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

[Chris Richard Adams]

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?

Thanks

> -----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 
> 
> 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
> Burnaby, British Columbia                          email: 
> siegert at sfu.ca
> Canada  V5A 1S6
> ==============================================================
> ==========
>