last 50 yrs. A query from constantologist Steve Finch revived my interest in whether special value identities like EllipticK[(1 + Sqrt[2])^4] == ((1/4 - I/2)π^(3/2))/ ((2 + Sqrt[2]) Gamma[3/4]^2) and EllipticK[8 2^(1/4) (-1 + Sqrt[2])^3* E^(I ArcTan[(3 (3 + Sqrt[2]))/(4 2^(1/4))])] == (I (1 + Sqrt[2]) E^(-2 I ArcCot[2^(1/4)]) π^(3/2))/ (4 Sqrt[2] Gamma[3/4]^2) are already known, and where one might find tables much more extensive than, e.g., http://functions.wolfram.com/EllipticIntegrals/EllipticK/03/01/ or http://dlmf.nist.gov/19.6 . DLMF says "Exact values of K(k) and E(k) for various special values of k are given in Byrd and Friedman (1971, 111.10 and 111.11)" (1st English edition, 1954) Neil and his Mom were kind enough to fetch me this comprehensive (*Handbook* of *Elliptic* Integrals for Engineers and Physicists<http://www.amazon.com/Handbook-Elliptic-Integrals-Engineers-Physicists/dp/B…>), and to my surprise and disappointment, there was not a single identity of the above form, i.e. algebraics times Gammas(rationals). How could this be? My guess: In the 1950s K and E were considered closed forms while Gamma was not! It makes sense: K and E are of lower computational complexity. We Funsters once concluded that Gamma[k/24] was rapidly computable thanks to elliptic K identities. But now, in papers like "EVALUATION OF COMPLETE ELLIPTIC INTEGRALS OF THE FIRST KIND AT SINGULAR MODULI" (people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf*) "evaluation" means "determin[ation] explicitly in terms of Gamma values". Which is what I always thought. See, e.g., http://en.wikipedia.org/wiki/Elliptic_integral#Special_values --rwg

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[math-fun] (no subject)
by Warren D Smith 24 May '15

24 May '15

Bill Gosper: 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? WDS: Let s=Sqrt[1-t]. Then (*) is rewritten as EllipticK[ 1/2 - (1+s^2)/(4s) ] == EllipticK[ -(1-s)^2/(4s) ] == s^(1/2) EllipticK[1-s^2] Now let s=1-m. Then this is EllipticK[ -m^2/(4(1-m)) ] == (1-m)^(1/2) EllipticK[(2-m)m]. These look kind of like the known identities here http://functions.wolfram.com/EllipticIntegrals/EllipticK/17/01/ but not the same. Perhaps yours arises by composing one of them, with a quadratic transformation. The final form of it I wrote down makes it clear how the argument (if m is small) gets to be quadratically way small. Thus, this offers a very fast way to compute EllipticK. Also, to compute pi/2=EllipticK[0] if we happened to know some other EllipticK special value. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Revolting robots
by Fred Lunnon 24 May '15

24 May '15

I note that according to its URL the quite thoughtful article below will not appear for another 10 days! http://www.nybooks.com/articles/archives/2015/jun/04/robots-are-winning/ Aside from well-known difficulties associated with an ability to read tomorrow's news today, I am prompted to reflect that the human race needs constant reminding that however ingeniously we go about creating our slaves, in the same process we eventually create our equals (perhaps even our masters). I wonder indeed if some such insight might not be formulated in terms of Situation Theory --- or perhaps already has been? Fred Lunnon On 5/24/15, Douglas Rogers <dgrbgsu(a)gmail.com> wrote: > FYI > < > http://www.theguardian.com/us-news/2015/may/24/john-nash-mathematician-beau… >> > > < > http://www.nytimes.com/2015/05/25/science/john-nash-a-beautiful-mind-subjec… >> > > Robots > <http://www.nybooks.com/articles/archives/2015/jun/04/robots-are-winning/>. >

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[math-fun] Infinite time sharing
by Warren D Smith 24 May '15

24 May '15

Crank up your Turing machine -- it needs to perform an infinite number of tasks simultaneously. It's the Hilbert Hotel for Turing machines. Specifically: let us say the TM performs task n a fraction F(n) of the time [for each n=1,2,3,...] and task n uses as its memory "tape," a subset of the cells on the whole tape -- specifically the cells that task n uses, are the integers k>0 such that G(k)=n. Here SUM_{j>0} F(j)=1, F(j)>0 for all j>0, G(k) is positive-integer-valued, and for each n>0 there exist an infinite set of integers k>0 with G(k)=n. THE PUZZLE: Design a "good" choice of the magic functions F() and G(), such that this time sharing scheme is maximally efficient. Unfortunately there are several possible measures of "efficiency" and it is not obvious which to use... including (as function of n and T) max latency before task n does another step, the min number of steps performed on tasks 1,2,...,n (combined) after time T, and the min number of steps performed on tasks 1,2,...,n (not combined, just take the worst one) after time T. A (PROBABLY POOR) SOLUTION just to get the ball rolling, I will propose one fairly concrete scheme. Imagine the "tape" of the TM is the infinite square spiral fedc 432b 501a 6789 in the pixelated plane, and G(n)=|y|, where y is the y-coordinate of tape cell #n, if that tape location is reserved for use by task #n. (The x-axis, y=0, is used for bookkeeping by the operating system which is "task #0.") Also imagine "time" as spiralling around (a second copy of) the same kind of spiral. The operating system works on task G(t)+1 for t^P steps, then advances t to the next spot on the time spiral. If the constant P is chosen large enough, then I believe this will work, i.e. it will get an infinite amount of private-TM-steps done on every task. This particular solution, however, involves F(n)=0 for all n, i.e. each job is worked on a fraction asymptotically 0 of the time. As opposed to, say, F(n)=zeta(2)/n^2. I do not know what the "best" solution to this puzzle is, and it is not even clear to me what "best" should mean exactly... but it nevertheless seems to me to be a pretty interesting puzzle. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] algebraic fcn simplification (Bill Gosper)
by Axel Vogt 24 May '15

24 May '15

> Date: Sun, 24 May 2015 03:39:38 -0700 > From: Bill Gosper <billgosper(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] algebraic fcn simplification > Content-Type: text/plain; charset=UTF-8 > > I just spent hours simplifying > (64 t^2 - 16 Sqrt[-32 t^2 + t^2/Sqrt[1 - t] + 31 Sqrt[1 - t] t^2 + 32 t^3 - > 17 Sqrt[1 - t] t^3 + 10 t^4 + Sqrt[1 - t] t^4])/(128 t^2) > > to > 1/2 + (4 - 5 t)/(8 (1 - t)^(1/4) t) - ((1 - t)^(1/4) (4 + t))/(8 t) > using Corey's denester on numeric special cases. > How should I have done this? > > Plotting the difference shows they only agree when arg t lies > within ?/4 of the real axis. Elsewhere it seems to be -1. Mma > gives 0 for Limit t->0, but Plot3D rather consistently spikes at > the origin. I don't know how to elicit that spike value except > by plotting. Bug? --rwg I think that is only a kind of numerical artefact of the plotting routine, which may not cancel out expressions in t=0 (or t=1) If I call a resp. b your expressions, then using Maple they are both roots of the polynomial P(y) = t^4 - 256*(-1+t)*(3*t-4)*(t-4)*y +256*(-1+t)*(19*t^2-16*t+16)*y^2 - 8192*t^2*(-1+t)*y^3 + 4096*t^2*(-1+t)*y^4 The limit for those roots in t=0 are -infinity*I, infinity*I, 0, 1 and one sees that a and b must coincide.

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[math-fun] CAS on smartphone
by Fred Lunnon 24 May '15

24 May '15

A couple of days ago my ten-year old mobe finally died, following which I reluctantly embraced the portable digital age in the form of a modest Android smartphone [8Gb Samsung Core Prime]. What's the prospects for installing a credible CAS on this device? Has anybody out there dipped a toe in this pond? With what outcome? The only realistic candidate so far seems to be Maxima: see https://sites.google.com/site/maximaonandroid/ or maybe JAS (but what happened to Yacas, eh?) http://krum.rz.uni-mannheim.de/jas/ What about SAGE (incl. Maxima & PARI/GP) or Magma? Fred Lunnon

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