Question

Question #d4992

Answer 1

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 `\alpha = -\frac{1}{2}` and `x=-0.01` for any `n\in \mathbb{N}`. You get

`\frac{1}{\sqrt{0.99}} = 1 - \frac{-0.01}{2} + \frac{3}{8}(-0.01)^2 +...`

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

You stop the development when you want.