How does (6.79) satisfy (6.70)?(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Verify the Rodrigues formula of the Legendre polynomials

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