How can I Laurent expand $\phi(x-t) = \sqrt{(x-t)^2 + c^2}$?

Question

Consider $$\phi(x-t) = \sqrt{(x-t)^2 + c^2},$$ for some $c \in \mathbb{R}$. This can be expanded as a Laurent series as \begin{align} \phi(x-t) = & \ \text{sign}(x)( x- t + \frac{1}{2}c^2x^{-1} \\ & + \frac{1}{2}tc^2x^{-2} + \frac{1}{8}(4t^2c^2 -c^4)x^{-3} \\ & + \dots). \end{align} I am unable to reproduce this expansion so I'm wondering if anyone know how it was done? It is from page 5 of this paper. The paper has some more details, and I understand that they careful to show the expansion is only valid for certain parameter values due to the branch cut for the square root. But I'm stuck on reproducing this Laurent expansion.

Answer 1

They are considering the expansion of $$A=\sqrt{1+\frac{a^2}{x^2}}$$ for infinitely large values of $x$.

So, let $x=\frac a {\sqrt t}$ to have $$A=\sqrt{1+t}$$ Now, using the binomial theorem or Taylor series around $t=0$ $$A=1+\frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}-\frac{5 t^4}{128}+O\left(t^5\right)$$ Now, replace $t$ by $\frac{a^2}{x^2}$ to get the expression in the paper.