Results 1 to 4 of 4

Math Help - Laurent Series

  1. #1
    Newbie
    Joined
    Aug 2011
    Posts
    11

    Laurent Series

    Hi, I have to compute the laurent series of this function \frac{z^2+1}{(z^3+1)^2} in |z|>1.
    Sigularities are: z=-1 and z=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}.
    Now I do not know how to continue, I wrote partial fractions \frac{z^2}{(z^3+1)^2}+\frac{1}{(z^3+1)^2} and I thought for the first fraction to use the derivative method, but there is z^{3}, so can't I use that method? While for the second fraction what I have to do... some ideas?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Laurent Series

    Quote Originally Posted by quidh View Post
    Now I do not know how to continue, I wrote partial fractions \frac{z^2}{(z^3+1)^2}+\frac{1}{(z^3+1)^2} and I thought for the first fraction to use the derivative method, but there is z^{3}, so can't I use that method? While for the second fraction what I have to do... some ideas?
    Using the derivative method you'll obtain 1/(u+1)^2}=\ldots=\sum_{n=0}^{+\infty} (-1)^n(n+1)u^{-n-2}\;\;(|u|>1) . If u=z^3 then, |u|>1\Leftrightarrow |z|>1 so 1/(z^3+1)^2=\ldots=\sum_{n=0}^{+\infty} (-1)^n(n+1)z^{-3n-6}\;\;(|z|>1) and z^2/(z^3+1)^2=\ldots=\sum_{n=0}^{+\infty} (-1)^n(n+1)u^{-3n-4}\;\;(|z|>1)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2011
    Posts
    11

    Re: Laurent Series

    Thanks for help!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2011
    Posts
    38

    Re: Laurent Series

    gives you the exact expansion
    \frac{1}{4}+\frac{x^2}{4}-\frac{x^3}{4}+\frac{5 x^4}{36 \left((-2)^{1/3}-x\right)^2}-\frac{7 (-1)^{1/3} x^4}{36\ 2^{2/3} \left((-2)^{1/3}-x\right)^2}+\frac{x^5}{12 \left((-2)^{1/3}-x\right)^2}+\frac{(-1)^{2/3} x^5}{9\ 2^{1/3} \left((-2)^{1/3}-x\right)^2}+\frac{5 x^4}{36 \left(2^{1/3}+x\right)^2}+\frac{7 x^4}{36\ 2^{2/3} \left(2^{1/3}+x\right)^2}+\frac{x^5}{12 \left(2^{1/3}+x\right)^2}+\frac{x^5}{9\ 2^{1/3} \left(2^{1/3}+x\right)^2}+\frac{5 x^4}{36 \left((-1)^{2/3} 2^{1/3}+x\right)^2}+\frac{7 \left(-\frac{1}{2}\right)^{2/3} x^4}{36 \left((-1)^{2/3} 2^{1/3}+x\right)^2}+\frac{x^5}{12 \left((-1)^{2/3} 2^{1/3}+x\right)^2}-\frac{\left(-\frac{1}{2}\right)^{1/3} x^5}{9 \left((-1)^{2/3} 2^{1/3}+x\right)^2}
    bye
    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: October 5th 2010, 08:41 PM
  2. Laurent Series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 22nd 2010, 10:13 AM
  3. Laurent Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 27th 2009, 10:03 PM
  4. Laurent Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 27th 2009, 07:38 PM
  5. Laurent Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 3rd 2007, 10:02 AM

Search Tags


/mathhelpforum @mathhelpforum