The generating function for legendre polynomials is $$\frac{1}{\sqrt(1-2ut+u^2)}=\sum_{i=0}^\infty u^iP_i(t)$$ where $P_i$ is $i^{th}$ legendre polynomial however for binomial expansion of left hand side with exponent –$\frac12$, we need to have $$|(u^2-2ut)|\le1$$. So how do we prove equality of two sides when this condition doesn't hold ? Or is it really a domain limitation ?
Domain limitations on generating function for Legendre polynomials
binomial theoremgenerating-functionslegendre polynomialsnegative binomialsequences-and-series
Best Answer
We consider the sequence $\{P_n(t)\}$ of Legendre polynomials. We describe how to construct a generating function \begin{align*} G(t,u)=\sum_{n=0}^\infty P_n(t)u^n \end{align*} and how to derive the region of convergence. We closely follow example 7.4 from Asymptotics and Special Functions by F.J.W. Olver.
We observe if $u\to 0$, then $z_1\to t$ and $|z_2|\to\infty$. Hence for sufficiently small $|u|$, $\mathcal{C}$ contains $z_1$ but not $z_2$. The residue theorem yields \begin{align*} G(t,u)=-\frac{2}{u}\frac{1}{z_1-z_2}=\frac{1}{\sqrt{1-2tu+u^2}} \end{align*}
We conclude, the desired expansion is given by \begin{align*} \color{blue}{\frac{1}{\sqrt{1-2tu+u^2}}=\sum_{n=0}^\infty P_n(t)u^n}\tag{1} \end{align*}
provided that $|u|$ is sufficiently small and the chosen branch of the square root tends to $1$ as $u\to 0$.
For $t\in[-1,1]$ the singularities of the left-hand side of (1) both lie on the circle $|u|=1$, hence in this case the radius of convergence of the series on the right-hand side is unity.