Solving Pi

[M.J.Shannon]

> In article <187 at ski.UUCP> eeg at ski.UUCP (eeg systems (bcx) writes:
> [ article deleted - program to calculate pi to X digits ]
> >**	      infinity                    infinity
> >**	      ____  16*(-1e(k+1))         ____  4*(-1e(k+1))
> >**	      \                           \
> >**	pi =   >    -------------    -     >    ------------
> >**	      /                           /
> >**	      ----  (2k-1)*5e(2k-1)       ----  (2k-1)*239e(2k-1)
> >**	      k = 1                       k = 1
> >**
> 
> Here is a more simple sum-evaluation of pi (thought I don't know if it will work
> with the original program):
> 
>                oo
>             ----         k-1
>             \        (-1)
>     pi = 4   >    ----------
> 	      /       2k-1
> 	      ----
> 	      k=1

The reason why the 2 infinite sums are used is that the single infinite sum
converges *VERY* slowly.
-- 
	Marty Shannon
UUCP:	ihnp4!eagle!mjs
Phone:	+1 201 522 6063