mersenneforum.org NEW YOU FORMULATE THAT THEY CALCULATE PI
 Register FAQ Search Today's Posts Mark Forums Read

 2007-04-09, 16:30 #1 ewmayer ∂2ω=0     Sep 2002 República de California 3·7·13·43 Posts NEW YOU FORMULATE THAT THEY CALCULATE PI I received the following in an e-mail over the weekend (during which time I was offline) - anybody care to see if there is anything new or at all interesting (e.g. in the sense of "yes, this is the well-known Fu-Bar series expansion...") here? I'm at work, no time to examine it in detail at this time. The integral formulae at least seem likely to be fairly trivial, possibly Fourier series (which often yield Pi-containing results by way of series expansions) in disguise. Anyway, might be some fun math exercises here. -Ernst p.s.: please try not to let laughter at the Borat-style mangled english subject line distract you from your work here. ;) p.p.s.: If someone would be kind enough to format the attached in nice LaTeX-style, that would be appreciated. Off to work... Code: ========================================================================== NEW YOU FORMULATE THAT THEY CALCULATE PI ========================================================================== FIRST FORMULA infinity n! PI 1 - SUM (--------------------------------) = ------ n=0 n 4 2 * PRODUCT ( 2 * p + 3 ) p=0 ========================================================================== SECOND FORMULA H[0] = 1 H[n] H[n + 1] = --------------------------- 1 + ( 1 + H[n]^2 )^(1/2) 2^(n + 2) * H[n] > PI ========================================================================== THIRD FORMULA 1 INTEGRAL ( ----------------------- ) between -1 and 0 X^2 1 + X + ------ 2 that he is equal to PI/2 developing in series the integral it is obtained formulates it 1 1 infinity 1 + ----- + ------- - SUM [ (-1)^(n+1) * 2 2*3 n=1 1 1 1 ( ------------------- + ------------------ + -------------------- ) ] 2^(2*n) * (4*n+1) 2^(2*n) * (4*n+2) 2^(2*n+1) * (4*n+3) that he is equal to PI/2 ========================================================================== FOURTH FORMULA 1 INTEGRAL (----------------------) between -1 and 0 x x^2 1 + ----- + ------ 2 4 that he is equal to (2 * SQRT(3) * PI ) / 9 developing in series the integral it is obtained formulates it 1 infinity 1 + ------ + SUM [ (-1)^n * 2*2 n=1 1 1 ( ------------------- + -------------------- ) ] 2^(3*n) * (3*n+1) 2^(3*n+1) * (3*n+2) that he is equal to ( 2 * SQRT(3) * PI ) / 9 ========================================================================== FIFTH FORMULA A[0] = 4 B[0] = SQRT( 1/2 ) 2 * A[n] * B[n] A[n + 1] = ------------------- 1 + B[n] 1 + B[n] B[n + 1] = SQRT ( ------------ ) 2 A[n] > A[n + 1] > PI ==========================================================================
 2007-04-09, 21:54 #2 T.Rex     Feb 2004 France 929 Posts First formula works PARI/gp: Code: G(N)= {x=0; for(n=0,N,y=1;for(p=0,n,y=y*(2*p+3));x=x+(factorial(n)/(2*y)));print((1-x)*4/Pi)} In TeX, the formula 1 is: $1-\sum_{n=0}^{\infty}\frac{n!}{\ \ \ \ 2 \prod_{p=0}^{n} (2 p+3)}=\frac{\pi}{4}$ Now time to go to bed ... Tony Last fiddled with by T.Rex on 2007-04-09 at 21:55

 Similar Threads Thread Thread Starter Forum Replies Last Post Stargate38 Factoring 7 2015-05-27 23:09 pepi37 Riesel Prime Search 8 2014-04-17 20:51 jmb1982 PrimeNet 1 2009-02-23 15:13 roemer2201 PrimeNet 2 2008-12-20 16:12 fropones Lone Mersenne Hunters 1 2003-05-27 23:01

All times are UTC. The time now is 04:44.

Wed Aug 17 04:44:48 UTC 2022 up 40 days, 23:32, 1 user, load averages: 0.66, 0.91, 1.02