# Clarify the singluarities of the function

1. Mar 9, 2006

### Wishbone "the function e^[(z/2)(z-1/z)] is the generating function for the bessel functions of the first kind of order n, Jn(s).

a) clarify the singluarities of the function."

I am wondering what is meant by clarifiying the singluarities. This is a term that I do not know, nor can find anywhere on the internet. He might have made this up. Any help would be greatly appreciated.

2. 3. Mar 9, 2006

### shmoe It's not mathematical terminology I've ever seen before. Maybe he means classify the singularities- is it removable, essential, etc.

Check your generating function again, those shouldn't all be z's in the exponent.

4. Mar 9, 2006

### Wishbone oh it might be e^[(s/2)(z-1/z)], he hand wrote the problem, and its kind of tought to see. Does that make more sense than the orignal function. If so, he didnt define it as f(z) or f(s), any idea on what the dependent variable is? also those z's aren't the same as in z=x+iy.

5. Mar 9, 2006

### shmoe That makes more sense. Consider it as a function of z. The Laurent series in z will have coefficients that depend on s, these are Bessel functions of the first kind.

6. Mar 9, 2006

### Wishbone hmm i am not aware of these terms, I thought the different types of singularities were either branch cuts or poles

7. Mar 9, 2006

### shmoe Use whatever terminology your course has introduced, you will almost surely have one that applies here.

8. Mar 9, 2006

### Wishbone hmm, I must admit after some effort, I cant seem to figure out how I should go about "clarifying" these singularities.

9. Mar 9, 2006

### shmoe Show what you've tried!

10. Mar 9, 2006

### Wishbone sry double post

Last edited: Mar 9, 2006
11. Mar 9, 2006

### Wishbone ya no prob.

First i tried plugging it in to the laurent series to see if I could get a function in which it would be apparent it has defined singularities.

$$f(z) = \Sigma a_n(z-z_0)^n$$

where

$$a_n = (1/2(\pi)i \oint \frac{f(z')dz'}{(z'-z_0)^{n+1}}$$

however that didn't really work out and got really messy, if you want i could put that up on the latex, it'll take a lil while tho. I don't think thats the correct way, I think I should have perhaps tried the same thing, but with bessel functions

Last edited: Mar 9, 2006
12. Mar 9, 2006

### shmoe Trying to work out the coefficients of the Laurent series using the integral for a general $$z_0$$, or even a specific one for that matter, isn't the best way to go.

You need to first locate the singularities- what values of z will make your function undefined?

13. Mar 9, 2006

### Wishbone well there's: z=0. That gives you a 0 in a denominator.

14. Mar 9, 2006

### shmoe Right. You may or may not be expected to consider what happens at infinity depending on your course.

Next go through your definitions for the different kinds of singularities for z=0, which does it fail to satisfy and what does it satsify?

15. Mar 9, 2006

### Wishbone well I only know about branch points, and poles. however neither would satify my equation at z=0, if thats what you are asking. (sorry if its not )

16. Mar 9, 2006

### shmoe You hopefully have different kinds of poles though? Something about the order of the pole relating the coefficients of the Laurent series? Anything when the Laurent series has an infinite number of powers of 1/z terms?

You should also have a way of determining the order of a pole without finding the Laurent series, something with limits.

17. Mar 9, 2006

### Wishbone oh yes I know that in

$$f(z) = \Sigma a_n(z-z_0)^n$$

the first nonvanishing term is a pole of order one. and the second non vanishing term is pole order two, and so one, right?

I do also know that the behavior of f(z) as z -> infintiy can be defind by f(1/t) as t->0. But I did not think of that since our singularity is at z=0

Last edited: Mar 9, 2006
18. Mar 9, 2006

### shmoe I'm not positive on your meaning, if your Laurent series at z=0 is:

$$\frac{a_{-1}}{z}+a_0+a_{1}z+a_{2}z^2+\ldots$$

with $$a_{-1}\neq 0$$, then z=0 is a pole of order 1 (aka a simple pole). If it's

$$\frac{a_{-2}}{z^2}+\frac{a_{-1}}{z}+a_0+a_{1}z+a_{2}z^2+\ldots$$

with $$a_{-2}\neq 0$$, then z=0 is a pole of order 2. Likewise for a pole of order 3, 4, etc.

You should have some way of determining the order of a pole (where something like the above applies, meaning your laurent series has a finite number of negative exponents with non-zero coefficients) without actually finding the Laurent series. e.g for a pole of order 1 you can say something about the limit of $$zf(z)$$ as z goes to zero.

edit-you mentioned the behavior at infinity so you have seen this concept. You know to examine this behavior you can instead look at f(1/z) near z=0. Work on f(z) at z=0 first, then come back to f(1/z) at z=0, the behavior will be similar.

edit#2 had problems with my latex.

Last edited: Mar 9, 2006
19. Mar 9, 2006

### Wishbone ya looking at f(z)=e^[(s/2)(z-1/z)] as it approaches z=0 from theleft, it blows up, and from the right it dies out. doing f(1/z) gives the opposite result, no?

20. Mar 9, 2006

### shmoe Yes. What about the limit of $$z^{n}f(z)$$ as z goes to zero where n is an integer?What does this say about this singularity?

21. Mar 10, 2006

### Wishbone well it goes to zero. What is says about the singularity...does it say that it is essential?