# Legendre polynomial of zero

Tags:
1. Mar 18, 2016

### watisphysics

1. The problem statement, all variables and given/known data
Using the Generating function for Legendre polynomials, show that:
$P_n(0)=\begin{cases}0 & n \ is \ odd\\\frac{(-1)^n (2n)!}{2^{2n} (n!)^2} & n \ is \ even\end{cases}$

2. Relevant equations
Generating function: $(1-2xt+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(x) t^n$

3. The attempt at a solution
I put $x=0$ in the generating function and got $(1+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(0) t^n$ Now, I think I should expand $(1+t^2)^{-1/2}$ using the binomial theorem, then I get an expression that's only valid for even values of $n$, but it's not the expression stated in the question. So, how can I solve this?

2. Mar 18, 2016

### Ray Vickson

So, what does that tell you about $P_n(0)$ for odd $n$?

3. Mar 22, 2016

### vela

Staff Emeritus
"Valid" isn't the right word here. The expansion only contains terms with even exponents, so what does that tell you about the coefficients of the odd terms?