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

Question Number 39291 by math khazana by abdo last updated on 04/Jul/18

l e t f (x) = \frac{e^{- x^{2}}}{1 + x^{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 abdo mathsup 649 cc last updated on 05/Jul/18

w e h a v e e^{- x^{2}} = \sum_{n = 0}^{\infty} \frac{{(- x^{2})}^{n}}{n!} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{n!}

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

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

= \sum_{n = 0}^{\infty} c_{n} x^{2 n} w i t h c_{n} = \sum_{i + j = n} a_{i} b_{j}

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

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

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

Σ

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

t h e r a d i u s o f c o n v e r g e n c e i s R = 1