# Find the Laurent Series of a function

• docnet
In summary, the problem involves decomposing the function f into a sum of fractions of the form ##\frac{c_1}{c_2-z}## and then using appropriate Taylor and Laurent series expansions to find the regions where the series converge. The goal is to choose the appropriate series expansions for each region to combine into a series that converges in that region. This involves using substitutions to switch between Taylor and Laurent series, and choosing the correct regions and singularities for each series.
docnet
Gold Member
Homework Statement
.
Relevant Equations
.

(a)
i tried to decompose the fracion as a sum of fractions of form ##\frac{1}{1-g}##

$$f=\frac{-z}{(1+z)(2-z)}=\frac{a}{1+z}+\frac{b}{2-z}$$
$$a=\frac{1}{3}, b=-\frac{2}{3}$$

$$f=\frac{1}{6}\frac{1}{1+z}-\frac{1}{3}\frac{1}{1-\frac{z}{2}}$$
$$f=\frac{1}{6}\sum_{n=0}^\infty (-z)^n-\frac{1}{3}\sum_{n=0}^\infty(\frac{z}{2})^n$$
sory i posted too early

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docnet said:
Homework Statement:: .
Relevant Equations:: .

View attachment 293612
(b)

$$f=-\frac{1}{3}\frac{1}{1-\frac{z}{2}}+\frac{1}{3z}\frac{1}{1+\frac{1}{z}}$$
$$f=-\frac{1}{3}\sum_{n=0}^\infty (\frac{z}{2})^n+\frac{1}{3z}\sum_{n=0}^\infty\frac{1}{(-z)^n}$$
then simplify...

(c) $$f=-\frac{2}{3z}\cdot \frac{1}{1-\frac{2}{z}}+\frac{1}{3z}\frac{1}{1+\frac{1}{z}}$$

$$f=-\frac{2}{3z}\sum_{n=0}^\infty (\frac{2}{z})^n+\frac{1}{3z}\sum_{n=0}^\infty\frac{1}{(-z)^n}$$

then simplify...aside from the need to simplify the expressions, is it a correct idea to re-write the fractions so the sum converges for each part of the problem?

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i don't understand the learning goal of this problem. anyone?

Give people more time to respond!

A Laurent series is an expansion of f into a series of terms ##(z-a)## for some fixed a. They ask you for three series so you probably need to pick three different values of ##a##. Does that help?

docnet
Office_Shredder said:
Give people more time to respond!

A Laurent series is an expansion of f into a series of terms ##(z-a)## for some fixed a. They ask you for three series so you probably need to pick three different values of ##a##. Does that help?
so even if the terms don't look like ##(z-a)##, it could be the correct expansion of f, right?

is there more than one correct way to expand f into a series over a fixed domain?

Is the Laurent series another form of the Taylor series?

can inequivalent terms have equivalent sums? like if you have two series expansions of ##f## over a shared domain, but one series expansion has a wider radius of convergence than the other. is there ever a benefit of choosing the series with the smaller radius of convergence? ( is this even possible?)

can one expand any function into a Laurent series?

docnet said:
aside from the need to simplify the expressions, is it a correct idea to re-write the fractions so the sum converges for each part of the problem?
Yes, you have the right idea.

docnet
docnet said:
Is the Laurent series another form of the Taylor series?

Yes, except a Laurent series can have terms with negative powers

can inequivalent terms have equivalent sums? like if you have two series expansions of ##f## over a shared domain, but one series expansion has a wider radius of convergence than the other. is there ever a benefit of choosing the series with the smaller radius of convergence? ( is this even possible?)

Yes, you could for example expand ##1/(z-1)## at ##a=0## and get something that converges when ##|z|<1##, or you could expand it around ##a=2/3## and get something that converges when ##1/3 < |z| < 1##. In general the radius of convergence is constrained to points around that are not singularities. On the other hand, if we expand it around ##a=1## we get something that converges for all ##z\neq 1##.

can one expand any function into a Laurent series?

As long as it's well behaved enough, yes.

docnet
For the function ##\frac{1}{1-z}##, you can have a Taylor series expanded at ##z_0=0## that converges for ##|z| \lt 1## and a Laurent series centered at ##z_0=0## that converges for ##|z| \gt 1## (since ##\frac{1}{1-z} = -a \frac{1}{1-a}##, where ##a=1/z)##. You can use this concept to combine appropriate series that converge in the regions specified in the problem.

docnet
FactChecker said:
For the function ##\frac{1}{1-z}##, you can have a Taylor series expanded at ##z_0=0## that converges for ##|z| \lt 1## and a Laurent series centered at ##z_0=0## that converges for ##|z| \gt 1## (since ##\frac{1}{1-z} = -a \frac{1}{1-a}##, where ##a=1/z)##. You can use this concept to combine appropriate series that converge in the regions specified in the problem.
Just to complete the explanation:
You already can express the function as the sum of two simple ##\frac{c_1}{c_2-z}## terms.
Each of those can be expanded into a Taylor series that converges in a disk out to its singularity. Call these ##T_1## and ##T_2##.
Also, using the substitution ##a=1/z## in each term, gives a Taylor series in ##a## that converges for ##a## in a disk to its singularity. Substituting the ##1/z## back in for ##a## gives a Laurent series with all negative powers of ##z## that converges for ##z## beyond the singularity. (Note: ##singularity_z = 1/singularity_a##.) Call these ##L_1## and ##L_2##.
For the region of part a of the problem, ##T_1+T_2## is a Taylor series that converges.
For the region of part b of the problem, ##L_1+T_2## is a Laurent series that converges.
For the region of part c of the problem, ##L_1+L_2## is a Laurent series that converges.

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docnet

## What is a Laurent series?

A Laurent series is a representation of a complex function as an infinite sum of terms, including both positive and negative powers of the variable. It is similar to a Taylor series, but it allows for terms with negative powers.

## Why is it important to find the Laurent series of a function?

The Laurent series allows us to extend the domain of a function beyond its usual region of convergence. This is useful in many applications, such as in complex analysis and in solving differential equations.

## How do you find the Laurent series of a function?

The Laurent series can be found by using the formula: f(z) = ∑(n= -∞ to ∞) cn(z-a)n, where cn is the coefficient of the (z-a)n term and a is the center of the series. The coefficients can be found by using the Cauchy integral formula or by using the Taylor series of the function.

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

The main difference between a Laurent series and a Taylor series is that a Taylor series only includes positive powers of the variable, while a Laurent series includes both positive and negative powers. Additionally, a Taylor series is centered at a point where the function is analytic, while a Laurent series can be centered at a point where the function has a singularity.

## Can a function have more than one Laurent series?

Yes, a function can have multiple Laurent series depending on the center chosen. The series may converge at different points and have different regions of convergence. However, the series will always be unique within its region of convergence.

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