# Math help: Calculate pi using Gregory's Series on Beowulf?

Martin Siegert siegert at sfu.ca
Fri Aug 3 10:49:04 PDT 2001

```On Fri, Aug 03, 2001 at 01:19:29PM -0300, Chris Richard Adams wrote:
> Calulating Pi.
>
> I found a link to compute pi using Gregory's Series. It states the form
> allowing quickest convergence
> is..
>
> 	pi = 2*sqrt(3)*[ 1 - (1/(3*3)) + (1/(5*3^2)) - (1/(7*3^3)) ... ]
>
> In the example for computing pi that comes with MPICH/MPI-Beowulf, I see
> something like...
>
> 	Sum = Sum + 4 * [ 1 / ( 1 + ( 1/n * ( i - 0.5 ) )^2 ) ]
>
> 	where we iterate n times and i increments from 1 to n.
>
> 	- then, pi = 1/n * sum
>
> ANyone see how to get from Gregory's series to this form, or does this
> form evolve from a different
> approach?  I'll keep trying... any feedback is appreciated.

This is different:

pi = 4 arctan(1) and arctan(x) = int_0^x[dy 1/(1 + y^2)]

If you want to know how to calculate pi check

http://www.cecm.sfu.ca/projects/pihex

Did you know that the quadrillionth bit of pi is '0' ?

Martin

========================================================================
Martin Siegert
Academic Computing Services                        phone: (604) 291-4691
Simon Fraser University                            fax:   (604) 291-4242
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