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
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Martin Siegert
Academic Computing Services phone: (604) 291-4691
Simon Fraser University fax: (604) 291-4242
Burnaby, British Columbia email: siegert at sfu.ca
Canada V5A 1S6
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