Legendre Functions

The Legendre equation

\[ \bigl(1-x^2\bigr)P''(x)-2xP'(x)+\lambda P(x)=0 . \] Its regular singular points are \(x=\pm1\) and \(x=\infty\), so the series solution about \(x=0\) converges for \(\abs x<1\) for every \(\lambda\). If \(\lambda=l(l+1)\) with \(l\) an integer, the series terminates after \(x^l\), leaving a polynomial of degree \(l\).

The generating function for the polynomial solutions of the Legendre ODE is \begin{equation*}\tag{11.1} g(x,t)=\frac1{\sqrt{1-2xt+t^2}}=\sum_{n=0}^\infty P_n(x)t^n . \end{equation*} To fix the scale of \(P_n\), set \(x=1\): \begin{equation*}\tag{11.2} g(1,t)=\frac1{\sqrt{1-2t+t^2}}=\frac1{1-t}=\sum_{n=0}^\infty t^n, \end{equation*} by the binomial theorem. Comparing with Eq. (11.1) gives \(P_n(1)=1\).

Replacing \(x\to-x\) and \(t\to-t\) leaves \(g(x,t)\) unchanged: \begin{equation*} \sum_{n=0}^\infty P_n(x)t^n=g(x,t)=g(-x,-t)=\sum_{n=0}^\infty P_n(-x)(-t)^n, \end{equation*} so \begin{equation*}\tag{11.4} P_n(-x)=(-1)^nP_n(x). \end{equation*} Hence \(P_n(-1)=(-1)^n\), and \(P_n(x)\) has the parity of \(x^n\).

At \(x=0\): \(P_{2n+1}\) is odd, so \(P_{2n+1}(0)=0\), while \begin{equation*}\tag{11.5} P_{2n}(0)=(-1)^n\frac{(2n-1)!!}{(2n)!!}. \end{equation*}