View Full Version : even function has a Laurent decomposition of even functions and even powers of z
caramello
Jul26-09, 02:03 PM
Hi,
if let's say that there's an even function f(z) then how do we know if its Laurent decomposition (i.e. f(z) = f0(z) + f1(z) ) will be even functions and have even powers of z?
Any help will be greatly appreciated.
Take a generic Laurent series and compute it's even and odd parts. What do you see?
(You know a formula for the even and odd part of any function, right?)
caramello
Jul26-09, 03:21 PM
i know that for even function f(z) = f(-z) which means that the sum from n=0 to infinity of a_n z^n is equals to the sum from n=0 to infinity of a_n (-z)^n.
(note: a_n is the coefficient of the series)
Then this will give me an equation of a_n = a_n (-1)^n ---> conclusion: a_n can't be equal to 0, am I right?
But then after this I don't know what else to do in order to prove that f0(z) and f1(z) are even functions that only has powers of z.
Thank you so much
g_edgar
Jul26-09, 04:35 PM
Must be Laurent series in powers of z.
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