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Residues of an essential singularity and a simple pole

by daoshay
Tags: essential, pole, residues, simple, singularity
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daoshay
#1
May11-10, 01:47 PM
P: 14
1. The problem statement, all variables and given/known data
Classify the isolated singularities and find the residues
[tex]
\frac {\sin(\frac {1}{z})}{1-z}
[/tex]


2. Relevant equations
I know the Taylor series expansion for 1/(1-z) when |z|<1
and I think I know the Taylor series for sin(1/z). The reciprocal of each term of the Taylor series of sin(z), right?

3. The attempt at a solution
znot = 0 is an essential singularity and znot = 1 is a simple pole.
I've tried using the limit approach to find the singularity at znot = 1, but I keep getting -sin(1) as an answer. I am thinking I should change sin(1/z) into (e^iz - e^-iz) / 2i,but I'm not sure if that is the right direction. If someone could just nudge me in the right direction, I'd be pumped.

P.S. I gave up on latex for now, it was driving me nuts...I'm learning from looking at other people's code. It kept throwing in a sin function into an expression I never coded a sin in. I have some learning to do.
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vela
#2
May11-10, 02:03 PM
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Quote Quote by daoshay View Post
1. The problem statement, all variables and given/known data
Classify the isolated singularities and find the residues
[tex]
\frac {\sin(\frac {1}{z})}{1-z}
[/tex]

2. Relevant equations
I know the Taylor series expansion for 1/(1-z) when |z|<1
and I think I know the Taylor series for sin(1/z). The reciprocal of each term of the Taylor series of sin(z), right?
No, that would be

[tex]\sum_{k=0}^\infty (-1)^k \frac{(2k+1)!}{z^{2k+1}}[/tex]

which is not the Taylor series for sin(1/z). I'm guessing that's not what you meant though.
3. The attempt at a solution
znot = 0 is an essential singularity and znot = 1 is a simple pole.
I've tried using the limit approach to find the singularity at znot = 1, but I keep getting -sin(1) as an answer. I am thinking I should change sin(1/z) into (e^iz - e^-iz) / 2i,but I'm not sure if that is the right direction. If someone could just nudge me in the right direction, I'd be pumped.
What's wrong with -sin(1)?
P.S. I gave up on latex for now, it was driving me nuts...I'm learning from looking at other people's code. It kept throwing in a sin function into an expression I never coded a sin in. I have some learning to do.
There's a problem with the forum not rendering equations on the page correctly. If you refresh the page, it usually fixes the problem.
daoshay
#3
May11-10, 04:03 PM
P: 14
Thanks, Vela. For now, I'm okay with the residue of 1 being -sin(1). I'm really uncertain about how to handle the 0 singularity. Do you suggest I expand the function into the Laurent series and find the -1 coefficient? If so, do I need to substitute

[tex]
\frac{e^{i1/z}-e^{-i1/z}}{2i}
[/tex]

for

[tex]
\sin\frac{1}{z}
[/tex]
?

jackmell
#4
May11-10, 07:05 PM
P: 1,666
Residues of an essential singularity and a simple pole

Whenever you have an essential singularity multiplied by an analytic function that's not a polynomial, the residue at zero is going to be an infinite sum:

[itex]\left(a_0+a_1 z+a_2 z^2+\cdots\right)\left(\frac{b_1}{z}+\frac{b_2}{z^2}+\cdots\right)[/itex]

And you have:

[itex]\sum_{n=0}^{\infty} \frac{(-1)^n}{z^{2n+1}(2n+1)!}\sum_{n=0}^{\infty} z^n[/itex]

Can you then form the Cauchy product of those sums and pick out all the terms with 1/z?
Count Iblis
#5
May11-10, 07:12 PM
P: 2,158
A simpler way that requires almost no computation is to consider the contour integral along a circle with center the origin of radius smaller than one.
daoshay
#6
May11-10, 07:18 PM
P: 14
So, I've got:

[tex]

\frac {1}{z}\left(1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\cdots\right)

[/tex]

where the alternating series is my residue and that has a limit of... (I don't recognize this one except for the coefficients in sin(x))
Count Iblis
#7
May11-10, 07:29 PM
P: 2,158
You should recognize this as sin(1). That's the weakness of this method.
daoshay
#8
May11-10, 07:36 PM
P: 14
So, I've got -sin(1) and sin(1) as the residues? That feels awkward, only because none of the examples we did had residues like that. But I suppose those are numbers too.
Count Iblis
#9
May11-10, 07:53 PM
P: 2,158
Quote Quote by daoshay View Post
So, I've got -sin(1) and sin(1) as the residues? That feels awkward, only because none of the examples we did had residues like that. But I suppose those are numbers too.
You should do more practice problems! Let me suggest one. Do the same problem again, but now with sin(1/z) replaced by
f(1/z) where f(z) is any arbitrary analytic function.
daoshay
#10
May12-10, 08:38 AM
P: 14
Thanks count. Will I end up getting -f(1), f(1/1)? I haven't tried it, but based on this problem and the behavior of the essential singularity times the analytic function, that's what I'm guessing I'll get. I -will- try it later this week.
HallsofIvy
#11
May12-10, 08:49 AM
Math
Emeritus
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P: 39,345
The Taylor's series for sin(z), about z= 0, is
[tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n+1}[/tex]

Therefore the Laurent's series for sin(1/z), about z= 0, is
[tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{-(2n+1)[/tex]

Az vela pointed out, that is not the same as "The reciprocal of each term of the Taylor series of sin(z)".

Since that has an infinite number of terms with negative exponent, z= 0 is an essential singularity.
Count Iblis
#12
May12-10, 09:28 AM
P: 2,158
Quote Quote by daoshay View Post
Thanks count. Will I end up getting -f(1), f(1/1)? I haven't tried it, but based on this problem and the behavior of the essential singularity times the analytic function, that's what I'm guessing I'll get. I -will- try it later this week.
You'll find that the residue at zero does not only involve f(1), so there was something special about the case f(z) = sin(z). You can also try to solve this more general problem: Find the residue at z = 0 of

[tex]\frac{f(\frac{1}{z})}{1-z^{n}}[/tex]

where f(z) is an arbitrary analytic function and n an arbitrary integer.
daoshay
#13
May12-10, 10:00 AM
P: 14
Quote Quote by HallsofIvy View Post
The Taylor's series for sin(z), about z= 0, is
[tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n+1}[/tex]

Therefore the Laurent's series for sin(1/z), about z= 0, is
[tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{-(2n+1)[/tex]

Az vela pointed out, that is not the same as "The reciprocal of each term of the Taylor series of sin(z)".
Thanks, Halls. I realized as soon as Vela pointed that out that I shouldn't have reciprocated because of the factorials. Only the z value is reciprocated. I'd been staring at this stuff for a while and wasn't thinking clearly.

You guys have been a great help.
benygh2002
#14
Jul8-11, 05:25 AM
P: 3
Hello friends,
I'm a student of mechanical engineering and I have a problem with computing residues of a complex function. I ve read your useful comments. Now I ve got some ideas about essential singularity and series expansion in computing the residue. However, I still can't find the solution to my problem.
I arrived at a complex function in the process of finding a solution to a mechanical problem.
Then I have to obtain the residues to proceed to the next steps. The function has the following form:

f(z)=exp(A*Z^N+B*Z^-N)/z

where A, B and N are real constants (N>=3).

I want to compute the resiude at z=0. I wrote the Laurent serie of f but got an infinite sum. I do not even know if I am at the right direction.
Engineering students like me have always problems with math, let alone complex analysis.
Anyway, I would be really thankful if someone could give me hint on this and put me back in the right direction.


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