# Verify the Rodrigues formula of the Legendre polynomials

1. Jan 11, 2016

### Happiness

How does (6.79) satisfy (6.70)?

After substitution, I get

$$(1-w^2)\frac{d^{l+2}}{dw^{l+2}}(w^2-1)^l-2w\frac{d^{l+1}}{dw^{l+1}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$

Using product rule in reverse on the first two terms,
$$(1-w^2)\frac{d^{l+1}}{dw^{l+1}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$
$$(1-w^2)\frac{d^{l+1}}{dw^{l+1}}(w^2-1)^l-2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$

Using product rule in reverse on the first two terms,
$$(1-w^2)\frac{d^{l}}{dw^{l}}(w^2-1)^l+2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$
$$\big[(1+2w-w^2)+l(l+1)\big]\frac{d^l}{dw^l}(w^2-1)^l$$

How do we prove it is equal to 0?

2. Jan 11, 2016

### blue_leaf77

I think it can be easily solved with the help of Leibniz rule for the differentiation of a product of two functions
$$\frac{d^n}{dx^n}(f(x)g(x)) = \sum_{k=0}^n C(n,k) \frac{d^{n-k}f}{dx^{n-k}} \frac{d^k g}{dx^n}$$
In the terms which contain differentiation $l+1$ or $l+2$ times, differentiate the function 1 or 2 times first.

3. Jan 12, 2016

### Happiness

I could't solve it.

Let $y=(w^2-1)^l$.
$\frac{dy}{dw}=2l(w^2-1)^{l-1}w$
$\frac{d^2y}{dw^2}=2l(w^2-1)^{l-2}\big[2(l-1)w+w^2-1\big]$

\begin{align}\frac{d^{l+1}y}{dw^{l+1}}&=\frac{d^{l}}{dw^{l}}2l(w^2-1)^{l-1}w\\&=2l\Big[w\frac{d^l}{dw^l}(w^2-1)^{l-1}+l\frac{d^{l-1}}{dw^{l-1}}(w^2-1)^{l-1}\Big]\end{align}

\begin{align}\frac{d^{l+2}y}{dw^{l+2}}&=\frac{d^{l}}{dw^{l}}2l(w^2-1)^{l-2}\big[2(l-1)w+w^2-1\big]\\&=4l(l-1)\Big[w\frac{d^l}{dw^l}(w^2-1)^{l-2}+l\frac{d^{l-1}}{dw^{l-1}}(w^2-1)^{l-2}\Big]+2l\Big[\frac{d^l}{dw^l}(w^2-1)^{l-1}\Big]\end{align}

After substituting (2) and (4) into the LHS of (6.70), it is not clear how the terms cancel out to give 0.

Last edited: Jan 12, 2016
4. Jan 12, 2016