1. ## Evaluate

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

find $\textup{f}(\frac{2\pi}{13})$

2. Originally Posted by banku12
$\textup{f(x)}=\prod_{r=0}^{5}\left ( 1+2\textup{cos}\left ( 2^r\textup{x} \right ) \right )$

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 $\prod_{k=1}^{12}\frac1{\omega^k} = 1.$