October 2014 - mathfun@mailman.xmission.com

[math-fun] Model of the Fano plane on the sphere?
by Mike Stay 03 Oct '14

03 Oct '14

I.e. can you assign each point in the Fano plane to antipodal points on the sphere and each line in the Fano plane to a great circle such that the incidence relation is preserved? -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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[math-fun] Proposed resolution of double-slit experiment paradox using Feynman path integral formalism
by Stuart Anderson 02 Oct '14

02 Oct '14

http://phys.org/news/2014-10-superposition-revisited-resolution-double-slit…

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[math-fun] A.Unzicker, physics curmudgeon
by Warren D Smith 02 Oct '14

02 Oct '14

Two books Alexander Unzicker: The Higgs Fake, How Particle Physicists Fooled the Nobel Committee Alexander Unzicker & Sheilla Jones: Bankrupting physics, Palgrave Macmillan 2013 I was having a blast reading these books, because the author is a huge curmudgeon, blasting merrily away at theoretical physics as a "bankrupt" field filled with frauds and hype artists intent on confusing the gullible media, general public, and each other. Not saying these books will do you any good, objectively, but they'll probably at least make you laugh. He also wrote this (which I also found highly comedic): http://arxiv.org/abs/0912.0443

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[math-fun] graphs on spheres in dimension D
by Warren D Smith 02 Oct '14

02 Oct '14

There is a well known theory of distance regular graphs and their embeddings with points on D dimensional spheres via eigenvectors of their matrices. So you can just use it and all your sphere/Fano questions are done with.

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Re: [math-fun] Multiplying repeating decimals
by Bill Gosper 01 Oct '14

01 Oct '14

I wonder how much of this stuff Mathematica already knows. E.g.: In[265]:= digitCycleLength[r_Rational, b_Integer?Positive] := MultiplicativeOrder[b, FixedPoint[Quotient[#, GCD[#, b]] &, Denominator[r]]] In[270]:= digitCycleLength[1/(10^13 - 1), 10] Out[270]= 13 In[271]:= digitCycleLength[1/(10^13 - 1)^2, 10] Out[271]= 129999999999987 VictorM>Let m be a positive integer relatively prime to 10. Then the period length of 1/m is the multiplicative order of 10 modulo m -- i.e. we look in the group (Z/mZ)*. Note that if p is an odd prime, that even though the multiplicative,a =order of 10 mod p might be small, the multiplicative order of 10 mod p^2 is overwhelmingly likely to be p*a (except when we have something like a Wieferich prime). If m,n are relatively prime, then by the Chinese remainder theorem the multiplicative order of 10 mod mn is the lcm of the multiplicative orders of 10 mod each of m and n. Victor On Mon, Sep 29, 2014 at 5:53 PM, James Propp <jamespropp(a)gmail.com> wrote: I had assumed that if you multiply an eventually repeating decimal of period m and an eventually repeating decimal of period n, you get an eventually repeating decimal whose period is bounded by some polynomial function of m and n. But today I learned from Henry Cohn that that's not true: the period length can be exponentially large in m and n. More specifically, 1/(10^n-1) repeats with period n, but its square repeats with period 10^n-1 or thereabouts. I'm sure someone has written about the repeat-length for 1/(10^n-1)^2; can anyone provide a reference? Jim Propp Rich> Your intuition isn't completely off base. If the two periods M & N are relatively prime, then the product period should be a divisor of 9MN. This is because 10^M - 1 and 10^N - 1 both divide 10^MN - 1, and if gcd(M,N)=1, then gcd(10^M-1,10^N-1) is 9. The 9 in 9MN provides the required factor of 9, so that (10^M-1)*(10^N-1) divides 10^9MN - 1. This suggests the general period formula is (a divisor of) lcm(M,N) * (10^gcd(M,N) - 1). Rich

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[math-fun] 11 Beautiful, Psychedelic GIFs Created by a Math Whiz
by Henry Baker 01 Oct '14

01 Oct '14

FYI -- Cool! http://www.wired.com/2014/10/11-beautiful-psychedelic-gifs-created-math-whi…

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[math-fun] Multiplying repeating decimals
by James Propp 01 Oct '14

01 Oct '14

I had assumed that if you multiply an eventually repeating decimal of period m and an eventually repeating decimal of period n, you get an eventually repeating decimal whose period is bounded by some polynomial function of m and n. But today I learned from Henry Cohn that that's not true: the period length can be exponentially large in m and n. More specifically, 1/(10^n-1) repeats with period n, but its square repeats with period 10^n-1 or thereabouts. I'm sure someone has written about the repeat-length for 1/(10^n-1)^2; can anyone provide a reference? Jim Propp

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[math-fun] about the volume of the debt.
by Simon Plouffe 01 Oct '14

01 Oct '14

Hello, yesterday, the amount of debt of France came out, which is over 2000 billion euros, actually the total external debt is 4542 billion euros. For the USA, the amout is almost 18 000 billion dollars. Well, that's a lot of money, I know that the amount has already been compared to other amounts or equivalently in VOLUME of the debt. I could not resist to compute that : For France the total volume of the debt is 5677500 cubic meters in coins of 1 euro. For USA, in quarters : a perfact cube of 414 meters side filled with 25 cents. In pennies (us one cent), the volume is roughly 1 cubic KILOMETER or 981960611 cubic meters. The calculation I used is a shortcut, I simply pile the coins in a square grid which is (I believe) the same amount of space than a random pile of coins, on this I am certain that some of the members can say if this rough model is accurate. For France it is the equivalent of 2 times the pyramid of Gizeh + about 500000 cubic meters. or 1 pyramid + 2 times the volume of the Eiffel Tower in 1 euro coins. The pyramid of Gizeh is about 2.6 million cubic meters and the Eiffel tower about 1.7 million cubic meters (enclosed space) which is by the way lightier than the enclosed air, the Eiffel tower is lightier than air (enclosed volume). I like that idea of 1 cubic kilometer of debt for the usa. Here are some references. http://en.wikipedia.org/wiki/Cent_%28United_States_coin%29 http://en.wikipedia.org/wiki/1_euro_coin http://en.wikipedia.org/wiki/List_of_countries_by_external_debt Best regards, Simon Plouffe

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[math-fun] Microsoft announces move to nonary numbers
by Henry Baker 01 Oct '14

01 Oct '14

Today Microsoft incremented "8" to "10". http://blogs.windows.com/bloggingwindows/2014/09/30/announcing-windows-10/

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[math-fun] calculator math
by Dave Dyer 01 Oct '14

01 Oct '14

This question on stackexchange may interest some of you. The essence is, where is there a source of specified precision algorithms for calculators? http://programmers.stackexchange.com/questions/257674/accuracy-of-calculato…

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