Results 1 to 8 of 8

Math Help - Another Laurent expansion

  1. #1
    Senior Member
    Joined
    Jul 2006
    Posts
    364

    Another Laurent expansion

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

    I've tried many things, but the |z-2|>0 thing gets me. Tips?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Ok, I did a substitution w=z-2 and got

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

    Does that look right?
    Last edited by scorpion007; May 15th 2010 at 03:56 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by scorpion007 View Post
    Ok, I did a substitution w=z-2 and got

    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<br /> <br />
\\ \displaystyle \frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^{\color{red}2}, &\text{for}~|z-2|>2<br />
\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 z-2, or \frac{1}{z-2}, respectively.

    For 0<|z-2|<2:

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

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

    \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
    =\frac{1}{z-2}-\frac{1}{z-2}\cdot\sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^n
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    I have a related question though -- we found the series for |z-2|>2 and |z-2|< 2, but what about |z-2| = 2?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    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 |z-2|=2 is satisfied by 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 z.

    But it might be that one of the two series that you have already found converges for some values other than z = 0 that satisfy |z-2|=2.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    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)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laurent expansion
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: August 26th 2010, 08:18 AM
  2. All Laurent series expansion around 1.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: July 14th 2010, 03:34 AM
  3. Help with a Laurent expansion
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 9th 2010, 03:48 AM
  4. Laurent series expansion
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: December 3rd 2009, 10:30 AM
  5. Laurent Expansion
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 11th 2008, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum