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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 ?

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  • $\begingroup$ The ininite series on the right is a Formal power series in which $u^n\to 0$ is all that is required for convergence. This is also why dividing by $0$ on the left is not an issue. $\endgroup$
    – Somos
    Mar 21, 2019 at 11:44
  • $\begingroup$ So that's really a limitation on domain of u ? $\endgroup$
    – Kutsit
    Mar 22, 2019 at 16:51
  • $\begingroup$ You are misunderstanding "formal" power series. There is no domain of $u$. Think of $u$ as an infinitesimal non-zero number if that helps. $\endgroup$
    – Somos
    Mar 22, 2019 at 17:36
  • $\begingroup$ Ok, read on wikipedia and got some basic understanding of the difference. Thanks $\endgroup$
    – Kutsit
    Mar 22, 2019 at 17:54
  • $\begingroup$ @Somos: I've added a post which determines the region of convergence of the generating function when the Legendre polynomials are treated as analytical functions. $\endgroup$
    – epi163sqrt
    Mar 23, 2019 at 8:33

1 Answer 1

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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 recall Rodrigues' formula
\begin{align*} P_n(t)=\frac{(-1)^n}{2^nn!}\frac{d^n}{dt^n}\left\{\left(1-t^2\right)^n\right\} \end{align*} and get using Cauchy's integral formula for the $n$-th derivative of an analytic function Schläfli's integral

\begin{align*} P_n(t)=\frac{1}{2^{n+1}\pi i}\int_{\mathcal{C}}\frac{(z^2-1)^n}{(z-t)^{n+1}}dz \end{align*}

in which $\mathcal{C}$ is any simple closed contour that encircles $z=t$; here $t$ may be real or complex. For fixed $\mathcal{C}$ and sufficiently small $|u|$, the series \begin{align*} \sum_{n=0}^\infty\frac{(z^2-1)^nu^n}{2^{n+1}\pi i(z-t)^{n+1}} \end{align*} converges uniformly with respect to $z\in\mathcal{C}$, by the M-test. By integration and summation we obtain \begin{align*} \frac{1}{2\pi i}\int_{\mathcal{C}}\left\{1-\frac{(z^2-1)u}{2(z-t)}\right\}^{-1}\frac{dz}{z-t}=\sum_{n=0}^\infty P_n(t)u^n=G(t,u). \end{align*} It follows \begin{align*} G(t,u)=-\frac{1}{\pi i}\int_{\mathcal{C}}\frac{dz}{uz^2-2z+(2t-u)}=-\frac{1}{u\pi i}\int_{\mathcal{C}}\frac{dz}{(z-z_1)(z-z_2)} \end{align*} where \begin{align*} z_1=\frac{1-\sqrt{1-2tu+u^2}}{u},\qquad z_2=\frac{1+\sqrt{1-2tu+u^2}}{u}, \end{align*}

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.

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