January 2012 - mathfun@mailman.xmission.com

[math-fun] Gosperian PaIn physics?
by Warren Smith 27 Jan '12

27 Jan '12

Although I have no idea what "matrix product physics" might mean, the problem of rolling a unit sphere (in R^3) along a piecewise smooth path in the plane -- and figuring out its net rotation -- is essentially taking a limit of the product of more and more smaller and smaller matrices . . . and this limit very definitely depends on the path the sphere is rolled along. --Dan Asimov Marc LeBrun wrote: One more ravelet: seeing path invariance as a kind of symmetry, what form of conservation laws might that imply for a matrix product physics? --WDS: In the Feynman path integral (FPI) formulation of quantum physics, everything is the continuum (i.e. "integral" not "sum") version of a matrix (actually operator) product, summed over all possible paths. In most situations, though, the FPI is very dependent upon the path. However, supposing some kind of quantum physics were living in some land where FPI did NOT depend on the path -- either because something fundamentally is that way, or because it's some cheesy model scenario you cooked up -- then you'd have Gosperian PaIn. Or PIMP. Whichever of these two wonderful acronyms you prefer. (How it can be that Gosper's ideas have escaped great attention in spite of having not one, but two, amazing acronyms is truly pathetic.) In which case the FPI would become trivial to evaluate, I guess. Which would be good, since situations so far in which physicists have been able to evaluate FPIs in close form, have been quite rare. A usual rubric in physics ("Noether's theorem") is that "to every symmetry there corresponds a conservation law." What would that tell us in the case? Well, PaIn would really be a humongous symmetry, much larger than the usual symmetries like translation invariance (which corresponds to momentum and energy conservation). I'm not sure what the heck conservation law it would correspond to (if any); it does not seem to obey the usual monkey-see-monkee-do pigeonholing that allows idiots like me to operate? I guess it would constitute some kind of unification, where that sort of physics really does not care what the environment is. It is also possible, Feynman later saw, to formulae statistical physics via a different kind of path-integral.

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Re: [math-fun] [EXTERNAL] Re: Fun with math: Dividing one by 998001 yields a surprising result
by Ray Tayek 27 Jan '12

27 Jan '12

At 11:34 AM 1/27/2012, you wrote: >SetAccuracy[1/998001, 4000] gives the same problem (no 998) for me. i used java. import java.math.*; public class Main { public static void main(String[] args) { BigDecimal n = BigDecimal.valueOf(998001); MathContext mc = new MathContext(3010); BigDecimal x = BigDecimal.ONE.divide(n, mc); String s = x.toString(); for (int i = 0; i < 10; i++) System.out.println(s.substring(2 + 300 * i, 2 + 300 * (i + 1))); } } 000001002003004005006007008009010011012013014015016017018019020021022023024025026027028029030031032033034035036037038039040041042043044045046047048049050051052053054055056057058059060061062063064065066067068069070071072073074075076077078079080081082083084085086087088089090091092093094095096097098099 100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199 200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299 300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399 400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499 500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599 600601602603604605606607608609610611612613614615616617618619620621622623624625626627628629630631632633634635636637638639640641642643644645646647648649650651652653654655656657658659660661662663664665666667668669670671672673674675676677678679680681682683684685686687688689690691692693694695696697698699 700701702703704705706707708709710711712713714715716717718719720721722723724725726727728729730731732733734735736737738739740741742743744745746747748749750751752753754755756757758759760761762763764765766767768769770771772773774775776777778779780781782783784785786787788789790791792793794795796797798799 800801802803804805806807808809810811812813814815816817818819820821822823824825826827828829830831832833834835836837838839840841842843844845846847848849850851852853854855856857858859860861862863864865866867868869870871872873874875876877878879880881882883884885886887888889890891892893894895896897898899 900901902903904905906907908909910911912913914915916917918919920921922923924925926927928929930931932933934935936937938939940941942943944945946947948949950951952953954955956957958959960961962963964965966967968969970971972973974975976977978979980981982983984985986987988989990991992993994995996997999000 --- co-chair http://ocjug.org/

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Re: [math-fun] Fun with math: Dividing one by 998001 yields a surprising result
by Dan Asimov 27 Jan '12

27 Jan '12

What am I doing wrong? (seriously): << In[3]:= N[1/99801,500] Out[3]= 0.0000100199396799631266219777356940311219326459654712878628470656606 . . . >> --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Fun with math: Dividing one by 998001 yields a surprising result
by Ray Tayek 27 Jan '12

27 Jan '12

http://www.iheartchaos.com/post/16393143676/fun-with-math-dividing-one-by-9… --- co-chair http://ocjug.org/

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[math-fun] surprises #4 & 5
by Bill Gosper 27 Jan '12

27 Jan '12

I described my inability to "undouble" k in a 2x2 system whose n matrix was of the form N(2 k, n), which requires finding K(k, n) given K(2 k, n).K(2 k + 1, n). In the case of upper triangular K, this is almost the same problem as reconstructing a series after its terms have been grouped pairwise, and requires solving linear recurrences with polynomial coefficients. (Does anybody know if there's such a solver in Koutschan's *HolonomicFunctions*.*m*?)* *Back when I was smarter, I found how to "untriple" k in the Rosetta system, getting diagonal N matrices of the form N(h,i,j,k/3,(k+1)/3,(k+2)/3,n), which induced interesting K matrices. So what happens when you untriple a staggered sum? In order to impose Dixon's condition that each numerator factor pair with a denominator factor to make equal sums, the N matrix must look like [ k + 1 k k - 1 k j ] [ (n + -----) (n + -) (n + -----) - + - - 1 ] [ 3 3 3 3 3 ] [ ------------------------------- 0 n + --------- ] [ j - 1 j j + 1 2 ] [ (n + -----) (n + -) (n + -----) ] [ 3 3 3 ] [ ] [ k + 1 k k - 1 ] [ (n + -----) (n + -) (n + -----) ] [ 3 3 3 ] [ 0 ------------------------------- 1 ] [ j - 1 j j + 1 ] [ (n + -----) (n + -) (n + -----) ] [ 3 3 3 ] [ ] [ 0 0 1 ] I.e., we must untriple wrt j as well as k. But I forget how to untriple. Is there some way to trick the *N(h,i,j,k/3,(k+1)/3,(k+2)/3,n)* system to give * N(j/3,(j+1)/3,(j+2)/3,k/3,(k+1)/3,(k+2)/3,n)*, even though k is in the numerator and j in the denominator? Yes, you can reciprocate the term ratio by inverting all the matrices with the transformation h->-h-1, i->-1-i, j->-j-1, k->-k-1, n->-n-1, and then cobble together the two systems by renaming variables. Interestingly, for q matrices, (where you have 1-q^n instead of n), you also need to reciprocate q. This reflects the existence of a seventh dimension (unfortunately, the next letter is e) in the q-Rosetta grid providing four-term recurrences relating 3φ2[...,q^e], 3φ2[...,q^(e+1)], etc. I.e., we also need e->-e-1. (That was surprise #4.) (Hypergeometrician geeks will recognize the 1,3 element in this Dixon N matrix as the fudge from "well-poised" to "very well-poised".) So when we stagger *and* untriple, we get surprise #5, a simple, rapidly convergent, but nontriangular J product [ j - k (j - k) (- k + j + 1) (j - 2) (j - 1) ] inf [ ----- - --------------------- --------------- ] /===\ [ 2 j 36 j 9 (- k + j - 1) ] | | [ ] | | [ 3 - k + j + 1 ] | | [ - ----------- 0 ] j = 1 [ j 2 j ] [ ] [ 0 0 1 ] (which I have in closed form as a pile of Gamma fns) which only becomes upper- triangular (and messy!) when sextupled (grouped six at a time)! Try it in Mma: K[j_, k_] :> {{(j - k)/(2*j), -((j - k)*((-k + j + 1)/(36*j))), (j - 2)*((j - 1)/(9*(-k + j - 1)))}, {3/j, (-k + j + 1)/(2*j), 0}, {0, 0, 1}} FullSimplify[Dot @@ Table[K[a + i, b] /. %, {i, 3}]] FullSimplify[Dot @@ Table[K[a + i, b] /. %%, {i, 6}]] This is the clearest evidence I've seen of what we're missing by insisting on summation and pFq notation. They don't even work well for sums. --rwg BtW, PaIn is a Warrenism. I don't know how he resisted Path Invariant Matrix Products.

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[math-fun] Can we make fast divisibility tests for binary computers?
by Warren Smith 26 Jan '12

26 Jan '12

After all that about how we humans working in base 10 can quickly test divisibility by some small primes by some ingenious tricks... who cares about base 10? Think binary so you can be cool. * N divisible by 2 <==> has a zero at the end <==> (N&1)==0. * Note that 63 = 3*3*7. If you add the consecutive 6-bit blocks of N together, result is congruent to N mod 63. Similarly for other 2^k-1 such as 3, 7, 15=5*3, 31=prime, 63=3*3*7, 127=prime, 255=5*3*17, 511=7*73 etc. * Note that 65=5*13. If you add and subtract alternately the 6-bit blocks of N then you get something congruent to N mod 65. Similarly for other 2^k+1 such as 33=3*11, 129=3*43, 257=prime, 513=3*3*3*19 etc. The numbers 2^k +- 2^j +- 1 would seem to be particularly important for this kind of game. The numbers 2^(2^k)-1 are especially important since computer wordlengths are powers of 2, and they have a lot of nice factors. Related question to "Can we make fast divisibility tests for binary computers?" is "...and should anybody really want to?"

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Re: [math-fun] Strong pseudoprimes [Jaeschke, Gilchrist, Feitsma]
by Phil Carmody 26 Jan '12

26 Jan '12

From: Warren Smith <warren.wds(a)gmail.com> > Gerhard Jaeschke: On strong pseudoprimes to several bases, > Maths. of Comput. 61,204 (1993) 915-926. > http://cr.yp.to/bib/1993/jaeschke.pdf > > Jaeschke also claims that there are no strong pseudoprimes below 10^12, > common to the four prime bases A=2, 13, 23, and 1662803. Several people have independently looked at pushing those limits. It's been a very long time since I looked at this field, but I think that Jason Moxham was the first to hit several of the limits, but used a slightly stronger combination of tests that Jaeschke did to reduce the number of composites which could slip through, and published his (as usual completely unreadable) code so that his techniques could be verified. A pair (or more) from perhaps China had done similar searches, but used even stronger sets of conditions that further reduced the number of composites that would slip through. They did publish the theoretical basis and findings in an international journal, but didn't supply their code. IIRC, Jason used that quartic character to, where possible, capture the square root of -1. I notice that Jaeschke mentions the quartic character in his paper, but doesn't seem to make use of its value. Any two witnesses to pseudoprimality which provide such a square root must, up to sign, provide the same primitive 4th root of unity. (and if they don't, you've factored the number.) Unfortunately this isn't such a big win as it first seems as you don't always end up with a primitive root. So you only gain when you're already in the cases where many tests are needed. However, when you get the square root you get it for free, so it seems wasteful not to do a simple comparision once you get two of them. Also IIRC, the Chinese pair generalised this to, where they exist, primitive 3rd roots of unity too. This only works when p==1(3), but that's half of the time. Again, you capture the 3rd root, and any two witnesses which provide a primitive one must, up to sign, supply the same one. (ditto re. factoring.) This can be generalised to 5th roots which, up to sign or squaring be identical, but that only aplies to p==1(5). Phil

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Re: [math-fun] Gosper's "path invariance" in matrix product land - symmetry & conservation
by Dan Asimov 26 Jan '12

26 Jan '12

Although I have no idea what "matrix product physics" might mean, the problem of rolling a unit sphere (in R^3) along a piecewise smooth path in the plane -- and figuring out its net rotation -- is essentially taking a limit of the product of more and more smaller and smaller matrices . . . and this limit very definitely depends on the path the sphere is rolled along. --Dan Marc wrote: << One more ravelet: seeing path invariance as a kind of symmetry, what form of conservation laws might that imply for a matrix product physics? >> ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Gosper's "path invariance" in matrix product land
by Warren Smith 26 Jan '12

26 Jan '12

as I've told Gosper & co, I had invented in an unfinished paper (that I've occasionally been fiddling with for many years) my own ideas about what I called the "matprod framework" for function approximation based on matrix products... but had kept getting the feeling the Gosper had somehow been there ahead of me. Now that I finally see Gosper's ideas, wow! In a lot of ways he WAS there ahead of me, plus his whole "path invariance" PaIn idea had never occurred to me and is really something much vaster than what I'd been working on (which was merely function approximation and algorithms). I can hardly comprehend the implications of PaIn, he has evidently found something incredibly vast here, but it seems plausible that only the tiniest fraction of it has yet been explored. For one example, he illustrates the path invariance idea on a "square grid"; but what about in the "continuum limit" as the grid side --> 0? Wouldn't that be a whole new kind of analysis? This is serious. This really could be the missing setting people have been wanting to do theoretical physics, and heaven knows what else. I had once seen the version in the Chudnovsky book, but had dismissed it as basically just random ravings. It's an outrage they printed that version. > The arboricidal version is > > Gosper, R. Wm., Strip Mining in the Abandoned Orefields of Nineteenth > Century Mathematics, Computers in Mathematics, (D. Chudnovsky & R. Jenks, > eds.), Lecture Notes in Pure and Applied Mathematics, Vol 125 (1990), p282. > but they printed my earliest draft instead of one of the many revisions I > sent > during the long publication delay. Better: > http://www.tweedledum.com/rwg/stanfordn1.pdf > http://www.tweedledum.com/rwg/stanfordn2.pdf > http://www.tweedledum.com/rwg/stanfordn3.pdf > http://www.tweedledum.com/rwg/stanfordn4.pdf > (The machine I TeXed them on was too microcephalic > to typeset more than a handful of pages per file! > And it couldn't import graphics--there are some > missing lines and a curve that were drawn by hand.) > --rwg

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Re: [math-fun] Approximating the Gamma function via odd-symmetric series without use of log or exp / Maple
by Axel Vogt 25 Jan '12

25 Jan '12

On 24.01.2012 22:52, math-fun-request(a)mailman.xmission.com wrote: ... > As you can see, the speed of convergence is pretty spectacular; 12 > terms in each series > will give you error<10^(-37). > > Were Gosper to Rational-Remez these it should become slightly better still. > In MAPLE, you allegedly can get it to find rational > Remez best approximation of degree 7/degree 5 by > with(numapprox); > wght := (x) -> 1; > minimax(q1, -1/4 - 10^(-87)..1/4 + 15.4632*10^(-88), [7, 5], wght, > 'maxerror' ); > #notice I have used a slight random enlargement of the > #interval -1/4, 1/4 to prevent MAPLE > #from activating some bullshit... but then it just gives different > kind of failure > > but in fact, I have never managed to make MAPLE actually succeed after > several tries, > gave up. MAPLE has a lot of bugs. side remark (for q1(x) = (1/GAMMA(x+1/2)-1/GAMMA(-x+1/2))/x) and Maple) concerning Maple: it can be made to work, but needs high working precision (and I lost overview what is the one in the current discussion), Digits=8*16 will give an error of ~ 0.4*1e-20 using the command on - 1e-87 ... 1/4 + tiny and now using symmetry in zero gives it. I guess one reason for troubles is degree = (7,5), while the function is one in x^2. If writing it as r1(x):= q1( sqrt(abs(x)) ), i.e. q1(x) = r1(x^2) then the same command but for r1 with degree = (7,5) gives an error of 0.1*1e-31. With degree = (7,6), thus even degree as denominator, error is ~ 0.5*1-e36.

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