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Something wrong with a proof relating to Laurent coefficients

  1. Oct 8, 2013 #1
    Something wrong with a "proof" relating to Laurent coefficients

    1. The problem statement, all variables and given/known data
    Let f(z) have an isolated singularity at z_0 and suppose that f(z) is bounded in some punctured neighborhood of z_0. Prove directly from the integral formula for the Laurent coefficients that a_(-j) = 0 for all j > 0. That is, f(z) must have a removable singularity at z_0

    3. The attempt at a solution
    Right I know one correct solution but I can't find my fault in this one:

    [tex]a_j = \frac{1}{2\pi i}\int \frac{f(z)}{(z-z_0)^{j+1}}[/tex]

    As f(z) is bounded let M be the maximum value of f(z) in the punctured disk.

    [tex]|\frac{1}{2\pi i}\int \frac{f(z)}{(z-z_0)^{j+1}}| \leq |\frac{M}{2\pi i} \int \frac{1}{(z-z_0)^{j+1}} |[/tex]

    But I also know (except I suspect this is where problems arise) that

    [tex]\int \frac{1}{(z-z_0)^n}dz [/tex]

    Around any circle centered at z_0 will have the value 2∏i if n = 1, and 0 otherwise. That would imply the only Laurent coefficient that isn't 0 is a_0 but I'm pretty sure that's false.
    I suspected that the last 'formula' is only valid simply connected domains, which I still believe to be the case (can be derived using Cauchy's integral formula for example).

    Now here's my problem. I got to the chapter Residue theory and in the first page they, it seems to be, does exactly what I did above. I'll quote

    Now we are applying the formula to a Laurent series expansion and as far as I understand the Laurent expansion is defined for an annulus around the singularity. Thus it's not a simply connected domain. But if I can use the formula on a domain that isn't simply connected how come my "proof" is false?

    Sorry, it was a bit long winded but explained the issue as thoroughly as I could. Any input would be appreciated, really at a loss here.

    Edit: apparently \bigg{|} won't show at physicsforums?
     
  2. jcsd
  3. Oct 9, 2013 #2
    Besides that f(z) has a singularity at ##z_0## is it analytic everywhere in the punctured disk?

    Your inequality is certainly correct, and it is further true that the Laurent expansion of f(z) = 1 will have only an ##a_0##.

    But this has little to do with ##\oint \frac{f(z)}{(z-z_0)^n}dz##

    You have to use the boundedness of f somewhere, but I don't think that is the place.
     
  4. Oct 10, 2013 #3
    I presumed it was implied f(z) was analytic in the whole neighborhood of z_0, and my disk is just a disk in the neighborhood. The correct proof utilizes the boundedness |f(z)|<M for some |z-z_0|<p<R, with C = |z-z_0|, then you use the ML inequality letting p -> 0 which shows a_(-j) = 0 for all positive j.

    Still, if my inequality was correct then my faulty result would be too. Or there's something else I'm misunderstanding. I'm using an altered version of the ML inequality which intuitively makes sense to me, but that's a trap I've fallen into before.

    The key would be "but this has little to do with the integral", could you explain how? Pretty sure that's the step I'm messing up on.
     
  5. Oct 10, 2013 #4
    Yes, I kind of assumed f would be analytic; just checking.

    You are integrating f(z) = 1, so the only Laurent coefficient is ##a_0## = 1, which is as we hope.

    You may conclude from this that ##| \oint \frac{f(z)}{z-z_0}dz | < M##, but what does that tell you about the Laurent coefficients of f? For example, suppose f(z) = ##z-z_0##, certainly bounded in an annulus around ##z_0##. We have ##a_1## = 1 and all the other coefficients are 0.

    How would the observation that ##| \oint \frac{f(z)}{z-z_0}dz | < M## tell you that ##a_1## = 1 and all the other coefficients are 0?
     
  6. Oct 10, 2013 #5
    Ah, I see now. I simply stopped looking at the function I wanted a Laurent coefficient of when I played around with the inequality, 100% without realizing. I'd say silly me but honestly the thought didn't even strike me until you pointed it out.

    I assume you meant a_0 everywhere after the second line as well? If so I'd say I can apply the same idea but now the absolute value of the integral is simply 0 and I ended up getting the Laurent series for f(z) = 1 in a slightly awkward fashion.

    Many thanks for your help, been puzzling me for days.
     
  7. Oct 10, 2013 #6
    Can't tell you how many times I have done the same.

    For f(z) = ##z-z_0## the ##a_0## coefficient is -##z_0## and the ##a_1## coefficient is 1 (no integration required). Probably was just thinking f(z) = z when I implied ##a_0## = 0 .
     
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