Proving Bonnets' Recursion with Rodrigues' Formula

Question

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.

Answer 1

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}

Answer 2

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*}