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Laurent Series-Finding Contour Integrals

  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫f(z)dz around the unit circle where f(z) is given by the following:

    a) [itex]\frac{e^{z}}{z^{3}}[/itex]
    b) [itex]\frac{1}{z^{2}sinz}[/itex]
    c) [itex] tanh(z) [/itex]
    d) [itex]\frac{1}{cos2z}[/itex]
    e) [itex]e^{\frac{1}{z}}[/itex]

    2. Relevant equations
    This is the chapter on Laurent Series, so I'm pretty sure:
    [itex]C_{n}=\frac{1}{2πi}\oint\frac{f(z)}{(z-z_{0})^{n+1}}dz[/itex]
    Is importnat

    3. The attempt at a solution
    My teacher has also given the hint to 'isolate the singulatiry' and expand the remaining function. For example: [itex]\frac{1}{z^{2}sinz}=\frac{1}{z^{3}}\frac{z}{sinz}[/itex]
    Then I expand [itex]\frac{z}{sinz}=\frac{z}{\sum\frac{(-1)^{j}}{(2j+1)!}z^{2j+1}}[/itex]


    My problem is that I'm not sure what to do from here. I'm having trouble reciprocating the sum. I want to use the binomial theorem but am not sure how to apply it for an infinite sum. I also don't know what to do once I get it in summation notation.
    Could I use uniform convergence to swap summation and integration and note that the only contribution to the sum is the integral with [itex]\frac{1}{z}[/itex] by Cauchy's integral formula??
    [itex]f^{(k)}(z_{0})=\frac{k!}{2πi}\oint\frac{f(z)}{(z-z_{0})^{n+1}}dz[/itex]
     
  2. jcsd
  3. Oct 21, 2012 #2
    This is no good; you should use a regular Taylor expansion on sin(z)/z instead. Or you don't even need an expansion, all you need is the limit z→0.

    Did you manage to do the first one? Start by using the integral theorem
    [tex] f''(z_0) = \frac{2}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^3} [/tex]
    Then choose appropriate f(z), evaluate it at the point you're interested z0 and solve for the integral.
     
  4. Oct 21, 2012 #3
    I could choose f(z)=exp(z), but then I'm just using Cauchy Integral Theorem. This is the chapter on Laurent series I'm trying to figure out how Laurent series can help me evaluate this.

    And the taylor series for sinz/z is the same as its Laurent series no???? And that is a regular taylor expansion. I just don't know how to 'reciporicate' the sum. Or I don't know how to find the series for z/sinz

    What do you mean when you say "all you need is the limit as z-->0"? We haven't done residue calculus yet by the way.
     
    Last edited: Oct 21, 2012
  5. Oct 21, 2012 #4
    You can expand all those in Laurent series then use the coefficient on the 1/z term to compute the integral. For example:

    [tex]\frac{1}{z^2} \frac{1}{\sin(z)}[/tex]

    use long division. Divide [itex]z-z^3/6+z^5/120+\cdots[/itex] into 1. You can do that right? Lemme try and find the latex for that . . .
     
  6. Oct 21, 2012 #5
    This part makes sense.

    ____
    Still a little unclear on this. I just wrote (z-z^3/6+z^5/120+...)| 1

    And I don't see how to compute this. I'll keep trying.

    Thanks for the help!
     
  7. Oct 21, 2012 #6
    Best if I show you what it looks like but unfortunately the Latex code doesn't seem to work. I'll start a thread down there where you ask latex questions to see if I can get it right.
     
  8. Oct 21, 2012 #7
    You can also deal with it using geometric series after writing it into: [itex] \frac{z}{\sin(z)} = \frac{1}{1-z^2/6 + ...} [/itex]
     
  9. Oct 21, 2012 #8
    I don't really see how geometric series would help. That would give a mess: I'd be a sum of a series plus a series squared plus...
    yuck
    what I need is the method of long division for an infinite series
     
  10. Oct 22, 2012 #9
    So the trick here is that you only need to extract one number. You don't need to care what the actual series is, just that single number.
     
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