November 2015 - mathfun@mailman.xmission.com

Re: [math-fun] Raghavendra's algorithm for solving linear systems over finite fields
by Henry Baker 06 Jun '16

06 Jun '16

I can see applications in distributed/parallel implementations. You can do most of the checking & probably a lot of the randomization in parallel. At 09:02 AM 8/15/2012, Warren Smith wrote: >>Preliminary draft: >> http://www.eecs.berkeley.edu/~prasad/linsystems.pdf > >--I don't understand why anybody would want to use this algorithm >for any purpose. It's slower than everything else I can think of and also >randomized i.e. can fail. > >However... perhaps it might be useful if changed to be about finite >RINGS not fields. In rings, you cannot divide, hence gaussian elimination is >not usable. Hence this algorithm then has less competition. >(Even then I'm dubious, but then it'd have at least a hope.)

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[math-fun] 5 and 7 outcome fair dice
by Bill Gosper 06 May '16

06 May '16

Odd barrel dice have the problem that two faces are always uppermost, so how do you label them? Alternative to face-transitivity, there are fair dice which cannot rest on any face, and yet have one face uppermost. Spoiler: gosper.org/5&7dice.png --rwg I assume these are old ideas. Check out Wikipedia if you haven't seen spherical, 6 outcome dice.

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[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] phi(N)*sigma(N)<N^2, phi(N)+sigma(N)>=2*N
by Warren D Smith 02 Dec '15

02 Dec '15

The formula for phi(N) is N * product (1 - 1/P) product taken over primes P that divide N. The formula for sigma(N) is N * product (1 + 1/P + 1/P^2 + ... 1/P^E) with P dividing N and P^E being the highest power of P that divides N. So obviously sigma(N) < N * product(1 + 1/P + 1/P^2 + ... ) = N^2 / phi(N). Q.E.D. Also obviously sigma(N) >= N * product ( 1 + 1/P ) so phi(N) + sigma(N) >= 2*N since the cross terms in the products all either cancel out or come out to something positive. [The avoidance of phi+sigma=2*N+5 is not so trivial, though.] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] "Q-carbon"
by Warren D Smith 30 Nov '15

30 Nov '15

New solid phase of carbon found, allegedly harder than diamond (!), supposedly may occur naturally inside some planetary cores. It is ferromagnetic (!). Created by heating a thin film of amorphous carbon to over 4000K with a huge power 200 nanosecond laser flash. "Novel Phase of Carbon, Ferromagnetism and Conversion into Diamond" will be published online Nov. 30 in the Journal of Applied Physics. "Direct conversion of amorphous carbon into diamond at ambient pressures and temperatures in air" Oct. 7 in the journal APL Materials. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Chess Engine Superfinal
by Hans Havermann 30 Nov '15

30 Nov '15

The Top Chess Engine Championship (TCEC) has just concluded its 100-game, season-8 superfinal between Komodo and Stockfish, each with a rating of over 3200. The 100 games averaged at just under 5.6 hours per game. Komodo won with a convincing +9 =89 -2. http://tcec.chessdom.com/archive.php

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[math-fun] Pfaff is still underrated!
by Bill Gosper 30 Nov '15

30 Nov '15

Page xii of Basic Hypergeometric Series says that Pfaff's 3F2[1] (long misattributed to Saalschütz) preceded Gauss's 2F1[1] limiting case by 16 years! (Did Pfaff ever think to take the n→∞ limit? He must have. But then why didn't he notice that the 2F1 was more useful? He was Gauss's research supervisor and housemate(?). Maybe he deliberately gave Gauss the credit.) Yikes, he also discovered Hypergeometric2F1[a_, b_, c_, z_] :> Hypergeometric2F1[c - a, b, c, z/(z - 1)]/(1 - z)^b ! --rwg Mathematica 10.3 still fails Pfaff: In[613]:= Assuming[n \[Element] Integers && n >= 0, FunctionExpand[HypergeometricPFQ[{a, b, -n}, {d, a + b - n - d + 1}, 1]]] Out[613]= HypergeometricPFQ[{a, b, -n}, {d, 1 + a + b - d - n}, 1] Reminder: If you don't know which parameter is the nonpositive integer, (c67) hypersimp(hyper_f[3,2]([a,b,c],[d,a+b+c-d+1])) Is any of a, b, or c a nonpositive integer? yes; (d67) - ((%pi)^2 * Gamma( - d + c + b + a + 1) * Gamma(d)/(Gamma( - d + b + a + 1) * Gamma( - d + c + a + 1) * Gamma( - d + c + b + 1) * Gamma(d - a) * Gamma(d - b) * Gamma(d - c) * (cos(%pi * d) * cos(%pi * (d - c - b - a)) - cos(%pi * a) * cos(%pi * b) * cos(%pi * c)))) (c68) neg_a_part(%,1,2,-1) (d68) ((%pi)^2 * Gamma( - d + c + b + a + 1) * Gamma(d)/(Gamma( - d + b + a + 1) * Gamma( - d + c + a + 1) * Gamma( - d + c + b + 1) * Gamma(d - a) * Gamma(d - b) * Gamma(d - c) * (cos(%pi * a) * cos(%pi * b) * cos(%pi * c) - cos(%pi * d) * cos(%pi * (d - c - b - a))))) (Also works: standardize_signs(%,d). Boy, does Mathematica ever need neg_a_part and standardize_signs.)

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