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Help showing bound for magnitude of complex log fcn

  1. Feb 25, 2006 #1


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    I'm working through an example problem wherein this bound is used:

    [tex]\left| \log \left( 1-\frac{1}{L^s}\right) \right| \leq L^{-\sigma},[/tex]

    where [tex]s:=\sigma +it[/tex] and it is known that [tex]\sigma >1.[/tex] How do I prove this? Should I assume the principle brach is taken?
  2. jcsd
  3. Feb 25, 2006 #2


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    It is perhaps important to note that in the context, [tex]L\rightarrow\infty[/tex]

    I think I got a start:

    For [tex]\left| z\right| <1,[/tex] the Taylor series about z=0 is

    [tex]\log (1-z)=-\sum_{j=1}^{\infty}\frac{z^j}{j}[/tex]

    so that

    [tex]\left| \log \left( 1-\frac{1}{L^s}\right) \right| =\left| -\sum_{j=1}^{\infty}\frac{L^{-js}}{j} \right| \leq \sum_{j=1}^{\infty}\left| \frac{L^{-js}}{j}\right| = \sum_{j=1}^{\infty} \frac{L^{-j\sigma}}{j} [/tex]
    Last edited: Feb 25, 2006
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