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$\qquad$正十七边形尺规作图代数推导<z2u
©Elinkage数学论坛 -- Elinkage极酷超级论坛 (V^zTK
令$\,\zeta=\exp\big(\frac{2\pi}{17}i\big),\;$则$\underset{\,}{\;}\zeta^k\,\small(k=\overline{0,16})\,$是$\,z^{17}-1=0\,$的全部根.1i:sJ
记$\,\lambda=\zeta+\zeta^{-1},\,\lambda_k=\zeta^k+\zeta^{-k},\,$由$\;\zeta^{17}=1\underset{\,}{\,}$得EsWc%$
$(\dagger)\underset{\,}{\quad}\lambda_m = \lambda_{17k\pm m},\;\;\lambda_m\lambda_n=\lambda_{m+n}+\lambda_{m-n}\,\;(k,m,n\in\mathbb{Z})$E_7m>
$(\ddagger)\underset{\,}{\quad}z^{16}+z^{15}+\cdots+z+1=0\,$是$\,\zeta\,$的最小多项式.o1
令$\quad\underset{\,}{\;} a=\lambda_1+\lambda_2+\lambda_4+\lambda_8,\;\,a'=\lambda_3+\lambda_5+\lambda_6+\lambda_7,$?\Nd
由$\underset{\,}{\,}(\dagger),(\ddagger)\,$可得$\,\;a+a'=-1,\;\;aa'=4(a+a')=-4.$E
$\therefore\underset{\,}{\quad}a\overset{>}{,}a'\,$是$\,z^2+z-4=0\,$的根${\small\,\dfrac{-1\pm\sqrt{17}}{2}.\;}\boxed{a=\small\frac{\sqrt{17}-1}{2}}$^
令$\underset{\,}{\quad} a=b+b',\;\,b=\lambda_1+\lambda_4\;(=\zeta+\zeta^4+\zeta^{13}+\zeta^{16}),\,\;$则l}
$\underset{\,}{\qquad}b\,b'=(\lambda_1+\lambda_4)(\lambda_2+\lambda_8)$X_{O0
$\underset{\,}{\qquad}\quad\,=\lambda_3+\lambda_1+\lambda_9+\lambda_7+\lambda_2+\lambda_6+\lambda_{12}+\lambda_4$Y
$\underset{\,}{\qquad}\quad\,=\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5+\lambda_6+\lambda_7+\lambda_8=-1.$tCyl%
$\therefore\underset{\,}{\quad}b\;\overset{>}{,}\;b'\,$是方程$\,z^2-az-1=0\,$的根$\;\boxed{b=\small\frac{a+\sqrt{a^2+4}}{2}}$mDCP
最后令$\underset{\,}{\;}b=c+c',\;c=\lambda=\zeta+\zeta^{-1},\;c'=\lambda_4,\;c\,c'=\lambda_1\lambda_4,$>#;VA
$\therefore\underset{\,}{\quad}c\;\overset{>}{,}\;c'\,$是$\,z^2-bz+\lambda_1\lambda_4\,$的根. 据$\,(\dagger),\;\lambda_k^2=\lambda_{2k}+2,$XzW
$\therefore\underset{\,}{\quad}b^2-b'-4=(\lambda_1+\lambda_4)^2-(\lambda_2+\lambda_8)=2\lambda_1\lambda_4$+q2X
$\therefore\underset{\,}{\quad}{\small\lambda=}\frac{b+\sqrt{-b^2+2b'+8}}{2}=\frac{a+\sqrt{a^2+4}+\sqrt{2a(2-a)-2(2+a)\sqrt{a^2+4}+28}}{4}$98[
$\therefore\small\;\;\boxed{\cos{\small\frac{2\pi}{\scriptsize 17}}=\scriptsize\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}{16}}$7l@c
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2015/11/04 03:59am 
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