Expanding brackets to power of -1/2 How do you expand this?
$$ \left(16-\frac{x^2}{4}\right)^{-\frac12} $$
And generally how would you expand any $ (a+b)^n $ including fractional and negative powers.

First we simplify
$$ f(x)=\left(16-\frac{x^2}{4}\right)^{-\frac12}= \left(\frac{64-x^2}{4}\right)^{-\frac12} = 2\left(64-x^2\right)^{-\frac12} $$
Then we expand
$$
f(x)=f(0)+\frac{x}{1!}\frac{d}{dx}f(0) + \frac{x^2}{2!}\frac{d^2}{dx^2}f(0) + \cdots 
$$
Note that
$$ \frac{d}{dx}f(x) = 2x\left(64-x^2\right)^{-\frac32}$$
$$ \frac{d^2}{dx^2}f(x) = 4\left(x^2+32\right)\left(64-x^2\right)^{-\frac52}$$
So now we have
$$ f(0) = 2\left(64\right)^{-\frac12} = \frac14 $$
$$ \frac{d}{dx}f(0) = 0 $$
$$ \frac{d^2}{dx^2}f(0) = 4\left(32\right)\left(64\right)^{-\frac52} = \frac1{256}$$
Therefore, by Maclaurin series expansion we have
$$ f(x)=2\left(64-x^2\right)^{-\frac12} = \frac14+ \frac{x^2}{512} + O(x^3)  $$
You are free to expand to any degree that you wish. 
