Editing Laurent series (section)

== Convergence ==
[[Image:Expinvsqlau GIF.gif|right|thumb|''e''<sup>−1/''x''<sup>2</sup></sup> and its Laurent approximations (labeled) with the negative degree rising. The neighborhood around the zero singularity can never be approximated.]]
[[Image:Expinvsqlau SVG.svg|right|thumb|''e''<sup>−1/''x''<sup>2</sup></sup> and its Laurent approximations. As the negative degree of the Laurent series rises, it approaches the correct function.]]

Laurent series with complex coefficients are an important tool in [[complex analysis]], especially to investigate the behavior of functions near [[mathematical singularity|singularities]].

Consider for instance the function <math>f(x) = e^{-1/x^2}</math> with <math>f(0) = 0</math>. As a real function, it is infinitely differentiable everywhere; as a complex function however it is not differentiable at <math>x = 0</math>. The Laurent series of <math>f(x)</math> is obtained via the [[Exponential_function#Definitions_and_fundamental_properties|power series representation]],
<math display="block">e^{-1/x^2} = \sum_{n=0}^{\infty} (-1)^n \, {x^{-2n}\over n!},</math>
which converges to <math>f(x)</math> for all <math>x \in \mathbb{C}</math> except at the singularity <math>x = 0</math>. The graph on the right shows <math>f(x)</math> in black and its Laurent approximations
<math display="block">\sum_{n=0}^N (-1)^n \, {x^{-2n}\over n!}, \quad \forall N \in \mathbb{N}^{+}.</math>
As <math>N\to\infty</math>, the approximation becomes exact for all (complex) numbers <math>x</math> except at the singularity <math>x=0</math>.

More generally, Laurent series can be used to express [[holomorphic function]]s defined on an [[Annulus (mathematics)|annulus]], much as [[power series]] are used to express holomorphic functions defined on a [[Disk (mathematics)|disc]].

Suppose 
<math display="block">\sum_{n=-\infty}^\infty a_n ( z - c )^n</math>
is a given Laurent series with complex coefficients <math>a_n</math> and a complex center <math>c</math>. Then there exists a [[unique (mathematics)|unique]] inner radius <math>r</math> and outer radius <math>R</math> such that:
* The Laurent series converges on the open annulus <math>A=\{z:r<|z-c|<R\}</math>. That is, both the positive- and negative degree power series converge. Furthermore, this convergence will be [[uniform convergence|uniform]] on [[compact set]]s. Finally, the convergent series defines a [[holomorphic function]] <math>f(z)</math> on <math>A</math>.
* Outside the annulus, the Laurent series diverges. That is, at each point in the [[exterior (topology)|exterior]] of <math>A</math>, either the positive- or negative degree power series diverges.
* On the [[boundary (topology)|boundary]] of the annulus, one cannot make a general statement, except that there is at least one point on the inner boundary and one point on the outer boundary such that <math>f(z)</math> cannot be holomorphically extended to those points; giving rise to a [[Riemann-Hilbert problem]].{{sfn|Ablowitz|Fokas|2003|p=152|ps=none}}

It is possible that <math>r</math> may be zero or <math>R</math> may be infinite; at the other extreme, it's not necessarily true that <math>r</math> is less than <math>R</math>.
These radii can be computed by taking the [[limit superior]] of the coefficients <math>a_n</math> such that:
<math display="block">\begin{align}
            r &= \limsup_{n\to\infty} |a_{-n}|^\frac{1}{n}, \\
  \frac{1}{R} &= \limsup_{n\to\infty} |a_n|^\frac{1}{n}.
\end{align}</math>

When <math>r=0</math>, the coefficient <math>a_{-1}</math> of the Laurent expansion is called the '''[[residue (complex analysis)|residue]]''' of <math>f(z)</math> at the singularity <math>c</math>.{{sfn|Ablowitz|Fokas|2003|p=130|ps=none}} For example, the function
<math display="block">f(z) = {e^z \over z} + e^{{1}/{z}},</math>
is holomorphic everywhere except at <math>z=0</math>. The Laurent expansion about <math>c=0</math> can then be obtained from the power series representation:
<math display="block">f(z) = \cdots + \left( {1 \over 3!} \right) z^{-3} + \left( {1 \over 2!} \right) z^{-2} + 2z^{-1} + 2 + \left( {1 \over 2!} \right) z + \left( {1 \over 3!} \right) z^2 + \left( {1 \over 4!} \right) z^3 + \cdots,</math>
hence, the residue is given by <math>a_{-1}=2</math>.

Conversely, for a holomorphic function <math>f(z)</math> defined on the annulus <math>A=\{z:r<|z-c|<R\}</math>, there always exists a unique Laurent series with center <math>c</math> which converges (at least on <math>A</math>) to <math>f(z)</math>.

For example, consider the following rational function, along with its [[partial fraction]] expansion:
<math display="block">
  f(z) = \frac{1}{(z - 1)(z - 2i)} 
       = \frac{1 + 2i}{5}\left(\frac{1}{z - 1} - \frac{1}{z - 2i}\right)
.</math>

This function has singularities at <math>z=1</math> and <math>z=2i</math>, where the denominator is zero and the expression is therefore undefined.
A [[Taylor series]] about <math>z=0</math> (which yields a power series) will only converge in a disc of [[radius]] 1, since it "hits" the singularity at <math>z=1</math>.

However, there are three possible Laurent expansions about 0, depending on the radius of <math>z</math>:

* One series is defined on the inner disc where {{math|{{!}}''z''{{!}} < 1}}; it is the same as the Taylor series, <math display="block">f(z) = \frac{1 + 2i}{5} \sum_{n=0}^\infty \left(\frac{1}{(2i)^{n + 1}} - 1\right)z^n.</math>  This follows from the partial fraction form of the function, along with the formula for the sum  of a [[geometric series]], <math display="block">\frac{1}{z-a} = - \frac{1}{a} \sum_{n=0}^\infty \left( \tfrac{z}{a} \right)^n </math> for <math>|z| < |a| </math>.
* The second series is defined on the middle annulus where <math>1<z<2</math> is caught between the two singularities: <math display="block">f(z) = \frac{1 + 2i}{5} \left(\sum_{n=1}^\infty z^{-n} + \sum_{n=0}^\infty \frac{1}{(2i)^{n + 1}} z^n\right).</math> Here, we use the alternative form of the geometric series summation, <math display="block"> \frac{1}{z - a} = \frac{1}{z}\sum_{n=0}^\infty \left(\frac{a}{z}\right)^n </math> for <math>|z| > |a|</math>.
* The third series is defined on the infinite outer annulus where <math>2<z<\infty</math>, (which is also the Laurent expansion at <math>z = \infty</math>) <math display="block"> f(z) = \frac{1 + 2i}{5} \sum_{n=1}^\infty \left(1 - (2i)^{n - 1}\right) z^{-n}.</math> This series can be derived using geometric series as before, or by performing [[polynomial long division]] of 1 by <math>(x-1)(x-2i)</math>, not stopping with a remainder but continuing into <math>x^{-n}</math> terms; indeed, the "outer" Laurent series of a rational function is analogous to the decimal form of a fraction. (The "inner" Taylor series expansion can be obtained similarly, just by reversing the [[monomial order|term order]] in the division algorithm.)