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Thread: Laurent expansions

  1. #1
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    Laurent expansions

    In my textbook, as part of an example, I am given that "$\displaystyle \frac{1}{{{{\left( {z - 1} \right)}^2}}}$ is its own Laurent expansion about z = 1, where it has a double pole."

    My reasoning is this:

    If we define a punctured disc centred at 1 with radius r > 0, then we can express it as a Laurent series $\displaystyle \sum\limits_{n = - \infty }^\infty {{c_n}{{\left( {z - 1} \right)}^n}} $, where $\displaystyle {c_n} = \frac{1}{{2\pi i}}\int_\gamma {\frac{{f\left( w \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w$. Having a double pole means that $\displaystyle {c_{ - 2}} \ne 0$, while $\displaystyle {c_i} = 0$, for all i < -2.

    How am I supposed to compute the coefficients of the Laurent series? Am I supposed to use the formula for $\displaystyle {c_n}$ described above? All I seem to be able to do is the following:

    $\displaystyle {c_n} = \frac{1}{{2\pi i}}\int_\gamma {\frac{{f\left( w \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\frac{{\left( {\frac{1}{{{{\left( {w - 1} \right)}^2}}}} \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{n + 3}}}}{\rm{ }}} {\rm{d}}w$, so

    $\displaystyle {c_{ - 2}} = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{ - 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\left( {w - 1} \right){\rm{ }}} {\rm{d}}w$, which does not equal zero (what I am trying to show), and, for example, $\displaystyle {c_{ - 3}} = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{ - 2}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {{{\left( {w - 1} \right)}^2}{\rm{ }}} {\rm{d}}w$, which does equal zero (what I am trying to show).

    Can anyone shed some light on this?
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  2. #2
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    Quote Originally Posted by abrak View Post
    In my textbook, as part of an example, I am given that "$\displaystyle \frac{1}{{{{\left( {z - 1} \right)}^2}}}$ is its own Laurent expansion about z = 1, where it has a double pole."

    My reasoning is this:

    If we define a punctured disc centred at 1 with radius r > 0, then we can express it as a Laurent series $\displaystyle \sum\limits_{n = - \infty }^\infty {{c_n}{{\left( {z - 1} \right)}^n}} $, where $\displaystyle {c_n} = \frac{1}{{2\pi i}}\int_\gamma {\frac{{f\left( w \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w$. Having a double pole means that $\displaystyle {c_{ - 2}} \ne 0$, while $\displaystyle {c_i} = 0$, for all i < -2.

    How am I supposed to compute the coefficients of the Laurent series? Am I supposed to use the formula for $\displaystyle {c_n}$ described above? All I seem to be able to do is the following:

    $\displaystyle {c_n} = \frac{1}{{2\pi i}}\int_\gamma {\frac{{f\left( w \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\frac{{\left( {\frac{1}{{{{\left( {w - 1} \right)}^2}}}} \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{n + 3}}}}{\rm{ }}} {\rm{d}}w$, so

    $\displaystyle {c_{ - 2}} = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{ - 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\left( {w - 1} \right){\rm{ }}} {\rm{d}}w$, which does not equal zero (what I am trying to show), and, for example, $\displaystyle {c_{ - 3}} = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{ - 2}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {{{\left( {w - 1} \right)}^2}{\rm{ }}} {\rm{d}}w$, which does equal zero (what I am trying to show).

    Can anyone shed some light on this?

    Of course $\displaystyle c_{-2}\neq 0$ : you're given that it is 1...!

    $\displaystyle \frac{1}{(z-1)^2} =...0\cdot (x-1)^{-3}+1\cdot (z-1)^{-2} + 0\cdot (z-1)^{-1} + ....$

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Of course $\displaystyle c_{-2}\neq 0$ : you're given that it is 1...!

    $\displaystyle \frac{1}{(z-1)^2} =...0\cdot (x-1)^{-3}+1\cdot (z-1)^{-2} + 0\cdot (z-1)^{-1} + ....$

    Tonio
    Oh, of course!

    So it's much easier than I thought: my function's Laurent expansion has only one non-zero term - that being the (-2)th term of the power series. That makes sense, thank you!
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  4. #4
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    Notice that $\displaystyle \int_{\gamma } \frac{1}{(z-1)^{n+3}}dz = 0$ if $\displaystyle n\neq -2$ and $\displaystyle 2\pi i$ if $\displaystyle n=-2$.

    On your work, you've made some mistakes:

    1)$\displaystyle -2+3\neq -1$
    2) $\displaystyle \int_{\gamma } (z-1)dz =0$ (Cauchy's theorem)
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