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Re: product

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  • Werner D. Sand
    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)).
    Message 1 of 11 , Dec 2, 2005
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      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
    • Bob Gilson
      Werner D. Sand wrote: Hi grostoon, my beginning of the product 2/3 * 7/5 * ... is quite correct. I thought I could avoid the ugly
      Message 2 of 11 , Dec 2, 2005
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        "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

        Since 2/3*11/13*29/31 converges to zero (I suspect), surely your 7/5*23/17*41/37*zero, will also converge to zero
        Cheers

        Bob




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      • Wojciech.Florek@amu.edu.pl
        ... All factors 7/5, 23/17,...,p(n+3)/p(n+2),... are larger than 1 so how their products can converge to zero? 5/7*17/23* ... seems to converge to 0, so its
        Message 3 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
          >
          > Since 2/3*11/13*29/31 converges to zero (I suspect),
          > surely your 7/5*23/17*41/37*zero, will also converge to zero
          > Cheers
          >
          > Bob

          All factors 7/5, 23/17,...,p(n+3)/p(n+2),... are larger than 1 so how
          their products can converge to zero?
          5/7*17/23* ... seems to converge to 0, so its inverse to tends to
          infinity, and Werner's product is of type 0*infinty, I suspect.
          But there are also other possiblilites, e.g.
          2; (1.5)*(1,5); (1.33)*(1.33)*(1.33); ...
          All numbers are larger than 1 (and 1+1/n > 1 for n<infinty) but their
          product is..., you know, I'm sure.


          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@...
        • 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 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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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