Clarify the singluarities of the function

[Wishbone]

The problem reads:

"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.

 

[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.

 

[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.

 

[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.

 

[Wishbone]

Maybe he means classify the singularities- is it removable, essential, etc.

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

 

[shmoe]

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

 

[Wishbone]

hmm, I must admit after some effort, I can't seem to figure out how I should go about "clarifying" these singularities.

 

[shmoe]

Show what you've tried!

 

[Wishbone]

sry double post

 

[Wishbone]

ya no prob.

First i tried plugging it into 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 that's the correct way, I think I should have perhaps tried the same thing, but with bessel functions

 

[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?

 

[Wishbone]

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

 

[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?

 

[Wishbone]

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

 

[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.

 

[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

 

[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.

 

[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?

 

[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?

 

[Wishbone]

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

 

[shmoe]

If z^nf(z) goes to zero as z goes to zero, ask what this means in terms of the laurent expansion, specifically the a_{-n} term and lower.

For your problem, let's try n=1 for now. Carefully find the left and right hand limits of zf(z), then try higher n's.

It's not critical, but the blowing up of e^[(s/2)(z-1/z)] from the left needed the assumption s>0, if s<0 it would have been reversed. What was important is the limit did not exist! This on it's own will tell you what kind of singularity it is, but do try to work out the limit in the last paragraph for higher n's and try to understand what they mean in general in terms of the laurent series and the order of the pole.

 

[Wishbone]

One problem I am having, is seeing how different n's will change the left and right hand limits. Am I totally missing something, or will all n's give the same behavior?

 

[Wishbone]

Anything when the Laurent series has an infinite number of powers of 1/z terms?

doesnt that mean its essential? (im not sure)

 

[shmoe]

One problem I am having, is seeing how different n's will change the left and right hand limits. Am I totally missing something, or will all n's give the same behavior?

They will flop the left and right hand behavior back and forth, but otherwise essentially the same- the limit doesn't exist. This should tell you something about the Laurent series.

doesnt that mean its essential? (im not sure)

Yes.

 

[Wishbone]

well I'm back with a bunch more of these problems, i think I have a good handle on the method, but I want to make sure.

The first one is

\frac{1}{z^2+a^2}

I know there are singularities at z=a and z=-a,

and to determine the order of the pole, I look at

z^n f(z) and look at which n will make that entire thing a constant, and that will be the order of the pole?

 

[shmoe]

It's a rational function (polynomial over a polynomial), you can read off the orders of poles and zeros by factoring it into linear factors.

 

[HallsofIvy]

well I'm back with a bunch more of these problems, i think I have a good handle on the method, but I want to make sure.

The first one is

\frac{1}{z^2+a^2}

I know there are singularities at z=a and z=-a,

No, you do not know that there are singularities at a and -a! Taking z= a or negative a gives \frac{1}{2a^2}- not a singularity. a and -a would be singularities for
\frac{1}{z^2- a^2}

If you really mean
\frac{1}{z^2+ a^2}
then the singularities are at ai and -ai.

and to determine the order of the pole, I look at

z^n f(z) and look at which n will make that entire thing a constant, and that will be the order of the pole?

First, there is no n which "make that entire thing a constant"- that should be obvious. Taking the limit of that as z goes to 0 and seeing that the limit existed would tell you that there was no singularity at z= 0.

To determine the order of the pole at ai, you should multiply by (z- ai)ⁿ and take the limit as z goes to ai, to determine the order at -ai, you should multiply by (z+ ai)ⁿ and take the limit as z goes to -ai. The order of the pole is the n that such that that limit exists and is non-zero.

As shmoe said, factoring the denominator should make the correct value of n obvious.