January 2016 - 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 16 Feb '16

16 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] The Wittless Duck
by Fred Lunnon 03 Feb '16

03 Feb '16

I have been struggling to recall a rather surprising theorem, encountered in passing while searching for something unrelated, concerning the structure of quadratic spaces for which the Witt index exceeds 2 . Not only can I not find the theorem; I can't even locate a definition of the Witt index --- that is |p - q| + r , where the (possibly degenerate) quadratic form signature involves p positive, q negative, r absent squares. Can somebody out there please unscramble my brain (or my surfing technique) for me? Dribble, mutter ... Fred Lunnon

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[math-fun] Pegg's crashing computer
by Warren D Smith 03 Feb '16

03 Feb '16

How did you find out what the problem was and how to fix it? (I have a similar crashing problem on my computer. I suspected it was a bad graphics chip.)

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[math-fun] Fwd: Pic: Clifford's 1st thm [Was: 2nd intersection, circles through points]
by Bill Gosper 31 Jan '16

31 Jan '16

gosper.org/cliff.png Anticipated by https://en.wikipedia.org/wiki/Clifford%27s_circle_theorems except for the (terse) drawing commands, for which you need circumboth[z1_, z2_, z3_] := {((z1 - z2)*Abs[z3]^2 + Abs[z2]^2*(z3 - z1) + Abs[z1]^2*(z2 - z3))/I, Abs[(z2 - z1)*(z2 - z3)*(z3 - z1)]}/(2*Im[Conjugate[z2 - z1]*(z3 - z1)]) second[z1_, z2_] := (Conjugate[z1] z2 - Conjugate[z2] z1)/Conjugate[z1 - z2] for {circumcenter, circumradius} and second intersection. ---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Sat, Jan 30, 2016 at 1:43 AM Subject: Pic: Clifford's 1st thm To: mathfuneavesdroppers(a)googlegroups.com, rcs(a)xmission.com Coxeter says: Put four arbitrary black circles through a point. Set aside each in turn. For each of the four remaining triads, put a red circle through their three second intersections. The four red circles meet at a point, and red and black can be interchanged. --rwg --------------- Related stuff: clifford circle chains <https://www.google.com/search?q=clifford+circle+chains&client=safari&rls=en…> images . Clifford's 2nd thm might require animation to be visually intelligible. A difficulty is curvatures changing sign (radii blowing up).

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[math-fun] Ancient Babylonians Figured Out Forerunner of Calculus
by Ray Tayek 30 Jan '16

30 Jan '16

http://science.slashdot.org/story/16/01/29/1847240/ancient-babylonians-figu… Posted by timothy <http://www.monkey.org/%7Etimothy/> on Friday January 29, 2016 @01:39PM from the clay-tablets-are-hard dept. sciencehabit <http://slashdot.org/%7Esciencehabit> writes: /Tracking and recording the motion of the sun, the moon, and the planets as they paraded across the desert sky, ancient Babylonian astronomers used simple arithmetic to predict the positions of celestial bodies. Now, new evidence reveals that these astronomers, working several centuries B.C.E., also employed sophisticated geometric methods that foreshadow the development of calculus <http://www.sciencemag.org/news/2016/01/math-whizzes-ancient-babylon-figured…>. Historians had thought such techniques did not emerge until more than 1400 years later, in 14th century Europe./ -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/

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[math-fun] Matrix question
by David Wilson 30 Jan '16

30 Jan '16

Which 2x2 matrices M satisfy M^2 = I? Is there an approach to solving this that doesn't involve algebra on the matrix elements?

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[math-fun] 2nd intersection
by Bill Gosper 30 Jan '16

30 Jan '16

I hope Henry or someone is collecting our various complex number geometricks e.g., PolygonArea[L_List] := Total[MapThread[Im[#1\[Conjugate]*#2] &, {L, RotateLeft[L]}]]/2 line segment intersect Interseg[z1_, z2_, z3_, z4_] := Block[{z12 = z1 - z2, z34 = z3 - z4}, (z34*Im[z2*z1\[Conjugate]] + z12*Im[z3*z4\[Conjugate]])/Im[z34*z12\[Conjugate]]] incenter, circumcenter, inradius, ... . For the Clifford circle theorems I needed second intersection: Given centers of circles though the origin z1, z2, secondintersection[z1_, z2_] := (Conjugate[z1] z2 - Conjugate[z2] z1)/Conjugate[z1 - z2] Is this obvious to anybody? I had to derive it with 8th grade analytic geometry. --rwg

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