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Laurent series of exp(1/z)/(z-1)

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the Laurent series of f(z) = exp(1/z)/(z-1) around 0, and find Res{f(z), 0}.


    2. Relevant equations



    3. The attempt at a solution
    To find the Laurent series, I wrote exp(1/z) = [itex]\sum_{n=0}^∞[/itex] z-n/n!, and 1/(z-1) = -[itex]\sum_{n=0}^∞[/itex] zn.

    Then, using Cauchy's product, and rearranging some terms, f(z) = -[itex]\sum_{n=0}^∞[/itex] z-n [itex]\sum_{k=0}^n[/itex] z2k/(n-k)!

    But I am unable to express this as a proper power series, with a closed expression for each coefficient, and thus find c-1.

    Since I'm unable to find c-1 directly, I found that Res(f, 1) = e, and Res(f, ∞) = -1, from which Res(f, 0) = 1-e. Is this correct?

    How do I find the correct Laurent series?
     
  2. jcsd
  3. Dec 15, 2011 #2

    vela

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    When you multiply the two series together, you get
    [tex]-\sum_{j=0}^\infty \frac{z^{-j}}{j!} \sum_{k=0}^\infty z^k = -\sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{k-j}}{j!}[/tex]You want to collect terms, so you need to find which terms satisfy n=k-j. It may help to picture points on the jk-plane and consider what n=k-j geometrically corresponds to.
     
  4. Dec 15, 2011 #3
    I'm not sure if I understand what you are saying. Could you provide some example? I don't find it intuitive since I've never worked with double sums that way.

    I was trying to use cauchy's product, but was unable to find a general expression for c_n.
     
  5. Dec 15, 2011 #4

    vela

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    The contributions to the coefficient to zn are from the terms that satisfy k-j = n. I'm just suggesting a way you can pictorially see exactly which terms from the product will contribute to the zn term.
     
  6. Dec 16, 2011 #5


    That's correct.

    How about though, I compute that using the power series and then you figure out how to do the rest?

    So when you form the Cauchy product, you can do it two ways. I'll write it as:

    [tex]f(z)=-\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{1}{(n-k)!} z^{2k-n}=-\sum_{n=-\infty}^{\infty}a_n z^n[/tex]

    right since the expression 2k-n will range from -infty to infty and also keep in mind the function has an essential singularity so we'd expect to have an infinite singular term in the sum. So that's the series at zero which means each coefficient [itex]a_n[/itex] is the sum of an infinite set of terms right? Just do a few manually ok, say let n=0, 1, 2,3 and see how it looks. Now if I wanted to find the coefficient for say the [itex]\frac{1}{z}[/itex] term, then that must mean [itex]2k-n=-1[/itex] right or [itex]n=2k+1[/itex] and keep in mind both n and k are non-negative integers when you do the others but in this case, k can range from zero to infinity. Now the inner sum, as n goes to infinity, k goes to infinity as well so substituting n=2k+1 into that inner sum and letting it range from 0 to infinity, I could write for the 1/z term:

    [tex]a_{-1}=-\sum_{k=0}^{\infty} \frac{1}{(2k+1-k)!}=-\sum_{k=0}^{\infty}\frac{1}{(k+1)!}=-(e-1)=1-e[/tex]

    Ok, now how about you figure out what the sum has to be for say the z or z^(-2) term? Also, try and look into checking your work in Mathematica:

    Code (Text):

    In[17]:=
    r = 0.75;
    i1 = (1/(2*Pi*I))*NIntegrate[(Exp[1/z]/(z - 1))*r*I*Exp[I*t] /. z -> r*Exp[I*t], {t, 0, 2*Pi}]

    Out[18]=
    -1.7182818284624202 - 6.106664988409952*^-13*I
     
    so if I did compute the a_1 term, how would I have to change that code to verify it?
     
    Last edited: Dec 16, 2011
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