<s>[INST] Proving Bonnets' Recursion with Rodrigues' Formula 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.
 [/INST] Using Rodrigues' formula $\displaystyle P_n(x)=\frac{1}{2^n n!}D^n\left(x^2-1\right)^n$, we can write
\begin{align*}
&\color{blue}{\left(n+1\right)P_{n+1}(x)}-\color{green}{\left(2n+1\right)xP_n(x)}+\color{red}{nP_{n-1}(x)}=\\
=\;&\frac{1}{2^{n+1}n!}D^{n-1}\biggl\{\color{blue}{D^2\left(x^2-1\right)^{n+1}}+\color{red}{4n^2\left(x^2-1\right)^{n-1}}\biggr\}-\color{green}{\left(2n+1\right)xP_n(x)}=\\
=\;&\frac{1}{2^{n+1}n!}D^{n-1}\biggl\{\left(x^2-1\right)^{n-1}\biggl[
\color{blue}{4n\left(n+1\right)x^2+2\left(n+1\right)\left(x^2-1\right)}+\color{red}{4n^2}\biggr]\biggr\}-\color{green}{\left(2n+1\right)xP_n(x)}=\\
=\;&\frac{2n+1}{2^{n}n!}D^{n-1}\biggl\{\left(x^2-1\right)^{n-1}\biggl[
\left(n+1\right)x^2+n-1\biggr]\biggr\}-\color{green}{\frac{2n+1}{2^{n}n!}xD^{n}\left(x^2-1\right)^n}.
\end{align*}
Now observe that $[D,x]=1$ implies $[D^n,x]=nD^{n-1}$. Using this to replace $xD^n$ in the above expression by $D^n\circ x-nD^{n-1}$, we can rewrite it as
\begin{align}
&\frac{2n+1}{2^{n}n!}D^{n-1}\biggl\{\left(x^2-1\right)^{n-1}\biggl[
\left(n+1\right)x^2+n-1\biggr]\color{green}{-D\left[x\left(x^2-1\right)^n\right]+n\left(x^2-1\right)^{n}}\biggr\}=\\
=\;&\frac{2n+1}{2^{n}n!}D^{n-1}\biggl\{\left(x^2-1\right)^{n-1}\cdot 0\biggr\}=0.
\end{align}
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