Laurent Series Expansion for Complex Functions

[rick1138]

Does anyone know of any examples of the explicit calculation of the Laurent series of a complex function? Any information would be appreciated.

 

[HallsofIvy]

The Laurent series is simply a power series that includes a finite number of negative powers. If a function is analytic at x= a, then its Taylor's series IS its Laurent series- there are no negative powers. If a function has an essential singularity at x= a, then it does not have a Laurent series. If a function has a pole of order n at x= a, then (x-a)ⁿf(x) is analytic at x= a. Construct the Taylor's series for (x-a)ⁿf(x) and multiply each term by (x-a)^-n.

As simple example: f(x)= e^x is analytic at x= 0 so its Laurent series there is the same as its Taylor's series: 1+ x+ (1/2)x²+ ...+ (1/n!)xⁿ+ ...

f(x)= e^x/x³ has a pole of order three at x= 0 (since x³f(x)= e^x is analytic at x= 0 but no lower power will give an analytic function). It's Laurent series is
x^-3(1+ x+ (1/2)x²+ (1/6)x³+ (1/24)x⁴+...+ (1/n!)xⁿ+...)= x^-3+ x^-2+ (1/2)x^-1+ (1/6)+ (1/24)x+ ...+ (1/n!)x^n-3+...

 

[rick1138]

Excellent. Exactly what I was looking for. Thanks.

 

[roman_1]

Construct the Taylor's series for (x-a)ⁿf(x) and multiply each term by (x-a)^-n.

I know this was six years ago, but would you believe it is the clearest explanation of Laurent series on the internet.

 

[Hurkyl]

And inaccurate for essential singularities. Complex analysis permits infinitely many negative powers as well.

In pure algebra, though, they usually limit Laurent series to ones that only have finitely many negative powers.

 

[zetafunction]

i posted a similar problem in this forum

my question about Laurent is this

let be the Taylor series $$f(1/x)= \sum_{n=0}^{\infty}c_{n}x^{n}$$ valid for |x| <1

then , if i make a change of variable $$x=1/y$$

$$f(y)= \sum_{n=0}^{\infty}c_{n}y^{-n}$$ is a LAURENT series for the function f(y) valid for |x| >1 ??