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Math Help - Laurent Series

  1. #1
    Junior Member CuriosityCabinet's Avatar
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    Laurent Series

    Find the laurent series of \frac{e^z}{z+1} about z=-1.

    I wrote out the Taylor expansion of e^z and \frac{1}{z+1} and now I think I need to do Cauchy product of the two series but stuck on how to do that. Help please?
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  2. #2
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    Re: Laurent Series

    The Taylor Series for \displaystyle \begin{align*} e^z \end{align*} centred at \displaystyle \begin{align*} z = -1 \end{align*} is \displaystyle \begin{align*} \sum_{n = 0}^{\infty} \frac{e^{-1}}{n!} \left( z + 1 \right) ^n  \end{align*}, so that means

    \displaystyle \begin{align*} \frac{e^z}{z + 1} &= \frac{\sum_{n=0}^{\infty} \frac{e^{-1}}{n!} \left( z + 1 \right) ^n}{z + 1} \\ &= \sum_{n = 0}^{\infty} \frac{e^{-1}}{n!} \left( z + 1 \right) ^{n - 1} \\ &= \frac{e^{-1}}{z + 1} + \sum_{n = 1}^{\infty} \frac{e^{-1}}{n!} \left( z + 1 \right) ^{n-1} \\ &= e^{-1} \sum_{n = 0}^{\infty} \left( -z \right) ^n + e^{-1}\sum_{n = 0}^{\infty} \frac{ \left( z + 1 \right) ^n}{ \left( n + 1 \right) ! } \textrm{ provided } \left| -z \right| < 1 \\ &= e^{-1} \sum_{n = 0}^{\infty} \left[ \left( -z \right) ^n + \frac{ \left( z + 1 \right) ^n}{ \left( n + 1 \right) !} \right] \textrm{ provided } |z| < 1 \end{align*}
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