Find the Laurent Series of a function

[docnet]

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

 

[docnet]

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?

 

[docnet]

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

 

[Office_Shredder]

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]

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?

 

[docnet]

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?

 

[vela]

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.

 

[Office_Shredder]

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.

 

[FactChecker]

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.

 

[FactChecker]

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.