What Determines the Values of Legendre Polynomials at Zero?

[watisphysics]

Homework Statement

Using the Generating function for Legendre polynomials, show that:
$P_n(0)=\begin{cases}0 & n \ is \ odd\\\frac{(-1)^n (2n)!}{2^{2n} (n!)^2} & n \ is \ even\end{cases}$

Homework Equations

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

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?

 

[Ray Vickson]

Homework Statement

Using the Generating function for Legendre polynomials, show that:
$P_n(0)=\begin{cases}0 & n \ is \ odd\\\frac{(-1)^n (2n)!}{2^{2n} (n!)^2} & n \ is \ even\end{cases}$

Homework Equations

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

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?

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

 

[vela]

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?

"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?