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

  1. Apr 18, 2007 #1
    1. The problem statement, all variables and given/known data

    f(z) = sin(z)/[cos(z^3)-1]

    Show that z=0 is a non-essential singularity with an ORD(f:0)=-5 and determine the co-eff. a_-1

    2. Relevant equations



    3. The attempt at a solution

    So, after expanding the power series z^5f(z) I showed that h(0) = 2, ie removable and has order -5. My question comes when I'm finding the co-eff. I cant seem to split the function into explicit h(1/z) + g(z) form so its somewhat hard to find the a_-1 coeff. Any tips would be appreciated.

    My power expansion thus far: (Sigma is always from n=0 to infinity for this)

    f(z) = Sigma[(-1)^n/(2n+1)! * z^(2n+1)] / Sigma[(-1)^(n+1)/(2n+2)! * z^(6n+6) ]

    How do I get a_-1 from this ?!:surprised

    Alright, I think I may have gotten it:

    Just written out the series goes something like:

    [z-z^3/3!+z^5/5!-z^7/7!.....] / [-z^6/2!+z^12/4!.......]

    The pattern I saw for the co-eff's of 1/z is as follows, tell me if you see the same thing :P.

    -Sigma[(3n+5)!/(2n+2)!]
     
    Last edited: Apr 18, 2007
  2. jcsd
  3. Apr 18, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I have no idea what you are trying to do with the 'Sigmas' - but I don't think you have to do it. Just work with the explicit series you've written out. Convert the factorials to numbers. Now try to write out the first few terms of the quotient series. Hint: start by multiplying top and bottom by -2/z^6. If you keep enough terms you'll find one of the form c/z.
     
  4. Apr 18, 2007 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    BTW, it's not a removable singularity. It's a pole of order 5.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Laurent Series
  1. Laurent series (Replies: 2)

  2. Laurent series (Replies: 12)

  3. Laurent Series (Replies: 2)

  4. Laurent series (Replies: 2)

  5. Laurent series (Replies: 6)

Loading...