Question

Question #5af5d

Answer 1

See below.

Explanation:

You can use the Binomial/ Taylor Series for `(1 + u)^\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 `(1+ x^2)^(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 + 1/2 x^2 - 1/8 x^4 + \cdots`

This will converge for `abs x < 1`.