I've poked around Mathematics Stack Exchange for a while, and while I'm sure this is an elementary problem to you guys, I cannot figure this out. I have found similar solved problem prompts on here, but the issue is that I cannot use Rodrigues' Formula or the generating function, I have to use the definition of a Legendre polynomial itself. What I'm trying to do is show the following (here is the original prompt):
$\text{From the definition of a Legendre polynomial show that} $
$$ P_{2n}(0)=(-1)^{n}\frac{(2n)!}{2^{2n}(n!)^2} $$
Here was my attempt so far:
$$ \text{Definition:} \ \ \ P_n(x)=\frac{1}{2^n} \sum_{k=0}^{M}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}x^{n-2k}
$$
$\\$
$$ P_{2n}(x)=\frac{1}{2^{2n}}\sum_{k=0}^{M}\frac{(-1)^k(4n-2k)!}{k!(2n-k)!(2n-2k)!}x^{2n-2k}
$$
$$
P_{2n}(0)=\frac{1}{2^{2n}}\sum_{k=0}^{M}\frac{(-1)^k(4n-2k)!}{k!(2n-k)!(2n-2k)!}(0)^{2n-2k}
$$
Here my confusion is that it looks like evaluating x at zero would make every single term disappear ($x^{2n-2k}=0$). Have I made a mistake, or how do I deal with this?
Best Answer
Remember the definition of $0^0$ (in series): $0^0 = 1$. Thus your $k=n$ term doesn't vanish. (I'm presuming you meant to have an $n$ in your upper limit, not $M$.) Making the appropriate substitution of $k=n$ in the Legendre polynomial coefficient yields the desired result.