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Math Help - Evaluate

  1. #1
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    Evaluate

    \textup{f(x)}=\prod_{r=0}^{5}\left ( 1+2\textup{cos}\left ( 2^r\textup{x} \right ) \right )

    find \textup{f}(\frac{2\pi}{13})
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  2. #2
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    Quote Originally Posted by banku12 View Post
    \textup{f(x)}=\prod_{r=0}^{5}\left ( 1+2\textup{cos}\left ( 2^r\textup{x} \right ) \right )

    find \textup{f}(\frac{2\pi}{13})
    The answer is 1. To see that, notice first that as r goes from 0 to 5, the set of values of \cos(2^{r+1}\pi/13) is the same as the set of values of \cos(2k\pi/13) as k goes from 1 to 6. In fact, modulo 13 we have 2^0\equiv1, 2^1\equiv2, 2^2\equiv4, 2^3\equiv8, 2^4\equiv3, 2^5\equiv6. So for r=0,1,2,4,5, \cos(2^{r+1}\pi/13) = \cos(2k\pi/13) for k = 1,2,4,3,6 respectively; and using the fact that \cos x = \cos(2\pi-x), we see that the remaining value \cos\bigl(8\tfrac{2\pi}{13}\bigr) is equal to \cos\bigl(5\tfrac{2\pi}{13}\bigr).

    Next, notice that 1+2\cos(2k\pi/13) is positive for k=1,2,3,4 and negative for k=5,6. So the product \prod_{k=1}^6(1+2\cos(2k\pi/13)) is positive. Also, \prod_{k=1}^6(1+2\cos(2k\pi/13)) = \prod_{k=7}^{12}(1+2\cos(2k\pi/13)) (because of that same fact \cos x = \cos(2\pi-x) as in the previous paragraph). So the prduct we are looking for is the positive square root of \prod_{k=1}^{12}(1+2\cos(2k\pi/13)).

    Let \omega = \exp(2\pi i/13). Then 1+2\cos(2k\pi/13) = 1+\omega^k+\omega^{-k}. The numbers \omega^k\ (1\leqslant k\leqslant12), together with 1, are the roots of the equation \omega^{13}=1, so their product is 1. Also, \prod_{k=1}^{12}(1+2\cos(2k\pi/13)) = \prod_{k=1}^{12}(1+\omega^k+\omega^{-k}) = \prod_{k=1}^{12}\frac{1-\omega^{3k}}{\omega^k(1-\omega^k)}. But as k goes from 1 to 12, the complex numbers \omega^{3k} run through the same set of values as the numbers \omega^k. So the terms 1-\omega^{3k} in the numerator all cancel with terms 1-\omega^k in the denominator, and we are left with \prod_{k=1}^{12}\frac1{\omega^k} = 1.
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