prove $P_{2n}(0)=(-1)^n\frac{(2n)!}{2^{2n}(n!)^2}$ of Legendre Polynomials.

Question

Use: $P_n(x)=\frac{1}{2^n}\sum_{m=0}^M(-1)^m\frac{(2n-2m)!}{m!(n-m)!(n-2m)!}x^{n-2m}$ where M=n/2 if even, (n-1)/2 if n odd, to prove $P_{2n}(0)=(-1)^n\frac{(2n)!}{2^{2n}(n!)^2}$ and $P_{2n+1}(0)=0$.

There's something I'm missing. If x=0, only the case where n=0 ($a_0$) will be non zero and equal 1. In case I didn't reproduce whatever it is that I'm missing, this problem is from Asmar Partial differential equations chpt 5.5 problem 2b.

Answer 1

Note that if we work with a even number the sum is $$P_{2n}\left(x\right)=\frac{1}{2^{2n}}\sum_{m=0}^{n}\left(-1\right)^{m} \frac{\left(4n-2m\right)!}{m!\left(2n-m\right)!\left(2n-2m\right)!}x^{2n-2m} $$ $$ =\frac{1}{2^{2n}}\sum_{m=0}^{n-1}\left(-1\right)^{m} \frac{\left(4n-2m\right)!}{m!\left(2n-m\right)!\left(2n-2m\right)!}x^{2n-2m}+\left(-1\right)^{n} \frac{1}{2^{2n}}\frac{\left(2n\right)!}{n!^{2}} $$ and so if take $x=0 $ we have $$P_{2n}\left(0\right)=\left(-1\right)^{n} \frac{1}{2^{2n}}\frac{\left(2n\right)!}{n!^{2}} $$ since every other addends depend on $x$ (the exponent is positive). If we work with a odd number we have $$P_{2n+1}\left(x\right)=\frac{1}{2^{2n+1}}\sum_{m=0}^{n}\left(-1\right)^{m} \frac{\left(4n+2-2m\right)!}{m!\left(2n+1-m\right)!\left(2n+1-2m\right)!}x^{2n+1-2m} $$ but in this case, for every $m=0,\dots,n$, $x$ has a positive exponent, so $$P_{2n+1}\left(0\right)=0.$$