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Math Help - Laurent series z0=infinity

  1. #1
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    Laurent series z0=infinity

    Hi there. I have to find the Laurent series for \frac{1}{z-2},z_0=\infty

    So what I did was calling z=\frac{1}{\omega}, I don't know if this is right.

    Then:

    f(\omega)=\frac{1}{\frac{1}{\omega}-2}=\frac{-1}{2-\frac{1}{\omega}}=\frac{-1}{2}\frac{1}{1-\frac{1}{2\omega}}=\frac{-1}{2}\sum_0^{\infty}\left ( \frac{1}{2\omega} \right )^n,|\omega|>\frac{1}{2}

    And then back to z:
    f(z)=\frac{-1}{2}\sum_0^{\infty}\left ( \frac{1}{2} \right )^n z^n,|z|<2

    I have to make the series for |z|>2, but I see that it gives the same as when I centered the series in zero, so I think I'm doing something wrong.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Laurent series z0=infinity

    We have f(z)=\frac{1/z}{1-2/z}=\dfrac{1}{z}\sum_{n=0}^{\infty}\left(\frac{2}{z  }\right)^n=\sum_{n=0}^{\infty}\frac{2^n}{z^{n+1}} , expansion valid for |2/z|<1 or equivalently for 2<|z|<\infty .
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    Re: Laurent series z0=infinity

    Yes, thats the other part for the Laurent series, but is it centered at z_0=\infty? because I've made the expansion centered at zero, and it gives exactly the same it gives when centered at infinity.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Laurent series z0=infinity

    Quote Originally Posted by Ulysses View Post
    Yes, thats the other part for the Laurent series, but is it centered at z_0=\infty?
    Of course, 2<|z|<\infty is a neighborhood of \infty .

    because I've made the expansion centered at zero, and it gives exactly the same it gives when centered at infinity.
    What? I see different expansions.
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    Re: Laurent series z0=infinity

    What's the difference for z_0=0 I just don't see it. I've proceeded exactly as you did when finding the Laurent series for this same function centered at zero, and I get the same series that we've found here, with exactly the same expressions for the modulus.

    If I take z_0=0 then I get:
    f(z)=\frac{1}{z-2}=\frac{-1}{2-z}=\frac{-1}{2}\frac{1}{1-\frac{z}{2}}=\frac{-1}{2}\sum_0^{\infty}\left( \frac{z}{2} \right )^n,|z|<2
    That's the Taylor part, the other gives the same that you found.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Laurent series z0=infinity

    Let us see, you have completed the Laurent expansion valid for |z|<2 that is, centered at z_0=0 which is the same than the Taylor expansion (there is no principal part). On the other hand we have completed the Laurent expansion valid for |z|>2 that is , in another region centered at z_0=0. Right? Then, in the second case we also say that we have the expansion centered at z_0=\infty because 2<|z|<\infty is a neighborhood of \infty .
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    Re: Laurent series z0=infinity

    Thank you Fernando I was reading some of your works by the way. That poly-dimensional time that you've proposed, and the movement of movements it sounds like a really interesting thing. It's not something that we should discuss right here, I can't discuss it anyway, but I wanted to mention it :P

    Bye bye! see you around.
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