Question #5af5d

1 Answer
Feb 18, 2017

Answer:

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#.