Puiseux Series? WolframAlpha says that $$\sqrt{x^2-1}$$ expanded in Puiseux series near 1 is $\sqrt 2 \sqrt{x-1}$.
I don't know what a Puiseux series is; I have searched on the net but I haven't understood much... can you briefly explain it to me and how I can obtain this result?
 A: If you are familiar with Taylor series, in this case you easily can get the same expansion "for free," without having to sweat too much. Set $y=x-1$, so that you are looking at 
$$
\sqrt{x^2-1} = \sqrt{(x+1)(x-1)} = \sqrt{(y+2)y} = \sqrt{2y}\sqrt{1+\frac{y}{2}}
$$
when $y\to 0$ (i.e., $x\to 1$). Recalling the Taylor expansion of $t\mapsto \sqrt{1+t}$ around $0$, you get
$$
\sqrt{1+\frac{y}{2}} = 1+\frac{y}{4}-\frac{y^2}{32} + o(y^3)
$$
(I only went to order $3$, but you can go much further) so that
$$\begin{align}
\sqrt{x^2-1} &= \sqrt{2y}\left(1+\frac{y}{4}-\frac{y^2}{32} + o(y^3)\right) = \sqrt{2y}+\frac{\sqrt{2}}{4}y^{3/2}-\frac{\sqrt{2}}{32}y^{5/2} + o\!\left(y^{7/2}\right) \\ 
&=\sqrt{2}\sqrt{x-1}+\frac{\sqrt{2}}{4}(x-1)^{3/2}-\frac{\sqrt{2}}{32}(x-1)^{5/2} + o\!\left((x-1)^{7/2}\right).
\end{align}$$
A: For understand more about Puiseux Series, you can consult Wikipedia or this lectures notes of Commutative Algebra and Algebraic Geometry by Franz Winkler, chapter 9.
A: A Puiseux series about $x=a$ is similar to a Taylor (or more generally Laurent) series, but allowing fractional powers of $x-a$ rather than just integer powers.  
I won't try to explain the whole theory, but here's a useful piece.
For simplicity, let's say the base point is $0$ (we can always arrange this by translation).  If for some positive integer $k$, a suitable branch of $f(w^k)$ is analytic in a neighbourhood of $w=0$, then we can write
$$f(w^k) = \sum_{j=0}^\infty a_j w^j$$
and then taking $w = z^{1/k}$ (for a suitable branch of this), we have the Puiseux series
$$ f(z) = \sum_{j=0}^\infty a_j z^{j/k} $$
In the case at hand, consider $f(\zeta) = \sqrt{(1+\zeta)^2 - 1} = \sqrt{2\zeta + \zeta^2}$ (I'm translating $z$ to $1+\zeta$ so the base point $z=1$ becomes $\zeta=0$).  This is not analytic at $\zeta=0$, but (for a suitable branch of the square root) $f(w^2) = w \sqrt{2 + w^2}$ is, and in fact 
$$ f(w^2) = \sqrt{2} \sum_{j=0}^\infty {1/2 \choose j} 2^{-j} w^{2j+1}$$
so that
$$ f(\zeta) = \sqrt{2} \sum_{j=0}^\infty {1/2 \choose j} 2^{-j} \zeta^{(2j+1)/2} $$
i.e. (translating back)
$$ \sqrt{z^2 - 1} = \sqrt{2} \sum_{j=0}^\infty {1/2 \choose j} 2^{-j} (z-1)^{(2j+1)/2}$$
