August 2010 - mathfun@mailman.xmission.com

Re: [math-fun] pi cf record
by Bill Gosper 06 Aug '10

06 Aug '10

YOW! 878783625 is still the largest know term?? Bill Gosper: > Has anyone listed the largest terms in Hans's data? I have. :) http://chesswanks.com/seq/cfpi/

1 0

[math-fun] Cantor, Pick and Fibonacci
by Victor Miller 06 Aug '10

06 Aug '10

I just got the new (August-September 2010) of the American Math Monthly. In it is the following delightful article "On Cantor's First Uncountability Proof, Pick's Theorem, and the irrationality of the Golden Ratio" http://www.math.jhu.edu/~wright/Cantor_Pick_Phi.pdf The authors (Mike Krebs and Thomas Wright) first discuss Cantor's first proof (given two years before his famous diagonalization argument) of the uncountability of the reals, which is based on an algorithm which takes as input a countable sequence of reals (a_n) with the "intermediate value property" (every term in the sequence lies strictly between two others) and produces two other sequences whose elements come from the sequence (a_n), called (b_n) and (c_n) which are both monotone of opposite growth (i.e. if the first in increasing then the second is decreasing, etc.) and on complementary sides (i.e. if the first is increasing then every term of the second is bigger than every term of the first, etc.). Cantor then shows that the LUB of the increasing sequence can't be in the the original sequence. The authors apply this algorithm to the sequence of rationals in (0,1) ordered by denominator, and then within constant denominator by numerator. What they first find, experimentally, is that the elements of (b_n) and (c_n) are all ratios of consecutive Fibonacci numbers. They prove this result by induction and an appeal to Pick's theorem (which gives the number of lattice points in the interior of a lattice polygon in terms of the area and the boundary), and thus use Cantor's original argument to prove the irrationality of the golden ratio. Victor

1 0

Re: [math-fun] 5000 billion digits of Pi
by rcs＠xmission.com 06 Aug '10

06 Aug '10

SBG replies ... ------- Date: Thu, 05 Aug 2010 20:42:41 -0700 Subject: Re: [math-fun] 5000 billion digits of Pi From: "Stephen B. Gray" <stevebg(a)roadrunner.com> To: math-fun <math-fun(a)mailman.xmission.com> 2.1 unnamed [text/html] 1.19 KB Download On 8/5/2010 3:27 PM, Eugene Salamin wrote: > The most evident objection to this kind of calculation is that base > 10 is just an arbitrary choice. A continued fraction result would > avoid this objection, and might actually be mathematically useful. > The only such calculation I'm aware of was the one by RWG some 30 > years ago that found 17 million terms. It appeared that pi has > large denominators appearing earlier than would be expected for a > random real number. > > Has anyone ever used these gazillion digits of pi for some purpose > besides setting a new record? Here's a goal for the new decade: > calculate pi to a quadrillion places, on a cell phone, while riding > a unicycle. > > -- Gene > > > That would be an impressive stunt but it will NOT help me calculate the circumference of my wading pool. Please, let's be practical. Steve Gray

1 0

[math-fun] pi cf record
by Bill Gosper 06 Aug '10

06 Aug '10

H.Havermann>In 1985. I, for one, was inspired by his effort. In June 1999, Mathematica spewed out 20 million terms for me on my personal computer and in October 2000, 53 million terms. And if I could up that number (alas, my current machine is seven years old), I would. ;) _______________________________________________ es>What pi algorithm did you use, arctan, AGM, Ramanujan's 1/pi, etc? Or, did you just use whatever was built into Mathematica? -- Gene --------------------------- The relative cost of the pi definition is insignificant compared to the efficiency of the c.f. conversion algorithm. Straightforward algorithms are hopeless. I am impressed that Mathematica has solved this, presumably with a divide-and-conquer approach designed around FFT multiply. Has anyone listed the largest terms in Hans's data? The GM?. The "density equivalent radix"? E.g., the lg(1+r) distribution predicts the information in a typical term to be about one digit in base 11. More testimony to the irrelevance of decimal! I actually checked my value by converting to the 8.5M/8.5M rational approximant (grouping terms pairwise--the opposite of divide-and-conquer). This was on a Lisp machine with "unlimited" precision rational arithmetic. What actually came out of the Ramanujan series was round(pi*2^58000000). (And before that, an enormous rational with a multimillion factorial gcd.) I've probably bored you all before with the tale of how desk-checking managed to uncover a bug that eluded my programmed error-checking with a (naive) probability of 2^-63. --rwg If pi bore any simple relationship to e, its c.f. would exhibit regular growth. But maybe we just haven't looked far enough.-)

3 2

[math-fun] Correction and confluent quadratic/linear cf.
by Bill Gosper 06 Aug '10

06 Aug '10

DanA points out that my last c.f. msg should have begun The cf with nth tail a[n]+b[n]/(a[n+1]+b[n+1]/ is computed by n a[n] b[n] P[n] p[n] prod ( ) = ( ) 1 0 Q[n] q[n] so that P[n+1] p[n+1] P[n] p[n] a[n+1] b[n+1] ( ) = ( ) ( ). Q[n+1] q[n+1] Q[n] q[n] 1 0 Also, my tutor Julian managed to reduce the quadratic/linear identity to Out[555]= K[e + n*(d - c^2*n), b + 2*c*n, {n, 0, oo}] -> (e*U[1 - (d + Sqrt[d^2 + 4*c^2*e])/(2*c^2), 1 - Sqrt[d^2 + 4*c^2*e]/c^2, -1 + b/c + d/c^2])/ (c*U[-((d + Sqrt[d^2 + 4*c^2*e])/(2*c^2)), 1 - Sqrt[d^2 + 4*c^2*e]/c^2, -1 + b/c + d/c^2]) In[560]:= %555 /. N[{b -> 69/10, c -> 105/100, d -> 288/10, e -> 239/10, K -> ContinuedFractionK, oo -> 19, U -> HypergeometricU}, 22] Out[560]= 2.2406188150339304802 -> 2.240618815033931406 In[561]:= %555 /. N[{b -> 69/10, c -> 105/100, d -> 288/10, e -> 239/10, K-> ContinuedFractionK, oo -> 22, U -> HypergeometricU}, 22] Out[561]= 2.2406188150339314056 -> 2.240618815033931406 where HypergeometricU is Kummer's U, defined in A&S 13.1.3 as the sum of two 1F1s weighted by various Gammas, a trig, and a power. It is probably safe not to learn the details and rely on FunctionExpand to know all the special values. Which seem to scrupulously exclude (for rational parameters) any simple function of e:=2.718, preserving the conjecture that there is no confluent series or cf for pi, and no nonconfluent formula for e. --rwg

1 0

[math-fun] CF convergence rate
by Bill Gosper 06 Aug '10

06 Aug '10

(This must be in Wall or Perron.) The cf with nth tail a[n]+b[n]/(a[n+1]+b[n+1]/ is computed by n a[n] b[n] P[n] p[n] prod ( ) = ( ) 1 0 Q[n] q[n] so that P[n+1] p[n+1] P[n] p[n] a[n] b[n] ( ) = ( ) ( ) Q[n+1] q[n+1] Q[n] q[n] 1 0 and P[n]/Q[n] - p[n]/q[n] estimates the uncertainty so far. Therefore P[n+1]/Q[n+1] - p[n+1]/q[n+1] r[n] := ----------------------------- P[n]/Q[n] - p[n]/q[n] as a measure of convergence is analogous to the term ratio of an ordinary series. I.e., -lg|rn| bits/term. Eliminating P, Q, p, and q, we get the peculiar result r[n] = a[n]*ccf[n+1]/b[n]-1, where ccf[n] is the "complementary cf" := -a[n]+b[n]/(-a[n+1] + b[n+1]/... . Approximating r[n+1] by r[n], r[n] ~ (sqrt(1+4*c[n])-1)/2/c[n] - 1 where c[n]:= b[n]/a[n]/a[n+1]. For a[n] := a*n^k+..., b[n] := c*n^(2*k)+..., r[n] -> (|a|*sqrt(4*c+a^2)-a^2)/2/c - 1 for large n. E.g., for Apery's continued fraction, k=3, a=34,and c=-1, giving r[oo] = 577-408*sqrt(2) = (1+sqrt(2))^-8. Note that for c->0 (e.g. linear/linear when k=1), r[n]->0. This is analogous to the term ratio -> 0 for e=sum(1/n!), which says that taking this limit won't tell us how many terms we need. At this point I planned to use the above to determine the optimal y for the twin cf Gamma formula but, experimentally, y=z kicks butt, and I'm too lazy to prove why. --rwg

1 0

[math-fun] 5000 billion digits of Pi
by Simon Plouffe 06 Aug '10

06 Aug '10

Apparently, a certain Kondo from Japan in collaboration with Alexander Yee over internet successfuly computed pi to 5000 decimal digits, There is no mention of method, timing, what computer , etc. here is the news in french http://www.starwizz.com/japon-nombre-pi-calculs-67780 they mention it somewhere on the 'Le Monde' in Paris, best regards and have a nice day, simon plouffe

4 7

Re: [math-fun] 5000 billion digits of Pi
by Dan Asimov 05 Aug '10

05 Aug '10

Steve G. wrote: << Would five trillion digits be enough to compute the circumference of my wading pool? Its diameter is five feet, more or less. >> It depends on how accurate you want to be. Since you already introduced significant inaccuracy with the very rough estimate of your pool's diameter, you wouldn't want any additional inaccuracy to creep in by using only 5x10^12 digits of pi. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

1 0

Re: [math-fun] 5000 billion digits of Pi
by rcs＠xmission.com 05 Aug '10

05 Aug '10

from SBG ... ------------- Date: Thu, 05 Aug 2010 10:40:48 -0700 Subject: Re: [math-fun] 5000 billion digits of Pi From: "Stephen B. Gray" <stevebg(a)roadrunner.com> On 8/5/2010 10:33 AM, Hans Havermann wrote: >> There is no mention of method, timing, what computer , etc > > > > http://www.numberworld.org/misc_runs/pi-5t/details.html > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > Would five trillion digits be enough to compute the circumference of my wading pool? Its diameter is five feet, more or less. Steve Gray

1 0

[math-fun] Number Theory 101
by Hans Havermann 04 Aug '10

04 Aug '10

The smallest decimal-digits prime composed of the concatenation of 7 distinct decimal-digit primes is 1113257317, composed of 11, 13, 2, 5, 7, 3, and 17. What are the first three digits of the smallest decimal- digits prime composed of the concatenation of 26 distinct decimal- digit primes? (Bonus marks: Allowing probable primes, what is the entire number?)

3 4