How is this Taylor expansion computed? I am reading a paper, where we consider the following function,
$$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)$$
where $b>0$ is a constant such that $r>2/b.$ After stating this definition the author writes that the taylor expression of $F$ is,
$$F(r) = b\frac{r^2}{4}-\frac{\ln(r)}{b} -\frac{1}{2b} -\frac{\ln(2b)}{b} + \mathcal{O}(b^{-3}r^{-2}).$$
I know how to compute taylor series, but I am not sure what the taylor series is centered around. I have computed the derivatives upto order $2$. 
$$F'(r) = \frac{1}{2}\sqrt{b^2r^2-4}, \quad F''(r) = \frac{b^2r}{2\sqrt{b^2r^2-4}}.$$ 
Any explanations regarding the expansion would be much appreciated. 
 A: Here you need the following expansions centered at $x=0$:
$$\sqrt{1+x}=1+\frac{x}{2}+O(x^2)\quad,\quad\ln(1+x)=x+O(x^2).$$
Hence, for $r$ large enough and $x=-\frac{4}{b^2r^2}$,
$$\begin{align*}F(r)&=\frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)\\
&=\frac{br^2}{4}\sqrt{1-\frac{4}{b^2r^2}}-\frac{1}{b}\ln(br(\sqrt{1-\frac{4}{b^2r^2}} + 1))\\
&=\frac{br^2}{4}\sqrt{1-\frac{4}{b^2r^2}}-\frac{\ln(br)}{b}-\frac{1}{b}\ln(\sqrt{1-\frac{4}{b^2r^2}}+1)\\
&=\frac{br^2}{4}\left(1-\frac{4}{2b^2r^2}\right)+O(b^{-3}r^{-2})-\frac{\ln(br)}{b}-\frac{1}{b}\ln(2-\frac{4}{2b^2r^2}+O(b^{-4}r^{-4}))\\
&=\frac{br^2}{4}-\frac{1}{2b}+O(b^{-3}r^{-2})-\frac{\ln(br)}{b}-\frac{\ln(2)}{b}-\frac{1}{b}\ln(1-\frac{1}{b^2r^2}+O(b^{-4}r^{-4}))\\
&=\frac{br^2}{4}-\frac{1}{2b}-\frac{\ln(2b)}{b}+\frac{\ln(r)}{b}+O(b^{-3}r^{-2})
\end{align*}$$
A: $$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)\tag{1}$$
Expand the square-root term with $1/(b^2r^2)$ as the small variable,
$$\sqrt{b^2r^2-4}=br\sqrt{1-\frac{4}{b^2r^2}}=br \left( 1-\frac{2}{b^2r^2} \right) \tag{2}=br - \frac{2}{br} $$
Similarly, expand the log term as
$$\ln(\sqrt{b^2r^2-4} + br) =\ln \left[br \left(1+ \sqrt{1-\frac{4}{b^2r^2}} \right) \right]$$
$$=\ln \left[br \left( 2-\frac{2}{b^2r^2}\right) \right] = \ln(2br) + \ln \left( 1-\frac{1}{b^2r^2}\right) = \ln(2b) + \ln r - \frac{1}{b^2r^2}\tag{3}$$
Plug the two Taylor expansions (2) and (3) into (1) to get
$$F(r) = b\frac{r^2}{4}-\frac{\ln(r)}{b} -\frac{1}{2b} -\frac{\ln(2b)}{b} $$
