January 2012 - mathfun@mailman.xmission.com

Re: [math-fun] Infinite product expansion for 1
by Phil Carmody 15 Jan '12

15 Jan '12

From: "Adam P. Goucher" <apgoucher(a)gmx.com> > > Consider the infinite product: > > > > (24/16) * (48/64) * (120/144) * ... * ((p^2 - 1)/(16*[p/4]^2)) * ... > > > > In Sloane notation, this is: > > > > Product (n >= 1) of (A061066(n+1) / (2*A024698(n+1)^2)) > > I've just realised that it has a much simpler representation: > > Product (p >= 5) of: > > (p+1)/(p-1) iff p = 4k + 1 > (p-1)/(p+1) iff p = 4k - 1 How about: Product[p >= 5] (p+Chi4(p))/(p-Chi4(p)) Where Chi4() is the character mod 4 with Chi4(3)=-1, or equivalently the Legendre symbol (-1|p). Phil

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[math-fun] Japanese supercomputer says: String theory says: universe is 3D expanding
by Warren Smith 14 Jan '12

14 Jan '12

Japanese supercomputer says: String theory says: universe is 3D expanding Normally I view string theory with great skepticism and lack of understanding. And now to up the ante on that, here is an amazing claim which I suppose one could consider to be an "experimental confirmation" of string theory. Press release: http://www.kek.jp/intra-e/press/2011/122209/ Paper: Sang-Woo Kim, Jun Nishimura, Asato Tsuchiya: Expanding (3+1)-dimensional universe from a Lorentzian matrix model for superstring theory in (9+1)-dimensions, http://arxiv.org/abs/1108.1540 (4 pages) This gist seems to be as follows. String theory says that particles are tiny 1-dimensional "strings" not points, and the universe is 9+1 dimensional not 3+1 dimensional, with the 6 extra dimensions curled up in some microscopic fashion for some reason so we do not perceive them -- which leaves it somewhat of a mystery why the universe is 3D and expanding. The "IKKT matrix model" is a way supposedly to simulate nonperturbative string theory using monte carlo on a supercomputer. So they put some strings in some kind of 9D box with periodic boundary conditions, simulated, and voila, the box automagically wanted to have 6 dimensions shrink small, and the other 3 dimensions expand. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] build table of multiplicative inverses without dying of embarrassment
by Warren Smith 14 Jan '12

14 Jan '12

Some internet & literature searching revealed a bunch of embarrassingly stupid schemes for making a table telling you for each x, what is y so that x*y (mod p) = 1 where p is some fixed prime modulus. There are less-stupid ways to do it. Here is one; maybe you can do even better. Step 1. find a generator mod p. That is, find g so that g^(p-1) mod p=1 but g^((p-1)/d) mod p is not 1 for any divisor d of p-1. Then g's inverse is f where f=g^(p-2). [Simply random trials will find such a g quickly, i.e. way sublinear(p) time.] Step 2. for k=0,1,...,p-1: write f^k (mod p) into entry g^k (mod p) of your table. The end. Another approach is to do this for some random g which might not be a generator, then while the table of inverses you get is not filled up, go to a random unfilled entry, make that location be your new g, and do it again. --Warren D. Smith

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[math-fun] build table of multiplicative inverses without dying of embarrassment
by Bill Gosper 14 Jan '12

14 Jan '12

HUH? The inverse of 42 mod 239: In[1131]:= Convergents[42/239] Out[1131]= {0, 1/5, 1/6, 3/17, 13/74, 42/239} In[1132]:= Mod[74*42, 239] Out[1132]= 1 --rwg

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Re: [math-fun] Gamma function (formulas collection)
by Bill Gosper 14 Jan '12

14 Jan '12

What about Remezing Spouge? This unfinished, three term (toy) example<http://gosper.org/rspouge3.png> looks pretty promising (on all of z>0!). But we can squeeze six fudgefactors into three terms <http://gosper.org/rspouge3+3.png>. Does this cost closer to three or six? --Bill Are you using E^x -> (1 + Tanh[x/2])/(1 - Tanh[x/2]) ? Robert>I have gotten the 10-term "munafacmu34" version performing to near the limits of what's possible in a 32-digit floating-point (Dekker style double-double) numerics library. To answer your question *"How many divides is a '+a5*z^2' worth?"*, I'd say it's probably about a one-to-one tradeoff. In other words, if you add one term to a polynomial to save one divide in the "Gamma[x+1]=x Gamma[x]" iteration, that would be about a break-even. So if you can reduce the minimum argument by 2 via adding one term, that's a win. The bigger issue I'm dealing with is the transcendental operations. My Sinh() is doing an 8-step Taylor series (for 1/z <= 1/14), Exp() is a 16 to 20 step Taylor series, and Log() is a little slower than Exp(). Since we need about 5 of these transcendentals to use the approximation formula, that's where most the computation time is being spent. - Robert [1] See bibliography at mrob.com/pub/math/f161.html On Tue, Jan 10, 2012 at 19:14, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote: > (Sorry about (no subject).)> Hmm, "only" 26 digits with seven b*ggerfactors on z>9:>> http://gosper.org/munafacmu26.png>> Note that the z^7 means we only need ~20 digit values for the seven> b*ggerfactors, but it looks like we'll need > eight, ior larger "9" for> quad (34digit)> precision. How many divides is a "+a5*z^2" worth?> --rwg> -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 -mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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Re: [math-fun] Infinite product expansion for 1
by Adam P. Goucher 13 Jan '12

13 Jan '12

> Consider the infinite product: > > (24/16) * (48/64) * (120/144) * ... * ((p^2 - 1)/(16*[p/4]^2)) * ... > > In Sloane notation, this is: > > Product (n >= 1) of (A061066(n+1) / (2*A024698(n+1)^2)) I've just realised that it has a much simpler representation: Product (p >= 5) of: (p+1)/(p-1) iff p = 4k + 1 (p-1)/(p+1) iff p = 4k - 1 = (6/4 . 6/8 . 10/12 . 14/12 . 18/16 . 18/20 . 22/24 . ...) = 1. Sincerely, Adam P. Goucher

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[math-fun] Irrationality of e
by Dan Asimov 13 Jan '12

13 Jan '12

I like that using the sum-of-reciprocal-factorials definition of e, one can immediately see why it must be irrational. (As made clear in that paper of Sondow's, at < http://arxiv.org/pdf/0704.1282v2 >.) For, any rational number is equal to L/N! for some integers L, N. But any partial sum Sum_{k=1..N} 1/k! of e is of the form J/N!, and so adding the next term 1/(N+1)! means dividing the interval [J/N!, (J+1)/N!] into N+1 equal intervals and ensuring that e must lie somewhere in the *second* one of them. Because second is neither the first nor the last, this immediately implies that e cannot be of the form L/N!, and since N is arbitrary, we're done. --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] patterns, lego and some tilings
by Simon Plouffe 13 Jan '12

13 Jan '12

Hello, this posting is not purely related to math. I stumbled on this site which has an impressive set of constructions in lego, one of them has some striking patterns : http://remocable.blogspot.com/2011/08/blog-post.html some other inventions are really cool, http://remocable.blogspot.com/2011/08/old-time-inventions-genius-is-one.html and many other contraptions. bonne journÃ©e and best regards, Simon Plouffe

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Re: [math-fun] Gamma function (formulas collection)
by Bill Gosper 13 Jan '12

13 Jan '12

(Sorry about (no subject).) Hmm, "only" 26 digits with seven b*ggerfactors on z>9: http://gosper.org/munafacmu26.png Note that the z^7 means we only need ~20 digit values for the seven b*ggerfactors, but it looks like we'll need > eight, ior larger "9" for quad (34digit) precision. How many divides is a "+a5*z^2" worth? --rwg On Tue, Jan 10, 2012 at 4:12 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > Robert> > > I only need it to work for real arguments. But if that were an issue I > could stick with Lanczos. > > Any given set of Lanczos parameters work equally well on the entire > half-plane Real[z]>0, and with the same guaranteed error bound. This > article describes it really well: > > http://www.haoli.org/nr/bookcpdf/c6-1.pdf > > which is from: William H. Press et al., Numerical recipes in C: the art > of scientific computing (2nd edition). Cambridge University Press, > (1992) ISBN 0521431085. > > On Mon, Jan 9, 2012 at 12:06, Warren Smith <warren.wds(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=warren.wds%40gmail.com>> wrote: > > Do you want it to work for complex or only for real argument?>> > I've been looking for a numeric Gamma method that gives 34 digits of> > precision suitable for quad-precision floating-point [2] although I would> > settle for 31 digits suitable for "double-double".> > -- > Robert Munafo -- mrob.com > Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 -mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com > --------------- > > Just two b*ggerfactors, da*n near double precision on z>9: > http://gosper.org/munafacmu16.png > Probably could be double prec for z>10. > (Manually Remezed coefficients aren't finished, but mu won't change much.) > > I'm guessing quad will take eight b*ggerfactors. > --rwg > >

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Re: [math-fun] Fracfac expansions (fractions in factorial base)
by Bill Gosper 13 Jan '12

13 Jan '12

This is a bit skew, but for a direct matrix product conversion of the (non regular!) CF of x/atan x to y/(1-y^2)/the taylor series for atan y, where atan x = 2 atan y, see bottom of p14, p15, http://www.tweedledum.com/rwg/stanfordn3.pdf . We should look for a direct matrix CF/series conversion for tanh. --rwg rcs> Googling for 'e irrational' turns up a 2006 MAA article by Ed Sandifer that credits Euler in 1737 for proving the irrationality of e and e^2. The method is somewhat roundabout, proving the continued fractions from the Ricatti equation, and noting that the infinite CF -> irrational. My (admittedly rusty) memory is that the CF for tanh(x), and I1/I0, can be established directly by doing GCD-like steps on the power series for sinh/cosh; the subsequent steps in the GCD process have relatively clean power series. For full rigor, you'd need to prove some convergence results. But this makes the results a little less mysterious. Rich

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