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Expand function in Laurent series

  1. Mar 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Expand ##f(z)=\frac{1}{z^2(z-1)}## in Laurent series for ##0<|z-1|<1##. Use binomial series.


    2. Relevant equations



    3. The attempt at a solution

    I am looking at this problem for quite some time now and still I got nothing.

    I do however think that this will come in handy ##\frac{1}{z^2(z-1)}=\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z-1}##. That's all I have and not even a clue on how to continue ... Please help.
     
  2. jcsd
  3. Mar 27, 2014 #2
    Update.

    Ok, I managed to find some examples http://sym.lboro.ac.uk/resources/Handout-Laurent.pdf and hopefully at least some of the things written below are ok.

    In ##f(z)=\frac{1}{z^2(z-1)}## I replace ##w=z-1## so that my open disk ##0<|z-1|=|w|<2## is centered around point ##0##.

    So ##f(w)=\frac{1}{w(w+1)^2}=\frac{1}{w}-\frac{1}{w+1}-\frac{1}{(w+1)^2}##.

    Now it makes sense to google binomial series and find out that ##\frac{1}{(1-z)^{\beta +1}}=\sum_{k=0}^{\infty}\binom{k+\beta }{k}z^k##.

    Now if I am not completely mistaken than ##\frac{1}{w}=\frac{-1}{-(1-1+w)}=-\frac{1}{1-(1+w)}=-\sum_{k=0}^{\infty}(1+w)^k##.

    And similar for ##-\frac{1}{w+1}## and ##-\frac{1}{(w+1)^2}##, which alltogether gives me ##f(w)=-\sum_{k=0}^{\infty}[(1+w)^k+(2+w)^k+(1+(1+w)^2)^k]##.

    And finally ##f(z)=-\sum_{k=0}^{\infty}[z^k+(z+1)^k+(1+(z)^2)^k]##.


    Yes or no?
     
  4. Mar 27, 2014 #3

    vela

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    No. You want to expand about z=1 or w=0. You expanded parts of the function about w=1. The series you end up with in the end should consist of powers of (z-1).

    Also, the disk is given by 0 < |z-1| < 1. If w=z-1, then you get 0 < |w| < 1, not 0 < |w| < 2.
     
  5. Mar 27, 2014 #4
    Ok, how do I do that?

    I agree, it was mistake.
     
  6. Mar 27, 2014 #5

    vela

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    For example, the term 1/w is already in the form of wn. You don't have to do anything with that term.

    Expand ##\frac{1}{1+w}## as a power series in ##w##. (You should memorize this series if you don't already know it.) Differentiate that and you'll get the series for ##{-\frac{1}{(1+w)^2}}##. Put it all together and you have the Laurent series in terms of ##w##.
     
  7. Mar 27, 2014 #6
    That's exactly how I wanted to do it, but the problem states that I have to use binomial series. So, of course my next question is what does binomial series have to do with all this? Where do I than use it if not here?
     
  8. Mar 27, 2014 #7

    vela

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    You can use the binomial theorem to expand (1+w)-1 and (1+w)-2.
     
  9. Mar 27, 2014 #8
    Remember: ##f(w)=\frac{1}{w(w+1)^2}=\frac{1}{w}-\frac{1}{w+1}-\frac{1}{(w+1)^2}##

    Hopefully this sounds a bit better:

    ##-\frac{1}{w+1}=-\frac{1}{1-(-w)}=-\sum_{k=0}^{\infty}\binom{k}{k}(-1)^kw^k=\sum_{k=0}^{\infty}(-1)^{k+1}w^k##

    and

    ##-\frac{1}{(w+1)^2}=-\frac{1}{(1-(-w))^2}=-\sum_{k=0}^{\infty}\binom{k+1}{k}(-1)^kw^k=\sum_{k=0}^{\infty}(k+1)(-1)^{k+1}w^k##
     
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