let-give-f-x-x-n-1-e-x-with-n-from-N-find-f-n-0-

Question Number 29349 by abdo imad last updated on 07/Feb/18

l e t g i v e f (x) = (x^{n} - 1) e^{- x} w i t h n f r o m N^{★}

f i n d f^{(n)} (0) .

Commented by abdo imad last updated on 09/Feb/18

t h e l e i b n i z f o r u l a g i v e f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(x^{n} - 1)}^{(k)} {(e^{- x})}^{(n - k)}

= (x^{n} - 1) {(e^{- x})}^{(n)} + \sum_{k = 1}^{n} C_{n}^{k} {(x^{n} - 1)}^{(k)} {(e^{- x})}^{(n - k)} b u t

{(e^{- x})}^{(1)} = - e^{- x}, {(e^{- x})}^{(2)} = {(- 1)}^{2} e^{- x} \dots {(e^{- x})}^{(p)} = {(- 1)}^{p} e^{- x} s o

f^{(n)} (x) = {(- 1)}^{n} (x^{n} - 1) e^{- x} + \sum_{k = 1}^{n} C_{n}^{k} {(- 1)}^{n - k} e^{- x} {(x^{n} - 1)}^{(k)}

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

{(x^{n} - 1)}^{(p)} = n (n - 1) \dots (n - p + 1) x^{n - p} = \frac{n!}{(n - p)!} x^{n - p} w i t h p ⩽ n

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

= {(- 1)}^{n} (x^{n} - 1) e^{- x} + {(- 1)}^{n} e^{- x} \sum_{k =}^{n - 1} (\dots) x^{n - p}

+ {(- 1)}^{n} e^{- x} {(- 1)}^{n} C_{n}^{n} n! \Rightarrow

f^{(n)} (0) = {(- 1)}^{n - 1} + n!