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?!?
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]
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!
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?
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....