TITLE: Taylor expansion of $\arccos(1-x)$ around $x=0$ to two terms
QUESTION [7 upvotes]: Doing a normal Taylor expansion of $\arccos(1-x)$ around $x=0$ to two terms by taking derivatives doesn't work because of division by zero.
I've put this into wolfram alpha: http://www.wolframalpha.com/input/?i=taylor+series+arccos%281-x%29. It's nice but I need to show that I can do it myself. This isn't an analysis class so I have never seen square roots in a series expansion before.
I found this in my search: Some approximations for $\arccos(1/(1+x))$ but I need the expansion to two terms.

REPLY [2 votes]: Although addressed indirectly in the other answers here, it seems relevant to point out that, strictly speaking, the resulting series expansion in all cases (so far) is -not- a Taylor series. 
The Taylor series of a function $f(x)$ about $x = c$ is given by:
$$
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n,
$$
i.e. it is an expansion in terms of positive, integer powers of $\mathbf{(x-c)}$, which the series given thus far are not. On the WolframAlpha link in the OP, the result is called a Puiseux series.
Personally, I don’t see any way of getting a Taylor series for this function, about $x = 0$. WolframAlpha certainly doesn’t list any such form.