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!

Contour integral, taylor and residue theory question

  1. Aug 17, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img243.imageshack.us/img243/4339/69855059.jpg [Broken]
    I can't seem to get far. It makes use of the Exponentional Taylor Series:

    2. Relevant equations

    http://img31.imageshack.us/img31/6163/37267605.jpg [Broken]


    3. The attempt at a solution
    taylor series expansions for cos and 1/z i assume - and stick it into the residual theorem, but i need to get it into the laurent series form first so that i can find out b1. I think it's like pole order 4 or 5 so it's going to be a pain
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 18, 2009 #2
    my attempt at a solution
    is this on the right track?
    http://img219.imageshack.us/img219/2425/image123t.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Aug 18, 2009 #3

    Cyosis

    User Avatar
    Homework Helper

    It has been a while but it does look correct to me. You can do it a little bit easier though if you know the series expansion of the cosine.

    [tex]
    z^5 \cos(\frac{1}{z})=z^5 \sum_{k=0}^{\infty} (-1)^k \frac{\left(\frac{1}{z}\right)^{2k}}{(2k)!}=...+\frac{-1}{6!z}+...
    [/tex]
     
    Last edited: Aug 18, 2009
  5. Aug 18, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why is the Taylor's series for ez a "relevant" equation? Use the Taylor's series for cos(z), replace z by 1/z so that you get a power series in negative powers of z (a "Laurent series") and then multiply by [itex]z^5[/itex] so you have a power series with highest power z5 and decreasing. It should be easy to see that, integrating term by term, every term gives 0 except the z-1 term.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook