f-x-x-n-e-nx-Determinate-f-n-x-

Question Number 166814 by mathocean1 last updated on 28/Feb/22

f (x) = x^{n} e^{- n x}

D e t e r m i n a t e f^{(n)} (x) .

Answered by TheSupreme last updated on 28/Feb/22

f (x) = {({x e}^{- x})}^{n} =

f^{n} (x) = D^{n} (t^{n}) ∣_{t = {x e}^{- x}} D^{n} ({x e}^{- x})

D^{1} = (1 - x) e^{- x}

D^{2} = (- 1 - 1 + x) e^{- x} = (- 2 + x) e^{- x}

D^{3} = (1 + 2 - x) e^{- x} = (3 - x) e^{- x}

D^{n} = {(- 1)}^{n} (x - n) e^{- x}

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

Answered by Mathspace last updated on 28/Feb/22

f^{(p)} (x) = \sum_{k = 0}^{p} C_{p}^{k} {(x^{n})}^{(k)} {(e^{- n x})}^{(p - k)}

b u t {(x^{n})}^{(k)} = n (n - 1) \dots) (

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

{(e^{- n x})}^{(p - k)} = {(- n)}^{p - k} e^{- n x} \Rightarrow

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

\Rightarrow f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} \frac{n!}{(n - k)!} {(- n)}^{n - k} e^{- n x}