1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laurent Series Expansion Centered on z=1

  1. Jul 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Find a Laurent Series of f(z)=[itex]\frac{1}{(2z-1)(z-3)}[/itex] about the point z=1 in the annular domain [itex]\frac{1}{2}<|z-1|<2[/itex].

    2. Relevant equations

    3. The attempt at a solution
    By partial fraction decomposition, [itex]f(z)=\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\frac{1}{z-3}-\frac{2}{5}\frac{1}{2z-1}[/itex].

    When [itex]\frac{1}{2}<|z-1|<2, \frac{(z-1)}{2}<1[/itex].

    Hence: [itex]\frac{1}{z-3}=-\frac{1}{3-z}=-\frac{1}{3-z+1-1}=-\frac{1}{2-(z-1)}=-\frac{1}{2}\frac{1}{1-\frac{(z-1)}{2}}[/itex]

    Therefore: [itex]\frac{1}{z-3}=-\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}[/itex]

    And: [itex]f(z)=-\frac{1}{5}\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}-\frac{2}{5(2z-1)}.[/itex]

    Is this complete? Is there something I could similarly do with the other term? Ive tried but havent been able to find anything that works.
     
    Last edited: Jul 16, 2012
  2. jcsd
  3. Jul 16, 2012 #2
    I would try explicitly calculating a few terms of the taylor series of the second term and try seeing a pattern.
     
  4. Jul 16, 2012 #3
    It's this, right? [itex]\frac{1}{2z-1}=\sum^{\infty}_{n=0}(-1)^{n+1}(z-1)^{n}*2^{n}[/itex]

    So the expansion would become: [itex]f(z)=-\frac{1}{5}\sum^{\infty}_{n=0}(z-1)^{n}[(-2)^{n+1}-2^{-n-1}][/itex]
     
  5. Jul 16, 2012 #4
    I'm not too sure about the n+1 on top of the -1. I'm also not too sure whether there should be a factorial there somewhere.
     
  6. Jul 16, 2012 #5
    You're right about the -1. It should be raise to n, not n+1. I mistakingly thought the index started at 1 instead of 0.
    I'm pretty sure the coefficient of the (z-1) term is a multiple of 2 in this case.
     
  7. Jul 16, 2012 #6
    You're correct, sorry for that. I was forgetting to divide by n! in the formula for a_n of a taylor series.
     
  8. Jul 16, 2012 #7
    No biggie. So is the last expression the answer then?
     
  9. Jul 16, 2012 #8
    I still think there are a couple of sign and index errors. Try going through your algebra again to find them (or to prove me wrong).
     
  10. Jul 16, 2012 #9
    I think I figured the other fraction out:

    [itex]\frac{1}{2}<|z−1|<2,\frac{1}{2(z-1)}<1.[/itex]

    Hence: [itex]\frac{1}{2z-1}=\frac{1}{2z-1-1+1}=\frac{1}{2(z-1)+1}=\frac{1}{2(z-1)}\frac{1}{1+\frac{1}{2(z-1)}}=\frac{1}{2(z-1)}\frac{1}{1-(-\frac{1}{2(z-1)})}=\frac{1}{2(z-1)}\sum^{\infty}_{n=0}(-1)^{n}[2(z-1)]^{-n}=\sum^{\infty}_{n=0}(-1)^{n}[2(z-1)]^{-n-1}[/itex]

    That looks much better. I hope.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laurent Series Expansion Centered on z=1
Loading...