Results 1 to 3 of 3

Thread: Laurent Series Problem

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Laurent Series Problem

    Hi

    Can someone tell me what are the steps in finding the Laurent series expansion of $\displaystyle f(z) = \frac{1}{z(z-1)^2}$ about z=0 and z=1

    i found the binomial expansion when z=0 is $\displaystyle \frac{1}{z}-2-3z-4z^2$

    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    Re: Laurent Series Problem

    Quote Originally Posted by Paymemoney View Post
    i found the binomial expansion when z=0 is $\displaystyle \frac{1}{z}-2-3z-4z^2$
    That is not correct. Consider $\displaystyle \frac{1}{1-z}=\sum_{n=0}^{+\infty}z^n$ and take derivatives in both sides. Around $\displaystyle z=1$ , use the substitution $\displaystyle u=z-1$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Re: Laurent Series Problem

    after doing the problem again i got $\displaystyle \frac{1}{z}-2+3z-4z^2$, however answers are $\displaystyle \frac{-1}{z}-2-3z-4z^2$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laurent Series/ Laurent Series Expansion
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 5th 2010, 08:41 PM
  2. [SOLVED] please help me with this Laurent series :D
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Sep 11th 2010, 11:26 PM
  3. laurent series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 7th 2010, 03:43 AM
  4. Laurent Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 27th 2009, 07:38 PM
  5. Laurent Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 30th 2007, 12:48 PM

Search Tags


/mathhelpforum @mathhelpforum