I'm attempting to express the following function (let $\xi$ be a parameter)
$$f(z) = \dfrac{1}{(1 - 2\xi z + z^2)^2} $$
So far, I have thought about getting the expression for $g(z) = \dfrac{1}{1 - 2\xi z + z^2} $ first, then multiply the series. However, I don't know how to effectively expand this in a neighbor of $0$, as I can only use the expansion $\dfrac{1}{1-a}$ when $|a| < 1$.
Is there any way to expand the series without taking derivatives? Thank you.
Note that it's $$ \frac{1}{2z} \frac{d}{d\xi} \frac{1}{1-2\xi+z^2}. $$ In fact we do have an expression for the latter fraction, although not really in closed form: $$ \frac{1}{1-2\xi+z^2} = \sum_{n=0}^{\infty} U_n(\xi) z^n, $$ where $U_n$ is the Chebyshev polynomial of the second kind. So we find that $$ \frac{1}{(1-2\xi z+z^2)^2} = \sum_{n=0}^{\infty} \frac{1}{2}U_{n+1}'(\xi) z^n. $$ These are also a special case of the ultraspherical/Gegenbauer polynomials, which have generating function $(1-2\xi z +z^2)^{-\alpha}$.