let-f-x-e-2x-ln-1-x-developp-f-at-integr-serie-

Question Number 37269 by abdo.msup.com last updated on 11/Jun/18

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

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

Commented by prof Abdo imad last updated on 24/Jun/18

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

{(e^{- 2 x})}^{(p)} = {(- 2)}^{p} e^{- 2 x}

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

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

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

f^{(n)} (0) = \sum_{k = 1}^{n} {(- 1)}^{k - 1} (k - 1)! {(- 2)}^{n - k}

= {(- 2)}^{n} \sum_{k = 1}^{n} {(- 1)}^{k - 1} (k - 1)! {(- 2)}^{- k} .

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

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

Commented by prof Abdo imad last updated on 24/Jun/18

a n o t h e r w a y b u t e a s y w e h a v e

e^{- 2 x} = \sum_{n = 0}^{\infty} \frac{{(- 2 x)}^{n}}{n!} a n d

{l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} w i t h ∣ x ∣< 1 \Rightarrow

l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n + 1} + c (c = 0)

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

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

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

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} + \sum_{n = 1}^{\infty} c_{n} x^{n} w i t h

c_{n} = \sum_{i + j = n} a_{i} b_{j} = \sum_{i + j = n} \frac{{(- 2)}^{i}}{i!} \frac{{(- 1)}^{j - 1}}{j}

= \sum_{i = 1}^{n - 1} a_{i} b_{n - i} = \sum_{i = 1}^{n - 1} \frac{{(- 2)}^{i}}{i!} \frac{{(- 1)}^{n - i - 1}}{n - i}