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Math Help - Series

  1. #1
    Senior Member slevvio's Avatar
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    Series

    Show that, for all  \theta \in \mathbb{R} ,

     \frac{cos\theta}{2} + \frac{cos3\theta}{2^2} + \frac{cos5\theta}{2^3} + \ldots = \frac{cos\theta}{1 + 8sin^2\theta} .

    I understand that this has something to do with a geometric series however taking  \frac{a_2}{a_1} and  \frac{a_3}{a_2} gave me different values... any help with this would be appreciated... thanks a lot and happy new year to all.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slevvio View Post
    Show that, for all  \theta \in \mathbb{R} ,

     \frac{cos\theta}{2} + \frac{cos3\theta}{2^2} + \frac{cos5\theta}{2^3} + \ldots = \frac{cos\theta}{1 + 8sin^2\theta} .

    I understand that this has something to do with a geometric series however taking  \frac{a_2}{a_1} and  \frac{a_3}{a_2} gave me different values... any help with this would be appreciated... thanks a lot and happy new year to all.
    Have you looked at a Fourier series for \frac{cos(\theta)}{1 + 8sin^2(\theta)}? (Use the sine and cosine form, not the complex exponential.)

    -Dan
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  3. #3
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    Quote Originally Posted by slevvio View Post
    Show that, for all  \theta \in \mathbb{R} ,

     \frac{\cos\theta}{2} + \frac{\cos3\theta}{2^2} + \frac{\cos5\theta}{2^3} + \ldots = \frac{\cos\theta}{1 + 8\sin^2\theta} .
    Why is this in the algebra sub-forum??

    Note that \cos z = \frac{1}{2} \left( e^{iz} + e^{-iz} \right).

    That means the series becomes,
    \sum_{n=0}^{\infty} \frac{\frac{1}{2}\left( e^{i(2n+1)\theta} + e^{-i(2n+1)\theta} \right)}{2^{n+1}} = \sum_{n=0}^{\infty} \frac{e^{i(2n+1)\theta}}{2^{n+2}} + \sum_{n=0}^{\infty}\frac{e^{-i(2n+1)\theta}}{2^{n+2}}

    Now, \sum_{n=0}^{\infty} \frac{e^{i(2n+1)\theta}}{2^{n+2}} = \frac{e^{i\theta}}{2^2} \sum_{n=0}^{\infty} \frac{e^{2ni\theta}}{2^n} = \frac{e^{i\theta}}{2^2} \sum_{n=0}^{\infty} \left( \frac{e^{2i\theta}}{2} \right)^n = \frac{e^{i\theta}}{2^2} \cdot \frac{2}{2 - e^{2i\theta}}

    Similarly, \sum_{n=0}^{\infty} \frac{e^{-i(2n+1)\theta}}{2^{n+2}} = \frac{e^{-i\theta}}{2^2}\cdot \frac{2}{2-e^{-2i\theta}}

    Thus series sums to, \frac{e^{i\theta}}{4 - 2e^{2i\theta}} + \frac{e^{-i\theta}}{4- 2e^{-2i\theta}}

    Now, \frac{e^{i\theta}}{4 - 2e^{2i\theta}} + \frac{e^{-i\theta}}{4- 2e^{-2i\theta}} = \frac{(4 - 2e^{-2i\theta})e^{i\theta} + (4 - 2e^{2i\theta})e^{-i\theta}}{(4-2e^{2i\theta})(4-2e^{-2i\theta})}

    Which becomes, \frac{2e^{i\theta} + 2e^{-i\theta}}{2^2(2-e^{2i\theta})(2-e^{-2i\theta})} = \frac{e^{i\theta}+e^{-i\theta}}{2[5 - 2(e^{2i\theta}+e^{-2i\theta})]}

    Use identity above, \frac{2\cos \theta}{10 - 8\cos 2\theta} = \frac{\cos \theta}{5 - 4\cos 2\theta}

    Double angle, \frac{\cos \theta}{5 - 4[1 - 2\sin^2 \theta]}

    And we finally get, \frac{\cos \theta}{1 + 8\sin^2 \theta}.
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