let-f-x-x-n-1-e-x-determine-f-n-x-with-n-integr-natural-

Question Number 52673 by maxmathsup by imad last updated on 11/Jan/19

l e t f (x) = (x^{n} - 1) e^{x} d e t e r m i n e f^{(n)} (x) w i t h n i n t e g r n a t u r a l

Commented by maxmathsup by imad last updated on 13/Jan/19

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

= \sum_{k = 0}^{n} C_{n}^{k} {(x^{n} - 1)}^{(k)} e^{x} = (x^{n} - 1) e^{x} + e^{x} \sum_{k = 1}^{n} C_{n}^{k} {(x^{n})}^{k} b u t w e h a v e

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

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

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

Answered by Smail last updated on 11/Jan/19

f^{'} (x) = ({n x}^{n - 1} + x^{n} - 1) e^{x}

f^{″} (x) = (n (n - 1) x^{n - 2} + 2 {n x}^{n - 1} + x^{n} - 1) e^{x}

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

f^{(4)} (x) = (n \to (n - 3) x^{n - 4} + 4 n \to (n - 2) x^{n - 3} + 6 n (n - 1) x^{n - 2} + 4 {n x}^{n - 1}) e^{x} + f (x)

f^{(n)} (x) = (P_{n - 0}^{n} C_{0}^{n} x^{0} + P_{n - 1}^{n} C_{1}^{n} x^{1} + P_{n - 2}^{n} C_{2}^{n} x^{2} + \dots + P_{0}^{n} C_{n}^{n} x^{n} - 1) e^{x}

f^{(n)} (x) = e^{x} (\underset{k = 0}{\sum^{n}} P_{n - k}^{n} C_{k}^{n} x^{k} - 1)

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