# Laurent Expansions

1. Mar 21, 2009

### latentcorpse

ok. i'm really confused by these.

(i) what's the difference between a Laurent expansion and Taylor expansion?

(ii) as i understand it, the point of these is to examine singularities of the complex function we're dealing with but, how do we actually go about doing it?

my notes on this were pretty rushed so aren't paricularly useful and i can't understand the wikipedia entry. anybody help me out?

2. Mar 21, 2009

### xepma

A taylor expansion is the "representation" of a function in terms of an (infinite) sum of powers of x, with coefficients evaluated at some point x=a.

However, this expansion breaks down as soon as the function has some singularity. For example, the function 1/x doesn't admit a Taylor expansion around x = a = 0. We can try to circumvent this by choosing some a unequal to zero, but the Taylor expansion would still not be valid beyond the point x = 0.

The solution is that we can still achieve a series expansion by admitting also negative exponents for x in the expansion. As a result, we are able to classify a lot more functions than before (which is good!). So all in all, a Laurent expansion admits for the possibility to come up with a series expansion when the function contains singularities. It's sort of a "better" version of the Taylor expansion, since by definition the Taylor expansion is contained in the Laurent expansion (check it).

The most practical use of a Laurent expansion is in the application to contour integral. There is a famous lemma that states that the integral along a closed contour over the complex plane for some analytic function is equal to the sum of the residues contained within this contour.

But to be fair, you gotta be a bit more specific on what details you want to hear...

3. Mar 21, 2009

### yyat

(i) A Taylor series, centered at a point $$a$$, has the form

$$\sum_{n=0}^{\infty}c_n(z-a)^n$$

and converges on a disk-shaped region (or just at $$a$$)
A Laurent series, on the other hand, can also have negative powers:

$$\sum_{n=-\infty}^{\infty}c_n(z-a)^n$$.

and converges on an annulus in general.

There is a formula for computing the coefficients $$c_n$$ in terms of path integrals which can be found in the wikipedia article.

(ii) Assume f is a holomorphic functions with a singularity at $$a$$. Then the Laurent series converges on a punctured disk centered at $$a$$. The type of singularity can be read off from the coefficients $$c_n$$ with $$n<0$$.

-If they are all 0, then f has a removable singularity at $$a$$

-If only finitely many are nonzero, then f has a pole at $$a$$

-If infinitely many are nonzero, then f has an essential singularity at $$a$$

The residue of f at $$a$$, important for computing path integrals (see residue theorem), is the coefficient $$c_{-1}$$.