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Math Help - Evaluate sum(requires use of complex numbers)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Evaluate sum(requires use of complex numbers)?

    Evaluate:
    \sin x + 3\sin 3x + 5\sin 5x + ... + (2n - 1)\sin \{(2n - 1)x\}

    Answer:
    Spoiler:
    \frac{2n\sin(2nx) - (2n - 1)\sin\{(2n + 1)x\} - 1}{4\sin^2 \frac{x}{2}}


    I do not know how the problem is done. Any ideas?
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  2. #2
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    Hi

    Are you sure about the answer ?
    I do not find the same and your answer seems not to work with values like \frac{\pi}{4} for instance ...
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  3. #3
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by running-gag View Post
    Hi

    Are you sure about the answer ?
    I do not find the same and your answer seems not to work with values like \frac{\pi}{4} for instance ...
    Hello running-gag,
    I checked the answer in the text again and it is the apparent answer. But it doesn't work with \frac{\pi}{4} as you said. Can you post the correct solution?
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  4. #4
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    Let f(x) = \sin x + 3\sin 3x + 5\sin 5x + ... + (2n - 1)\sin \{(2n - 1)x\} = \sum_{k=1}^{n} (2k-1) \sin{(2k-1)x}

    Let F(x) one antiderivative of f

    Then F(x) = -\sum_{k=1}^{n} \cos{(2k-1)x}

    F(x) = -Re\left(\sum_{k=1}^{n} e^{i(2k-1)x}\right)

    After some calculations I can find

    f(x) = \frac{\sin(2nx) \cos x - 2n \cos(2nx) \sin x}{2 \sin^2x}

    But there are many other possible expressions for f(x)
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  5. #5
    Super Member fardeen_gen's Avatar
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    We cannot get a single unique answer?
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  6. #6
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    If you use \sin{(2n+1)x} = \sin(2nx) \cos x + \cos(2nx) \sin x

    you can substitute \sin(2nx) \cos x = \sin{(2n+1)x} - \cos(2nx) \sin x

    or even \cos(2nx) \sin x = \sin{(2n+1)x} - \sin(2nx) \cos x

    in the below expression of f

    f(x) = \frac{\sin(2nx) \cos x - 2n \cos(2nx) \sin x}{2 \sin^2x}

    At first I thought that this was the reason why I did not find the same result as yours but then I realized that your solution is not correct
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