Laurent Series doubt about the Sum limits

  • #1

Main Question or Discussion Point

Hi everyone!

I'm studying Laurent Series, and I thought I understood it but after solving one exercises I get confused. The problem was to get the Laurent series for the following function:

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

I do it this way:

[tex]\frac{1}{\left(z-1\right)^2}\cdot \frac{1}{z+3+1-1}=[/tex]
[tex]\frac{1}{\left(z-1\right)^2}\cdot \frac{1}{4-\left(-\left(z-1\right)\right)}=[/tex]
[tex]\frac{1}{\left(z-1\right)^3}\cdot \frac{1}{1-\left(-\left(\frac{z-1}{4}\right)\right)}=[/tex]
[tex]\frac{1}{\left(z-1\right)^3}\cdot \sum_{n=0}^\infty \left(-\frac{4}{z-1}\right)^n=[/tex]
[tex]\sum_{n=0}^\infty \left(-4\right)^n\cdot \frac{1}{\left(z-1\right)^{n+3}}[/tex]

But when I saw the solution they say that is:

[tex]
\frac{1}{4\left(z-1\right)^2} - \frac{1}{4^2\left(z-1\right)}+\frac{1}{4^3}-\frac{\left(z-1\right)}{4^4}+\frac{\left(z-1\right)^2}{4^5}+\dots
[/tex]

which is different from what I get. However I noticed that if I change the limits of the sum to -inf,0 the expression is equal. Why do this happens or what do I did wrong?
 

Answers and Replies

  • #2
22,097
3,278
Writing

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

is correct. But when you changed that fraction in a sum, that is incorrect. The correct sum is

[tex]
\frac{1}{\left(z-1\right)^3}\sum_{n=0}^{+\infty}{\left(-\frac{z-1}{4}\right)^n}
[/tex]
 
  • #3
22,097
3,278
Also, that 3 in my above post should be a 2.
 
  • #4
I didn't notice that. After all it was just an annoying error. Thanks for your help!

Just one more thing: Is the Laurent serie unique?
I'm asking this because I tried different approaches to get the Laurent serie, and each one gave me a different expression.
 
  • #5
22,097
3,278
Yes, the coefficients of the Laurent series are unique. Strange that you got different expressions. I think you've made a mistake somewhere...
 
  • #6
For example if I do:

[tex]\frac{1}{(z-1)^{2}}\cdot\frac{1}{z+3-2+2}=[/tex]
[tex]\frac{1}{(z-1)^{2}}\cdot\frac{1}{z-1+2}=[/tex]
[tex]-\frac{1}{(z-1)^{3}}\cdot\frac{1}{1-\frac{2}{(z-1)}}=[/tex]
[tex]-\frac{1}{(z-1)^{3}}{\displaystyle \sum_{n=0}^{\infty}(\frac{2}{z-1})^{n}=\sum_{n=0}^{\infty}\frac{2^{n}}{(z-1)^{n+3}}}[/tex]

or:

[tex]\frac{1}{\left(z-1\right)^{2}}\cdot\frac{1}{4-\left(-\left(z-1\right)\right)}=[/tex]
[tex]-\frac{1}{\left(z-1\right)^{3}}\cdot\frac{1}{1-\frac{4}{(z-1)}}=[/tex]
[tex]\sum_{n=0}^{\infty}-\frac{4^{n}}{(z-1)^{n+3}}[/tex]

The expressions are different.
 
  • #7
Sorry forget the first case. I have already seen the mistake. The second I still don't understand.
 

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