# Expansion of $x^{-1/2}$ at $0$

Consider the function $f(x) = x^{-1/2}$ on the non-negative real line.

The point $z=0$ is 'distinguished' because it is the boundary of the domain of $f$, and because $f$ has a pole there. It seems natural to ask whether there exists a expansion of $f$ near $0$. What is the natural candidate for this? As far as I know, most series (like Laurent) are only developed in the interior of a domain.

The motivation is compute an integral that involves $f$ (inside a more complicated expression) by computing the terms of a series, where the terms are assumed to be simpler (rational functions at most) and the method converges.

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It has a very simple expression as a Puiseux series, see en.wikipedia.org/wiki/Puiseux_series – Gerry Myerson Feb 1 '13 at 11:33

There is no expansion of a function about a branch point singularity. Terms such as $x^{\alpha}$ for noninteger $\alpha$ and $\log{x}$ are examples of such singularities in which the argument of the term is undefined. For example, functions resulting from differential equations having a regular singular point at $x=0$ may be expressed, by the Froebenius method, in a series involving powers of $x$ and terms in $x^{\alpha}$ and $\log{x}$. (See, for example, many of the solutions of the second kind of the orthogonal basis functions, e.g., Bessel, Legendre, etc., at regular singular points of their generating differential equations.)