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Re: product convergence

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  • djbroadhurst
    ... See Hardy and Wright Sections 22.7 and 22.8 for the Mertens product prod(p
    Message 1 of 8 , May 4, 2010
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      --- In primenumbers@yahoogroups.com,
      "tsr21" <timro21@...> wrote:

      > Is there a simple proof of this?

      See Hardy and Wright Sections 22.7 and 22.8
      for the Mertens product

      prod(p<x, (p-1)/p) ~ exp(-Euler)/log(x)

      where the product runs over primes.

      As Andrey remarked, your product tends to zero faster:

      prod(2<p<x, (p-2)/p) = O(1/log(x)^2)

      David
    • djbroadhurst
      ... So let s work out the constant, say K, in prod(2
      Message 2 of 8 , May 4, 2010
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        --- In primenumbers@yahoogroups.com,
        "djbroadhurst" <d.broadhurst@...> wrote:

        > prod(2<p<x, (p-2)/p) = O(1/log(x)^2)

        So let's work out the constant, say K, in

        prod(2<p<x, (p-2)/p) ~ K/log(x)^2

        We should use the twin-prime constant

        C2 = prod (2<p, p*(p-2)/(p-1)^2) = 0.6601618158...

        and then use the square of Mertens' formula,
        remembering that the latter includes p = 2.

        K = C2*(exp(-Euler)*2)^2 =
        0.832429065661945278030805943531465575045445318077417053240894...

        Sanity check:

        default(primelimit,10^8);
        \p5
        P=1.;x=10^8;forprime(p=3,x,P*=1-2/p);print(P*log(x)^2);

        0.83242

        Looks OK to me...

        David
      • Kermit Rose
        ... Hmm.... And I thought I had proven that the product converged to zero. David, what is wrong with my Proof below which I had already sent to Tim? Kermit
        Message 3 of 8 , May 5, 2010
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          primenumbers@yahoogroups.com wrote:
          >
          >
          >
          > Messages in this topic (5)
          > ________________________________________________________________________
          > 1b. Re: product convergence
          > Posted by: "djbroadhurst" d.broadhurst@... djbroadhurst
          > Date: Tue May 4, 2010 1:17 pm ((PDT))
          >
          >
          >
          > --- In primenumbers@yahoogroups.com,
          > "djbroadhurst" <d.broadhurst@...> wrote:
          >
          >
          >> prod(2<p<x, (p-2)/p) = O(1/log(x)^2)
          >>
          >
          > So let's work out the constant, say K, in
          >
          > prod(2<p<x, (p-2)/p) ~ K/log(x)^2
          >
          > We should use the twin-prime constant
          >
          > C2 = prod (2<p, p*(p-2)/(p-1)^2) = 0.6601618158...
          >
          > and then use the square of Mertens' formula,
          > remembering that the latter includes p = 2.
          >
          > K = C2*(exp(-Euler)*2)^2 =
          > 0.832429065661945278030805943531465575045445318077417053240894...
          >
          > Sanity check:
          >
          > default(primelimit,10^8);
          > \p5
          > P=1.;x=10^8;forprime(p=3,x,P*=1-2/p);print(P*log(x)^2);
          >
          > 0.83242
          >
          > Looks OK to me...
          >
          > David
          >
          >
          >


          Hmm.... And I thought I had proven that the product converged to zero.

          David, what is wrong with my "Proof" below which I had already sent to Tim?

          Kermit


          > *****************************************************
          >
          >
          >
          > Hello Tim.
          >
          >
          >
          > http://en.wikipedia.org/wiki/Infinite_product
          >
          >
          >
          > The infinite product 3/5 5/7 9/11 ...
          >
          > converges to zero.
          >
          >
          >
          > would converge to a positive number between 0 and 1 only if
          >
          > the infinite product
          >
          > 5/3 7/5 11/9 ..... converged to a positive number > 1.
          >
          >
          >
          > 5/3 7/5 11/9 ..... = (1 + 2/3) (1 + 2/5) (1 + 2/9) (1 + 2/11) (1 +
          > 2/15) (1 + 2/17) ...
          >
          >
          > 1 + 2/3 + 2/5 + 2/9 + 2 / 11 + .... =< (1 + 2/3) (1 + 2/5) (1 + 2/9)
          > (1 + 2/11) (1 + 2/15) (1 + 2/17) ... =< exp( 2/3 + 2/5 + 2/9 + 2/11
          >
        • djbroadhurst
          ... Now let x tend to infinity and I think that you will see that K/log(x)^2 vanishes, not so :-? David
          Message 4 of 8 , May 5, 2010
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            --- In primenumbers@yahoogroups.com,
            Kermit Rose <kermit@...> wrote:

            > I thought I had proven that the product converged to zero.

            It does. I wrote:

            > prod(2<p<x, (p-2)/p) ~ K/log(x)^2
            > K = 0.832429065661945278030805943531465575045445318077417053240894...

            Now let x tend to infinity and I think that you
            will see that K/log(x)^2 vanishes, not so :-?

            David
          • djbroadhurst
            ... It began OK and then fizzled out. Here is an elementary proof. Proposition: The product prod(p 2, (p-2)/p) over primes p 2 vanishes. Proof: It suffices to
            Message 5 of 8 , May 8, 2010
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              --- In primenumbers@yahoogroups.com,
              Kermit Rose <kermit@...> wrote:

              > what is wrong with my "Proof"

              It began OK and then fizzled out.
              Here is an elementary proof.

              Proposition: The product prod(p>2, (p-2)/p) over primes p>2 vanishes.

              Proof: It suffices to show that the contrary proposition is absurd.
              Suppose that prod(p>2, (p-2)/p) did not vanish.
              Then, by taking logs, we would conclude that
              sum(p>2, log(p) - log(p-2)) is finite. But
              log(p) - log(p-2) > 2/p. Thus
              sum(p>2, 1/p) would also be finite.
              Yet that is easily proven to be absurd:
              http://primes.utm.edu/infinity.shtml#punchline
              Hence the product prod(p>2, (p-2)/p) does indeed vanish.

              Comment: By the same argument, it follows that
              the product prod(p>x, (p-x)/p) over primes p > x
              vanishes for all real x > 0.

              David
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