I would like to show that $(n+1)P_{n+1}(x)+nP_{n-1}(x)=(2n+1)xP_{n}(x)$ using Rodrigues' formula, not the generating function.
I got to this point, but have not been able to progress further.
$$(n+1)P_{n+1}(x)=\frac{x(n+1)(D^{n}(x^{2}-1)^{n})+n(n+1)(D^{n-1}(x^{2}-1)^{n})}{2^{n}n!}$$
$D^{n}$ denotes the $n^{th}$ derivative with respect to $x$.
I see that part of the $(2n+1)xP_{n}(x)$ term is formed in the first term of the fraction, but do not see how to recover the rest of Bonnets' recursion.
Any help would be appreciated.
Thanks.
Best Answer
Note that by Rodrigues' formula, $P_{n+1}=\frac{1}{2^{n+1}(n+1)!}D^{n+1}[(x^{2}-1)^{n+1}]$. Multiplying by $(n+1)$, we have: \begin{align*} &{\left(n+1\right)P_{n+1}(x)}={\left(n+1\right)\frac{1}{2^{n+1}(n+1)!}D^{n+1}[(x^{2}-1)^{n+1}]}\\ =\;&{\left(n+1\right)\frac{1}{2^{n}n!}D^{n}[x(x^{2}-1)^{n}]}\\ \end{align*} Using the Leibniz rule: \begin{align*} =\;&{\left(n+1\right)\frac{xD^{n}[(x^{2}-1)^{n}]+nD^{n-1}[(x^{2}-1)^{n}]}{2^{n}n!}}\\ =\;&x(n+1)P^{n}(x)+\frac{n(n+1)}{2^{n}n!}D^{n-1}[(x^2-1)^n]\\ \end{align*} Adding and subtracting $\frac{1}{2^{n}(n-1)!}xD^{n}[(x^2-1)^n]$ we get: \begin{align*} =\;&x(2n+1)P^{n}(x)+\frac{n(n+1)}{2^{n}n!}D^{n-1}[(x^2-1)^n]-\frac{1}{2^{n}(n-1)!}xD^{n}[(x^2-1)^n]\\ \end{align*} Using L.G.'s observation (or deriving it with the Leibniz rule) we have: \begin{align*} =\;&x(2n+1)P^{n}(x)+\frac{(n+1)}{2^{n}(n-1)!}D^{n-1}[(x^2-1)^n]-\frac{1}{2^{n}(n-1)!}(D^{n}[x(x^2-1)^n]-nD^{n-1}[(x^2-1)^n])\\ =\;&x(2n+1)P^{n}(x)+\frac{2n}{2^{n}(n-1)!}D^{n-1}[(x^2-1)^{n}-x^{2}(x^{2}-1)^{n-1}]\\ =\;&x(2n+1)P^{n}(x)+\frac{n}{2^{n-1}(n-1)!}D^{n-1}[(x^2-1)^{n-1}(-1)]\\ =\;&x(2n+1)P^{n}(x)-nP^{n-1}(x). \end{align*}