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Laurent series of a logarithm

  1. Dec 18, 2011 #1
    Hello, I am having difficulty matching one term in my Laurent series to that which mathematica tells me is the correct answer. For the function

    [itex]f(z)=log\frac{1+z}{1-z}[/itex]

    we know that there exists a k such that

    [itex]Log|1+z|-Log|1-z|+i2\pi k[/itex]

    Now, we know that the Taylor series of f is as below

    [itex] f(z)=log\frac{1+z}{1-z}=\int \frac{1}{1+z}dz+\int \frac{1}{1-z} dz
    =\int \sum_{n=0}^{\infty} (-z)^n dz+\int \sum_{n=0}^{\infty} dz=2 \sum_{n=0}^{\infty}\frac{z^(2n+1)}{2n+1} [/itex]

    Now we would like to find the Laurent series. I used a similar approach resulting in

    [itex] f(z)=log\frac{1+z}{1-z}=\int \frac{1}{z}\frac{1}{1+1/z}dz-\int \frac{1}{z}\frac{1}{1-1/z} dz =2 \sum_{n=1}^{\infty} \frac{z^{-(2n+1)}}{2n+1}[/itex]

    Notice that since I used integration to find the coefficients, I was unable to determine the coefficient of the power z^-1 which according to mathematica should give me the following term in the series

    [itex] \frac{2}{z}+i\pi [/itex]

    I know that we can use the following definition to find this coefficient

    [itex] b_{-1}=\frac{1}{2 \pi i} \oint_c f(z) dz [/itex]

    However I do not know how to integrate the closed contour with a branch cut.
    Also, it seems that the branch cut we choose is important to make our Laurent
    series expansion consistent with the function, however what tells me what the
    correct branch cut is?
     
  2. jcsd
  3. Dec 18, 2011 #2
    Misunderstood the problem initially so if you saw my first one, it was wrong. Ok, you want the Laurent series for [itex]|z|>1[/itex]. When I go through the algebra by first combining both series then integrating, I get:

    [tex]-2\sum_{n=1}^{\infty}\frac{z^{1-2n}}{1-2n}[/tex]

    I'm unclear about the addition of the [itex]\pi i[/itex] term but I believe the general expression can be written as:

    [tex]f(z)=\pi i+2n\pi i-2\sum_{n=1}^{\infty}\frac{z^{1-2n}}{1-2n},\quad |z|>1[/tex]
     
    Last edited: Dec 18, 2011
  4. Dec 18, 2011 #3
    I agree with what you found for |z|>1, however where are you getting the 2nπi eqpression from? That may help me to figure out where I have gone wrong.
     
  5. Dec 19, 2011 #4
    If it's just a log(h(z)) where h(z) is single-valued, then the branches differ by [itex]2n\pi i[/itex]. I probably should not have used n though and written it as:

    [tex]f(z)=\pi i+2k\pi i-2\sum_{n=1}^{\infty}\frac{z^{1-2n}}{1-2n},\quad |z|>1[/tex]

    where k is the k'th branch and you know what, I'm starting to think that [itex]\pi i[/itex] thing is just an artifact of Mathematica because the function [itex]\displaystyle \frac{1+z}{1-z}[/itex] for [itex]|z|>1[/itex] lies entirely in the left half-plane, a linear fractional transformation, so the analytic continuation of the arg function in that range would not agree with Mathematica's "principal" arg function which is [itex]-\pi<\theta\leq \pi[/itex] and therefore to represent the analytic continuation of the function in Mathematica, we need to add the [itex]\pi i[/itex] to adjust the argument so that it is analytically continuous there. Not sure though. Anyone can clarify that for me please?
     
    Last edited: Dec 19, 2011
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