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? – paul garrett Jun 11 '15 at 22:40
• From Wolfram Alpha, this appears to be at $x=1$. – Clement C. Jun 11 '15 at 22:41
• More terms to clarify: $$\sqrt{x^2-1} = \sqrt {2} (x-1)^{1/2} +\frac{1}{4}\,\sqrt {2} \left( x-1 \right) ^{3/2}-\frac{1}{32}\, \sqrt {2} \left( x-1 \right) ^{5/2}+{\frac {\sqrt {2} \left( x-1 \right) ^{7/2}}{128}}-{\frac {5\,\sqrt {2} \left( x-1 \right) ^{9/2} }{2048}}+{\frac {7\,\sqrt {2} \left( x-1 \right) ^{11/2}}{8192}}+ \ldots$$ – Robert Israel Jun 11 '15 at 22:41
• @paulgarrett yes, it is near 1. I have edited the question. Where is the typo? – sunrise 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.) – Clement C. 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. – Robert Israel Oct 22 '17 at 8:17

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.

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}