How can I find the Laurent series for Cos(1/z) at z=0?

[Crush1986]

Homework Statement

I need to find the Laurent Series of Cos[\frac{1}{z}] at z=0

Homework Equations

None

The Attempt at a Solution

I've gone through a lot of these problems and this is one of the last on the problem set. With all the other trig functions it's been just computing their Taylor series, then just simplifying the terms of the problem. Usually, I'd have to do like a geometric series in there as well.

This function though I'm having trouble with because f(0) doesn't exist. I'm not sure what else I can do. Can anyone point me in the right direction?

Thanks.

 

[RUber]

What if you were asked to solve for the series for $\cos(u)$? Then let $u = z^{-1}$?

 

[Crush1986]

What if you were asked to solve for the series for $\cos(u)$? Then let $u = z^{-1}$?

Ok, I was thinking of doing that. I just dunno, it seems to weird. Obviously then the problem is just the simplest Taylor series in the world just about.

Thanks.

 

[RUber]

Right...the Laurent series allows you to build into the negative exponents which can capture the information near the singularity of 1/z.
So, if the Taylor series of f(u) is $\sum_{k = 0}^\infty a_k u^k$ then the equivalent Laurent series for f(1/u) is $\sum_{k = 0}^{\infty} a_k u^{-k}$