by serchinnho
Tags: laurent, series
 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?!?
 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{}{} $$f(z) = \frac{1}{(z-c)^2}$$ or $$f(z) = \sum_{-\infty}^{\infty} a_n (z-c)^n}$$
 HW Helper P: 3,309 welcome to pf by the way
P: 3

Yeah, I'm sorry, I'm asked to find the laurent series of
$$f(z) = \frac{1}{(2-z)^2(1-z)^2}$$
in two rings: 1<|z|<2 and |z|>2. Using partial fractions I got
$$f(z) = \frac{-2}{1-z} + \frac{1}{(1-z)^2} + \frac{2}{2-z} + \frac{1}{(2-z)^2}$$
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!
 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?
 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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