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 is the Puiseux series, I have search on the net but I don't have understood so much... Can you briefly explain me it and how can I obtain the result?

• There must be a typo in what you wrote... Also, you'd need to clarify where (a base point) you were expanding near. I suspect it's "near $1$", but what you've written is just the first-order term there, since the value of $\sqrt{x+1}$ there is that $\sqrt{2}$. Check your source? Clarify? Jun 11 '15 at 22:40
• From Wolfram Alpha, this appears to be at $x=1$. Jun 11 '15 at 22:41
• More terms to clarify: $$\sqrt{x^2-1} = \sqrt {2} (x-1)^{1/2} +\frac{1}{4}\,\sqrt {2} \left( x-1 \right) ^{3/2}-\frac{1}{32}\, \sqrt {2} \left( x-1 \right) ^{5/2}+{\frac {\sqrt {2} \left( x-1 \right) ^{7/2}}{128}}-{\frac {5\,\sqrt {2} \left( x-1 \right) ^{9/2} }{2048}}+{\frac {7\,\sqrt {2} \left( x-1 \right) ^{11/2}}{8192}}+ \ldots$$ Jun 11 '15 at 22:41
• @paulgarrett yes, it is near 1. I have edited the question. Where is the typo? Jun 11 '15 at 22:46
• @sunrise: Rewrite $\sqrt{x^2-1}=\sqrt{(x-1)(x+1)}=\sqrt{x+1}\sqrt{x-1}$. By continuity, when $x\to 1$ you have $\sqrt{x+1} = \sqrt{2} + o(1)$, so $\sqrt{x^2-1} = \sqrt{2}\sqrt{x-1} + o\left(\sqrt{x-1}\right)$. (You don't actually need Puiseux series for the first term.) Jun 11 '15 at 22:52

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) $$\sqrt{z^2 - 1} = \sqrt{2} \sum_{j=0}^\infty {1/2 \choose j} 2^{-j} (z-1)^{(2j+1)/2}$$

• Oops. I'll edit. Oct 22 '17 at 8:17

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}

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.