Results 1 to 2 of 2

Math Help - Laurent Series

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    1

    Laurent Series

    Hi guys,

    I need help with a problem. we need to find the Laurent series in this problem.

    we have the partial fraction

    f(z) = 4/z((1/1+z) + (1/2-z). Now it says, for 0 < |z| < 1, we expand each of the fractions in the parenthesis in powers of z. This gives:

    f(z) = -3 + 9z/2 - 15z^2/4 + 33z^3/8 + ...... + 6/z

    can you please tell me how are we getting this expansion?? i really need your help. i have a final coming up.

    thanks a lot for your support.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by sonaiarko View Post
    Hi guys,

    I need help with a problem. we need to find the Laurent series in this problem.

    we have the partial fraction

    f(z) = 4/z((1/1+z) + (1/2-z). Now it says, for 0 < |z| < 1, we expand each of the fractions in the parenthesis in powers of z. This gives:

    f(z) = -3 + 9z/2 - 15z^2/4 + 33z^3/8 + ...... + 6/z

    can you please tell me how are we getting this expansion?? i really need your help. i have a final coming up.

    thanks a lot for your support.
    Use a geometric series expansion like this

    \frac{4}{z}\left[\frac{1}{1+z}+\frac{1}{2-z}\right]=\frac{4}{z}\left[\sum_{k=0}^\infty (-1)^k z^k+\frac{1}{2}\sum_{k=0}^\infty \left(\frac{z}{2}\right)^k\right]

    =\sum_{k=0}^\infty 4\left[(-1)^k+\frac{1}{2^{k+1}}\right]z^{k-1}=\sum_{k=-1}^\infty 4\left[(-1)^{k+1}+\frac{1}{2^{k+2}}\right]z^k
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laurent Series
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 17th 2011, 01:46 AM
  2. Laurent Series/ Laurent Series Expansion
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 5th 2010, 09:41 PM
  3. [SOLVED] please help me with this Laurent series :D
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 12th 2010, 12:26 AM
  4. Laurent series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 13th 2009, 01:10 PM
  5. [SOLVED] Help! Laurent series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 5th 2007, 09:29 AM

Search Tags


/mathhelpforum @mathhelpforum