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About LinkBacksI'm trying to construct Laurent series aroundand
for
with
the only pole of this function.
I have = \sum_{n=-\infty}^{\infty}a_{n}(z-c)^n)
with
and
and
so am I correct in assuming that around z=c:
 = \sum_{n=-\infty}^{\infty}a_{n}(z-c)^n = \sum_{n=-\infty}^{\infty} (z - c)^{n}\lim_{z \to z_{0}}f(z)\frac{z - z_{0}}{(z - c)^{n+1}})
or did I make a mistake?
And how the hell do I evaluate the function around z=infinity?
I'm trying to construct Laurent series around
I have
with
and
and
so am I correct in assuming that around z=c:
or did I make a mistake?
And how the hell do I evaluate the function around z=infinity?
I am confused about what you are doing here.has a pole at z= 2, as you say, but is analytic at z= 0. A function's "Laurent series" about any point at which it is analytic is just its Taylor series.
In this case,which is the sum of a geometric series with first term a= 1/2 and common ratio r= z/2.
As for "at z=infinity", the Laurent series of any function, f(z), about
Ifthen
.
Thank you.
Yeah, I was kinda wondering why only this problem had different values for c and z0 so it makes sense this is a special case. But if I were told to look for the Laurent Series near c=2=z0 would my last equation be correct?
I looked up the answer using an online calculator (for z0 = 2 and c=0) and it corresponded to what I found using my method above, except for the sign: the calculator found: -1/2 -z/4 -(z^2)/8 etc.. while I found 1/2 z/4 (z^2)/8 etc... is it a coincidence that I got so close?
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