cos(2π/17)の計算

この記事では${\displaystyle \cos \frac{2\pi }{17}}$の計算をします。かなり計算量が多いので計算好きの方は読んでみてください。

${\displaystyle \zeta =e^{\frac{2\pi i}{17}}}$とおきます。
すると、${\zeta }^{17}=1$なので、$\zeta +{\zeta }^2+...+{\zeta }^{16}=-1$となります。
ここで、$3$は$\mathrm{mod}17$における原始根となっているので${\zeta }^{3^k}$の形に並び替えます。

$$\begin{gathered} \zeta +{\zeta }^3+{\zeta }^9+{\zeta }^{10}+{\zeta }^{13}+{\zeta }^5+{\zeta }^{15}+{\zeta }^{11} \\ +{\zeta }^{16}+{\zeta }^{14}+{\zeta }^8+{\zeta }^7+{\zeta }^4+{\zeta }^{12}+{\zeta }^2+{\zeta }^6 \\ =-1 \end{gathered}$$

これを奇数番目と偶数番目に分けて、
$\zeta +{\zeta }^9+{\zeta }^{13}+{\zeta }^{15}+{\zeta }^{16}+{\zeta }^8+{\zeta }^4+{\zeta }^2=\alpha$
${\zeta }^3+{\zeta }^{10}+{\zeta }^5+{\zeta }^{11}+{\zeta }^{14}+{\zeta }^7+{\zeta }^{12}+{\zeta }^6=\beta$
と置きます。

すると、$\alpha +\beta =-1$であり、
$$\begin{gathered} \alpha \beta \\ =(\zeta +{\zeta }^9+{\zeta }^{13}+{\zeta }^{15}+{\zeta }^{16}+{\zeta }^8+{\zeta }^4+{\zeta }^2) \\ ({\zeta }^3+{\zeta }^{10}+{\zeta }^5+{\zeta }^{11}+{\zeta }^{14}+{\zeta }^7+{\zeta }^{12}+{\zeta }^6) \\ ={\zeta }^4+{\zeta }^{11}+{\zeta }^6+{\zeta }^{12}+{\zeta }^{15}+{\zeta }^8+{\zeta }^{13}+{\zeta }^7 \\ +{\zeta }^{12}+{\zeta }^2+{\zeta }^{14}+{\zeta }^3+{\zeta }^6+{\zeta }^{16}+{\zeta }^4+{\zeta }^{15} \\ +{\zeta }^{16}+{\zeta }^6+\zeta +{\zeta }^7+{\zeta }^{10}+{\zeta }^3+{\zeta }^8+{\zeta }^2 \\ +\zeta +{\zeta }^8+{\zeta }^3+{\zeta }^9+{\zeta }^{12}+{\zeta }^5+{\zeta }^{10}+{\zeta }^4 \\ +{\zeta }^2+{\zeta }^9+{\zeta }^4+{\zeta }^{10}+{\zeta }^{13}+{\zeta }^6+{\zeta }^{11}+{\zeta }^5 \\ +{\zeta }^{11}+\zeta +{\zeta }^{13}+{\zeta }^2+{\zeta }^5+{\zeta }^{15}+{\zeta }^3+{\zeta }^{14} \\ +{\zeta }^7+{\zeta }^{14}+{\zeta }^9+{\zeta }^{15}+\zeta +{\zeta }^{11}+{\zeta }^{16}+{\zeta }^{10} \\ +{\zeta }^5+{\zeta }^{12}+{\zeta }^7+{\zeta }^{13}+{\zeta }^{16}+{\zeta }^9+{\zeta }^{14}+{\zeta }^8 \\ =4(\zeta +{\zeta }^2+...+{\zeta }^{16})=-4 \end{gathered}$$

従って、${\displaystyle \alpha ,\beta =\frac{-1\pm \sqrt{17}}{2}}$
となるが、${\displaystyle \alpha =2(\cos \frac{2\pi }{17}+\cos \frac{4\pi }{17}+\cos \frac{8\pi }{17}+\cos \frac{16\pi }{17})>0}$より、${\displaystyle \alpha =\frac{-1+\sqrt{17}}{2},\beta =\frac{-1-\sqrt{17}}{2}}$です。

$\alpha =\zeta +{\zeta }^9+{\zeta }^{13}+{\zeta }^{15}+{\zeta }^{16}+{\zeta }^8+{\zeta }^4+{\zeta }^2$を奇数番目と偶数番目に分けて
${\alpha }_1=\zeta +{\zeta }^{13}+{\zeta }^{16}+{\zeta }^4$
${\alpha }_2={\zeta }^9+{\zeta }^{15}+{\zeta }^8+{\zeta }^2$
とします。

${\alpha }_1+{\alpha }_2=\alpha$
$$\begin{gathered} {\alpha }_1{\alpha }_2 \\ =(\zeta +{\zeta }^{13}+{\zeta }^{16}+{\zeta }^4)({\zeta }^9+{\zeta }^{15}+{\zeta }^8+{\zeta }^2) \\ ={\zeta }^{10}+{\zeta }^{16}+{\zeta }^9+{\zeta }^3+{\zeta }^5+{\zeta }^{11}+{\zeta }^4+{\zeta }^{15} \\ +{\zeta }^8+{\zeta }^{14}+{\zeta }^7+\zeta +{\zeta }^{13}+{\zeta }^2+{\zeta }^{12}+{\zeta }^6 \\ =-1 \end{gathered}$$

従って、${\displaystyle {\alpha }_1,{\alpha }_2=\frac{\alpha \pm \sqrt{{\alpha }^2+4}}{2}}$

また、${\displaystyle {\alpha }_1=2(\cos \frac{2\pi }{17}+\cos \frac{8\pi }{17})>0}$より、${\displaystyle {\alpha }_1=\frac{\alpha +\sqrt{{\alpha }^2+4}}{2},{\alpha }_2=\frac{\alpha -\sqrt{{\alpha }^2+4}}{2}}$

$\beta ={\zeta }^3+{\zeta }^{10}+{\zeta }^5+{\zeta }^{11}+{\zeta }^{14}+{\zeta }^7+{\zeta }^{12}+{\zeta }^6$も同様に奇数番目と偶数番目に分けて
${\beta }_1={\zeta }^3+{\zeta }^5+{\zeta }^{14}+{\zeta }^{12}$
${\beta }_2={\zeta }^{10}+{\zeta }^{11}+{\zeta }^7+{\zeta }^6$
とします。

${\beta }_1+{\beta }_2=\beta$
$$\begin{gathered} {\beta }_1{\beta }_2 \\ =({\zeta }^3+{\zeta }^5+{\zeta }^{14}+{\zeta }^{12})({\zeta }^{10}+{\zeta }^{11}+{\zeta }^7+{\zeta }^6) \\ ={\zeta }^{13}+{\zeta }^{14}+{\zeta }^{10}+{\zeta }^9+{\zeta }^{15}+{\zeta }^{16}+{\zeta }^{12}+{\zeta }^{11} \\ +{\zeta }^7+{\zeta }^8+{\zeta }^4+{\zeta }^3+{\zeta }^5+{\zeta }^6+{\zeta }^2+\zeta \\ =-1 \end{gathered}$$
${\displaystyle {\beta }_1=2(\cos \frac{6\pi }{17}+\cos \frac{10\pi }{17})>0}$
より、
${\displaystyle {\beta }_1=\frac{\beta +\sqrt{{\beta }^2+4}}{2},{\beta }_2=\frac{\beta -\sqrt{{\beta }^2+4}}{2}}$

そして、${\alpha }_1=\zeta +{\zeta }^{13}+{\zeta }^{16}+{\zeta }^4$も奇数番目と偶数番目に分けてみます。
$(\zeta +{\zeta }^{16})+({\zeta }^{13}+{\zeta }^4)={\alpha }_1$
$(\zeta +{\zeta }^{16})({\zeta }^{13}+{\zeta }^4)={\zeta }^{14}+{\zeta }^5+{\zeta }^{12}+{\zeta }^3={\beta }_1$

ここで、${\displaystyle \zeta +{\zeta }^{16}=2\cos \frac{2\pi }{17},{\zeta }^{13}+{\zeta }^4=2\cos \frac{8\pi }{17}}$
より、$\zeta +{\zeta }^{16}>{\zeta }^{13}+{\zeta }^4$となるので、
$$\begin{gathered} {\displaystyle 2\cos \frac{2\pi }{17}=\zeta +{\zeta }^{16}} \\ {\displaystyle =\frac{{\alpha }_1+\sqrt{{{\alpha }_1}^2-4{\beta }_1}}{2}} \end{gathered}$$

従って、${\displaystyle \cos \frac{2\pi }{17}=\frac{{\alpha }_1+\sqrt{{{\alpha }_1}^2-4{\beta }_1}}{4}}$と求まりましたね。
仕上げに、
${\displaystyle {\alpha }_1=\frac{\alpha +\sqrt{{\alpha }^2+4}}{2}}$
${\displaystyle {\beta }_1=\frac{\beta +\sqrt{{\beta }^2+4}}{2}}$
${\displaystyle \alpha =\frac{-1+\sqrt{17}}{2}}$
${\displaystyle \beta =\frac{-1-\sqrt{17}}{2}}$
を代入してみます。
$$\begin{gathered} {\displaystyle {\alpha }_1=\frac{\frac{-1+\sqrt{17}}{2}+\sqrt{{\left(\frac{-1+\sqrt{17}}{2}\right)}^2+4}}{2}} \\ {\displaystyle =\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}}{4}} \end{gathered}$$
$$\begin{gathered} {\displaystyle {\beta }_1=\frac{\frac{-1-\sqrt{17}}{2}+\sqrt{{\left(\frac{-1-\sqrt{17}}{2}\right)}^2+4}}{2}} \\ {\displaystyle =\frac{-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}}{4}} \end{gathered}$$
$$\begin{gathered} {\displaystyle \cos \frac{2\pi }{17}} \\ {\displaystyle =\frac{\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}}{4}+\sqrt{{\left(\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}}{4}\right)}^2-(-1-\sqrt{17}+\sqrt{34+2\sqrt{17}})}}{4}} \\ {\displaystyle =\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}} \end{gathered}$$

計算できましたね。
お疲れ様でした。