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Thread: Geometric Series

  1. November 18th 2007, 05:38 AM #1
    slevvio
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    Geometric Series

    Use complex numbers and geometric series to sum:

    1 + cos2\theta + cos4\theta +...+ cos(2n-2)\theta .

    I don't really have a clue what to do... any help would be appreciated... thanks. The answer is \frac{\cos{(n-1)\theta} \sin\: {n\theta}}{\sin{\theta}}.
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  2. November 18th 2007, 07:42 AM #2
    Simba
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    Consider \Re (1 + e^{2i\theta}\ + e^{4i\theta}\ + ... + e^{2(n - 1)i\theta})\ .
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  3. November 18th 2007, 07:49 AM #3
    PaulRS
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    Hello

    The key point is to use de Moivre's formula: [\cos(x)+i\cdot{\sin(x)}]^k=\cos(kx)+i\cdot{\sin(kx)}
    And mix it with the geometrical sum


    http://en.wikipedia.org/wiki/De_Moivre's_formula

    PS. It can also be done using telescoping sums
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  4. November 18th 2007, 09:04 AM #4
    ThePerfectHacker
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    Just be careful what Simba posted does not work for \theta = \pi n. (Because that is the special case of geometric series which fails).
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