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Re: [PrimeNumbers] Demichel

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  • Phil Carmody
    ... In 1914, numerical evidence proved that π(x)
    Message 1 of 8 , Jan 2, 2011
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      --- On Sun, 1/2/11, Andrey Kulsha wrote:
      > There are also two new (very similar, huh)
      > papers analyzing and improving these results:
      > http://eprints.ma.man.ac.uk/1541/01/Munibah2010.pdf

      "In 1914, numerical evidence proved that π(x) < li(x) for all x. "

      Ewww...

      Phil
    • Phil Carmody
      ... It gets worse. Looking at the start of Chapter 3, Numerical Results (i), we have the assertion: Now we know that (gonna simplify glyphs, sorry) e^(iwy)
      Message 2 of 8 , Jan 2, 2011
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        --- On Sun, 1/2/11, Phil Carmody wrote:
        > --- On Sun, 1/2/11, Andrey Kulsha wrote:
        > > There are also two new (very similar, huh)
        > > papers analyzing and improving these results:
        > > http://eprints.ma.man.ac.uk/1541/01/Munibah2010.pdf
        >
        > "In 1914, numerical evidence proved that π(x) < li(x)
        > for all x. "
        >
        > Ewww...

        It gets worse. Looking at the start of Chapter 3, Numerical Results (i), we have the assertion:

        Now we know that (gonna simplify glyphs, sorry)

        e^(iwy) e^(iwy) e^(-iwy)
        ------- = ------- + --------
        p B + iy B - iy

        where none of the terms are defined. It seems chapter 2 most recently defined p = B + iy.

        So his assertion seems to be that:

        e^(iwy) e^(iwy) e^(-iwy)
        ------- = ------- + --------
        B + iy B + iy B - iy

        Or:

        e^(-iwy)
        0 = --------
        B - iy

        Or are my eyes playing tricks with me?

        Phil
      • Andrey Kulsha
        ... I guess they mean the conjugated pair of zeta zeros. Best regards, Andrey
        Message 3 of 8 , Jan 2, 2011
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          > It gets worse. Looking at the start of Chapter 3, Numerical Results (i),
          > we have the assertion:
          >
          > Now we know that (gonna simplify glyphs, sorry)
          >
          > e^(iwy) e^(iwy) e^(-iwy)
          > ------- = ------- + --------
          > p B + iy B - iy

          I guess they mean the conjugated pair of zeta zeros.

          Best regards,

          Andrey
        • Phil Carmody
          ... Maths by guesswork? What happened to rigour? Then again, seeing how long those two take to get from 24/8 to 3, I suspect that the papers will be closely
          Message 4 of 8 , Jan 2, 2011
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            --- On Sun, 1/2/11, Andrey Kulsha wrote:
            > > e^(iwy)    e^(iwy)   e^(-iwy)
            > > ------- =  ------- + --------
            > >    p        B + iy    B - iy
            >
            >     I guess they mean the conjugated pair of zeta zeros.

            Maths by guesswork? What happened to rigour? Then again, seeing how long those two take to get from 24/8 to 3, I suspect that the papers will be closely associated with rigor mortis.

            Phil
          • Kermit Rose
            _ ... Shows that that dissertation did not have careful proofreading. http://primes.utm.edu/howmany.shtml However in 1914 Littlewood proved that pi(x)-Li(x)
            Message 5 of 8 , Jan 2, 2011
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              _
              > 3d. Re: Demichel
              > Posted by: "Phil Carmody" thefatphil@... thefatphil
              > Date: Sun Jan 2, 2011 4:32 am ((PST))
              >
              > --- On Sun, 1/2/11, Andrey Kulsha wrote:
              >> There are also two new (very similar, huh)
              >> papers analyzing and improving these results:
              >> http://eprints.ma.man.ac.uk/1541/01/Munibah2010.pdf
              >
              > "In 1914, numerical evidence proved that π(x)< li(x) for all x. "
              >
              > Ewww...
              >
              > Phil
              >

              Shows that that dissertation did not have careful proofreading.

              http://primes.utm.edu/howmany.shtml

              However in 1914 Littlewood proved that pi(x)-Li(x) assumes both positive
              and negative values infinitely often.

              Kermit
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