S.O.S. Mathematics CyberBoard

I was looking at the function .

I see that this has a removable singularity at z = 0 by looking at the power series expansions (centered at 0) of the functions.

For the poles of the function, I know that where k is an integer.

So if I want to look at the power series expansions centered at , is there a shortcut to change the original power series?

Apart from this I also know that the cos function has a zero of order 1. Does this mean that the order of the pole will also be 1?

Given f(z) = I see that the denominator has a zero of degree 2 at z = 0. Hence the function has a pole of degree 2 at z = 0. How Should I go about finding the power series expansion so that I can find the singular part of the function? That is, the first two terms of its Laurent series expansion.

Any suggestions?

"the cos function has a zero of order 1. Does this mean that the order of the pole will also be 1?"

Yes. If you have not encountered this lemma, it's probably in the next section. In one text I have, it appears right before Picard's Theorem. You'll find it.

"zero of degree 2"

Are you sure?

So as you point out, it is in fact a pole of degree one.

How do I go about calculating the singular part of the this function?

I am sorely tempted to notice that it looks an awful lot like a simple geometric series expansion, but that seems to overlook the criteria for doing that sort of thing.

Why not just calculate it directly? The long division algorithm certainly is tedious, but you can get couple of terms quickly enough.

Of course, if I actually remembered anything about this, I would be more help. You're about at my limit on this one.