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Riemann Zeta function zeros

  1. Nov 3, 2009 #1
    Hi:
    ____________________________________________________________________
    Added Nov.3, 2009

    (For anyone who can't read the formula below (probably everyone) and who
    might have an interest in the subject: - the derivation of two simple equations
    that locate all the zeros of the zeta function on the imaginary (critical) line can be downloaded from

    http://www.magma.ca/~gmtrcs/papers/zeta.pdf )

    ___________________________________________________________________

    Can anyone tell me if the formula below is already known?

    The zeros of the Zeta function along the imaginary (critical) line coincide with the zeros of
    the following equation:

    {D_{R}\,{\zeta _{I}^{\prime} } } + {N}\,{{\zeta _{R}^{\prime} } =0 }.

    where N =
    \mathit{N} = {\displaystyle \frac {{C_{m}}\,\mathrm{cos}({\rho
    _{\pi }})}{\sqrt{\pi }}} - {\displaystyle \frac {{C_{p}}\,
    \mathrm{sin}({\rho _{\pi }})}{\sqrt{\pi }}}

    and D is

    \mathit{D_{R}} = {\displaystyle \frac {1}{2}} - {\displaystyle
    \frac {1}{2}} \,{\displaystyle \frac { {C_{p}}\,\mathrm{cos}({
    \rho _{\pi }}) + {C_{m}}\,\mathrm{sin}({\rho _{\pi }})}{\sqrt{\pi
    }}}

    {C_{p}}& =& \mathrm{cosh}({\displaystyle \frac {\pi \,\rho }{2}} )\,{
    \Gamma _{R}} + \mathrm{sinh}({\displaystyle \frac {\pi \,\rho }{2
    }} )\,{\Gamma _{I}}\\
    {C_{m}}& = & - \mathrm{sinh}({\displaystyle \frac {\pi \,\rho }{2}} )
    \,{\Gamma _{R}} + \mathrm{cosh}({\displaystyle \frac {\pi \,\rho
    }{2}} )\,{\Gamma _{I}}

    Gamma _{I} is the imaginary part of Gamma(1/2+I*rho)
    Gamma _{R} is the Real part of Gamma(1/2+I*rho)
    and similarly for Zeta\prime, the first derivative of Zeta (s)
    with s=1/2+I*rho


    I am new to this forum and it does not seem possible to attach a file to this message.
    Or of it is, it doesn't seem to work for me.

    If someone would like a copy of the derivation of this formula, please send a message with
    an email address and I will send a copy of the full derivation.

    Thank you

    Mike
     
    Last edited: Nov 3, 2009
  2. jcsd
  3. Nov 3, 2009 #2
    OP's post with LaTex fixed:

    ----------------------------------------------------------------

    Hi:
    __________________________________________________ __________________
    Added Nov.3, 2009

    (For anyone who can't read the formula below (probably everyone) and who
    might have an interest in the subject: - the derivation of two simple equations
    that locate all the zeros of the zeta function on the imaginary (critical) line can be downloaded from

    http://www.magma.ca/~gmtrcs/papers/zeta.pdf )

    __________________________________________________ _________________

    Can anyone tell me if the formula below is already known?

    The zeros of the Zeta function along the imaginary (critical) line coincide with the zeros of
    the following equation:

    [tex]
    {D_{R}\,{\zeta _{I}^{\prime} } } + {N}\,{{\zeta _{R}^{\prime} } =0 } [/tex].

    where [tex] \mathit{N} = {\displaystyle \frac {{C_{m}}\,\mathrm{cos}({\rho
    _{\pi }})}{\sqrt{\pi }}} - {\displaystyle \frac {{C_{p}}\,
    \mathrm{sin}({\rho _{\pi }})}{\sqrt{\pi }}} [/tex]


    and D is


    [tex] \mathit{D_{R}} = {\displaystyle \frac {1}{2}} - {\displaystyle
    \frac {1}{2}} \,{\displaystyle \frac { {C_{p}}\,\mathrm{cos}({
    \rho _{\pi }}) + {C_{m}}\,\mathrm{sin}({\rho _{\pi }})}{\sqrt{\pi
    }}}

    {C_{p}}& =& \mathrm{cosh}({\displaystyle \frac {\pi \,\rho }{2}} )\,{
    \Gamma _{R}} + \mathrm{sinh}({\displaystyle \frac {\pi \,\rho }{2
    }} )\,{\Gamma _{I}}\\
    {C_{m}}& = & - \mathrm{sinh}({\displaystyle \frac {\pi \,\rho }{2}} )
    \,{\Gamma _{R}} + \mathrm{cosh}({\displaystyle \frac {\pi \,\rho
    }{2}} )\,{\Gamma _{I}} [/tex]

    [tex] \Gamma _{I} [/tex] is the imaginary part of [tex] \Gamma(1/2+I \rho) [/tex].

    [tex] \Gamma _{R} [/tex] is the Real part of [tex] \Gamma(1/2+I \rho) [/tex] and similarly for [tex] \zeta^{\prime} [/tex] with [tex] s=1/2+I \rho [/tex].


    I am new to this forum and it does not seem possible to attach a file to this message.
    Or of it is, it doesn't seem to work for me.

    If someone would like a copy of the derivation of this formula, please send a message with
    an email address and I will send a copy of the full derivation.

    Thank you

    Mike
     
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