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Math Help - [SOLVED] Laurent series

  1. #1
    Bop
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    [SOLVED] Laurent series

    Hello, I want to calculate laurent series and its convergence radius of:

    f(z)=\frac{1}{sin({\pi z})}

    Here it is what i've done, I'm not sure if we can do it in this way:


    sin z =\sum \frac{(-1)^{n}}{(2n+1)!}z^{2n+1}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-O(z^7) for -1<z<1

    so:

    sin (\pi z) =\sum \frac{(-1)^{n}}{(2n+1)!}(\pi z)^{2n+1}=\pi z-\frac{(\pi z)^3}{3!}+\frac{(\pi z)^5}{5!}-O((\pi z)^7) for -1<z<1

    so:


    \frac{1}{sin (\pi z)} =\sum \frac{(2n+1)!}{(-1)^{n}}\frac{1}{(\pi z)^{2n+1}}=\frac{1}{\pi z}-\frac{3!}{(\pi z)^3}+\frac{5!}{(\pi z)^5}-\frac{1}{O((\pi z)^7)} for -1<z<1


    Is it right?


    Thank you.
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  2. #2
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    warning
    \sum \frac{A}{B} \neq (\sum \frac{B}{A})^(-1)
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  3. #3
    MHF Contributor chisigma's Avatar
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    We can start from the series expansion...

    \frac{\sin (\pi z)}{\pi z} = 1 - \frac{(\pi z)^{2}}{3!} + \frac{(\pi z)^{4}}{5!} - \dots (1)

    ... and then, setting...

    f(z) = \frac{\pi z} {\sin (\pi z)} = a_{0} + a_{1} z + a_{2} z^{2} + \dots (2)

    ... find the a_{n} imposing...

    f(z) \frac{\sin (\pi z)}{\pi z} = (a_{0} + a_{1} z + a_{2} z^{2} + \dots) \{ 1 - \frac{(\pi z)^{2}}{3!} + \frac{(\pi z)^{4}}{5!} - \dots\} = 1 (3)

    From (3) now we derive directly...

    a_{0} = 1

    a_{1} =0

    a_{2} - a_{0} \frac{\pi ^{2}}{3!} = 0 \rightarrow a_{2} = \frac{\pi^{2}}{6}

    a_{3} =0

    a_{4} - a_{2} \frac{\pi^{2}}{3!} + a_{0} \frac{\pi^{4}}{5!}=0 \rightarrow a_{4} = \frac{7 \pi^{4}}{360} (4)

    ... and so one, so that is...

    \frac{\pi z}{\sin (\pi z)} = 1 + \frac {\pi^{2} z^{2}} {6} + \frac{7 \pi ^{4} z^{4}}{360} + \dots (5)

    ... and now, deviding (5) by  \pi z, we obtain...

    \frac{1}{\sin (\pi z)} = \frac{1}{\pi z} + \frac {\pi z} {6} + \frac{7 \pi ^{3} z^{3}}{360} + \dots (6)

    The Laurent series (6) converges for 0 < |z| < 1 ...

    Kind regards

    \chi \sigma
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