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!

Complex Analysis - Radius of convergence of a Taylor series

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the radius of convergence of the Taylor series at 0 of this function

    f(z) = [itex]\frac{e^{z}}{2cosz-1}[/itex]


    2. Relevant equations



    3. The attempt at a solution
    Hi everyone,

    Here's what I've done so far:

    First, I tried to re-write it as a Laurent series to find where the closest singularity to 0 is.

    e^z = Ʃ(x^n)/n!

    cos z = Ʃ (-1)^n . x^2n / (2n)!

    However, I'm a little unsure how to combine these into a single laurent series. Is this even going about the problem in the correct way?

    Thanks
     
  2. jcsd
  3. May 18, 2012 #2
    Actually, I have had a wave of inspiration since - is this correct?

    The singularities occur for 2.cos(z)-1 = 0 i.e. cos(z) = 1/2

    This happens for z = pi/3 (+ 2k.pi, but this z is the smallest one)

    So then

    the distance from z=1 to z=pi/3 is:

    √(1 - pi/3)^2) = 2pi/3

    which is then the radius of convergence?
     
  4. May 18, 2012 #3
    If you're given the function, then the radius of convergence is the distance to the nearest singular point. That's different if you're only given the series and you don't know what function the series represents. In that second case, you have to compute the radius of convergence. So you got the function, from zero, how far is the nearest singular point?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex Analysis - Radius of convergence of a Taylor series
Loading...