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Laurent series - where does f(z) need to be holomorphic?

  1. May 27, 2013 #1

    BruceW

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    Hey everyone,
    I've got a question that might be interesting to those of you that enjoy maths. Note: I only did a physics degree, so I have never really done analysis in the proper way a mathematician has. But I am eager to learn about more rigorous maths.
    Right, so for a Laurent series about a point c, we have:
    [tex]f_{(c)} = \sum_{n=-\infty}^{\infty} \ \frac{(z-c)^n}{2\pi i} \ \oint \frac{f_{(z)}}{(z-c)^{n+1}} \ dz[/tex]
    And from looking at wikipedia, they say that f(z) only needs to be holomorphic in an annulus centred on c, which contains the curve I integrate over. So in other words, this is very useful, because even if there is a singularity at c, we can still get the Laurent series about point c.
    Also, we have the Cauchy integral formula:
    [tex]f_{(c)}^{(n)} = \frac{n!}{2\pi i} \oint \ \frac{f_{(z)}}{(z-c)^{n+1}} \ dz[/tex]
    And wikipedia seems to say that in this case, f(z) needs to be holomorphic for the equation to work. (i.e. can't be a singularity at c). Interestingly, the equation for the Laurent series seems to contain the Cauchy integral formula, and if we make the substitution, then the Laurent series now looks like this:
    [tex]f_{(c)} = \sum_{n=-\infty}^{\infty} \ \frac{(z-c)^n}{n!} \ f_{(c)}^{(n)} [/tex]
    Which makes sense to me, since it looks a bit like the Taylor series, but with negative coefficients also included. But now comes my problem. The Cauchy integral formula needs f(z) holomorphic at c, but the Laurent series does not need f(z) holomorphic at c. So for my final equation (the one that looks like a Taylor series with negative coefficients), does this require f(z) to be holomorphic at c ? (since I have made use of the Cauchy integral formula, which does require f(z) holomorphic at c).

    I hope I have made it clear what my question is. Please ask me if I have not explained what I mean clearly enough. Also, if the answer to my question is yes, then I guess this means that the Laurent series is 'more general' than the final equation which I wrote? Thank-you in advance!

    edit: I guess that if f(z) is not holomorphic at c, then we can't define the n'th differential at c. Therefore, the Cauchy integral formula clearly cannot work if there is a singularity at c. Is this basically the right way to think about it? I feel like there might be a deeper reason.

    second edit: really, I was hoping that the 1st equation can be derived from the 3rd equation. But now I am starting to think that it is actually the other way around?
     
    Last edited: May 27, 2013
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  3. May 27, 2013 #2

    micromass

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    I think you already identified your problem. There is no way to define ##f^{(n)}(c)## unless ##f## is defined at ##c## and is holomorphic there (and thus actually has ##n##th derivatives). So yes, your third equation would require ##f## to be holomorphic.

    The Laurent series is certainly more general than the Taylor series. In fact, the integrals

    [tex]\frac{n!}{2\pi i}\oint \frac{f(z)}{(z-c)^n}dz[/tex]

    should be seen as generalizations of the ordinary derivative. The Cauchy integral formula then shows that this generalization coincides with the ordinary derivative in the usual cases (that is: if the anulus has the form ##B(c,\varepsilon)\setminus \{c\}## and ##c## is removable singularity). In fact, the above integral makes sense too for all ##n\in \mathbb{R}## instead of just integers. This induces the notion of a fractional derivative.
     
  4. May 28, 2013 #3

    BruceW

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    ah, cool. that makes sense. I'll remember that about fractional derivatives, that seems pretty interesting too. thanks for the reply :)
     
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