Lambert series

  • Thread starter lokofer
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  • #1
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Main Question or Discussion Point

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]
 

Answers and Replies

  • #2
saltydog
Science Advisor
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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:

[tex]S(x)=\frac{x}{(1-x)^2}[/tex]
 

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