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Math Help - Laurent expansions

  1. #1
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    Laurent expansions

    In my textbook, as part of an example, I am given that " \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 \sum\limits_{n =  - \infty }^\infty  {{c_n}{{\left( {z - 1} \right)}^n}} , where {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 {c_{ - 2}} \ne 0, while {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 {c_n} described above? All I seem to be able to do is the following:

    {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

    {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, {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 " \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 \sum\limits_{n = - \infty }^\infty {{c_n}{{\left( {z - 1} \right)}^n}} , where {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 {c_{ - 2}} \ne 0, while {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 {c_n} described above? All I seem to be able to do is the following:

    {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

    {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, {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 c_{-2}\neq 0 : you're given that it is 1...!

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

    \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 \int_{\gamma } \frac{1}{(z-1)^{n+3}}dz = 0 if n\neq -2 and 2\pi i if n=-2.

    On your work, you've made some mistakes:

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