Math help: Calculate pi using Gregory's Series on Beowulf?
Chris Richard Adams
chrisa at ASPATECH.COM.BR
Fri Aug 3 15:55:28 PDT 2001
Thanks, but I don't understand your notation:
arctan(x) = int_0^x[dy 1/(1 + y^2)]
how is a function of x dependent on y? I'll look this up, but is this
where you are saying the expression in mpi-beowulf/examples/ came from?
> -----Original Message-----
> From: Martin Siegert [mailto:siegert at sfu.ca]
> Sent: Friday, August 03, 2001 2:49 PM
> To: Chris Richard Adams
> Cc: beowulf at beowulf.org
> Subject: Re: Math help: Calculate pi using Gregory's Series
> on Beowulf?
> 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
> Did you know that the quadrillionth bit of pi is '0' ?
> 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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