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MathGroup Archive 2000

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Re: formula for Pi

  • To: mathgroup at smc.vnet.net
  • Subject: [mg22036] Re: formula for Pi
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Fri, 11 Feb 2000 02:38:17 -0500 (EST)
  • Organization: Universitaet Leipzig
  • References: <87tq26$5ji@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

go to the next book store and buy/order 

"Mathematica in Action"
S. Wagon
TELOS/Springer 1999
Second edition

it has a whole chapter about Pi and the ways to 
compute it with Mathematica. 

But

Sum[(1/16)^k (4/(8k + 1) - 2/(8k + 4) - 1/(8k + 5) - 1/(8k + 6)),
    {k, 0, Infinity}] // FullSimplify

will give Pi.

Regards
  Jens

Arnold wrote:
> 
> The following remarkable identity for Pi can be used to calculate the
> nth hexadecimal
> digit of Pi without calculating first the earlier digits.
> Mathematica 4.0 simplifies the sum in terms of hypergeometric functions
> 
> In[2]:=
> pi = Sum[(1/16)^k (4/(8k + 1) - 2/(8k + 4) - 1/(8k + 5) - 1/(8k + 6)),
> {k, 0,
>       Infinity}]
> Out[2]=
> \!\(\(-2\)\ ArcTanh[1\/4] + 4\ Hypergeometric2F1[1, 1\/8, 9\/8, 1\/16] -
> 
>     1\/5\ Hypergeometric2F1[1, 5\/8, 13\/8, 1\/16] -
>     1\/6\ Hypergeometric2F1[1, 3\/4, 7\/4, 1\/16]\)
> 
> Can one use Mathematica to show that this last expression equals Pi?
> 
> (In the December 1999 issue of the American Mathematical Monthly p.903
> it is shown how to prove the sum equals Pi using another system.)
> 
> Arnold Knopfmacher
> Witwatersrand University
> South Africa


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