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Math Help - Sum of this geometric series

  1. #1
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    Sum of this geometric series

    Hi all,

    I'm trying to find:
    \sum_{k=0}^{n} (e^{i\Theta})^{k}

    I manipulate this the following way:

    = 1 + \sum_{k=1}^{n} (e^{i\Theta})^{k} = 1 + \frac{1-e^{in\Theta}}{1-e^{i\Theta}}

    Is this right? Thanks
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  2. #2
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    Re: Sum of this geometric series

    Yes you can derive that result directly in the same way that the general formula is derived, I just tried it myself and the peculiar common ratio in the series does not change the derivation.
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  3. #3
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    Re: Sum of this geometric series

    I see, thanks. Now, If I want to obtain the sum from just the real part of (e^{i\Theta})^{k},
    that is, \sum_{k=0}^{n} cos(k\Theta)

    Could I start from:

    \sum_{k=0}^{n} (e^{i\Theta})^{k} = \sum_{k=0}^{n} cos(k\Theta) + i\sum_{k=0}^{n} sin(k\Theta)

    and then continue this way:

    \sum_{k=0}^{n} cos(k\Theta) = \sum_{k=0}^{n} (e^{i\Theta})^{k} - i\sum_{k=0}^{n} sin(k\Theta)

    Or is there an easier method?
    Last edited by director; February 2nd 2013 at 02:21 PM.
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  4. #4
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    Re: Sum of this geometric series

    The expression for sum of n terms of a GP is
    Sum of this geometric series-gp.png Now I am sure you can make out where did you go wrong
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  5. #5
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    Re: Sum of this geometric series

    Actually it's \displaystyle \begin{align*} S_n = \frac{a\left( 1 - r^n \right)}{1 - r} \end{align*}...
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