Whats a Laurent series? And how do I use one to represent a function?

In summary: Then put them back together to get the Laurent series for z= i. In summary, to represent a function with a Laurent series, you must identify the poles, pick one, and use that point as the center of your series. You must expand everything in powers of (z-c)^n as you said, where c is the location of the pole. The formula to use is the geometric series.
  • #1
laura_a
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0

Homework Statement


Write TWO laurent series in powers of z that represent the function

f(z)= \frac{1}{z(1+z^2)}

In certain domains, and specify the domains


Homework Equations



Well that's my prob, not sure what the terms in the Laurent series are

The formula I'm looking at is

\sum^{infty}_{n=0} a_n * (z-c)^n

for a complex function f(z) about a point c and a_n is a constant.

I sort of understand that, but do I use it to represent a function? Thats the part I'm not sure about

The Attempt at a Solution




My lame guess is that sub in the f(z) I'm given into the equation. I've seen one example where the fraction is split into two and then the Laurent expansion is applied. I have the answers (because it's a textbook question)

They are

\sum^{\infty}_{n=0} (-1)^{n+1} * z^{2n+1} + 1/z (0 < |z| < 1)

and

\sum^{\infty}_{n=0} [(-1)^{n+1}] / z^{2n+1} (1 < |z| < \infty)

All I need to know is what to I plug into where and I'll work on the rest :) any suggestions will make my day (night actually, but who's counting)

Thanks
Laura
 
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  • #2
laura_a said:

Homework Statement


Write TWO laurent series in powers of z that represent the function

f(z)= \frac{1}{z(1+z^2)}

In certain domains, and specify the domains


Homework Equations



Well that's my prob, not sure what the terms in the Laurent series are

The formula I'm looking at is

\sum^{infty}_{n=0} a_n * (z-c)^n

for a complex function f(z) about a point c and a_n is a constant.

I sort of understand that, but do I use it to represent a function? Thats the part I'm not sure about

The Attempt at a Solution




My lame guess is that sub in the f(z) I'm given into the equation. I've seen one example where the fraction is split into two and then the Laurent expansion is applied. I have the answers (because it's a textbook question)

They are

\sum^{\infty}_{n=0} (-1)^{n+1} * z^{2n+1} + 1/z (0 < |z| < 1)

and

\sum^{\infty}_{n=0} [(-1)^{n+1}] / z^{2n+1} (1 < |z| < \infty)

All I need to know is what to I plug into where and I'll work on the rest :) any suggestions will make my day (night actually, but who's counting)

Thanks
Laura
Btw, put your TeX code between [ itex ] and [ /itex ] (with no spaces...I left spaces there just so that they would show up in my post.)

First you must identify the poles of your function.

Once you have the poles, pick one and use that point as the center of your Laurent series. Then you must expand everything in powers of (z-c)^n as you said, where c is the location of the pole.
The formula to use is the geometric series
[tex] \frac{1}{1-q} = \sum_{n=0}^\infty q^n [/tex]
at the condition that q < 1. (in your case, it will be a condition on the magnitude of a complex quantity).


For the domain, you will have to check if you have any restriction from using the geometric series and you must laos make sure that you don't go far enough from your pole that you run into one of the other poles.
 
Last edited:
  • #3
To answer the question in your title, a Laurent series is a power series that includes negative powers. When your original function is a rational function, like the one you give, [tex]f(x)= \frac{1}{z(1+z^2)}[/tex], a good method is to take out one of the factors in the denominator, say 1/z, and expand the remaining function, 1/(1+z2), in a Taylor's series. Here we can write [tex]\frac{1}{1+z^2}= \frac{1}{1- (-z^2)}[/tex] and recognize that as a geometric series with "common factor" -z2 (I just noticed that is what nrqed said): [itex]1- z^2+ z^4- z^6+ \cdot\cdot\cdot[/itex]. Finally, put the 1/z back in: [itex]z^{-1}- z+ z^3- z^5+ \cdot\cdot\cdot[/itex]. Obviously, that cannot converge at z= 0. Further more the original power series only converges for the absolute value "common factor" less than 1. Here that is z2< 1 so the Laurent series converges for 0< |z|< 1.

For a different Laurent series, notice that 1/(1+ z^2) is discontinuous at z= i. Factor out 1/(z- i), leaving [tex]\frac{1}{z(z+ i)}[/tex] and find a Taylor's series for that, about z= i. A good method is to separate into two different fractions using "partial fractions" and write each as a geometric series.
 

1. What is a Laurent series?

A Laurent series is a mathematical representation of a complex function in the form of an infinite sum of terms. It is named after mathematician Pierre Alphonse Laurent and is used to represent functions that have both positive and negative powers of the variable.

2. How is a Laurent series different from a Taylor series?

A Taylor series is a special case of a Laurent series where all the terms have non-negative powers of the variable. This means that a Taylor series is valid only for functions that are analytic (infinitely differentiable) at a single point. In contrast, a Laurent series can represent functions that are not analytic at a point, but are analytic in an annulus (a region between two concentric circles).

3. How do I find the coefficients of a Laurent series?

The coefficients of a Laurent series can be found using the formula: cn = (1/2πi) ∫C f(z) (z-z0)-n-1 dz, where cn is the coefficient of z-n-1, C is a closed curve around the point z0, and f(z) is the given function. Alternatively, the coefficients can also be found by expanding the function as a power series and identifying the coefficients of the negative powers of the variable.

4. How do I use a Laurent series to represent a function?

To use a Laurent series to represent a function, first determine the domain of convergence (where the series is valid) by finding the singularities of the function. Then, write the Laurent series in the form ∑n=-∞ cn (z-z0)n where cn are the coefficients found using the methods mentioned above. Finally, substitute in the values of z and z0 to evaluate the series for a given function.

5. What are some applications of Laurent series?

Laurent series have various applications in mathematics, physics, and engineering. They are used to solve differential equations, evaluate integrals, and approximate functions. They also have applications in the field of complex analysis, where they are used to study the behavior of functions near their singularities.

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