Solving Pi
M.J.Shannon
mjs at eagle.UUCP
Mon Jul 29 02:39:13 AEST 1985
> 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
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