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Math Help - Help with Laurent series

  1. #1
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    Help with Laurent series

    Hi all.

    So, I've got my complex analysis exam coming up on Tuesday, and I am having a bit of difficulty with finding Laurent series (among other things too )

    Could someone give me a quick tutorial on how to do it, and possibly an example or two

    Thanks in advanced for the help
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  2. #2
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    Write what as a Laurent series? How you do that depends strongly on the individual series. For an analytic function, of course, the Laurent series is just its Taylor series- there are no negative powers of x. A lot of the time, if you have some analytic function divided by powers of x, you can write the Laurent series for the whole function by starting with the Taylor series for the analytic "part" and then dividing each term by the the powers of x.

    For example, if f(x)= \frac{e^x}{x^2}, start with the Taylor series e^x= \sum_{n=0}^\infty \frac{1}{n!}x^n and divide each term by x^2: \sum_{n=0}^\infty \frac{1}{n!}x^{n-2} which is the same (letting k= n-2) as \sum_{k=-2}^\infty \frac{1}{(k+2)!}x^k.

    On the other hand, if you have other functions, like logarithm or tangent, that have worse singularities than just powers of x, the Laurent series is much, much harder to find.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Write what as a Laurent series? How you do that depends strongly on the individual series. For an analytic function, of course, the Laurent series is just its Taylor series- there are no negative powers of x. A lot of the time, if you have some analytic function divided by powers of x, you can write the Laurent series for the whole function by starting with the Taylor series for the analytic "part" and then dividing each term by the the powers of x.

    For example, if f(x)= \frac{e^x}{x^2}, start with the Taylor series e^x= \sum_{n=0}^\infty \frac{1}{n!}x^n and divide each term by x^2: \sum_{n=0}^\infty \frac{1}{n!}x^{n-2} which is the same (letting k= n-2) as \sum_{k=-2}^\infty \frac{1}{(k+2)!}x^k.

    On the other hand, if you have other functions, like logarithm or tangent, that have worse singularities than just powers of x, the Laurent series is much, much harder to find.
    OK, so what if we have something along the lines of {z-1}/{z+1}? How would I go about doing that?

    Also how do I determine the radius of convergence and domain of a given series?
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