How to proof that $(n+1)P_{n}(x)-(2n+1)x P_{n}(x)+n P_{n-1}(x)=0$?

Question

How to proof that $(n+1)P_{n+1}(x)-(2n+1)x P_{n}(x)+n P_{n-1}(x)=0$

where

$P_{n}(x)=\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}$.

Answer 1

You may use Rodrigues' formula to prove that Legendre polynomials have to following generating function: $$ \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n\geq 0}P_n(x)\,t^n = g(t)\tag{1}$$ The relation $(n+1) P_{n+1}(x)-(2n+1) x P_n(x) + n P_{n-1}(x)=0 $ translates into:

$$ g'(t) - \left( x g(t) + 2tx g'(t)\right)+t\cdot\frac{d}{dt}\left( t g(t)\right) =0 \tag{2}$$

or:

$$(1-2tx+t^2)\, g'(t) = (x-t)\, g(t) \tag{3} $$

or:

$$ \frac{d}{dt}\log g(t) = \frac{x-t}{1-2tx+t^2} \tag{4}$$

that is straightforward to prove, given the LHS of $(1)$.