let-f-x-x-n-e-2nx-with-n-integr-natural-calculate-f-n-0-

Question Number 40105 by maxmathsup by imad last updated on 15/Jul/18

l e t f (x) = x^{n} e^{- 2 n x} w i t h n i n t e g r n a t u r a l

c a l c u l a t e f^{(n)} (0) .

Commented by maxmathsup by imad last updated on 17/Jul/18

l e i b n i z f o r m u l a g i v e f^{(p)} (x) = \sum_{k = 0}^{p} C_{p}^{k} {(x^{n})}^{(k)} {(e^{- 2 n x})}^{(p - k)} b u t

{(x^{n})}^{(k)} = n (n - 1) \dots . (n - k + 1) x^{n - k} i f k ⩽ n a n d {(x^{n})}^{(k)} = 0 i f k > n \Rightarrow

{(x^{n})}^{(k)} = \frac{n!}{(n - k)!} x^{n - k} a l s o w e h a v e {e^{- 2 n}}^{(p - k)} = {(- 2 n)}^{p - k} e^{- 2 n x} \Rightarrow

f^{(p)} (x) = \sum_{k = 0}^{p} C_{p}^{k} \frac{n!}{(n - k)!} x^{n - k} {(- 2 n)}^{p - k} e^{- 2 n x} l e t s u p p o s e n ⩽ p

f^{(p)} (x) = \sum_{k = 0}^{n - 1} C_{p}^{k} \frac{n!}{(n - k)!} x^{n - k} {(- 2 n)}^{p - k} e^{- 2 n x} + C_{p}^{n} \frac{n!}{0!} {(- 2 n)}^{p - n} e^{- 2 n x}

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

f^{(p)} (0) = C_{p}^{n} n! {(- 2 n)}^{p - n} \Rightarrow f^{(n)} (0) = n! C_{n}^{n} = n!