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Lambert series

  1. Aug 22, 2006 #1
    If we have the next Lambert series:

    [tex] S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n } [/tex]

    my question is..if you know what S(x) is then..could you obtain the value of the a(n) ?..or simply working with the Dirichlet series version:

    [tex] \sum_{n=0}^{\infty}a(n)n^{-s}= \zeta (s) \sum _{n=0}^{\infty}b(n)n^{-s} [/tex]

    Where [tex] S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n }=\sum_{n=0}^{\infty} \frac{ b(n) x^n } [/tex]
  2. jcsd
  3. Aug 25, 2006 #2


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    Notice this one still at zero so I'll say my two cents:

    If you can represent S(x) as a taylor series, then the coefficients of the taylor series are related to the coefficients of the Lambert series; you can obtain the a(n) from those of the Taylor series through iteration. Check out Lambert series on Mathworld to see this and then try it for:

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