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Thread: Taylor Series

  1. #1
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    Taylor Series

    Use taylor series to verify $\displaystyle e^{-iz}= cos z - isinz$

    i know cos z =$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{2n!}$

    and sin z = $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{2n+1!}$

    and $\displaystyle e^{-iz} =\sum_{n=0}^{\infty}\frac{(-iz)^n}{n!} $

    but i cant get the steps to get there.
    Last edited by linalg123; Sep 21st 2012 at 10:03 AM.
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  2. #2
    Super Member girdav's Avatar
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    Re: Taylor Series

    The sum for $\displaystyle \cos$ runs over even terms, and the corresponding one for $\displaystyle \sin$ of odd terms. So you just have to check that the coefficients of $\displaystyle e^{-iz}$ for odd/even indexes match.
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  3. #3
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    Re: Taylor Series

    Start with the $\displaystyle e^{-iz}$ series.
    Break it down by what happens to that $\displaystyle (-i)^n$ factor.
    That will break the $\displaystyle e^{-iz}$ series into the sum of four series via the values of n mod 4.
    Collect real and imaginary terms.
    Simpify for the real and imaginary parts (this takes some care - gotta manipulate the series and their indices and such carefully).
    That'll do it.
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