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Non-zero zeta function on plane Re(z)>1

  1. Dec 28, 2008 #1
    Riemann says that the zeta function doesn't have zeros on the half plane [itex]\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\}[/itex], because the sum

    [tex]
    \log(\zeta(z)) = \log\Big(\frac{1}{\underset{p\in\mathbb{P}}{\prod}\big(1 - \frac{1}{p^z}\big)}\Big) = -\sum_{p\in\mathbb{P}}\log\big(1 - \frac{1}{p^z}\big)
    [/tex]

    remains finite. Well, why does that sum remain finite? Doesn't look quite obvious to me.

    hmhmh... what kind of set is the set

    [tex]
    \zeta(\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\})\subset\mathbb{C}?
    [/tex]

    Does it fit in a domain of some logarithm, or does it wind around the origo?
     
    Last edited: Dec 28, 2008
  2. jcsd
  3. Dec 28, 2008 #2

    Hurkyl

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    Well, [itex]\log(1-x) \approx -x[/itex] (for small x), so it looks plausible.
     
  4. Dec 28, 2008 #3
    hmhmhm... ok

    Possible winding problems can probably be avoided by using a real logarithm.

    [tex]
    \log(|\zeta(z)|) = -\sum_{p\in\mathbb{P}}\log\big(|1 - p^{-z}|\big) = -\frac{1}{2}\sum_{p\in\mathbb{P}} \log\big(1 + p^{-2x} - 2p^{-x} \cos(y\log(p))\big)
    [/tex]
     
    Last edited: Dec 28, 2008
  5. Dec 28, 2008 #4
    Well doesn't the Euler product suffice to show that there are no zeroes in re z > 1?
     
  6. Dec 28, 2008 #5
    It can happen that [itex]x_n\neq 0[/itex] for all [itex]n=1,2,3,\ldots[/itex], but still

    [tex]
    \prod_{n=1}^{\infty} x_n = 0,
    [/tex]

    so [itex]\zeta(z)\neq 0[/itex] is not clear at least merely because the factors [itex](1 - p^{-z})^{-1}[/itex] are non-zero.
     
  7. Dec 28, 2008 #6
    Yes but [tex]\lim_{n\to\infty} x_n=1[/tex] in this case so I think it should hold.

    I must however confess I have very small knowledge on the area, and might well be mistaken.

    (edit: ugly tex =o )
     
  8. Dec 29, 2008 #7
    If [itex]a_1,a_2,a_3,\ldots[/itex] is a sequence such that

    [tex]
    a_n\to 0
    [/tex]

    but

    [tex]
    \sum_{k=1}^{n} a_k \to \pm\infty,
    [/tex]

    then [itex]e^{a_1}, e^{a_2}, e^{a_3},\ldots [/itex] is a sequence such that

    [tex]
    e^{a_n} \to 1
    [/tex]

    but

    [tex]
    \prod_{k=1}^{n} e^{a_k} \to 0\;\textrm{or}\;\infty.
    [/tex]
     
  9. Dec 29, 2008 #8
    Well that clears that up then :tongue:.

    (edit: a moment of clarity)

    in that case,

    [tex]\log \zeta(z) = -\sum_p \log(1-p^{-z}) = \sum_p \sum_{n=1}^\infty \frac{p^{-nz}}{n}[/tex]

    using the taylor series for log(1-x), we see that the sum remains finite.

    Does that hold?
     
    Last edited: Dec 29, 2008
  10. Jan 16, 2009 #9
    As I realise now that what I said, although looking plausible does not prove anything.

    However I looked into absolute convergence for products and it seems that the product,

    [tex]\prod_i 1+a_i[/tex] converges absolutely iff [tex]\sum_i a_i[/tex] does so.

    Apply with [tex]\prod_p \left( 1+\frac{1}{p^s-1} \right)[/tex] and you end up with the convergence of [tex]\sum_p \frac{1}{p^s-1}[/tex]
     
  11. Jan 18, 2009 #10
    I agree that the idea in the post #8 is the one that leads into the proof, but there are some details that have to be worked out. Do you have a proof for the claim in the post #9?

    Anyway, my proof of the fact [itex]\zeta(z)\neq 0[/itex] for [itex]\textrm{Re}(z)>1[/itex] goes like this now. Firstly, we aim for the proof of

    [tex]
    \log |\zeta(z)| > -\infty
    [/tex]

    The real logarithm must be used unless somebody knows with certainty that a complex logarithm exists in the image of [itex]\zeta(\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\})[/itex]. So the task becomes a proof of the claim

    [tex]
    \sum_{p\in\mathbb{P}} \log\big(1 + p^{-2x} - 2p^{-x}\cos(y\log(p))\big) < \infty
    [/tex]

    The Taylor series of the logarithm cannot be substituted right a way, because many of the terms inside logarithm may not be in domain of convergence [itex]]0,2[[/itex]. In order to prove that the series converges, it suffices to prove that it converges when the sum is restricted to values [itex]p>P[/itex] with some [itex]P[/itex]. When [itex]z[/itex] is fixed,

    [tex]
    \lim_{p\to\infty}\big( p^{-2x} - 2p^{-x}\cos(y\log(p)) \big) = 0
    [/tex]

    so we can choose sufficiently big [itex]P[/itex] so that the terms in the logarithm are inside the Taylor series' domain of convergence, and then the task is to prove that

    [tex]
    \underset{p>P}{\sum_{p\in\mathbb{P}}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \big(
    p^{-2x} - 2p^{-x}\cos(y\log(p))\big)^n
    [/tex]

    converges.

    [tex]
    \big| p^{-2x} - 2p^{-x}\cos(y\log(p))\big| \leq 3 p^{-x}
    [/tex]

    so actually it suffices to prove

    [tex]
    \underset{p>P}{\sum_{p\in\mathbb{P}}} \sum_{n=1}^{\infty} \frac{1}{n} p^{-xn} < \infty.
    [/tex]

    This can be done with Fubini's theorem and some approximations.

    [tex]
    \underset{p>P}{\sum_{p\in\mathbb{P}}} p^{-xn} \leq \int\limits_{P}^{\infty} p^{-xn} dp = \frac{P^{1-xn}}{xn-1}
    [/tex]

    [tex]
    \sum_{n=1}^{\infty} \underset{p>P}{\sum_{p\in\mathbb{P}}} \frac{1}{n}p^{-xn} \leq P \sum_{n=1}^{\infty} \frac{1}{n^2} \frac{1}{x - \frac{1}{n}} P^{-xn} < \infty
    [/tex]
     
  12. Jan 18, 2009 #11
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