# Find the Laurent Series of a function

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Homework Statement:
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Relevant Equations:
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(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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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?

Staff Emeritus
Gold Member
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?

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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?

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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?

Staff Emeritus
Homework Helper
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.

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Staff Emeritus
Gold Member
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.

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Homework Helper
Gold Member
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.

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