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 ?

  • $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.