# Question d4992

Feb 18, 2015

Hello !

Use the formula

(1+x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha -1)}{2}x^2 + \frac{\alpha(\alpha -1)(\alpha -2)}{6}x^3 + \ldots + \frac{\alpha(\alpha -1)\ldots (\alpha -n+1)}{n!}x^n + o(x^n)#

with $\setminus \alpha = - \setminus \frac{1}{2}$ and $x = - 0.01$ for any $n \setminus \in \setminus m a t h \boldsymbol{N}$. You get

$\setminus \frac{1}{\setminus \sqrt{0.99}} = 1 - \setminus \frac{- 0.01}{2} + \setminus \frac{3}{8} {\left(- 0.01\right)}^{2} + \ldots$

These three terms give 1,0050375. The real values is 1,005037815....

You stop the development when you want.