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Thread: Another Laurent expansion

  1. #1
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    Another Laurent expansion

    Expand $\displaystyle f(z) = \frac{2}{z(z-2)}$ in a Laurent series valid for $\displaystyle |z-2|>0$.

    I've tried many things, but the |z-2|>0 thing gets me. Tips?
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  2. #2
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    Ok, I did a substitution $\displaystyle w=z-2$ and got

    $\displaystyle f(z)=\begin{cases}\displaystyle -\frac{1}{z-2}\sum_{n=0}^\infty \left(-\frac{z-2}{2}\right)^n, & \text{for}~ |z-2|<2

    \\ \displaystyle \frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^n, &\text{for}~|z-2|>2
    \end{cases}$

    Does that look right?
    Last edited by scorpion007; May 15th 2010 at 03:56 AM.
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  3. #3
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    Quote Originally Posted by scorpion007 View Post
    Ok, I did a substitution $\displaystyle w=z-2$ and got

    $\displaystyle f(z)=\begin{cases}\displaystyle -\frac{1}{z-2}\sum_{n=0}^\infty \left(\frac{z-2}{2}\right)^{\color{red}2}, & \text{for}~ |z-2|<2

    \\ \displaystyle \frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^{\color{red}2}, &\text{for}~|z-2|>2
    \end{cases}$

    Does that look right?
    To me it looks somewhat right, but not quite. Let me see: I'd do a partial fraction decomposition first, then massage the thing into something I can expand in terms of $\displaystyle z-2$, or $\displaystyle \frac{1}{z-2}$, respectively.

    For $\displaystyle 0<|z-2|<2$:

    $\displaystyle \frac{2}{z(z-2)}=\frac{1}{z-2}-\frac{1}{z}=\frac{1}{z-2}-\frac{1}{2+(z-2)}=\frac{1}{z-2}-\frac{1}{2}\cdot\frac{1}{1+\frac{z-2}{2}}$

    which gives
    $\displaystyle =\frac{1}{z-2}-\frac{1}{2}\sum_{n=0}^\infty \left(-\frac{z-2}{2}\right)^n$

    For $\displaystyle 2<|z-2|$, on the other hand:

    $\displaystyle \frac{2}{z(z-2)}=\frac{1}{z-2}-\frac{1}{z}=\frac{1}{z-2}-\frac{1}{2+(z-2)}=\frac{1}{z-2}-\frac{1}{z-2}\cdot\frac{1}{1+\frac{2}{z-2}}$

    which gives
    $\displaystyle =\frac{1}{z-2}-\frac{1}{z-2}\cdot\sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^n$
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  4. #4
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    Thanks, I edited my post, I had 2 typos. The powers of two were supposed to be n in the series. I'll look over your working and see why it's different.

    EDIT: It seems that your answer is equivalent! Cool
    Last edited by scorpion007; May 15th 2010 at 03:43 AM.
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  5. #5
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    I have a related question though -- we found the series for |z-2|>2 and |z-2|< 2, but what about |z-2| = 2?
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  6. #6
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    Quote Originally Posted by scorpion007 View Post
    I have a related question though -- we found the series for |z-2|>2 and |z-2|< 2, but what about |z-2| = 2?
    Well, the condition that $\displaystyle |z-2|=2$ is satisfied by $\displaystyle z=0$ for which the original function has a pole. Hence you cannot expect to get a Laurent-Series expansion that converges for all such values of $\displaystyle z$.

    But it might be that one of the two series that you have already found converges for some values other than $\displaystyle z = 0$ that satisfy $\displaystyle |z-2|=2$.
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  7. #7
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    Ah, I see!

    Also, would it be correct to say that while there is a pole of order 1 for the first case (|z-2|<2), that there is an essential (unremovable) singularity for the second case? (Since we have infinitely many negative power terms)
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  8. #8
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    Quote Originally Posted by scorpion007 View Post
    Ah, I see!

    Also, would it be correct to say that while there is a pole of order 1 for the first case (|z-2|<2), that there is an essential (unremovable) singularity for the second case? (Since we have infinitely many negative power terms)
    Let me put it this way: you can never have an essential singularity inside the disc of convergence of a power series, or inside the annulus of convergence of a Laurent series.

    But you can (and typically do have such a singularity) right on the boundrary of that disc / that annulus of convergence. Stronger: what limits the size of the disc / annulus of convergence is the existence of such a singularity on its boundary.

    Convergence or divergence of those power / Laurent series expansions for points that lie right on the boundary of the disc / annulus of convergence is a separate problem that might need some further trickery to decide.
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