공대생 공부노트

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^{..

$$\frac{d^{n}}{dx^{n}}\left[x(x^{2}-1)^{n}\right]=x\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}+n\frac{d^{n-1}}{dx^{n-1}}(x^{2}-1)^{n}$$ 우리에게 알맞게 정리해주면 $$x\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}={\color{blue}{\frac{d^{n}}{dx^{n}}x(x^{2}-1)^{n}-n\frac{d^{n-1}}{dx^{n-1}}(x^{2}-1)^{n}}}$$ 우리가 전개하던 식에 대입해주면 $$\frac{(n+1)}{2^{n}(n-1)!}\frac{d^{n-1}}{dx^{n-1}}(x^{2}-1)^{n}-\frac{1}{2^{n}(n-1)!}\left(\frac{d^{n}}{dx^{n..