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About Laurent series... 
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#1
Apr1410, 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/(cz)^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/(cz) 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?!? 


#2
Apr1410, 01:37 AM

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P: 3,307

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}{(zc)^2} [/tex] or [tex] f(z) = \sum_{\infty}^{\infty} a_n (zc)^n} [/tex] 


#3
Apr1410, 01:41 AM

HW Helper
P: 3,307

welcome to pf by the way



#4
Apr1410, 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}{(2z)^2(1z)^2} [/tex] in two rings: 1<z<2 and z>2. Using partial fractions I got [tex] f(z) = \frac{2}{1z} + \frac{1}{(1z)^2} + \frac{2}{2z} + \frac{1}{(2z)^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! 


#5
Apr1510, 10:01 AM

HW Helper
P: 3,307

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? 


#6
Apr1510, 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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