Laurent series

The Laurent series for a complex function $f(z)$ about an arbitrary point $c$ is given by{{sfn|Ablowitz|Fokas|2003|p=128|ps=none}}{{citation | last=Folland | first=Gerald B. | author-link=Gerald Folland | title=Fourier analysis and its applications | publisher=Wadsworth & Brooks/Cole | publication-place=Pacific Grove, Calif | date=1992 | isbn=978-0-534-17094-3 |page=395}}

$$f(z)\; =\; \backslash sum\_\{n=-\backslash infty\}^\backslash infty\; a\_n(z-c)^n,$$

where the coefficients $a\_n$ are defined by a contour integral that generalizes Cauchy's integral formula:

$$a\_n\; =\backslash frac\{1\}\{2\backslash pi\; i\}\backslash oint\_\backslash gamma\; \backslash frac\{f(z)\}\{(z-c)^\{n+1\}\}\; \backslash ,\; dz.$$

The path of integration $\backslash gamma$ is counterclockwise around a Jordan curve enclosing $c$ and lying in an annulus $A$ in which $f(z)$ is holomorphic (analytic). The expansion for $f(z)$ will then be valid anywhere inside the annulus. The annulus is shown in red in the figure on the right, along with an example of a suitable path of integration labeled $\backslash gamma$. When $\backslash gamma$ is defined as the circle $|z-c|\; =\; \backslash varrho$, where , this amounts

to computing the complex Fourier coefficients of the restriction of $f$ to $\backslash gamma$.{{sfn|Ablowitz|Fokas|2003|pp=196-197|ps=none}} The fact that these integrals are unchanged by a deformation of the contour $\backslash gamma$ is an immediate consequence of Green's theorem.

One may also obtain the Laurent series for a complex function $f(z)$ at $z\; =\; \backslash infty$. However, this is the same as when $R\; \backslash rightarrow\; \backslash infty$.

In practice, the above integral formula may not offer the most practical method for computing the coefficients

$a\_n$ for a given function $f(z)$; instead, one often pieces together the Laurent series by combining known Taylor expansions. Because the Laurent expansion of a function is unique whenever

it exists, any expression of this form that equals the given function

$f(z)$ in some annulus must actually be the Laurent expansion of $f(z)$.

Image:Expinvsqlau GIF.gif

Image:Expinvsqlau SVG.svg

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

Consider for instance the function $f(x)\; =\; e^\{-1/x^2\}$ with $f(0)\; =\; 0$. As a real function, it is infinitely differentiable everywhere; as a complex function however it is not differentiable at $x\; =\; 0$. The Laurent series of $f(x)$ is obtained via the power series representation,

$$e^\{-1/x^2\}\; =\; \backslash sum\_\{n=0\}^\{\backslash infty\}\; (-1)^n\; \backslash ,\; \{x^\{-2n\}\backslash over\; n!\},$$

which converges to $f(x)$ for all $x\; \backslash in\; \backslash mathbb\{C\}$ except at the singularity $x\; =\; 0$. The graph on the right shows $f(x)$ in black and its Laurent approximations

$$\backslash sum\_\{n=0\}^N\; (-1)^n\; \backslash ,\; \{x^\{-2n\}\backslash over\; n!\},\; \backslash quad\; \backslash forall\; N\; \backslash in\; \backslash mathbb\{N\}^\{+\}.$$

As $N\backslash to\backslash infty$, the approximation becomes exact for all (complex) numbers $x$ except at the singularity $x=0$.

More generally, Laurent series can be used to express holomorphic functions defined on an annulus, much as power series are used to express holomorphic functions defined on a disc.

Suppose

$$\backslash sum\_\{n=-\backslash infty\}^\backslash infty\; a\_n\; (\; z\; -\; c\; )^n$$

is a given Laurent series with complex coefficients $a\_n$ and a complex center $c$. Then there exists a unique inner radius $r$ and outer radius $R$ such that:

• The Laurent series converges on the open annulus
• Outside the annulus, the Laurent series diverges. That is, at each point in the exterior of $A$, either the positive- or negative degree power series diverges.
• On the 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 $f(z)$ 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 $r$ may be zero or $R$ may be infinite; at the other extreme, it's not necessarily true that $r$ is less than $R$.

These radii can be computed by taking the limit superior of the coefficients $a\_n$ such that:

$$\backslash begin\{align\}$$

r &= \limsup_{n\to\infty} |a_{-n}|^\frac{1}{n}, \\

\frac{1}{R} &= \limsup_{n\to\infty} |a_n|^\frac{1}{n}.

\end{align}

When $r=0$, the coefficient $a\_\{-1\}$ of the Laurent expansion is called the residue of $f(z)$ at the singularity $c$.{{sfn|Ablowitz|Fokas|2003|p=130|ps=none}} For example, the function

$$f(z)\; =\; \{e^z\; \backslash over\; z\}\; +\; e^\{\{1\}/\{z\}\},$$

is holomorphic everywhere except at $z=0$. The Laurent expansion about $c=0$ can then be obtained from the power series representation:

$$f(z)\; =\; \backslash cdots\; +\; \backslash left(\; \{1\; \backslash over\; 3!\}\; \backslash right)\; z^\{-3\}\; +\; \backslash left(\; \{1\; \backslash over\; 2!\}\; \backslash right)\; z^\{-2\}\; +\; 2z^\{-1\}\; +\; 2\; +\; \backslash left(\; \{1\; \backslash over\; 2!\}\; \backslash right)\; z\; +\; \backslash left(\; \{1\; \backslash over\; 3!\}\; \backslash right)\; z^2\; +\; \backslash left(\; \{1\; \backslash over\; 4!\}\; \backslash right)\; z^3\; +\; \backslash cdots,$$

hence, the residue is given by $a\_\{-1\}=2$.

Conversely, for a holomorphic function $f(z)$ defined on the annulus

For example, consider the following rational function, along with its partial fraction expansion:

f(z) = \frac{1}{(z - 1)(z - 2i)}

= \frac{1 + 2i}{5}\left(\frac{1}{z - 1} - \frac{1}{z - 2i}\right)

.

This function has singularities at $z=1$ and $z=2i$, where the denominator is zero and the expression is therefore undefined.

A Taylor series about $z=0$ (which yields a power series) will only converge in a disc of radius 1, since it "hits" the singularity at $z=1$.

However, there are three possible Laurent expansions about 0, depending on the radius of $z$:

• One series is defined on the inner disc where {{math|{{!}}z{{!}} < 1}}; it is the same as the Taylor series, $$f(z)\; =\; \backslash frac\{1\; +\; 2i\}\{5\}\; \backslash sum\_\{n=0\}^\backslash infty\; \backslash left(\backslash frac\{1\}\{(2i)^\{n\; +\; 1\}\}\; -\; 1\backslash right)z^n.$$ This follows from the partial fraction form of the function, along with the formula for the sum of a geometric series, $$\backslash frac\{1\}\{z-a\}\; =\; -\; \backslash frac\{1\}\{a\}\; \backslash sum\_\{n=0\}^\backslash infty\; \backslash left(\; \backslash tfrac\{z\}\{a\}\; \backslash right)^n$$ for .
• The second series is defined on the middle annulus where
• The third series is defined on the infinite outer annulus where

Suppose a function $f(z)$ holomorphic on the annulus

$$f(z)\; =\; \backslash sum\_\{n=-\backslash infty\}^\{\backslash infty\}\; a\_\{n\}\; (z-c)^n\; =\; \backslash sum\_\{n=-\backslash infty\}^\{\backslash infty\}\; b\_\{n\}\; (z-c)^n.$$

Multiply both sides by $(z-c)^\{-k-1\}$, where k is an arbitrary integer, and integrate on a path γ inside the annulus,

$$\backslash oint\_\{\backslash gamma\}\backslash ,\backslash sum\_\{n=-\backslash infty\}^\{\backslash infty\}\; a\_\{n\}\; (z-c)^\{n-k-1\}\backslash ,dz\; =\; \backslash oint\_\{\backslash gamma\}\backslash ,\backslash sum\_\{n=-\backslash infty\}^\{\backslash infty\}\; b\_\{n\}\; (z-c)^\{n-k-1\}\backslash ,dz.$$

The series converges uniformly on $r+\backslash varepsilon\; \backslash leq\; |z-c|\; \backslash leq\; R-\backslash varepsilon$, where ε is a positive number small enough for γ to be contained in the constricted closed annulus, so the integration and summation can be interchanged. Substituting the identity

$$\backslash oint\_\{\backslash gamma\}\backslash ,(z-c)^\{n-k-1\}\backslash ,dz\; =\; 2\backslash pi\; i\backslash delta\_\{nk\}$$

into the summation yields

$$a\_k\; =\; b\_k.$$

Hence the Laurent series is unique.

{{main|Laurent polynomial}}

A Laurent polynomial is a Laurent series in which only finitely many coefficients are non-zero. Laurent polynomials differ from ordinary polynomials in that they may have terms of negative degree.

{{broader|Principal part}}

The principal part of a Laurent series is the series of terms with negative degree, that is

$$\backslash sum\_\{k=-\backslash infty\}^\{-1\}\; a\_k\; (z-c)^k.$$

If the principal part of $f$ is a finite sum, then $f$ has a pole at $c$ of order equal to (negative) the degree of the highest term; on the other hand, if $f$ has an essential singularity at $c$, the principal part is an infinite sum (meaning it has infinitely many non-zero terms).

If the inner radius of convergence of the Laurent series for $f$ is 0, then $f$ has an essential singularity at $c$ if and only if the principal part is an infinite sum, and has a pole otherwise.

If the inner radius of convergence is positive, $f$ may have infinitely many negative terms but still be regular at $c$, as in the example above, in which case it is represented by a different Laurent series in a disk about $c$.

Laurent series with only finitely many negative terms are well-behaved—they are a power series divided by $z^k$, and can be analyzed similarly—while Laurent series with infinitely many negative terms have complicated behavior on the inner circle of convergence.

Laurent series cannot in general be multiplied.

Algebraically, the expression for the terms of the product may involve infinite sums which need not converge (one cannot take the convolution of integer sequences).

Geometrically, the two Laurent series may have non-overlapping annuli of convergence.

Two Laurent series with only finitely many negative terms can be multiplied: algebraically, the sums are all finite; geometrically, these have poles at $c$, and inner radius of convergence 0, so they both converge on an overlapping annulus.

Thus when defining formal Laurent series, one requires Laurent series with only finitely many negative terms.

Similarly, the sum of two convergent Laurent series need not converge, though it is always defined formally, but the sum of two bounded below Laurent series (or any Laurent series on a punctured disk) has a non-empty annulus of convergence.

Also, for a field $F$, by the sum and multiplication defined above, formal Laurent series would form a field $F((x))$ which is also the field of fractions of the ring $Fx$ of formal power series.

• Puiseux series
• Mittag-Leffler's theorem
• Formal Laurent series{{snd}} Laurent series considered formally, with coefficients from an arbitrary commutative ring, without regard for convergence, and with only finitely many negative terms, so that multiplication is always defined.
• Z-transform{{snd}} the special case where the Laurent series is taken about zero has much use in time-series analysis.
• Fourier series{{snd}} the substitution $z=e^\{\backslash pi\; i\; w\}$ transforms a Laurent series into a Fourier series, or conversely. This is used in the q-series expansion of the j-invariant.
• Padé approximant{{snd}} Another technique used when a Taylor series is not viable.

{{reflist}}

• {{citation | last=Ablowitz | first=Mark J. | author1-link = Mark Ablowitz | last2=Fokas | first2=Athanassios S. | author2-link=Athanassios Fokas | title=Complex Variables | publisher=Cambridge University Press | date=2003 | isbn=978-0-521-53429-1}}

• {{springer|title=Laurent series|id=p/l057690}}
• {{MacTutor Biography|id=Laurent_Pierre}}
• {{MathWorld | urlname=LaurentSeries | title=Laurent Series }}
• [http://www.mrob.com/pub/muency/laurentseries.html Laurent Series and Mandelbrot set by Robert Munafo]

{{series (mathematics)}}

{{DEFAULTSORT:Laurent Series}}

Category:Complex analysis

Category:Series expansions