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Re: product

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  • Werner D. Sand
    Hi Jacques and Wojtek, thanks for computing. No chance to reach limit=1? Good idea to do the log, I will try. Werner
    Message 1 of 11 , Dec 2, 2005
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      Hi Jacques and Wojtek,

      thanks for computing. No chance to reach limit=1?
      Good idea to do the log, I will try.

      Werner
    • Mark Underwood
      ... I was just averaging some figures around the n = 40,000 area and it worked out to be around .9105. Looking a Jacque s findings the product seems to be ever
      Message 2 of 11 , Dec 2, 2005
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        --- In primenumbers@yahoogroups.com, "Jacques Tramu"
        <jacques.tramu@e...> wrote:
        >
        > I have computed the product and found
        > n= 200 000 000 : 0.9048530178
        > n= 1 000 000 0000 : 0.9048042986
        > n= 1 500 000 000 : 0.9047987573
        >

        I was just averaging some figures around the n = 40,000 area and it
        worked out to be around .9105. Looking a Jacque's findings the
        product seems to be ever so slowly shrinking.

        Contrast this the product 1/2 * 4/3 * 5/6 * 8/7 * ....

        and this appears to be converging to .599070...

        Mark
      • Werner D. Sand
        I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge to 1, oscillating around 1.
        Message 3 of 11 , Dec 4, 2005
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          I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge
          to 1, oscillating around 1.
        • Kermit Rose
          From: Werner D. Sand Date: 12/04/05 04:58:46 To: primenumbers@yahoogroups.com Subject: [PrimeNumbers] Re: product I suspect the product 2/3 * 7/5 * 11/13 *
          Message 4 of 11 , Dec 4, 2005
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            From: Werner D. Sand
            Date: 12/04/05 04:58:46
            To: primenumbers@yahoogroups.com
            Subject: [PrimeNumbers] Re: product

            I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge
            to 1, oscillating around 1.


            Kermit says:
            With primes, anything might be possible.

            [Non-text portions of this message have been removed]
          • Dario Alpern
            ... This product is not very difficult to compute if you know about the function gamma (the extension of factorials to the complex domain). You are trying to
            Message 5 of 11 , Dec 5, 2005
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              --- In primenumbers@yahoogroups.com, "Mark Underwood"
              <mark.underwood@s...> wrote:
              >
              > --- In primenumbers@yahoogroups.com, "Jacques Tramu"
              > <jacques.tramu@e...> wrote:
              > >
              > > I have computed the product and found
              > > n= 200 000 000 : 0.9048530178
              > > n= 1 000 000 0000 : 0.9048042986
              > > n= 1 500 000 000 : 0.9047987573
              > >
              >
              > I was just averaging some figures around the n = 40,000 area and it
              > worked out to be around .9105. Looking a Jacque's findings the
              > product seems to be ever so slowly shrinking.
              >
              > Contrast this the product 1/2 * 4/3 * 5/6 * 8/7 * ....
              >
              > and this appears to be converging to .599070...
              >
              > Mark
              >

              This product is not very difficult to compute if you know about the
              function gamma (the extension of factorials to the complex domain).

              You are trying to find:

              k=inf (4k+1)(4k+4) k=inf 2
              Prod ------------ = Prod 1 - ------------
              k=0 (4k+2)(4k+3) k=0 (4k+2)(4k+3)

              This is clearly convergent. But what is the limit?

              k=n (4k+1)(4k+4) k=n (k+1/4)(k+1)
              Prod ------------ = Prod -------------- =
              k=0 (4k+2)(4k+3) k=0 (k+2/4)(k+3/4)

              gamma(k+5/4) gamma(k+2) gamma(2/4) gamma(3/4)
              = ------------------------- ---------------------
              gamma(k+6/4) gamma(k+7/4) gamma(1/4) gamma(1)


              = A(k) x B

              where A(k) is the first fraction and B the second.

              We are interested in the value of A(k) as k->inf.

              Fortunately it turns out that the limit is 1. This can be seen by
              using Stirling approximation.

              Since gamma(1) = 1 and gamma(1/2) = sqrt(pi) we finally get:

              k=inf (4k+1)(4k+4) sqrt(pi) * gamma(3/4)
              Prod ------------ = ----------------------
              k=0 (4k+2)(4k+3) gamma(1/4)

              This is about 0.5990701173677961037199612

              Best regards,

              Dario Alejandro Alpern
              Buenos Aires - Argentina
              http://www.alpertron.com.ar/ENGLISH.HTM
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