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Thread: Laurent Series

  1. #1
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    Laurent Series

    Find the Laurent series representation of $\displaystyle \frac{e^z}{(z+1)^2}$ for $\displaystyle (0<\left |z+1 \right |< \infty)$.

    I really don't understand the process of getting the series representation.
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  2. #2
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    Quote Originally Posted by davesface View Post
    Find the Laurent series representation of $\displaystyle \frac{e^z}{(z+1)^2}$ for $\displaystyle (0<\left |z+1 \right |< \infty)$.

    I really don't understand the process of getting the series representation.
    First get the Taylor's series for $\displaystyle e^z$ about the point z= -1. Can you do that?
    (The Taylor's series for f(z) about z= a is $\displaystyle \sum_{n=0}^\infty \frac{f^{(n)}(a)}{a!}(z- a)^n$ where $\displaystyle f^{(n)}$ indicates the nth derivative. It should be easy to find the nth derivative of $\displaystyle e^z$ at z= -1.

    Then divide each term by $\displaystyle (z+ 1)^2$
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  3. #3
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    Taking that approach, I get $\displaystyle L(f(z))=\frac{1}{(z+1)^2e}*\sum_{n=0}^{\infty}\fra c{(z+1)^n}{(-1)!}$. Shouldn't it be $\displaystyle n!$ instead of $\displaystyle a!$?
    Last edited by davesface; Apr 22nd 2010 at 10:40 AM.
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