Mastering Laurent Series Expansion: A Layman's Guide

In summary, the conversation discussed the steps for breaking a function into partial fractions and finding its Laurent series. It also addressed the difference between a Laurent series and a Taylor series, and the role of poles in the Laurent series.
  • #1
j-lee00
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Homework Statement



In attachment

The Attempt at a Solution



I break into partial fractions, then get stuck. Please help me in layman terms

f(z) = -1/3[3(z+1)] + 4/3[z+4]

Now I am stuck.
 

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  • #2
Hi!

I will help you with a full list of steps for one of the parts:

[tex]

\frac{4}{3(z + 4)} = \frac{4}{3(4(\frac{z}{4} + 1))} = \frac{1}{3(\frac{z}{4} + 1)} = \frac{1}{3(1 + \frac{z}{4})} = \frac{1}{3}\cdot \frac{1}{(1 + \frac{z}{4})} = \frac{1}{3}\cdot\frac{1}{(1-(-\frac{z}{4}))} = \frac{1}{3}\left ( 1 + (-\frac{z}{4}) + (-\frac{z}{4})^{2} + (-\frac{z}{4})^{3} +\cdots \right ) =

[/tex]

[tex]

= \frac{1}{3}\left ( 1 - \frac{z}{4} + \frac{z^{2}}{4^{2}} - \frac{z^{3}}{4^{3}} +\cdots \right ) = \frac{1}{3} - \frac{z}{3\cdot 4} + \frac{z^{2}}{3\cdot 4^{2}} - \frac{z^{3}}{3\cdot 4^{3}} +\cdots
[/tex]

Remember that for a given geometric series to converge |x| has to be less than 1. which in your case means:

[tex]
\left |\frac{z}{4} \right |<1
[/tex]I hope you will understand all the steps otherwise do ask.
P.S. I used the sum of the geometric series (which is also the Taylor and Laurent series): 1/(1 - x) = 1 + x + x^2 + x^3 + ...
|x|<1
 
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  • #3
Thanks for your help.

With the expansion that you made, how does this differ to the taylor expansion? or it the same?
 
  • #4
If a Laurent series contains only non-negative powers, it will be identical to the Taylor series for the function.
 
  • #5
Are the negative powers of the Laurent expansion only as big as the largest pole?
 
  • #6
What do you mean by "largest pole"?

Note that

[tex]\frac{4}{3(z + 4)} = \frac{1}{3} \left( 1 - \frac{z}{4} + \frac{z^2}{4^2} - \frac{z^3}{4^3} + \cdots \right)[/tex]

has a pole at z=-4 yet no negative-power terms. The Laurent series about z=0 of the function for |z|>4 is

[tex]\frac{4}{3(z + 4)} = \frac{4}{3z}\left( 1 - \frac{4}{z} + \frac{4^2}{z^2} - \frac{4^3}{z^3} + \cdots \right)[/tex]

has an infinite number of negative-power terms.
 
  • #7
What you wrote clarified my problem. tks
 

1. What is a Laurent series expansion?

A Laurent series expansion is a mathematical representation of a complex function as an infinite sum of terms, including both positive and negative powers of a complex variable, centered around a specific point within the function's domain.

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

A Taylor series expansion only includes positive powers of the complex variable, while a Laurent series expansion includes both positive and negative powers. Additionally, a Taylor series expansion is centered around a point where the function is defined, while a Laurent series expansion can be centered around a singularity or pole of the function.

3. Why is a Laurent series expansion useful?

A Laurent series expansion can be used to approximate complex functions, especially those with singularities or poles. It can also help in analyzing the behavior of a function near these points and determining the convergence or divergence of the series.

4. What is the difference between a principal part and a regular part in a Laurent series expansion?

The principal part of a Laurent series expansion includes the terms with negative powers of the complex variable, while the regular part includes the terms with positive powers. The principal part helps in understanding the behavior of the function near singularities, while the regular part represents the "smooth" part of the function.

5. How is a Laurent series expansion related to the residue theorem?

The residue theorem states that the integral of a function around a closed contour is equal to the summation of the residues of the function at its singularities inside the contour. A Laurent series expansion can be used to calculate these residues and thus, can be used in applications of the residue theorem.

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