I am given with a relation \begin{equation} \frac{d^2}{dx^2}(P_l(x))=\frac{1}{2}\sum_{n=(0,1),2}^{l-2}(l-n)(l+n+1)(2n+1)P_n(x). \end{equation} The above sum starts with 0 for even end point, and 1 for odd end point, and proceeds in a step of 2. Using the above equation, I get \begin{equation} \frac{d^4}{dx^4}(P_l(x))=\frac{1}{4}\sum_{n=(0,1),2}^{l-2}\sum_{m=(0,1),2}^{n-2}(l-n)(l+n+1)(2n+1)(n-m)(n+m+1)(2m+1)P_m(x). \end{equation} The problem is that I want the double sum in the above as a single sum, just in terms of $n, l$. Is there any idea how to simplify the above?
$\endgroup$ Reset to defaultDerivative of Legendre Polynomial
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