let-f-x-2-x-developp-f-at-integr-serie-and-give-the-radius-of-convergence-

Question Number 34696 by abdo imad last updated on 10/May/18

l e t f (x) = \sqrt{2 - x}

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

c o n v e r g e n c e .

Commented by math khazana by abdo last updated on 11/May/18

f (x) = \sqrt{2} \sqrt{1 - \frac{x}{2}} = \sqrt{2} {(1 - \frac{x}{2})}^{\frac{1}{2}} b u t w e k n o w t h a t

{(1 + u)}^{α} = 1 + α u \frac{α (α - 1)}{2!} u^{2} + \dots

= 1 + \sum_{n = 1}^{\infty} \frac{α (α - 1) \dots (α - n + 1)}{n!} u^{n}

a n d {(1 - u)}^{n} = 1 + \sum_{n = 1}^{\infty} α (α - 1) \dots (α - n + 1) \frac{{(- 1)}^{n}}{n!} u^{n}

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

= 1 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n! 2^{n + 1}} (- \frac{1}{2}) \dots (- \frac{3}{2}) \dots (- \frac{2 n - 3}{2}) x^{n}

= 1 + \sum_{n = 1}^{\infty} \frac{(- 1)}{n! 2^{n + 1}} . \frac{1.3.5 \dots . . (2 n - 3)}{2^{n - 1}} x^{n}

= 1 + \sum_{n = 1}^{\infty} \frac{- 1}{n! 2^{2 n}} (1.3.5 \dots . . (2 n - 3)) x^{n} \Rightarrow

f (x) = \sqrt{2} - \sqrt{2} \sum_{n = 1}^{\infty} \frac{1.3.5 \dots . (2 n - 3)}{n! 2^{2 n}} x^{n} .