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RE: [PrimeNumbers] pi=(2/1)x(3/2)x(5/6)x(7/6)x(11/10)x(13/14)x(17/18)x(19/18)...

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  • Tim Roberts
    Is this product mentioned in any of the literature? I haven t been able to find it anywhere previous to Eric s email! Tim ... From: mikeoakes2@aol.com
    Message 1 of 2 , Dec 21, 2003
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      Is this product mentioned in any of the literature? I haven't been able to find it anywhere previous to Eric's email!


      -----Original Message-----
      From: mikeoakes2@... [mailto:mikeoakes2@...]
      Sent: Sun 21/12/2003 8:07 PM
      To: primenumbers@yahoogroups.com
      Subject: Re: [PrimeNumbers] pi=(2/1)x(3/2)x(5/6)x(7/6)x(11/10)x(13/14)x(17/18)x(19/18)...

      In a message dated 21/12/03 02:05:29 GMT Standard Time, lh2072@...

      > My guess would be
      > pi =infiniteproduct(over all primes = 1 mod 4 of p/(p+1)) *
      > infiniteproduct(over all other primes of p/(p-1))
      > Next terms:
      > 29/30 * 31/30 * 37/38...

      You've got it!

      And here is the proof of why the sum = pi.

      Define the "character" chi(p) by:-
      chi(p) = +1 if p is a prime = +1 mod 4
      chi(p) = -1 if p is a prime = -1 mod 4
      chi(p) = 0 for all other integers p > 0.

      Eric's product is:-
      = 2/P, say
      where P=(2/3)*(6/5)*(6/7)*(10/11)*(14/13)*...
      = (1-1/3)*(1+1/5)*(1-1/7)*(1-1/11)*(1+1/13)*...
      = product{p odd prime}(1+chi(p)/p)
      = product{p odd prime}(1-1/p^2) / product{p odd prime}(1-chi(p)/p)
      = [(4/3)* product{p prime}(1-1/p^2)] / product{p odd prime}(1-chi(p)/p)
      = [(4/3)/zeta(2)] * L(chi,1)
      = [(4/3)/(pi^2/6)] * [pi/4] = 2/pi
      where zeta(s) is the Riemann zeta function and L(chi,s) is the Dirichlet
      L-function associated with chi().

      A useful reference for the background to all this is:

      -Mike Oakes

      [Non-text portions of this message have been removed]

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