let-f-x-e-2x-ln-1-2x-1-find-f-n-x-and-f-n-0-2-developp-f-at-integr-serie-

Question Number 83245 by mathmax by abdo last updated on 29/Feb/20

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

1) f i n d f^{(n)} (x) a n d f^{(n)} (0)

2) d e v e l o p p f a t i n t e g r s e r i e

Commented by mathmax by abdo last updated on 03/Mar/20

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

= l n (2 x + 1) {(- 2)}^{n} e^{- 2 x} + \sum_{k = 1}^{n} C_{n}^{k} {(l n (2 x + 1))}^{(k)} {(- 2)}^{n - k} e^{- 2 x}

w e h a v e {(l n (2 x + 1))}^{(1)} = \frac{2}{2 x + 1} = \frac{1}{x + \frac{1}{2}} \Rightarrow

{(l n (2 x + 1))}^{(k)} = {(\frac{1}{x + \frac{1}{2}})}^{(k - 1)} = \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + \frac{1}{2})}^{k}} \Rightarrow

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

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

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

f^{(n)} (0) = \sum_{k = 1}^{n} (k - 1)! C_{n}^{k} = \sum_{k = 1}^{n} (k - 1)! \times \frac{n!}{k! (n - k)!}

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

Commented by mathmax by abdo last updated on 03/Mar/20

2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = f (0) + \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

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