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Question Number 74501 by mathmax by abdo last updated on 25/Nov/19

$$letP\left(x\right)=\frac{1}{n!}{(x^2-1)}^n$$$$calculate{P}^{\left(n\right)}\left(x\right)and{P}^{\left(n\right)}\left(0\right)$$$$$$

Commented by mathmax by abdo last updated on 28/Nov/19

$$wehaveP\left(x\right)=\frac{1}{n!}\sum _{k=0}^n{C}_n^k{x}^{2k}{(-1)}^{n+k}$$$$=\frac{{(-1)}^n}{n!}\sum _{k=0}^n{(-1)}^k{C}_n^k{x}^{2k}\Rightarrow P^{\left(n\right)}\left(x\right)=\frac{{(-1)}^n}{n!}+\frac{{(-1)}^n}{n!}\sum _{k=1}^n{(-1)}^k{C}_n^k{\left(x^{2k}\right)}^{)n)}$$$${\left(x^{2k}\right)}^{\left(1\right)}=\left(2k\right)x^{2k-1},{\left(x^{2k}\right)}^{\left(2\right)}=\left(2k\right)(2k-1)x^{2k-2}....$$$${\left(x^{2k}\right)}^{\left(n\right)}=\left(2k\right)(2k-1)....(2k-n+1)x^{2k-1}$$$$=\frac{\left(2k\right)!}{(2k-n)!}{x}^{2k-1}\Rightarrow$$$$P^{\left(n\right)}\left(x\right)=\frac{{(-1)}^n}{n!}+\frac{{(-1)}^n}{n!}\sum _{k=1}^n{(-1)}^k{C}_n^k\frac{\left(2k\right)!}{(2k-n)!}{x}^{2k-1}$$

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