Laurent Series

**bakerconspiracy** · March 12th 2011, 01:04 AM

Hey guys, I'm having a bit of trouble with the creating Laurent series representations of functions. Specifically, I have no idea what the domain does to the series representation.

Anyways, this is what I'm working on...

$\frac{e^z}{(z+1)^2}$ for $0 < |z+1| < \infty$

This is where I go with this?...

$\frac{e^z}{(z+1)^2}= \frac{1}{(z+2)^2} \displaystyle\sum_{n=0}^{\infty}\frac{(z+1)^n}{(n+ 1)!}$

No idea where to go from here.

Also, if a function is analytic everywhere in a domain, is the Laurent series representation just the Taylor series?

Thanks for the help in advance.

**FernandoRevilla** · March 12th 2011, 02:34 AM

Use the substitution $u=z+1$ , then:

$f(z)=\dfrac{e^{u-1}}{u^2}=\dfrac{1}{e}\displaystyle\sum_{n\geq 0}\dfrac{u^{n-2}}{n!}=\ldots(0<|u|<+\infty)$

Originally Posted by bakerconspiracy

Also, if a function is analytic everywhere in a domain, is the Laurent series representation just the Taylor series?

Yes, it is.

**Opalg** · March 12th 2011, 02:36 AM

Originally Posted by bakerconspiracy

Hey guys, I'm having a bit of trouble with the creating Laurent series representations of functions. Specifically, I have no idea what the domain does to the series representation.

Anyways, this is what I'm working on...

$\frac{e^z}{(z+1)^2}$ for $0 < |z+1| < \infty$

This is where I go with this?...

$\frac{e^z}{(z+1)^2}= \frac{1}{(z+2)^2} \displaystyle\sum_{n=0}^{\infty}\frac{(z+1)^n}{(n+ 1)!}$

No idea where to go from here.

Write the function as $\displaystyle \frac{e^z}{(z+1)^2}= \frac{e^{-1}e^{z+1}}{(z+1)^2}= e^{-1}\sum_{n=0}^\infty\dfrac{(z+1)^{n-2}}{n!}$ . That is the Laurent expansion in powers of z+1.

Originally Posted by bakerconspiracy

Also, if a function is analytic everywhere in a domain, is the Laurent series representation just the Taylor series?

Yes, provided that you use the expansion about the same point in both cases.

Edit. Sorry, didn't notice that FernandoRevilla got there first.

**bakerconspiracy** · March 12th 2011, 03:23 AM

Thank you both. I'm starting to understand this better now.

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