let-f-x-ln-2-x-1-give-D-n-0-of-f-2-developp-f-at-integr-serie-

Question Number 40113 by maxmathsup by imad last updated on 15/Jul/18

l e t f (x) = l n (2 + x)

1) g i v e D_{n} (0) o f f

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 prof Abdo imad last updated on 17/Jul/18

1) w e h a v e f (x) = \sum_{k = 0}^{n} \frac{f^{(k)} (0)}{k!} x^{k} + \frac{x^{n + 1}}{(n + 1)!} θ (x)

w i t h {l i m}_{x \to 0} θ (x) = 0 b u t

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

\Rightarrow f^{(k)} (0) = \frac{{(- 1)}^{k - 1} (k - 1)!}{k! 2^{k}} f o r k ⩾ 1 \Rightarrow

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

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

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

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

Commented by prof Abdo imad last updated on 17/Jul/18

f^{(k)} (0) = \frac{{(- 1)}^{k - 1} (k - 1)!}{2^{k}} f o r k ⩾ 1