Proof the Recurrence formula with Rodrigues' Formula

Question

My University Professor gave the task to proof the recurrence formula without generating function of Legendre Polynomial , only with Rodrigues' Formula .

So far, I used :

$P_l\left(x\right)=\frac{1}{{2}^{l}l!}\frac{{d}^{l}}{{dx}^{l}}{\left({x}^{2}-1\right)}^{l}$

I want to proof the recurrence formula :

$\left({x}^{2}-1\right)\frac{d{P}_{l}}{dx}=lx{P}_{l}-l{P}_{l-1}$

I did : $\left({x}^{2}-1\right)\frac{d{P}_{l}}{dx}=\frac{2l}{{2}^{l}l!}\left[\left({x}^{2}-1\right)\frac{{d}^{l}}{{dx}^{l}}\right]x{\left({x}^{2}-1\right)}^{l-1}$

by far, I used $[D^n,x]=n{D}^{n-1}$.but still not sure the way to proof !

Any help would be appreciated.

Thanks :)

Answer 1

I am trying to answer my pull.

$$D{P}_{l+1}\left(x\right)=\frac{1}{\left(l+1\right)!{2}^{l+1}}{D}^{l+2}{\left({x}^{2}-1\right)}^{l+1}=\frac{1}{\left(l\right)!{2}^{l}}{D}^{l+1}x{\left({x}^{2}-1\right)}^{l}=\frac{1}{\left(l\right)!{2}^{l}}\left[x{D}^{l+1}+\left(l+1\right){D}^{l+1}\right]{\left({x}^{2}-1\right)}^{l}=xD{P}_{l}\left(x\right)+\left(l+1\right){P}_{l}\left(x\right)$$ $|1|$ $$\implies D{P}_{l+1}\left(x\right)=xD{P}_{l}\left(x\right)+\left(l+1\right){P}_{l}\left(x\right)$$ Now,

$$xD{P}_{l}\left(x\right)-D{P}_{l-1}\left(x\right)=\frac{1}{\left(l\right)!{2}^{l}}\left[x{D}^{l+1}{\left({x}^{2}-1\right)}^{l}-2l{D}^{l}{\left({x}^{2}-1\right)}^{l-1}\right]=\frac{1}{\left(l\right)!{2}^{l}}\left[\left({D}^{l}x-l{D}^{l-1}\right){\left({x}^{2}-1\right)}^{l-1}2xl-2l{D}^{l}{\left({x}^{2}-1\right)}^{l-1}\right]=\frac{1}{\left(l\right)!{2}^{l}}\left[2l{D}^{l}\left({x}^{2}-1\right){\left({x}^{2}-1\right)}^{l-1}-l{D}^{l-1}{\left({x}^{2}-1\right)}^{l-1}2xl\right]=\frac{1}{\left(l\right)!{2}^{l}}\left[2l{D}^{l}{\left({x}^{2}-1\right)}^{l}-l{D}^{l}{\left({x}^{2}-1\right)}^{l}-l{D}^{l}a\right]=\frac{1}{\left(l\right)!{2}^{l}}l{D}^{l}{\left({x}^{2}-1\right)}^{l}$$ $|2|$ $$\implies xD{P}_{l}\left(x\right)-D{P}_{l-1}\left(x\right)=l{P}_{l}\left(x\right)$$

$|2|-{|1|}_{l=l-1}\times x$ gives us: $$\left(1-{x}^{2}\right)D{P}_{l}\left(x\right)=l{P}_{l-1}\left(x\right)-lx{P}_{l}\left(x\right)$$ NB it is also valid for $x=0$.