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Infinite zeroes proof

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  • Jon Perry
    Anybody seen this before? http://www.geocities.com/rze17/zeros.pdf (home page http://www.geocities.com/rze17/math.html) Jon Perry perry@globalnet.co.uk
    Message 1 of 5 , Feb 28, 2003
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    • Jose Ramón Brox
      I m looking at the proof for if I could find a flaw. The only thing I have observed is that he took off the null roots of the polynomial, and I was wondering
      Message 2 of 5 , Feb 28, 2003
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        I'm looking at the proof for if I could find a flaw.

        The only thing I have observed is that he took off the null roots of the polynomial, and I was wondering if the zeta function had any zero root or not (if they were infinite, then the proof is not valid).

        I went to Mathworld and find this equality:

        zeta(1-s) = 2· (2pi)^(-s) · cos(1/2 · s · pi) · gamma(s) · zeta(s)

        I thought that "If s=1 then zeta(1-s) = zeta(0) = .... · cos(pi/2) = 0." but I have just realised that zeta(1) is the harmonic series! And then the value is in principle indetermined... can anyone say?

        What happens if we quit a zero root? zeta(1-s)/s = 2· (2pi)^(-s) · cos(1/2 · s · pi) · gamma(s) · zeta(s) /s ... and then evaluate in s=1 again...

        and I'd apply the L'Hôpital rule if I knew how to derivate zeta(1-s) - the s in the denominator is carried out -.

        If it can have an infinity of zero roots, and then the proof is not rigorous.

        Regards. Jose Brox.

        ----- Original Message -----
        From: Jon Perry
        To: Prime Numbers
        Sent: Friday, February 28, 2003 5:31 PM
        Subject: [PrimeNumbers] Infinite zeroes proof


        Anybody seen this before?

        http://www.geocities.com/rze17/zeros.pdf

        (home page http://www.geocities.com/rze17/math.html)

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry/maths/
        http://www.users.globalnet.co.uk/~perry/DIVMenu/
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        [Non-text portions of this message have been removed]
      • Jose Ramón Brox
        Well, in the same page of Mathworld (http://mathworld.wolfram.com/RiemannZetaFunction.html) we can see the graphic of zeta(x) and it isn t zero in x=0, so
        Message 3 of 5 , Feb 28, 2003
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          Well, in the same page of Mathworld (http://mathworld.wolfram.com/RiemannZetaFunction.html)
          we can see the graphic of zeta(x) and it isn't zero in x=0, so forget my previous message.

          Jose.


          ----- Original Message -----
          From: Jose Ramón Brox
          To: Prime Numbers
          Sent: Friday, February 28, 2003 6:30 PM
          Subject: Re: [PrimeNumbers] Infinite zeroes proof



          I'm looking at the proof for if I could find a flaw.

          The only thing I have observed is that he took off the null roots of the polynomial, and I was wondering if the zeta function had any zero root or not (if they were infinite, then the proof is not valid).

          I went to Mathworld and find this equality:

          zeta(1-s) = 2· (2pi)^(-s) · cos(1/2 · s · pi) · gamma(s) · zeta(s)

          I thought that "If s=1 then zeta(1-s) = zeta(0) = .... · cos(pi/2) = 0." but I have just realised that zeta(1) is the harmonic series! And then the value is in principle indetermined... can anyone say?

          What happens if we quit a zero root? zeta(1-s)/s = 2· (2pi)^(-s) · cos(1/2 · s · pi) · gamma(s) · zeta(s) /s ... and then evaluate in s=1 again...

          and I'd apply the L'Hôpital rule if I knew how to derivate zeta(1-s) - the s in the denominator is carried out -.

          If it can have an infinity of zero roots, and then the proof is not rigorous.

          Regards. Jose Brox.

          ----- Original Message -----
          From: Jon Perry
          To: Prime Numbers
          Sent: Friday, February 28, 2003 5:31 PM
          Subject: [PrimeNumbers] Infinite zeroes proof


          Anybody seen this before?

          http://www.geocities.com/rze17/zeros.pdf

          (home page http://www.geocities.com/rze17/math.html)

          Jon Perry
          perry@...
          http://www.users.globalnet.co.uk/~perry/maths/
          http://www.users.globalnet.co.uk/~perry/DIVMenu/
          BrainBench MVP for HTML and JavaScript
          http://www.brainbench.com

          Yahoo! Groups Sponsor
          ADVERTISEMENT




          Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
          The Prime Pages : http://www.primepages.org/



          Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.



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


          Yahoo! Groups Sponsor
          ADVERTISEMENT




          Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
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          [Non-text portions of this message have been removed]
        • David Broadhurst <d.broadhurst@open.ac.u
          I knew of the use of Euler-Maclaurin in Sondow, Jonathan The Riemann hypothesis, simple zeros and the asymptotic convergence degree of improper Riemann sums.
          Message 4 of 5 , Feb 28, 2003
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            I knew of the use of Euler-Maclaurin in

            Sondow, Jonathan
            The Riemann hypothesis, simple zeros and the asymptotic
            convergence degree of improper Riemann sums.
            Proc. Amer. Math. Soc. 126 (1998), no. 5, 1311--1314.

            It is not clear to me what is added to that by
            Nadim Restom web postings.

            David
          • Carl Devore
            ... zeta(0) = -1/2. lim_{s to 1} cos(pi/2*s)*zeta(s) = -pi/2.
            Message 5 of 5 , Feb 28, 2003
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              On Fri, 28 Feb 2003, [iso-8859-1] Jose Ramón Brox wrote:
              > Well, in the same page of Mathworld (http://mathworld.wolfram.com/RiemannZetaFunction.html)
              > we can see the graphic of zeta(x) and it isn't zero in x=0, so forget my previous message.

              zeta(0) = -1/2.

              lim_{s \to 1} cos(pi/2*s)*zeta(s) = -pi/2.
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