Laurent expansion of ##ze^{1/z}##

In summary, the conversation is about finding a Laurent series for the function f(z) = ze^(1/z) in powers of z-1. The original poster is struggling with this as it is not a typical expansion seen in textbooks, and their attempt at a solution is too complicated. Some suggestions and attempts at finding a solution are discussed, but it is ultimately suggested to use a Taylor series expansion by substituting z = u+1 and taking derivatives. However, it is mentioned that this method also proves to be difficult in finding a simple formula for the series.
  • #1
Terrell
317
26

Homework Statement


Find a Laurent series of ##f(z)=ze^{1/z}## in powers of ##z-1##. Is there an easier way to go about this as this is not a typical expansion I see on textbooks. It seems that my incomplete solution is too complicated. Please help, exam is in two days and I am working on past exams. Worked out solutions are welcome, also. P.S. it may seem like I'm cramming, but it's not under my control - :cry:

Homework Equations


Laurent's Theorem.png


The Attempt at a Solution


Since ##f(z)## is entire, then it is analytic in any annular domain. So let's pick the domain ##\frac{1}{2} \lt \vert z-1\vert \lt 3##. By Laurent's theorem the coefficients
\begin{align}a_k=\frac{1}{2\pi i}\int_{c}\frac{(1+e^{it})e^{\frac{1}{1+e^{it}}}}{e^{it\cdot k}}\end{align}
where ##C## is the contour ##z(t)=1+e^{it}## and hence, ##dz=ie^{it}dt##. Note that the contour ##C## is within our selected annular domain as it should be according to the theorem.
 

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  • #2
I think I see a trick to this one: ## f'(z)=(z-1)f(z) ##, and if you do the algebra, and I think you can show ## f^n(z)=(z-1)f^{n-1}+(n-1) f^{n-2}(z) ##. ## \\ ##That would make it suitable for a Taylor expansion about ## z=1 ##. Note that ##f^n(1)=(n-1)f^{n-2}(1) ##. ## \\ ## It looks like the odd number derivatives are zero and the even ones are ## (n-1)! =(n-1)(n-3)(n-5)...(1) ##. (I think I have the correct definition of n!=Edit: yes I googled it: see http://mathworld.wolfram.com/DoubleFactorial.html) And hopefully I don't get dinged for supplying a solution to this, because other than this approach, I wouldn't know what to do with this one. :wink:
 
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  • #3
Charles Link said:
f′(z)=(z−1)f(z)
I don't see why this is true. I got \begin{align}f'(z)=e^{1/z}-\frac{e^{1/z}}{z}\end{align}
But ##(z-1)f(z)=(z-1)ze^{1/z}=z^2e^{1/z}-ze^{1/z}##
 
  • #4
I made an algebraic error here: ## f'(z)=\frac{(z-1)}{z^2} f(z) ##. Oh well, back to the drawing board.
 
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  • #5
I think a similar method might still be possible: Using ## f'(z)=f(z)g(z) ## where ## g(z)=\frac{1}{z}-\frac{1}{z^2} ##, so that ## f'=fg ##, ## f''=f'g+fg' ##, etc, I think you should be able to get some kind of expression for ## f^n(z) ##.
 
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  • #6
Charles Link said:
I think a similar method might still be possible: Using f′(z)=f(z)g(z)f′(z)=f(z)g(z) f'(z)=f(z)g(z) where g(z)=1z−1z2g(z)=1z−1z2 g(z)=\frac{1}{z}-\frac{1}{z^2} , so that f′=fgf′=fg f'=fg , f′′=f′g+fg′f″=f′g+fg′ f''=f'g+fg' , etc, I think you should be able to get some kind of expression for fn(z)fn(z) f^n(z) .
Is there a standard general approach you are trying to follow? Thanks!
 
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  • #7
Terrell said:
Is there a standard general approach you are trying to follow? Thanks!
I am just trying to find something that works. Had I not made a simple algebraic error, the first solution that I had would have been a good one. Upon further inspection, I'm not having any luck with anything that works...
 
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  • #8
Charles Link said:
I am just trying to find something that works. Had I not made a simple algebraic error, the first solution that I had would have been a good one. Upon further inspection, I'm not having any luck with anything that works...
I find it hard to believe they put this on a past exam. It's suppose to be a test of understanding, not puzzles. I'll try my hand at this again later, I'm working on other problems for the moment. Thank you!
 
  • #9
Terrell said:
I find it hard to believe they put this on a past exam. It's suppose to be a test of understanding, not puzzles. I'll try my hand at this again later, I'm working on other problems for the moment. Thank you!
I hope the people who designed the problem didn't make the same mistake that I did in thinking there was a simple solution. It's unlikely, but certainly quite possible.:wink:
 
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  • #10
Charles Link said:
I hope the people who designed the problem didn't make the same mistake that I did in thinking there was a simple solution. It's unlikely, but certainly quite possible.:wink:
I would be outrage! These are comprehensive exams and they don't get returned to us. :mad:
 
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  • #11
@Orodruin You have much more math expertise. Does this one have a simple solution that we are missing?
 
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  • #12
Here's my short-cut way to an answer:

##f(z) = z e^\frac{1}{z}## has a singularity only at ##z=0##. Which means that it doesn't have any singularity at ##z=1##. If it doesn't have a pole at ##z=1##, that means that it can be expanded in an ordinary Taylor series, using just positive powers of ##z-1##.

Given that, I think I would just let ##z = u+1## and then our function becomes ##f(u) = (1+u) e^{\frac{1}{1+u}}##. You can find the Taylor series by just taking derivatives, rather than doing complex integrals. I guess it depends on whether the assignment is intended to give you practice at doing complex integration.
 
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  • #13
With a Taylor expansion, and taking derivatives, I was unable to succeed at getting a simple formula for the series. This only worked when I made an algebraic error that was clearly wrong. Perhaps the OP or someone else can find such a formula.
 

Related to Laurent expansion of ##ze^{1/z}##

What is the Laurent expansion of ##ze^{1/z}##?

The Laurent expansion of ##ze^{1/z}## is a mathematical series that represents the complex function ##ze^{1/z}## as a sum of infinitely many terms. It is used to approximate the behavior of the function near a singular point, such as when ##z = 0##.

What is the purpose of the Laurent expansion?

The purpose of the Laurent expansion is to provide a way to analyze the behavior of a complex function near singular points, where the function may not be defined or may have a singularity. It allows us to approximate the behavior of the function and make predictions about its behavior in a given region.

How is the Laurent expansion calculated?

The Laurent expansion is calculated using the Cauchy integral formula, which involves integrating the function along a contour surrounding the singular point. The series is then constructed by finding the coefficients of the terms in the power series expansion of the function.

What is the difference between a Laurent expansion and a Taylor series?

The main difference between a Laurent expansion and a Taylor series is that a Taylor series represents a function as a sum of powers of ##z##, while a Laurent expansion includes both positive and negative powers of ##z##. A Taylor series is used to approximate a function near a point, while a Laurent expansion is used to approximate a function near a singular point.

When is the Laurent expansion useful?

The Laurent expansion is useful in many areas of mathematics and science, such as in complex analysis, physics, and engineering. It is often used to approximate the behavior of functions in situations where singularities or branch cuts occur, such as in the study of complex functions with poles or essential singularities.

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