let f(x)= e^(−2x) ln(1+x) developp f at integr serie .


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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
$f^((n)) (x)=Σ_(k=0) ^n (ln(1+x))^((k)) (e^(−2x) )^((n−k)) but (e^(−2x) )^((p)) =(−2)^p e^(−2x) ln(1+x)^((1)) = (1/(1+x)) ⇒(ln(1+x))^((p)) = (((−1)^(p−1) (p−1)!)/((1+x)^p )) f^((n)) (x)=(−2)^n e^(−2x) ln(1+x) +Σ_(k=1) ^n (((−1)^(k−1) (k−1)!)/((1+x)^k )) (−2)^(n−k) e^(−2x) ⇒ f^((n)) (0) =Σ_(k=1) ^n (−1)^(k−1) (k−1)!(−2)^(n−k) =(−2)^n Σ_(k=1) ^n (−1)^(k−1) (k−1)!(−2)^(−k) . f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=0) ^∞ (((−2)^n )/(n!)){Σ_(k=1) ^n (−1)^(k−1) (k−1)!(−2)^(−k) }x^n$
$f^{(n)} (x) = \sum_{k = 0}^{n} {(l n (1 + x))}^{(k)} {(e^{- 2 x})}^{(n - k)} b u t$ ${(e^{- 2 x})}^{(p)} = {(- 2)}^{p} e^{- 2 x}$ $l n {(1 + x)}^{(1)} = \frac{1}{1 + x} \Rightarrow {(l n (1 + x))}^{(p)} = \frac{{(- 1)}^{p - 1} (p - 1)!}{{(1 + x)}^{p}}$ $f^{(n)} (x) = {(- 2)}^{n} e^{- 2 x} l n (1 + x)$ $+ \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} (k - 1)!}{{(1 + x)}^{k}} {(- 2)}^{n - k} e^{- 2 x} \Rightarrow$ $f^{(n)} (0) = \sum_{k = 1}^{n} {(- 1)}^{k - 1} (k - 1)! {(- 2)}^{n - k}$ $= {(- 2)}^{n} \sum_{k = 1}^{n} {(- 1)}^{k - 1} (k - 1)! {(- 2)}^{- k} .$ $f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$ $= \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{n!} {\sum_{k = 1}^{n} {(- 1)}^{k - 1} (k - 1)! {(- 2)}^{- k}} x^{n}$
Commented by prof Abdo imad last updated on 24/Jun/18

$a n o t h e r w a y b u t e a s y w e h a v e$ $e^{- 2 x} = \sum_{n = 0}^{\infty} \frac{{(- 2 x)}^{n}}{n!} a n d$ ${l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} w i t h ∣ x ∣< 1 \Rightarrow$ $l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n + 1} + c (c = 0)$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} \Rightarrow$ $f (x) = (\sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{n!} x^{n}) (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n})$ $= (1 + \sum_{n = 1}^{\infty} \frac{{(- 2)}^{n}}{n!} x^{n}) (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n})$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} + \sum_{n = 1}^{\infty} c_{n} x^{n} w i t h$ $c_{n} = \sum_{i + j = n} a_{i} b_{j} = \sum_{i + j = n} \frac{{(- 2)}^{i}}{i!} \frac{{(- 1)}^{j - 1}}{j}$ $= \sum_{i = 1}^{n - 1} a_{i} b_{n - i} = \sum_{i = 1}^{n - 1} \frac{{(- 2)}^{i}}{i!} \frac{{(- 1)}^{n - i - 1}}{n - i}$
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