I should prove that
$$P_n^n(\cos \theta)=(2n-1)!! \sin^n\theta$$
$$P_n^m(0)=\begin{Bmatrix} (-1)^{(m+n)/2}\displaystyle\frac{(n+m-1)!!}{(n-m)!!} & \mbox{ if }& n+m \text{ even}\\ 0 & \mbox{if}& n+m \text{ odd}\end{Bmatrix}$$
$P_n^m(x)$ is a associated Legendre functions
I don't know what way follow. I apreciatte any advice to solve it.
Thanks a lot!!!
The associated Legendre polynomials $P_l^m$ (which are actually not polynomials for odd m :-)) are given by $$P_l^m(x) = \frac{(-1)^m}{2^ll!}(1-x^2)^{m/2}[(x^2-1)^l]^{(l+m)}$$ Where $f^{(n)}$ stands for the $n^{th}$ derivative of $f$. In this formula, $x = cos \theta \in[-1,1]$. We derive \begin{equation} \begin{split} P_n^n(x)&=\frac{(-1)^n}{2^nn!}(1-x^2)^{n/2}[(x^2-1)^n]^{(2n)} \\ &=\frac{(-1)^n(2n)!}{2^nn!}(1-x^2)^{n/2} \\ &=(-1)^n(2n-1)!!(1-x^2)^{n/2} \end{split} \end{equation} or, in goniometric terms, $$P_n^n(cos \theta) = (-1)^n(2n-1)!! \cdot sin^n \theta$$ Next, \begin{equation} \begin{split} P_n^m(0)&=\frac{(-1)^m}{2^nn!}(1-x^2)^{m/2}[(x^2-1)^n]^{(n+m)} \left.\right|_{x=0} \\ &=\frac{(-1)^m}{2^nn!}[(x^2-1)^n]^{(n+m)} \left.\right|_{x=0} \\ &=\frac{(-1)^m}{2^nn!}\left[\sum_{k=0}^n \binom{n}{k}x^{2n-2k}(-1)^k\right]^{(n+m)} \left.\right|_{x=0} \\ &=\begin{cases} 0 &\text{ if $n+m$ odd} \\ \text {see below} &\text{ if $n+m$ even} \end{cases} \end{split} \end{equation} The latter is determined by the term with $2n-2k = n+m$, hence $k = \frac{n-m}{2}$, and can be evaluated as follows: \begin{equation} \begin{split} P_n^m(0)&=\frac{(-1)^m}{2^nn!}\binom{n}{(n-m)/2}(n+m)!(-1)^{(n-m)/2} \\ &= \frac{(-1)^{(n+m)/2}}{2^nn!}\frac{n!}{\frac{n-m}{2}!\frac{n+m}{2}!}(n+m)! \\ &= (-1)^{(n+m)/2}\frac{(n+m)!}{(n-m)!!(n+m)!!} \\ &= (-1)^{(n+m)/2}\frac{(n+m-1)!!}{(n-m)!!} \end{split} \end{equation}
$P_n^m(\cos \theta)=\displaystyle\frac{(-1)^n}{2^nn!}\sin^{m/2} \theta \displaystyle\frac{d^{n+m}}{dx^{n+m}}\sin^{n}\theta$
What should I do with $\displaystyle\frac{d^{n+m}}{dx^{n+m}}\sin^{n}\theta$?
– EQJ Apr 22 '14 at 03:02