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Math Help - Laurant Series

  1. #1
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    Laurant Series

    Forgive me if this takes a while to post, attempting to get the LaTeX right.

    Attempting to find a Laurent series expansion for the function:

    f (z)=\frac{4z^2}{1-z^2}

    Over 2 regions:

    0<|z|<1

    1<|z|< {\infty}

    I have seen similar problems like:

    f(z)=\frac{z+2}{z-2} where you can simplify by taking

    f(z)=\frac{z+(-2+2)+2}{z-2}=\frac{z-2}{z-2}+\frac{4}{z-2}=1+\frac{2}{\frac{z}{2}-1}=1-\frac{2}{1-\frac{z}{2}}

    Which is solved here

    I can't seem to find a way to simplify my equation in a similar manner.

    I know that (obviously):

    f(z)=\frac{4z^2}{(1+z)(1-z)}

    Not sure where to go from there. If I knew a Maclaurin series for something like \frac{z}{1+z} and \frac{z}{1-z} or how to find one, that would make things much simpler as I could just multiply the series together...?

    Thanks in advance to anyone who read all that.
    Last edited by Negativ; April 23rd 2010 at 02:49 PM.
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  2. #2
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    Quote Originally Posted by Negativ View Post
    Forgive me if this takes a while to post, attempting to get the LaTeX right.

    Attempting to find a Laurant series expansion for the function:

    f (z)=\frac{4z^2}{1-z^2}

    Over 2 regions:

    0<|z|<1

    1<|z|< {\infty}
    Consider the following: for |z|<1, we have |z^2|<1 hence \frac{1}{1-z^2}=\sum_{n=0}^\infty z^{2k} and \frac{z}{1-z^2}=\sum_{n=0}^\infty z^{2k+1}=z+z^3+z^5+\cdots. I let you consider the other case.

    By the way it is Laurent series, with an "e"!
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  3. #3
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    I can see, judging by your username, that I was wrong. Haha. Thanks for the reply. I'll see what I can work out.
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  4. #4
    MHF Contributor chisigma's Avatar
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    For |z|>1 is...

    f(z)= \frac{4\cdot z^{2}}{1-z^{2}} = \frac{-4}{1-(\frac{1}{z})^{2}} = -4 - \frac{4}{z^{2}} - \frac{4}{z^{4}} - \dots (1)

    For |z|<1 is...

    f(z)= \frac{4\cdot z^{2}}{1-z^{2}} = 4\cdot z^{2} \cdot (1 + z^{2} + z^{4} + \dots) (2)

    Kind regards

    \chi \sigma
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