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!!!

  • $\begingroup$ Did you try using the Rodrigues formula, inputting $x = \cos \theta$ and thus $(x^2-1)^l = (-1)^l \sin^{2l} \theta$? $\endgroup$ – Dead-End Apr 22 '14 at 2:33
  • $\begingroup$ Trying this way I get: $P_n^m(\cos \theta)=\displaystyle\frac{(-1)^n}{2^nn!}\sin^{m/2} \theta \displaystyle\frac{d^{n+m}}{dx^{n+m}}\sin^{n}\theta$ What should I do with $\displaystyle\frac{d^{n+m}}{dx^{n+m}}\sin^{n}\theta$? $\endgroup$ – YTS Apr 22 '14 at 3:02

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.