Problem. Use the following relationship for the Legendre polynomials
$$ P_n(\cos\theta) = \frac{1}{n!} \frac{\partial^n}{\partial t^n} (1 – 2t\cos\theta + t^2)^{(-1/2)} |_{t=0} $$
in order to prove that $P_n(1) = 1$ and $P_n(-1)=(-1)^n$.
My attempt. Given that the generating function for the Legendre polynomials is $\sum_{n=0}^\infty P_n (x) t^n = (1 – 2tx + t^2)^{(-1/2)}$, then
\begin{align*}
P_n(\cos\theta)
= \frac{1}{n!} \frac{\partial^n}{\partial t^n} \sum_{m=0}^\infty P_m (\cos\theta) t^m|_{t=0}
= \frac{1}{n!} \sum_{m=0}^\infty P_m (\cos\theta) \frac{\partial^n}{\partial t^n} \left( t^m \right) |_{t=0}
\end{align*}
Note that, $m \in [0, \infty)$ then at some point $m=n$. Thus, when $m<n$ the $n$-th partial derivative will be equal to 0, implying that the $m$-th term of the sum is 0. Henceforth,
\begin{align*}
P_n(\cos\theta)
&= \frac{1}{n!} \sum_{m=n}^\infty P_m (\cos\theta) \frac{\partial^n}{\partial t^n} \left( t^m \right) |_{t=0}\\
&= \frac{1}{n!} \left( P_n (\cos\theta) \frac{\partial^n}{\partial t^n} t^n + P_{n+1} (\cos\theta) \frac{\partial^n}{\partial t^n} t^{n+1} + … \right) |_{t=0}\\
&= \frac{1}{n!} \left( n! P_n (\cos\theta) + t (n+1)! P_{n+1} (\cos\theta) + … \right) |_{t=0}
\end{align*}
Since we will evaluate at $t=0$, all the terms containing $t$ will vanish and we will eventually arrive at $P_n(\cos\theta) = P_n(\cos\theta)$.
Any suggestions?
Best Answer
$P_n(1)=P_n(\cos 0)=\frac{1}{n!} \frac{\partial^n}{\partial t^n} (1-t)^{-1} |_{t=0}=1$.
$P_n(-1)=P_n(\cos \pi)=\frac{1}{n!} \frac{\partial^n}{\partial t^n} (1+t)^{-1} |_{t=0}=(-1)^n$
Note that the square root in the definition of Legendre polynomials through a generating function is chosen to be $1$ at $t=0$ so $((1-t)^2)^{-1/2}=(1-t)^{-1}$ and same for $1+t$