Laurent Series expansions

[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)= ex 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)= ex/x3 has a pole of order three at x= 0 (since x3f(x)= ex 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!)xn-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. :frown: 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 fucntion f(y) valid for |x| >1 ??