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Math Help - Laurent Series

  1. #1
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    Laurent Series

    Hi all

    I'm trying to get my head around the Laurent series for the following function :

    f(z) = zsinh(1/z+1)

    about -1 and trying to give the general term of the series for odd and even powers of (z+1).

    Also, how would I write down a punctured open disk D, containing the circle C={z:|z+1|=1} of which f is represented on this series, and what would be the nature of teh singularity of f at -1.

    Any help would be greatly appreciated.

    Many thanks
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  2. #2
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    Is that 1/(z+1) or 1/z + 1
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  3. #3
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    its 1/(z+1)
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  4. #4
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    Quote Originally Posted by moolimanj View Post
    f(z) = zsinh(1/z+1)
    f(z) = z \cdot \sinh \left( \frac{1}{z+1} \right)

    First we know that,
    \sinh w = w + \frac{w^3}{3!}+\frac{w^5}{5!}+... = \sum_{n=0}^{\infty} \frac{w^{2n+1}}{(2n+1)!}

    So if, w=\frac{1}{z+1} we have,

    \sinh \left( \frac{1}{z+1} \right) = (z+1)^{-1} + \frac{(z+1)^{-3}}{3!} + \frac{(z+1)^{-5}}{5!}+...

    Now think of z as (z+1) - 1

    Thus,
    [ (z+1) - 1 ] \left( (z+1)^{-1} + \frac{(z+1)^{-3}}{3!} + \frac{(z+1)^{-5}}{5!}+... \right)

    Factor,
    \left(1 + \frac{(z+1)^{-2}}{3!} + \frac{(z+1)^{-4}}{5!} + ... \right) - \left( (z+1)^{-1} + \frac{(z+1)^{-3}}{3!} + \frac{(z-1)^{-5}}{5!}+ ... \right)

    Combine,
    -(z+1)^{-1} + \frac{(z+1)^{-2}}{3!} - \frac{(z+1)^{-3}}{3!} + \frac{(z+1)^{-4}}{5!} - \frac{(z+1)^{-5}}{5!} + ...

    Now you can state the formula for the general term.
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