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  1. #1
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    Evaluate

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

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

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

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

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