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Re: [PrimeNumbers] Re: product

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  • Jacques Tramu
    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
    Message 1 of 11 , Dec 2, 2005
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      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


      > my beginning of the product 2/3 * 7/5 * ... is quite correct.
      > I thought I could avoid the ugly formula:
      > product(p(n)/p(n+1)) * (p(n+3)/p(n+2)).
      > It's an `alternating' product. Otherwise the limit would be simply = 0
      > (I suspect).
      >
    • Wojciech.Florek@amu.edu.pl
      ... Hi, I ve calculated very rough estimation of products p1=2/3*11/13*... and p2=7/5*19/17*... up to LIM primes (it is very rough since I used Nextprime from
      Message 2 of 11 , Dec 2, 2005
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        On Fri, 2 Dec 2005, Bob Gilson wrote:

        >
        >
        > "Werner D. Sand" <Theo.3.1415@...> wrote: Hi grostoon,
        >
        > my beginning of the product 2/3 * 7/5 * ... is quite correct.
        > I thought I could avoid the ugly formula:
        > product(p(n)/p(n+1)) * (p(n+3)/p(n+2)).
        > It's an `alternating' product. Otherwise the limit would be simply = 0
        > (I suspect).
        >
        > Werner
        >
        Hi,
        I've calculated very rough estimation of products p1=2/3*11/13*...
        and p2=7/5*19/17*... up to LIM primes (it is very rough since I used
        Nextprime from GMP lib):
        LIM= 1000 p1=0.131402 p2= 6.980669 p1p2=0.917275
        LIM= 2000 p1=0.108394 p2= 8.584098 p1p2=0.930461
        LIM= 4000 p1=0.088371 p2=10.297879 p1p2=0.910038
        LIM= 8000 p1=0.073380 p2=12.474514 p1p2=0.915374
        LIM= 16000 p1=0.060168 p2=15.118286 p1p2=0.909637
        LIM= 32000 p1=0.049730 p2=18.303402 p1p2=0.910235
        LIM= 64000 p1=0.041075 p2=22.084929 p1p2=0.907139
        LIM= 128000 p1=0.033984 p2=26.647697 p1p2=0.905591
        I have no time to analyze these series (e.g. quotients
        18.303/15.118, etc), but I have another idea:

        You investigate

        Prod_{n=0} p(n)/p(n+1) * p(n+3)/p(n+2)
        (I assume p(0)=2, for convenience). Another formulation is
        Prod_{n=0} a_n/a_{n+1}, where
        a_n=p(2n)/p(2n+1).
        What about log of this product? I'm not good in logs for infinite products
        but for finite one we have

        log(Prod_{n=0} a_n/a_{n+1}) = Sum_{n=0} log a_n - log a_{n+1},
        so it is an alternating series. Is it enough to apply Leibniz criterion?

        log a_n= log [ p(2n)/p(2n+1)] = log p(2n) - log p(2n+1).

        It is possible to say something about this difference? It is, of course,
        negative, but what more? |log (a_n)| > |log(a_{n+1}| ????
        There are large gaps in primes, so sometimes a_n is quite small, but for
        twin primes it tends to 1.

        Wojtek



        ===============================================
        Wojciech Florek (WsF)
        Adam Mickiewicz University, Faculty of Physics
        ul. Umultowska 85, 61-614 Poznan, Poland

        Phone: (++48-61) 8295033 fax: (++48-61) 8257758
        email: Wojciech.Florek@...
      • Werner D. Sand
        Hello Bob, you are not allowed to handle an infinite product like that. Did you already calculate step by step? Your argumentation is as if you would say: 1 -
        Message 3 of 11 , Dec 2, 2005
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          Hello Bob,

          you are not allowed to handle an infinite product like that. Did you
          already calculate step by step? Your argumentation is as if you would
          say:
          1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 +- ... = inf - inf = indefinite,
          actually it is pi/4.

          Werner
        • 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 4 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 5 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 6 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 7 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 8 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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