# Question 5af5d

Feb 18, 2017

See below.

#### Explanation:

You can use the Binomial/ Taylor Series for ${\left(1 + u\right)}^{\setminus} \alpha$:

 (1 + u)^\alpha = \sum_{k=0}^{\infty} ((alpha),( k)) u^k = 1 + \alpha u + \frac{\alpha(\alpha-1)}{2!} u^2 + \cdots 

Here we have ${\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$ so pattern match in $u = {x}^{2}$ to get:

 (1 + x^2)^(1/2) = 1 + 1/2 x^2 + \frac{1/2(-1/2)}{2!} x^4 + \cdots #

$= 1 + \frac{1}{2} {x}^{2} - \frac{1}{8} {x}^{4} + \setminus \cdots$

This will converge for $\left\mid x \right\mid < 1$.