How can we obtain the coefficients of S(x) from its Taylor series?

[lokofer]

If we have the next Lambert series:

$$S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n }$$

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:

$$\sum_{n=0}^{\infty}a(n)n^{-s}= \zeta (s) \sum _{n=0}^{\infty}b(n)n^{-s}$$

Where $$S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n }=\sum_{n=0}^{\infty} \frac{ b(n) x^n }$$

 

[saltydog]

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:

$$S(x)=\frac{x}{(1-x)^2}$$

 

[tacman]

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The coefficients of S(x) can be obtained from its Taylor series by using the formula for the nth derivative at x=0:

a(n) = f^n(0) / n!

Where f^n(0) is the nth derivative of S(x) evaluated at x=0 and n! is the factorial of n.

In the case of the Lambert series, we can use the formula above to obtain the coefficients a(n) by taking the nth derivative of S(x) and evaluating it at x=0. However, this may not always be a practical or feasible method, especially for more complicated functions.

Alternatively, we can use the Dirichlet series version of S(x) to obtain the coefficients a(n) by using the formula:

a(n) = b(n) / n

Where b(n) is the coefficient of the nth term in the Dirichlet series. This method may be more efficient and easier to use, especially for functions with complicated Taylor series.

In summary, both methods can be used to obtain the coefficients of S(x) from its Taylor series, but the choice of method may depend on the specific function and its complexity.