let give f_n (x)= ((x^2 −1)^n )^((n)) find f


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Question Number 28428 by abdo imad last updated on 25/Jan/18

$l e t g i v e f_{n} (x) = {({(x^{2} - 1)}^{n})}^{(n)} f i n d f_{n} .$
Commented by abdo imad last updated on 27/Jan/18

$w e h a v e f_{n} (x) = {(p (x))}^{(n)} w i t h p (x) = {(x^{2} - 1)}^{n}$ $p (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{2 k} {(- 1)}^{n - k} = {(- 1)}^{n} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} x^{2 k}$ ${(p (x))}^{(n)} = {(- 1)}^{n} \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} {(x^{2 k})}^{(n)} b u t i f$ $2 k < n {(x^{2 k})}^{(n)} = 0 s o$ $f_{n} (x) = {(- 1)}^{n} \sum_{k = [\frac{n - 1}{2}] + 1}^{n} C_{n}^{k} {(x^{2 k})}^{(n)} . l e t f i n d$ ${(x^{p})}^{(n)} f o r p ⩾ n w e h a v e {(x^{p})}^{(1)} = p x^{p - 1}, {(x^{p})}^{(2)} = p (p - 1) x^{p - 2}$ $s o {(x^{p})}^{(n)} = p (p - 1) . . . . (p - n + 1) x^{p - n}$ $= \frac{p (p - 1) . . . . (p - n + 1) (p - n)!}{(p - n)!} x^{p - n} = \frac{p!}{(p - n)!} x^{p - n}$ $\Rightarrow {(x^{2 k})}^{(n)} = \frac{(2 k)!}{(2 k - n)!} x^{2 k - n} s o$ $f_{n} (x) = {(- 1)}^{n} \sum_{k = [\frac{n - 1}{2}] + 1}^{n} C_{n}^{k} \frac{(2 k)!}{(2 k - n)!} x^{2 k - n} .$
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