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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
>



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