Math help: Calculate pi using Gregory's Series on Beowulf?
Carlos O'Donell Jr.
carlos at baldric.uwo.ca
Fri Aug 3 14:10:55 EDT 2001
> > 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 ) ]
> > - then, pi = 1/n * sum
> 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
> Did you know that the quadrillionth bit of pi is '0' ?
You can also check out:
The artcan calucation of pi has logarithmic convergance, and
thus is a very poor performer. See the AGM algorithms used today!
I think I'll have to go pickup a "calculate pi" program and see
how big I cang get it! :)
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