About Laurent series...


by serchinnho
Tags: laurent, series
serchinnho
serchinnho is offline
#1
Apr14-10, 12:43 AM
P: 3
1. The problem statement, all variables and given/known data
I'm asked to find the Laurent series of some rational function and using partial fractions I encounter something like 1/(c-z)^2 with c > 0.


2. Relevant equations



3. The attempt at a solution
I've tried several 'algebraic tricks' like multiplying for z^2 or just staring at it several hours without any results... besides a some red eyes! I know I just can't multiply the Laurent series of 1/(c-z) and I ran out of ideas... Please, a little help!
By the way, if you remember your middle school and just do the division the result seems like the Laurent series 1/z^2+2c/z^3+3c^2/z^4+4c^3/z^5+... (and I say 'seems like' because I don't know which is the Laurent series...), why that happens?! Does it has to do with the uniqueness of the Laurent series?!?
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lanedance
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#2
Apr14-10, 01:37 AM
HW Helper
P: 3,309
maybe it would help if you gave the rational function you're attempting to write a laurent series for & about which point & region you want it to be convergent... along with what you've tried, otherwise i don't know exactly what it is you're asking

click on tex code below to see how to write it, they open & close with the tags "tex" & "/tex" in [] brackets... eg to write a fraction use \frac{}{}
[tex] f(z) = \frac{1}{(z-c)^2} [/tex]
or
[tex] f(z) = \sum_{-\infty}^{\infty} a_n (z-c)^n} [/tex]
lanedance
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#3
Apr14-10, 01:41 AM
HW Helper
P: 3,309
welcome to pf by the way

serchinnho
serchinnho is offline
#4
Apr14-10, 04:44 PM
P: 3

About Laurent series...


Yeah, I'm sorry, I'm asked to find the laurent series of
[tex]
f(z) = \frac{1}{(2-z)^2(1-z)^2}
[/tex]
in two rings: 1<|z|<2 and |z|>2. Using partial fractions I got
[tex]
f(z) = \frac{-2}{1-z} + \frac{1}{(1-z)^2} + \frac{2}{2-z} + \frac{1}{(2-z)^2}
[/tex]
and I can easily obtain the laurent series in both rings of the first and third partial fraction, but I'm stuck in the other two!

By the way, thanks!
lanedance
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#5
Apr15-10, 10:01 AM
HW Helper
P: 3,309
should be a similar method

but as a shortcut, the 2nd & 4th are proprotional to square of 1st & 3rd resepctively, have you tried evaluating the square of each series?
serchinnho
serchinnho is offline
#6
Apr15-10, 08:17 PM
P: 3
The problem is I just can't multiply let's say the first by itself without knowing for sure if the result is a convergent series... I was told it had to do with the Binomial Series Theorem:

http://en.wikipedia.org/wiki/Binomial_theorem

And with that I got the 3rd and 4th in the first ring (but I already had the 3rd one) but to get the Laurent series of the 2nd in any of the two rings....


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