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Comples analysis - Radius of convergence of a Taylor series question

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the radius of convergence of the Taylor series at z = 1 of the function:

    [itex]\frac{1}{e^{z}-1}[/itex]


    2. Relevant equations



    3. The attempt at a solution
    Hi everyone,

    Here's what I've done so far.

    Multiply top and bottom by minus 1 to get:

    -1/(1-e^z)

    And then this is the expansion of an infinite geometric series:

    1/(1-z) = Ʃ(z^n)

    Then use the formula 1/R = lim sup (nth root of absolute value of a_n)

    where here a_n is constant, 1.

    So then you will get that the lim sup of this is 1.

    So then R = 1.


    However, I don't like this solution as I think that geometric formula only applies if |z|<1, which is not the case for e^z.
    I am a bit stuck as to how else to write out the Taylor series though. And how do I make use of the fact z=1? Does that mean I have to write out a Taylor series in the form Ʃa_n.(z-1)^n?

    I'd really appreciate if someone could please point me in the right direction.

    Thanks
     
  2. jcsd
  3. May 17, 2012 #2
    You want to find the radius of convergence of the series of the form [itex]\sum_{n=0}^{\infty} a_n (z-1)^n[/itex], yes. Don't bother looking at the series itself though. Just use the fact that the radius of convergence of a Taylor series is just the distance from its center to the nearest singularity. Where are the singularities of [itex]\frac{1}{e^z - 1}[/itex]?
     
  4. May 17, 2012 #3
    Thanks for your reply!

    Ah. So the singularity would be at z=0 then?
    So the distance from z=0 to z=1 would be 1, so the radius of convergence is then R=1 centered at the point z=1?
     
  5. May 17, 2012 #4
    Also, just out of interest, how exactly would you write that formula as a power/Laurent series?
     
  6. May 17, 2012 #5
    That's the closest singularity, yes. But for future reference, remember that there are also other singularities at [itex]2\pi ik[/itex] for every [itex]k\in \mathbb{Z}[/itex].

    Yep. You got it!

    It's easy enough to find the Taylor series of [itex]e^z - 1[/itex] directly, and then you can just use long division to get the terms of the Taylor series for [itex]\frac{1}{e^z-1}[/itex]. It's a pretty tedious calculation though.
     
  7. May 17, 2012 #6
    If you restrict values of z so that ##\left|\frac{1}{e^z}\right| < 1##,

    ##
    \begin{eqnarray*}
    \displaystyle \frac{1}{e^z - 1} &=& \left(\frac{1}{e^z}\right)\left(\frac{1}{1-\frac{1}{e^z}}\right)\\
    &=& \left(\frac{1}{e^z}\right) \cdot \sum_{n=0}^{\infty} \left(\frac{1}{e^z}\right)^n\\
    &=& \sum_{n=0}^{\infty} \left(\frac{1}{e^z}\right)^{n+1}\\
    &=& \sum_{n=0}^{\infty} (e^z)^{-n-1}\\
    &=& \sum_{n=-\infty}^{0} (e^z)^{n-1}\\
    &=& \sum_{n=-\infty}^{-1} (e^z)^n
    \end{eqnarray*}
    ##

    Edit: I just realized this isn't in power series form, you need a ##(z-a)^n##.. I'll leave this here because it's a good technique anyway. Yeah, it would be tough to put it into power series form.
     
  8. May 18, 2012 #7
    Thanks guys, I appreciate it :)
     
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