Legendre recurrence formula (2)

Rodrigues' formula

 

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

 

Change \(n\) to \((n+1)\)

 

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

 

Multiply \((n+1)\) on both side

 

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

$$=\frac{(n+1)}{2^{n+1}(n+1)!}\frac{d^{n}}{dx^{n}}\left[\frac{d}{dx}\left\{(x^{2}-1)^{n}(x^{2}-1)\right\}\right]$$

 

re-arrange the inner term

 

$$\frac{d}{dx}\left\{(x^{2}-1)^{n}(x^{2}-1)\right\}$$

$$=n(x^{n}-1)^{n-1}(2x)\cdot(x^{2}-1)+(x^{2}-1)^{n}(2x)$$

$$=2nx(x^{2}-1)^{n}+2x(x^{2}-1)^{n}$$

$$=(2nx+2x)(x^{2}-1)^{n}$$

$$=2x(n+1)(x^{2}-1)^{n}$$

 

Back to the original equation

 

$$(n+1)P_{n+1}(x)=\frac{(n+1)}{2^{n+1}(n+1)!}\frac{d^{n}}{dx^{n}}\left[2x(n+1)(x^{2}-1)^{n}\right]$$

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

$$={\color{blue}{\frac{(n+1)}{2^{n}n!}}}\frac{d^{n}}{dx^{n}}x(x^{2}-1)^{n}$$

 

re-arrange the right inner term

 

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

 

Then, insert in back to the original equation

 

$$\frac{(n+1)}{2^{n}n!}\left[x\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}+\frac{d^{n}}{dx^{n}}x(x^{2}-1)^{n}\right]\cdots(a)$$

 

Bring 'Rodrigues' formula' to make first term of equation \((a)\)

 

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

 

multiply \(x(n+1)\) on both side

 

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

 

Then, arrange the second term of equation \((a)\)

 

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

 

Bring 'Leibniz formula'

 

$$\frac{d^{n}}{dx^{n}}(uv)=\left(\frac{d^{n}}{dx^{n}}u\cdot v\right)+\frac{n}{1!}\frac{d^{n-1}}{dx^{n-1}}u\frac{dv}{dx}+\frac{n(n-1)}{2!}\frac{d^{n-2}}{dx^{n-2}}u\frac{d^{2}v}{dx^{2}}+\cdots + \frac{n}{1!}\frac{du}{dx}\frac{d^{n-1}v}{dx^{n-1}}+u\frac{d^{n}v}{dx^{n}}$$