Simplifying a Legendre polynomial

dwsmith

Well-known member
Given the following expression
$\mathcal{P}_{n}(0) = \left.\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}} \sum_{k = 0}^{n}\binom{n}{k}(x^2)^k(-1)^{n - k}\right|_{x = 0}, \qquad (*)$
I know for $$k$$ even we get
$\mathcal{P}_{n}(0) = \frac{1}{2^{n}n!}\binom{n}{\frac{n}{2}}n!(-1)^{n / 2}. \qquad (**)$
However, I don't see how this is done. Can someone explain how we go from $$(*)$$ to $$(**)$$?

girdav

Member
The key point is the computation of $\frac{d(x^p)}{dx^q}$: if $q>p$ this is $0$, if $p<q$ this gives, evaluated at $0$, again $0$. And the $p$-th derivative of $x^p$ is $p!$.

dwsmith

Well-known member
How does this simplify?
\begin{align*}
I_{n} &= \frac{1}{2n + 1}\left[\frac{1}{2^{n - 1}}
\binom{n - 1}{\frac{n - 1}{2}}(-1)^{(n - 1)/2} -
\frac{1}{2^{n + 1}}
\binom{n + 1}{\frac{n + 1}{2}}(-1)^{(n + 1)/2}\right]\\
&= \frac{1}{(2n + 1)2^{n - 1}\left[\left(\frac{n - 1}{2}\right)!\right]^2} \left[(n-1)!(-1)^ {(n - 1)/2} - \frac{4}{n^2}(-1)^{(n + 1)/2}\right]\\
&= ???\\
&= \frac{(-1)^{(n - 1)/2}}{2^{n - 1}}\frac{(n - 1)!}{
\left(\frac{n - 1}{2}\right)!\left(\frac{n - 1}{2}\right)!}
\frac{1}{n + 1}
\end{align*}