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Math Help - Laurent Series

  1. #1
    Newbie bakerconspiracy's Avatar
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    Laurent Series

    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.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    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)


    Quote Originally Posted by bakerconspiracy View Post
    Also, if a function is analytic everywhere in a domain, is the Laurent series representation just the Taylor series?

    Yes, it is.
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by bakerconspiracy View Post
    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.

    Quote Originally Posted by bakerconspiracy View Post
    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.
    Last edited by Opalg; March 12th 2011 at 04:03 AM.
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  4. #4
    Newbie bakerconspiracy's Avatar
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    Thank you both. I'm starting to understand this better now.
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