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

Question Number 34849 by math khazana by abdo last updated on 11/May/18

l e t f (x) = \frac{e^{- x}}{2 + 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 math khazana by abdo last updated on 13/May/18

w e h a v e f (x) = \frac{e^{- x}}{2 (1 + \frac{x}{2})} \Rightarrow f o r ∣ x ∣< 2

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

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

= \sum_{n = 0}^{\infty} c_{n} x^{n} w i t h c_{n} = \sum_{i + j = n} a_{i} b_{j}

= \sum_{i + j = n} \frac{{(- 1)}^{i}}{2^{i}} \frac{{(- 1)}^{j}}{j!} = \sum_{i = 0}^{n} a_{i} b_{n - i}

= \sum_{i = 0}^{n} \frac{{(- 1)}^{i}}{2^{i}} \frac{{(- 1)}^{n - i}}{(n - i)!} = \sum_{i = 0}^{n} \frac{{(- 1)}^{n}}{(n - i)! 2^{i}} \Rightarrow

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

Commented by math khazana by abdo last updated on 13/May/18

a n o t h e r m e t h o d

f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} l e t c a l c u l a t e f^{(n)} (0)

l e i b n i t z f o r m u l a g i v e f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\frac{1}{2 + x})}^{(k)} {(e^{- x})}^{(n - k)}

= \sum_{k = 0}^{n} C_{n}^{k} \frac{{(- 1)}^{k} k!}{{(x + 2)}^{k + 1}} {(- 1)}^{n - k} e^{- x} \Rightarrow

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

f (x) = \sum_{n = 0}^{\infty} (\sum_{k = 0}^{n} C_{n}^{k} \frac{k!}{n!} \frac{{(- 1)}^{n}}{2^{k + 1}}) x^{n} b u t

C_{n}^{k} . \frac{k!}{n!} = \frac{n!}{k! (n - k)!} \frac{k!}{n!} = \frac{1}{(n - k)!} \Rightarrow

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