Re: For loop problem in mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg90288] Re: For loop problem in mathematica
- From: "Dana DeLouis" <dana.del at gmail.com>
- Date: Fri, 4 Jul 2008 04:00:15 -0400 (EDT)
> x=0;For[i = 1, i < 10,000,000, i++, x = x + 1/(i^2)];N[Sqrt(6*x),25]
I know this is a question about looping, but just to mention...
Sum[1/r^2, {r, Infinity}]
Pi^2/6
HarmonicNumber[Infinity, 2]
Pi^2/6
Or...just use the '2 (ie i^2) in the Zeta function
Zeta[2]
Pi^2/6
Hence...
Sqrt[6*Zeta[2]]
Pi
--
HTH
Dana DeLouis
"PhysNova" <skhoshbinfar at gmail.com> wrote in message
news:g4d2qf$emo$1 at smc.vnet.net...
> Hi,
> i wrote a simple program of evaluating Pi number in M6 to test cpu
> computation timing, to do this a simple for loop
>
> was used:
>
> x=0;For[i = 1, i < 10,000,000, i++, x = x + 1/(i^2)];N[Sqrt(6*x),25]//
> Timing
>
> the result was catastrophe! it take few minuates. but i first expect
> to do this very simple job in few
>
> seconds.computation time is just satisfactory up to 100000 cycle.
>
> could anyone interperet this falut?
> our get an idea to improve the result?
>
> Tanks
>