Exercise 9.2.9

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
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In this exercise, you will use numerical computations and the previous exercise to find radical expressions for various $f$-periods when $p=17$.
\be
\item[(a)] Show that
\begin{align*}
(8,1) &=  2\cos(2\pi/17) + 2\cos(4\pi/17)+2\cos(8\pi/17)+2\cos(16\pi/17)\
(4,1) &= 2\cos(2\pi/17)+2\cos(8\pi/17)\
(4,3) &= 2\cos(6\pi/17)+2\cos(10\pi/17)\
(2,1) &= 2\cos(2\pi/17)
\end{align*}
Then compute each of these periods to five decimal places.
\item[(b)] Use the numerical computations of part (a) and the quadratic polynomials of Exercise 8 to show that
\begin{align*}
(8,1) &= \frac{1}{2}\left(-1+\sqrt{17}ight)\
(8,3) &= \frac{1}{2}\left(-1-\sqrt{17}ight)\
(4,1) &= \frac{1}{4} \left(-1+\sqrt{17} +\sqrt{34-2\sqrt{17}}ight)\
(4,2) &= \frac{1}{4} \left(-1+\sqrt{17} - \sqrt{34-2\sqrt{17}}ight)\
(4,3) &= \frac{1}{4} \left(-1-\sqrt{17} +\sqrt{34+2\sqrt{17}}ight)
\end{align*}

\item[(c)] Use the quadratic polynomial $x^2 - (4,1) x + (4,3)$ and part (b) to derive (9.19).
\ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\be} {\begin{enumerate}} ewcommand{\ee} {\end{enumerate}} ewcommand{\deb}{\begin{eqnarray*}} ewcommand{\fin}{\end{eqnarray*}} ewcommand{\ssi} {si et seulement si } ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} Recall (see Exercise 8) that \begin{align*} H_8 &= \{1,2,4,8,9,13,15,16\}\ H_4 &= \{1,4,13,16\}\ 3H_4 &= \{3,5,12,14\}\ H_2 &= \{1,16\} \end{align*} Write $\zeta=\zeta_{17}$. \begin{enumerate} \item[(a)] Using these results, and $\zeta^{-k} = \zeta^{17-k}, k=1,2,4,8$, and also $\zeta^{k} + \zeta^{-k} = 2 \cos(2k\pi/17)$, we obtain \begin{align*} (8,1) &= \sum_{[a]\in H_8} \zeta^a\ &=\zeta+ \zeta^2 + \zeta^4 + \zeta^8+\zeta^9 + \zeta^{13}+ \zeta^{15}+\zeta^{16}\ &=(\zeta+\zeta^{-1})+ (\zeta^2+\zeta^{-2})+(\zeta^4+\zeta^{-4})+ (\zeta^8+\zeta^{-8})\ &= 2\cos(2\pi/17) + 2\cos(4\pi/17)+2\cos(8\pi/17)+2\cos(16\pi/17)\ \ (4,1) &= \sum_{[a]\in H_4} \zeta^a\ &=\zeta+\zeta^4+\zeta^{13}+ \zeta^{16}\ &= (\zeta+ \zeta^{-1}) + (\zeta^4+\zeta^{-4})\ &= 2\cos(2\pi/17)+2\cos(8\pi/17) \ (4,3) &= \sum_{[a]\in 3H_4} \zeta^a\ &=\zeta^{3}+\zeta^5+\zeta^{12}+ \zeta^{14}\ &= (\zeta^{3}+ \zeta^{-3}) + (\zeta^5+\zeta^{-5})\ &= 2\cos(6\pi/17)+2\cos(10\pi/17)\ \ (2,1) &= \sum_{[a]\in H_2} \zeta^a\ &=\zeta+\zeta^{16}\ &=\zeta+ \zeta^{-1}\ &=2\cos(2\pi/17) \end{align*} $(2,1) = 2\cos(2\pi/17) \simeq 0.93247$, $(4,1) \simeq 2.04948, (4,3)\simeq 0.34415$, $(8,1) \simeq 1.56155$. As $(4,1)+(4,2) = (8,1)$, we obtain $(4,2) \simeq -0.48792<0$. \item[(b)] By Exercise 8, $(8,1),(8,3)$ are the roots of $x^2+x-4$, and by part (a) $(8,1)>0$. The only positive root of $x^2+x-4$ is $(-1+\sqrt{17})/2$, therefore \begin{align*} (8,1) &= \frac{1}{2}\left(-1+\sqrt{17} ight),\ (8,3) &= \frac{1}{2}\left(-1-\sqrt{17} ight).\ \end{align*} $(4,1),(4,2)$ are the roots of $x^2-(8,1)x-1$, with discriminant $$\Delta= \frac{1}{4}(-1+\sqrt{17})^2+4 = \frac{1}{4}(34 - 2\sqrt{17}),$$ therefore $$\{(4,1),(4,2)\} =\left \{\frac{1}{4} \left(-1+\sqrt{17} +\sqrt{34-2\sqrt{17}} ight), \frac{1}{4} \left(-1+\sqrt{17} -\sqrt{34-2\sqrt{17}} ight) ight\}.$$ By part (a) $(4,2)<0<(4,1)$, so \begin{align*} (4,1) &= \frac{1}{4} \left(-1+\sqrt{17} +\sqrt{34-2\sqrt{17}} ight),\ (4,2)&= \frac{1}{4} \left(-1+\sqrt{17} - \sqrt{34-2\sqrt{17}} ight). \end{align*} $(4,3),(4,6)$ are the roots of $x^2-(8,3)x-1$, with discriminant $$\Delta= \frac{1}{4}(-1-\sqrt{17})^2+4 = \frac{1}{4}(34 + 2\sqrt{17}),$$ therefore $$\{(4,3),(4,6)\} =\left \{\frac{1}{4} \left(-1-\sqrt{17} +\sqrt{34+2\sqrt{17}} ight), \frac{1}{4} \left(-1-\sqrt{17} -\sqrt{34+2\sqrt{17}} ight) ight\}.$$ As $(4,3)>0$, \begin{align*} (4,3) &= \frac{1}{4} \left(-1-\sqrt{17} +\sqrt{34+2\sqrt{17}} ight),\ (4,6) &=\frac{1}{4} \left(-1-\sqrt{17} -\sqrt{34+2\sqrt{17}} ight). \end{align*} \item[(c)] $(2,1) = 2\cos(2\pi/17)$, and also $(2,4)$, is root of $x^2 -(4,1)x+(4,3)$, with discriminant $$\Delta = (4,1)^2 - 4(4,3).$$ As \begin{align*} (4,1)^2 &= \sum\limits_{\lambda \in H_4} (4,\lambda+1)\ &=(4,2)+(4,5)+(4,14)+4\ &=(4,2)+2(4,3)+ 4, \end{align*} then \begin{align*} \Delta &= (4,2) -2(4,3) + 4\ &= \frac{1}{4} \left ( -1+\sqrt{17}- \sqrt{34-2\sqrt{17}} -2\left(-1-\sqrt{17} +\sqrt{34+2\sqrt{17}} ight) +16 ight)\ &= \frac{1}{4} \left ( 17 + 3\sqrt{17} - \sqrt{34-2\sqrt{17}} -2 \sqrt{34+2\sqrt{17}} ight). \end{align*} The roots of $x^2 -(4,1)x+(4,3)$ are so $\frac{1}{2}((4,1)\pm \sqrt{\Delta})$ $=\frac{1}{8} \left(-1+\sqrt{17} +\sqrt{34-2\sqrt{17}} ight) \pm \frac{1}{4} \sqrt{17 + 3\sqrt{17} - \sqrt{34-2\sqrt{17}} -2 \sqrt{34+2\sqrt{17}} }.$ As $(2,4) = 2\cos(8\pi/17) < 2 \cos(2\pi/17) = (2,1)$, we can conclude that \begin{align*} \cos\left(\frac{2\pi}{17} ight) =&-\frac{1}{16}+ \frac{1}{16}\sqrt{17} +\frac{1}{16}\sqrt{34-2\sqrt{17}}\ & + \frac{1}{8} \sqrt{17 + 3\sqrt{17} - \sqrt{34-2\sqrt{17}} -2 \sqrt{34+2\sqrt{17}} }. \end{align*} \end{enumerate} \end{proof}

Exercise 9.2.9

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Exercise 9.2.9

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