Worksheet10: Dickson 042
Current Work (editable)
Taking alternate terms in z_{1}, we obtain the periods v_{1} = R + R^{16}, v_{2} = R^{13} + R^{4}. Now, v_{1} + v_{2} = z_{1}, v_{1} v_{2} = w_{1}. Hence (25) v_{1}, v_{2} satisfy v^{2} - z_{1}v + w_{1} = 0, (26) R, R^{16} satisfy \rho^{2} - v_{1}\rho + 1 = 0. Hence we can find R by solving a series of quadratic equations. Which of the sixteen values of R we shall thus obtain depends upon which root of (22) is called y_{1} and which y_{2}, and similarly in (23)-(26). We shall now show what choice is to be made in each such case in order that we shall finally get the value of the particular root R = cos 2\pi/17 + i sin 2\pi/17. Then 1/R = cos 2\pi/17 - i sin 2\pi/17, v_{1} = R + 1/R = 2 cos 2\pi/17, R^{4} = cos 8\pi/17 + i sin 8\pi/17, v_{2} = R^{4} + 1/R^{4} = 2 cos 8\pi/17. Hence v_{1} > v_{2} > 0, and therefore z_{1} = v_{1} + v_{2} > 0. Similarly, w_{1} = R^{3} + 1/R^{3} + R^{5} + 1/R^{5} = 2 cos 6\pi/17 + 2 cos 10\pi/17 = 2 cos 6\pi/17 - 2 cos 7\pi/17 > 0, y_{2} = 2 cos 6\pi/17 + 2 cos 10\pi/17 + 2 cos 12\pi/17 + 2 cos 14\pi/17 < 0, since only the first cosine in y_{2} is positive and it is numerically less than the third. But y_{1} y_{2} = -4. Hence y_{1} > 0. Thus (22)-(24) give y_{1} = 1/2( $${17} - 1), y_{2} = 1/2(-$${17} - 1), z_{1} = 1/2y_{1} + $${1 + 1/4 y_{1}^{2}}, w_{1} = 1/2y_{2} + $${1 + 1/4 y_{2}^{2}}. We may readily construct segments of these lengths. Evidently $${17} is the length of the hypotenuse of a right triangle whose legs are of lengths 1 and 4, while for the radical in z_{1} we employ legs of lengths 1 and 1/2 y_{1}. We thus obtain segments representing the coefficients of the
Answer Key (non-editable)
Taking alternate terms in $z_{1}$, we obtain the periods \[ v_{1} = R + R^{16}, \qquad v_{2} = R^{13} + R^{4}. \] Now, $v_{1} + v_{2} = z_{1}$, $v_{1} v_{2} = w_{1}$. Hence \begin{align*} \Tag{(25)} v_{1},\ v_{2} &\quad\text{satisfy}\quad v^{2} - z_{1}v + w_{1} = 0, \\ \Tag{(26)} R,\ R^{16} &\quad\text{satisfy}\quad \rho^{2} - v_{1}\rho + 1 = 0. \end{align*} Hence we can find $R$ by solving a series of quadratic equations. Which of the sixteen values of $R$ we shall thus obtain depends upon which root of \Eq{(22)} is called $y_{1}$ and which $y_{2}$, and similarly in \Eq{(23)}--\Eq{(26)}. We shall now show what choice is to be made in each such case in order that we shall finally get the value of the particular root \[ R = \cos \frac{2\pi}{17} + i \sin \frac{2\pi}{17}. \] Then \begin{alignat*}{4} \frac{1}{R} &= \cos \frac{2\pi}{17} - i \sin \frac{2\pi}{17}, &\qquad v_{1} &= R &&+ \frac{1}{R} &&= 2 \cos \frac{2\pi}{17}, \\ % R^{4} &= \cos \frac{8\pi}{17} + i \sin \frac{8\pi}{17}, & v_{2} &= R^{4} &&+ \frac{1}{R^{4}} &&= 2 \cos \frac{8\pi}{17}. \end{alignat*} Hence $v_{1} > v_{2} > 0$, and therefore $z_{1} = v_{1} + v_{2} > 0$. Similarly, \begin{align*} w_{1} &= R^{3} + \frac{1}{R^{3}} + R^{5} + \frac{1}{R^{5}} = 2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17} = 2 \cos \frac{6\pi}{17} - 2 \cos \frac{7\pi}{17} > 0, \\ % y_{2} &= 2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17} + 2 \cos \frac{12\pi}{17} + 2 \cos \frac{14\pi}{17} < 0, \end{align*} since only the first cosine in $y_{2}$ is positive and it is numerically less than the third. But $y_{1} y_{2} = -4$. Hence $y_{1} > 0$. Thus \Eq{(22)}--\Eq{(24)} give \begin{align*} y_{1} &= \tfrac{1}{2}( \sqrt{17} - 1), & y_{2} &= \tfrac{1}{2}(-\sqrt{17} - 1), \\ % z_{1} &= \tfrac{1}{2}y_{1} + \sqrt{1 + \tfrac{1}{4}y_{1}^{2}}, & w_{1} &= \tfrac{1}{2}y_{2} + \sqrt{1 + \tfrac{1}{4}y_{2}^{2}}. \end{align*} We may readily construct segments of these lengths. Evidently $\sqrt{17}$ is the length of the hypotenuse of a right triangle whose legs are of lengths $1$ and $4$, while for the radical in $z_{1}$ we employ legs of lengths $1$ and $\frac{1}{2} y_{1}$. We thus obtain segments representing the coefficients of the
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