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Laurent series, deep confusion

  1. Nov 14, 2007 #1
    I'm so lost. I can follow the steps in the examples in the book, but all of the examples seem easier than this problem-- and in each case there is a point where you "spot" a similarity to the geometric series or something and then you can just patch it in... But, I just don't know how to make that happen here.

    Given the function [tex]f(z)=\frac{z}{a^{2}-z^{2}}[/tex] Expand f(z) in a laurent series in the regions: |z| < a and |z| > a.

    From another problem I know that:

    [tex]\frac{1}{1-z}=\displaystyle\sum_{n=0}^{\infty}z^{n}[/tex], |z| < 1 so:

    [tex]\frac{1}{a-z}=\displaystyle\sum_{n=0}^{\infty}z^{n}[/tex], |z| < a (right?)

    Now I want to use this fact so I write:

    [tex]f(z)=\frac{z}{(a+z)(a-z)}=\frac{z}{(a+z)}\left(\displaystyle\sum_{n=0}^{\infty}z^{n}\right)[/tex]

    um... and I don't know what to do next. I've looked at the example in the book and they used -1/z to repace z when working with [tex]\frac{z}{(a+z)}[/tex].

    But, if I do that it changes the boundaries for the whole problem. I tried writing:

    [tex]=\left(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{z^{n+1}}\right)\left(\displaystyle\sum_{n=0}^{\infty}z^{n}\right)[/tex]

    for my next step, but I know this is wrong. I can only do that replacement if I change the boundary... I'm sort of confused about how to go about doing these problems. Help?
     
    Last edited: Nov 14, 2007
  2. jcsd
  3. Nov 14, 2007 #2

    HallsofIvy

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    Why not use "partial fractions"? That is, write
    [tex]\frac{z}{a^2- z^2}= \frac{z}{(a-z)(a+z)= \frac{A}{a- z}+ \frac{B}{a+z}[/itex]
    Then write the two separate fractions as power series in z using the fact that
    [itex]\frac{1}{1- z}= \sum_{n=0}^{\infty}z^n[/itex]
    and add corresponding terms of the two series.
     
  4. Nov 14, 2007 #3
    Okay, I have

    [tex]\frac{z}{(a+z)(a-z)}=\frac{-\frac{1}{2}}{a+z}+\frac{\frac{1}{2}}{a-z}[/tex]

    Then,

    [tex]=\frac{1}{2}\left(\displaystyle\sum_{n=0}^{\infty}z^{n}-\displaystyle\sum_{n=0}^{\infty}(-1)^{n}z^{n}\right)[/tex]

    [tex]=\displaystyle\sum_{n=0}^{\infty}z^{2n+1}[/tex]

    for |z| <a

    okay, but what about |z|> a?

    Something about this makes no sense... where are the negative powers?
     
  5. Nov 14, 2007 #4
    It is important to consider the cases of [itex] |z|<a \text{ and } |z|>a [/itex] seperately.

    Let us first consider [itex] |z|>a[/itex].

    We want to be able to use the fact that

    [tex]\frac{1}{1-x} = \displaystyle\sum_{n=0}^{\infty}x^{n}[/tex]

    Thus factor a z out of the addition term to get

    [tex]\frac{1}{z\left( \frac{a}{z} +1 \right) } [/tex]

    Now since [itex] |z|>a[/itex], then we know that the sum will converge, and so we can proceed as before.

    For the record, it is not necessary to use partial fractions for either case.

    [tex]f(z)=\frac{z}{(a^2-z^2)}=\frac{z}{a^2 \left( 1- \frac{z^2}{a^2} \right) } [/tex]
     
    Last edited: Nov 14, 2007
  6. Nov 14, 2007 #5
    You'll note that using

    [tex]f(z)=\frac{z}{(a^2-z^2)}=\frac{z}{a^2 \left( 1- \frac{z^2}{a^2} \right) } [/tex]

    is a much faster way of getting your solution. Furthermore, when you get your closed solution, who says that the index variable must be a positive integer?
     
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