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

  • Thread starter lokofer
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In summary, if given the Lambert series S(x), the value of the coefficients a(n) can be obtained by representing S(x) as a Taylor series and using iteration to find the relationship between the coefficients. This can be seen on Mathworld's page on Lambert series and can be applied to a specific example like S(x)=\frac{x}{(1-x)^2}.
  • #1
lokofer
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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]
 
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  • #2
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]
 
  • #3
{n}

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.
 

What is a Lambert series?

A Lambert series is an infinite series of the form Σ(n=1 to ∞) an x^n, where a is a constant and x is the variable. It is named after mathematician Johann Heinrich Lambert.

What is the convergence of a Lambert series?

A Lambert series converges when the absolute value of x is less than the inverse of the limit of the sequence an as n approaches infinity.

What is the purpose of a Lambert series?

Lambert series are used in mathematics to represent certain functions, such as the logarithm function, as an infinite sum of powers. They are also used in number theory and combinatorics.

What is the relationship between Lambert series and the Riemann zeta function?

Lambert series can be used to express the Riemann zeta function as a special case. Specifically, the Riemann zeta function can be written as a Lambert series with a = 1 and x = 1/n.

What are some applications of Lambert series in physics?

Lambert series have applications in physics, particularly in quantum field theory and statistical mechanics. They are also used in the study of black holes and in cosmology to model gravitational lensing.

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