let-P-x-1-n-x-2-1-n-calculate-P-n-x-and-P-n-0-

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

l e t P (x) = \frac{1}{n!} {(x^{2} - 1)}^{n}

c a l c u l a t e P^{(n)} (x) a n d P^{(n)} (0)

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

w e h a v e P (x) = \frac{1}{n!} \sum_{k = 0}^{n} C_{n}^{k} x^{2 k} {(- 1)}^{n + k}

= \frac{{(- 1)}^{n}}{n!} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} x^{2 k} \Rightarrow P^{(n)} (x) = \frac{{(- 1)}^{n}}{n!} + \frac{{(- 1)}^{n}}{n!} \sum_{k = 1}^{n} {(- 1)}^{k} C_{n}^{k} {(x^{2 k})}^{) n)}

{(x^{2 k})}^{(1)} = (2 k) x^{2 k - 1}, {(x^{2 k})}^{(2)} = (2 k) (2 k - 1) x^{2 k - 2} \dots .

{(x^{2 k})}^{(n)} = (2 k) (2 k - 1) \dots . (2 k - n + 1) x^{2 k - 1}

= \frac{(2 k)!}{(2 k - n)!} x^{2 k - 1} \Rightarrow

P^{(n)} (x) = \frac{{(- 1)}^{n}}{n!} + \frac{{(- 1)}^{n}}{n!} \sum_{k = 1}^{n} {(- 1)}^{k} C_{n}^{k} \frac{(2 k)!}{(2 k - n)!} x^{2 k - 1}