摘要: 本文分別探討以 $\Big \{\big (2\cos\frac{2\pi}{7}\big )^n$, $\big (2\cos\frac{4\pi}{7}\big )^n$, $\big (2\cos\frac{6\pi}{7}\big )^n\Big \}$, $\Big \{\big (2\cos\frac{2\pi}{7} 2\cos\frac{4\pi}{7}\big )^n$, $\big (2\cos\frac{4\pi}{7} 2\cos\frac{6\pi}{7}\big )^n, \big (2\cos\frac{6\pi}{7} 2\cos\frac{2\pi}{7}\big )^n\Big \}$, $\Big \{2\cos\frac{2k\pi}{2n+1}\mid 1\le k\le n\Big \}$ 或 $\Big \{2\cos\frac{2k\pi}{2n}\mid 1\le k\le n\Big \}$ 等四組無理數為根的整係數多項式, 進而探討該諸多項式的係數間的關係。 這些研究顯示透過整係數多項式, 我們可以把由三角函數所形成的一些無理數一線牽起來。
引理 1: $\displaystyle\Big (x-2\cos\frac{2\pi}{7}\Big ) \Big (x-2\cos\frac{4\pi}{7}\Big ) \Big (x-2\cos\frac{6\pi}{7}\Big )=x^3+x^2-2x-1$
令 $T_n=\displaystyle\Big (2\cos\frac{2\pi}{7}\Big )^n + \Big (2\cos\frac{4\pi}{7}\Big )^n + \Big (2\cos\frac{6\pi}{7}\Big )^n$,
$Q_n=\displaystyle\Big (2\cos\frac{2\pi}{7} 2\cos\frac{4\pi}{7}\Big )^n +\Big (2\cos\frac{4\pi}{7} 2\cos\frac{6\pi}{7}\Big )^n +\Big (2\cos\frac{6\pi}{7} 2\cos\frac{2\pi}{7}\Big )^n$,
則 $T_n\in Z$, $(T_1, T_2, T_3)=(-1, 5, -4)$, $T_{n+3}=-T_{n+2}+2T_{n+1}+T_n$,
$Q_n=\displaystyle\frac{1}{2}[T_n^2-T_{2n}]$。
證: (1). $\displaystyle\alpha=2\cos\frac{2\pi}{7}$, $\displaystyle\beta=2\cos\frac{4\pi}{7}$, $\displaystyle\gamma=2\cos\frac{6\pi}{7}$ 是 $x^3=-x^2+2x+1$ 的三根 \begin{eqnarray*} \quad && \Rightarrow \alpha, \beta, \gamma \hbox{ 也是 } x^{n+3} = -x^{n+2} + 2x^{n+1} + x^n \hbox{ 的三根} \\ && \Rightarrow \left \{ \begin{array}{l} \alpha^{n+3} = -\alpha^{n+2} + 2\alpha^{n+1} + \alpha^n \\ \beta^{n+3} = -\beta^{n+2} + 2\beta^{n+1} + \beta^n \\ \gamma^{n+3} = -\gamma^{n+2} + 2\gamma^{n+1} + \gamma^n \end{array} \right. \\ && \Rightarrow (\alpha^{n+3}+\beta^{n+3}+\gamma^{n+3}) = - (\alpha^{n+2}+\beta^{n+2}+\gamma^{n+2}) + 2(\alpha^{n+1}+\beta^{n+1}+\gamma^{n+1}) \\ && + (\alpha^n+\beta^n+\gamma^n) \\ && \Rightarrow T_{n+3} = -T_{n+2} + 2T_{n+1} + T_n. \end{eqnarray*}
\begin{eqnarray*} (2). Q_n&&=\displaystyle\alpha^n\beta^n + \beta^n\gamma^n + \gamma^n\alpha^n =\frac{1}{2} \bigg [\Big (\alpha^n+\beta^n+\gamma^n\Big )^2 -\Big ((\alpha^n)^2+(\beta^n)^2+(\gamma^n)^2\Big )\bigg ]\\ &&=\displaystyle\frac{1}{2} \Big [T_n^2-T_{2n}\Big ]. \end{eqnarray*}
命題 2: 令 $f(x)=\displaystyle\prod_{j=1}^m (x-\alpha_j)$, 則 $x^{n+1} f'(x)\div f(x)$ 之商的常數項為 $\displaystyle\sum_{j=1}^m \alpha_j^n$,
特別是當 $\displaystyle(\alpha_1, \alpha_2, \alpha_3)=\Big (2\cos\frac{2\pi}{7}, 2\cos\frac{4\pi}{7}, 2\cos\frac{6\pi}{7}\Big )$ 時, $(3x^{n+3}+2x^{n+2}-2x^{n+1})\div$ $(x^3+x^2-2x-1)$ 之商的常數項為 $T_n$ 。
證: 觀察綜合除法 $x^{n+1}\div(x-\alpha_1)$:
表成 $\displaystyle\frac{x^{n+1}}{x-\alpha_1}=x^n+\sum_{i=1}^{n-1} \alpha_1^i x^{n-i}+\alpha_1^n+\frac{\alpha_1^{n+1}}{x-\alpha_1}$
同理 $\displaystyle\frac{x^{n+1}}{x-\alpha_2}=x^n+\sum_{i=1}^{n-1} \alpha_2^i x^{n-i}+\alpha_2^n+\frac{\alpha_2^{n+1}}{x-\alpha_2}$
$\vdots$
$\displaystyle\frac{x^{n+1}}{x-\alpha_m}=x^n+\sum_{i=1}^{n-1} \alpha_m^i x^{n-i}+\alpha_m^n+\frac{\alpha_m^{n+1}}{x-\alpha_m}$
此 $m$ 個式子相加: $\displaystyle\sum_{j=1}^m \frac{x^{n+1}}{x-\alpha_j}=mx^n+\sum_{j=1}^m \sum_{i=1}^{n-1} \alpha_j^i x^{n-i} +\sum_{j=1}^m \alpha_j^n+\sum_{j=1}^m \frac{\alpha_j^{n+1}}{x-\alpha_j}$
左式 $\displaystyle =x^{n+1} \sum_{j=1}^m \frac{1}{x-\alpha_j} =x^{n+1} \frac{\sum\limits_{k=1}^m \prod\limits_{j\not=k}^m (x-\alpha_j)}{\prod\limits_{j=1}^m (x-\alpha_j)} =x^{n+1} \frac{f'(x)}{f(x)}$ 。
命題 3: 令 $a_n\in Z$, $(a_1, a_2, a_3)\equiv (1, 1, 0)$ $(\hskip -.25cm\mod 2)$, 且 $a_{n+3}=a_{n+2}+a_n$,
則 $(a_{7m+1}, a_{7m+2}, \ldots, a_{7m+7})\equiv (1, 1, 0, 1, 0, 0, 1)$ $(\hskip -.25cm\mod 2)$
$(a_n, a_{2n})\equiv (1, 1)$ 或 $(0, 0)$ $(\hskip -.25cm\mod 2)$
特別是 $(T_{7m+1}, T_{7m+2}, \ldots, T_{7m+7})\equiv (1, 1, 0, 1, 0, 0, 1)$ $(\hskip -.25cm\mod 2)$
$(T_n, T_{2n})\equiv (1, 1)$ 或 $(0, 0)$ $(\hskip -.25cm\mod 2)$ $Q_n\in Z$ 。
證: (1).①當 $m=0$, $(a_1, a_2, \ldots, a_7)\equiv(1, 1, 0, a_3+a_1, a_4+a_2, a_5+a_3, a_6+a_4)$
$\equiv(1, 1, 0, 1, 0, 0, 1)$ $(\hskip -.25cm\mod 2)$
②設 $m=k$ 成立, 即設 $(a_{7k+1}, a_{7k+2}, \ldots, a_{7k+7})\equiv (1, 1, 0, 1, 0, 0, 1)$ $(\hskip -.25cm\mod 2)$
③當 $m=k+1$, $a_{7k+8}=a_{7k+7}+a_{7k+5}\equiv 1+0 \equiv 1$ $(\hskip -.25cm\mod 2)$
$a_{7k+9}=a_{7k+8}+a_{7k+6}\equiv 1+0 \equiv 1$ $(\hskip -.25cm\mod 2)$
$a_{7k+10}=a_{7k+9}+a_{7k+7}\equiv 1+1 \equiv 0$ $(\hskip -.25cm\mod 2)$
$a_{7k+11}=a_{7k+10}+a_{7k+8}\equiv 0+1 \equiv 1$ $(\hskip -.25cm\mod 2)$
$a_{7k+12}=a_{7k+11}+a_{7k+9}\equiv 1+1 \equiv 0$ $(\hskip -.25cm\mod 2)$
$a_{7k+13}=a_{7k+12}+a_{7k+10}\equiv 0+0 \equiv 0$ $(\hskip -.25cm\mod 2)$
$a_{7k+14}=a_{7k+13}+a_{7k+11}\equiv 0+1 \equiv 1$ $(\hskip -.25cm\mod 2)$,
由數學歸納法故得證。 \begin{eqnarray*} \hbox{(2).} \quad && \hskip -25pt (a_{2(7k+1)}, a_{2(7k+2)}, a_{2(7k+3)}, a_{2(7k+4)}, a_{2(7k+5)}, a_{2(7k+6)}, a_{2(7k+7)}) \hskip 2.6cm \\ &=& (a_{7h+2}, a_{7h+4}, a_{7h+6}, a_{7\ell+1}, a_{7\ell+3}, a_{7\ell+5}, a_{7\ell+7}) \\ &\equiv& (1, 1, 0, 1, 0, 0, 1) \equiv (a_{7k+1}, a_{7k+2}, \ldots, a_{7k+7}) (\hskip -.4cm\mod 2) \end{eqnarray*}
命題 4: $(T_{3m+1}, T_{3m+2}, T_{3m+3})\equiv (6, 5, 3)$ $(\hskip -.25cm\mod 7)$
證: (1). 當 $m=0$, $(T_1, T_2, T_3)\equiv (-1, 5, -4)\equiv (6, 5, 3)$ $(\hskip -.25cm\mod 7)$
(2). 設 $m=k$ 成立, 即設 $(T_{3k+1}, T_{3k+2}, T_{3k+3})\equiv (6, 5, 3)$ $(\hskip -.25cm\mod 7)$
(3). 當 $m=k+1$, $T_{3k+4}=-T_{3k+3}+2T_{3k+2}+T_{3k+1}\equiv -3+2\cdot 5+6\equiv 6$ $(\hskip -.25cm\mod 7)$
$T_{3k+5}=-T_{3k+4}+2T_{3k+3}+T_{3k+2}\equiv -6+2\cdot 3+5\equiv 5$ $(\hskip -.25cm\mod 7)$
$T_{3k+6}=-T_{3k+5}+2T_{3k+4}+T_{3k+3}\equiv -5+2\cdot 6+3\equiv 3$ $(\hskip -.25cm\mod 7)$
$\Rightarrow$ 由數學歸納法故得證。
定理 5: 1. 令 $f_n(x)=\displaystyle\Big (x-\Big (2\cos\frac{2\pi}{7}\Big )^n\Big ) \Big (x-\Big (2\cos\frac{4\pi}{7}\Big )^n\Big ) \Big (x-\Big (2\cos\frac{6\pi}{7}\Big )^n\Big )$, 則
$f_n(x)\in Z[x]$, 且 $f_n(x)=x^3-T_nx^2+Q_nx-1$。
2. 令 $g_n(x)=\displaystyle\Big (x-\Big (2\cos\frac{2\pi}{7}\Big )^{2^n}\Big ) \Big (x-\Big (2\cos\frac{4\pi}{7}\Big )^{2^n}\Big ) \Big (x-\Big (2\cos\frac{6\pi}{7}\Big )^{2^n}\Big )$,
則 $g_n(x)\in Z[x]$, 且 $g_n(x)=x^3-r_nx^2+s_nx-1$, 式中 $r_{n+1}=r_n^2-2s_n$, $s_{n+1}=s_n^2-2r_n$, 當 $n\ge 2$ 時 $r_n$, $s_n$ 皆為 13 的倍數。
證: (1). 由 $\displaystyle\alpha=2\cos\frac{2\pi}{7}$, $\displaystyle\beta=2\cos\frac{4\pi}{7}$, $\displaystyle\gamma=2\cos\frac{6\pi}{7}$ \begin{eqnarray*} && \Rightarrow \left \{ \begin{array}{l} \alpha^n + \beta^n + \gamma^n = T_n \\ \alpha^n \beta^n + \beta^n\gamma^n + \gamma^n\alpha^n =\displaystyle\frac{1}{2} \bigg [(\alpha^n+\beta^n+\gamma^n)^2 - \Big ((\alpha^n)^2+(\beta^n)^2+(\gamma^n)^2\Big )\bigg ] \\ \hskip 3.85cm =\displaystyle\frac{1}{2} [T_n^2-T_{2n}] = Q_n \in Z \\ \alpha^n \cdot \beta^n \cdot \gamma^n = (\alpha\beta\gamma)^n = 1 \end{array} \right. \end{eqnarray*}
(2). $r_{n+1}=\alpha^{2^{n+1}}+\beta^{2^{n+1}}+\gamma^{2^{n+1}}=(\alpha^{2^n}+\beta^{2^n}+\gamma^{2^n})^2 - 2(\alpha^{2^n}\beta^{2^n}+\beta^{2^n}\gamma^{2^n}+\gamma^{2^n}\alpha^{2^n})$
$=r_n^2-2s_n$ \begin{eqnarray*} s_{n+1} &=& \alpha^{2^{n+1}}\beta^{2^{n+1}} \!+\! \beta^{2^{n+1}}\gamma^{2^{n+1}} \!+\! \gamma^{2^{n+1}}\alpha^{2^{n+1}} \!=\! (\alpha^{2^n}\beta^{2^n})^2 \!+\! (\beta^{2^n}\gamma^{2^n})^2 \!+\! (\gamma^{2^n}\alpha^{2^n})^2 \\ &=& (\alpha^{2^n}\beta^{2^n}+\beta^{2^n}\gamma^{2^n}+\gamma^{2^n}\alpha^{2^n})^2 - 2(\alpha^{2^n}\beta^{2^n}\beta^{2^n}\gamma^{2^n} + \beta^{2^n}\gamma^{2^n}\gamma^{2^n}\alpha^{2^n} \\ && +\gamma^{2^n}\alpha^{2^n}\alpha^{2^n}\beta^{2^n}) \\ &=& (\alpha^{2^n}\beta^{2^n} + \beta^{2^n}\gamma^{2^n} + \gamma^{2^n}\alpha^{2^n})^2 - 2\alpha^{2^n} \beta^{2^n} \gamma^{2^n} (\beta^{2^n}+\gamma^{2^n}+\alpha^{2^n}) \\ &=& s_n^2 - 2 \cdot 1 \cdot r_n = s_n^2 - 2r_n. \end{eqnarray*}
(3). 由 $r_2=13$ 與 $s_2=26$, 知當 $n\ge 2$ 時 $r_n$ 與 $s_n$ 都是 13 的倍數。
定理 6: 令 \begin{eqnarray*} f_n(x)&=&\displaystyle\Big (x-\Big (2\cos\frac{2\pi}{7} 2\cos\frac{4\pi}{7}\Big )^n\Big ) \Big (x-\Big (2\cos\frac{4\pi}{7} 2\cos\frac{6\pi}{7}\Big )^n\Big ) \\ &&\times\displaystyle\Big (x-\Big (2\cos\frac{6\pi}{7} 2\cos\frac{2\pi}{7}\Big )^n\Big ), \end{eqnarray*} 則 $f_n(x)\in Z[x]$, 且 $f_n(x)=x^3-Q_nx^2+T_nx-1$。
證: 由 $\displaystyle\alpha=2\cos\frac{2\pi}{7}$, $\displaystyle\beta=2\cos\frac{4\pi}{7}$, $\displaystyle\gamma=2\cos\frac{6\pi}{7}$ \begin{eqnarray*} && \Rightarrow \left \{ \begin{array}{l} (\alpha\beta)^n + (\beta\gamma)^n + (\gamma\alpha)^n = \alpha^n\beta^n + \beta^n\gamma^n + \gamma^n\alpha^n = Q_n \\ (\alpha\beta)^n (\beta\gamma)^n + (\beta\gamma)^n (\gamma\alpha)^n + (\gamma\alpha)^n (\alpha\beta)^n = \alpha^n\beta^n\gamma^n \cdot (\beta^n+\gamma^n+\alpha^n) \\ \hskip 7.3cm = 1 \cdot T_n = T_n \\ (\alpha\beta)^n \cdot (\beta\gamma)^n \cdot (\gamma\alpha)^n = (\alpha^n \beta^n \gamma^n)^2 = 1^2 = 1 \end{array} \right. \end{eqnarray*}
引理 7: 1. $\displaystyle\sum_{k=1}^n \cos \frac{2k\pi}{2n+1}=-\frac{1}{2}$ 2. $\displaystyle\sum_{k=1}^n \cos \frac{2k\pi}{2n}=-1$
證: (1). 令 $\displaystyle\omega=\cos\frac{2\pi}{2n+1}+i\sin\frac{2\pi}{2n+1}$, 則 $\omega^{2n+1}=1$ 且 $\omega^{2n}+\omega^{2n-1}+\cdots+\omega^1+1=0$ \begin{eqnarray*} \hbox{左式 } &=& \frac{1}{2} (\omega^1+\omega^{-1}) + \frac{1}{2} (\omega^2+\omega^{-2}) + \cdots + \frac{1}{2} (\omega^n+\omega^{-n}) \\ &=& \frac{1}{2} (\omega^1 + \omega^{2n} + \omega^2 + \omega^{2n-1} + \cdots + \omega^n + \omega^{n+1}) = \frac{1}{2}(-1) = -\frac{1}{2} \end{eqnarray*}
(2). 令 $\displaystyle\omega=\cos\frac{2\pi}{2n}+i\sin\frac{2\pi}{2n}$, 則 $\omega^{2n}=1$ 且 $\omega^{2n-1}+\omega^{2n-2}+\cdots+\omega^1+1=0$ 且 $\omega^n=-1$ \begin{eqnarray*} \hbox{左式 } &=& \frac{1}{2} (\omega^1+\omega^{-1}) + \frac{1}{2} (\omega^2+\omega^{-2}) + \cdots + \frac{1}{2} (\omega^n+\omega^{-n}) \\ &=& \frac{1}{2} (\omega\!+\!\omega^{2n-1}\!+\!\omega^2\!+\!\omega^{2n-2}\!+\!\cdots\!+\!\omega^n\!+\!\omega^n) = \frac{1}{2} (-1+\omega^n) = \frac{1}{2} (-1-1) = -1 \end{eqnarray*}
接下來的證明中, $\displaystyle\prod_{k=1}^n \Big (x-2\cos\frac{2k\pi}{2n+1}\Big )=0$ 的根為 $\displaystyle\Big \{x_k=2\cos\frac{2k\pi}{2n+1}\Big| 1\le k\le n\Big \}$, 而 $z^{2n+1}=1$ 的前 $n$ 個根為 $\displaystyle\Big \{z_k=\omega^k=\cos\frac{2k\pi}{2n+1}+i\sin\frac{2k\pi}{2n+1}\Big| 1\le k\le n\Big \}$, 其關係為 $x_k=z_k+z_k^{-1}$, 即 $x=z+z^{-1}$ 。
定理 8: 1. 令 $f_n(x)=\displaystyle\prod_{k=1}^n \Big (x-2\cos\frac{2k\pi}{2n+1}\Big )$, 則 $f_n(x)\in Z[x]$, $f_{n+2}(x)=xf_{n+1}(x)-f_n(x)$, 且 $\displaystyle f_n(x)=x^n+x^{n-1}-(n-1)x^{n-2}-(n-2)x^{n-3} +\frac{1}{2}(n-2)(n-3)x^{n-4} + \frac{1}{2}(n-3)(n-4) x^{n-5}+\cdots+(-1)^{[\frac{n}{2}]}$
2. 令 $g_n(x)=\displaystyle\prod_{k=1}^n \Big (x-2\cos\frac{2k\pi}{2n}\Big )$, 則 $g_n(x)\in Z[x]$, $g_{n+2}(x)=xg_{n+1}(x)-g_n(x)$, 且 $\displaystyle g_n(x)=x^n+2x^{n-1}-(n-2)x^{n-2}-2(n-2)x^{n-3} +\frac{1}{2}(n-3)(n-4)x^{n-4}+\frac{2}{2}(n-3)(n-4)x^{n-5}+\cdots+\Big ((-1)^{[\frac{n}{2}]} + (-1)^{[\frac{n+3}{2}]}\Big )$
證: (1). ① 令 $\omega=\displaystyle\cos\frac{2\pi}{2n\!+\!1}+i\sin\frac{2\pi}{2n\!+\!1}$, 則 $\omega^{2n+1}=1$ 且$\omega^{2n}+\omega^{2n-1}+\cdots+\omega^1+1=0$ \begin{eqnarray*} f(x) &=& \prod_{k=1}^n \Big (x-2\cos\frac{2k\pi}{2n+1}\Big ) = \prod_{k=1}^n \Big (x - (\omega^k+\omega^{-k})\Big ) \\ &=& \prod_{k=1}^n \Big (x - (\omega^k+\omega^{2n+1-k})\Big ) = \prod_{k=1}^n \Big (z + z^{-1} - (\omega^k+\omega^{2n+1-k})\Big ) \\ &=& z^{-n} \prod_{k=1}^n \Big (z^2 - (\omega^k+\omega^{2n+1-k})z+1\Big ) = z^{-n} \prod_{k=1}^n (z-\omega^k)(z-\omega^{2n+1-k}) \\ &=& z^{-n} \prod_{k=1}^{2n} (z-\omega^k) = z^{-n} (z^{2n}+z^{2n-1}+\cdots+1) = z^n + z^{n-1} + \cdots + z^{-n}. \end{eqnarray*} \begin{eqnarray*} ② && x \cdot f_{n+1}(x) - f_n(x) \\ &=& (z+z^{-1})(z^{n+1}+z^n+\cdots+z^{-n-1}) - (z^n+z^{n-1}+\cdots+z^{-n}) \\ &=& (z^{n+2}\!+\!z^{n+1}\!+\!\cdots\!+\!z^{-n}) \!+\! (z^n\!+\!z^{n-1}\!+\!\cdots\!+\!z^{-n-2}) \!-\! (z^n\!+\!z^{n-1}\!+\!\cdots\!+\!z^{-n}) \\ &=& (z^{n+2}+z^{n+1}) + (z^n+z^{n-1}+\cdots+z^{-n-2}) = f_{n+2}(x). \end{eqnarray*}
(2).①當 $n\!=\!6$, $f_6(x)=x^6\!+\!x^5\!-\!5x^4\!-\!4x^3\!+\!6x^2\!+\!3x\!-\!1$ $\Rightarrow f_6(x)$ 降羃前 6 項成立
當 $n\!=\!7$, $f_7(x)=x^7\!+\!x^6\!-\!6x^5\!-\!5x^4\!+\!10x^3\!+\!6x^2\!-\!4x\!-\!1$ $\Rightarrow f_7(x)$ 降羃前 6 項成立
② 設 $n=k$ 成立, 即設 $f_k(x)$ 降羃前 6 項
為 $\displaystyle x^k + x^{k-1} - (k-1)x^{k-2} - (k-2)x^{k-3} + \frac{1}{2}(k-2)(k-3)x^{k-4} + \frac{1}{2}(k-3)(k-4)x^{k-5}$
設 $n=k+1$ 成立, 即設 $f_{k+1}(x)$ 降羃前 6 項
為 $\displaystyle x^{k+1} + x^k - kx^{k-1} - (k-1)x^{k-2} + \frac{1}{2}(k-1)(k-2)x^{k-3} + \frac{1}{2}(k-2)(k-3)x^{k-4}$
③當 $n=k+2$, $x\cdot f_{k+1}(x)-f_k(x)$ 降羃前『8』項 \begin{eqnarray*} \hbox{為} && x^{k+2} + x^{k+1} - kx^k - (k-1)x^{k-1} + \frac{1}{2}(k-1)(k-2)x^{k-2} \\ && + \frac{1}{2}(k-2)(k-3)x^{k-3} - \Big [x^k + x^{k-1} - (k-1)x^{k-2} - (k-2)x^{k-3} \\ && + \frac{1}{2}(k-2)(k-3)x^{k-4} + \frac{1}{2}(k-3)(k-4)x^{k-5}\Big ] \\ &=& x^{k+2} + x^{k+1} - (k+1)x^k - kx^{k-1} + \frac{1}{2}k(k-1)x^{k-2} \\ && + \frac{1}{2}(k-1)(k-2)x^{k-3} - \frac{1}{2}(k-2)(k-3)x^{k-4} - \frac{1}{2} (k-3)(k-4)x^{k-5} \end{eqnarray*} $\Rightarrow f_{k+2}(x)$ 降羃前 6 項成立, 由數學歸納法故得證。
(3). 設 $f_n(x)$ 常數項為 $c_n$, 由 $f_1(x)$ 與 $f_2(x)$, 知 $c_1=1$ 與 $c_2=-1$,
由 $f_{n+2}(x)=x\cdot f_{n+1}(x)-f_n(x)$, 知 $c_n=-c_{n-2}$,
故 $c_n$ 為 $1, -1, -1, 1, \ldots\ldots$ 的週期數列, 相同於 $(-1)^{[\frac{n}{2}]}$ 。
(以下為 $g_n(x)$ 的證明, 類似 $f_n(x)$ 的證明)
(4).①令 $\omega=\displaystyle\cos\frac{2\pi}{2n}\!+\!i\sin\frac{2\pi}{2n}$, 則 $\omega^{2n}=1$ 且 $\omega^{2n-1}\!+\!\omega^{2n-2}\!+\!\cdots\!+\!\omega^1\!+\!1=0$ 且 $\omega^n=-1$ \begin{eqnarray*} g(x) &=& \prod_{k=1}^n \Big (x-2\cos\frac{2k\pi}{2n}\Big ) = \prod_{k=1}^n \Big (x - (\omega^k+\omega^{-k})\Big ) \\ &=& \prod_{k=1}^n \Big (x - (\omega^k+\omega^{2n-k})\Big ) = \prod_{k=1}^n \Big (z + z^{-1} - (\omega^k+\omega^{2n-k})\Big ) \\ &=& z^{-n} \prod_{k=1}^n \Big (z^2 - (\omega^k+\omega^{2n-k})z+1\Big ) = z^{-n} \prod_{k-1}^n (z-\omega^k)(z-\omega^{2n-k}) \\ &=& z^{-n}(z-\omega^n) \prod_{k=1}^{2n-1} (z-\omega^k) = z^{-n} (z+1)(z^{2n-1}+z^{2n-2}+\cdots+1) \\ &=& (z+1)(z^{n-1}+z^{n-2}+\cdots+z^{-n}) \\[5pt] ② && x \cdot g_{n+1}(x) - g_n(x) \\ &=& (z\!+\!z^{-1})(z\!+\!1)(z^n+z^{n-1}+\cdots+z^{-n-1}) - (z\!+\!1)(z^{n-1}+z^{n-2}+\cdots+z^{-n}) \\ &=& (z+1) \Big [(z^{n+1}+z^n+\cdots+z^{-n}) + (z^{n-1}+z^{n-2}+\cdots+z^{-n-2}) \\ && - (z^{n-1}+z^{n-2}+\cdots+z^{-n})\Big ] \\ &=& (z+1) \Big [(z^{n+2}+z^{n+1}) + (z^n+z^{n-1}+\cdots+z^{-n-2})\Big ] = g_{n+2}(x) \end{eqnarray*}
(5).①當 $n=6$, $g_6(x)=x^6+2x^5-4x^4-8x^3+3x^2+6x$ $\Rightarrow g_6(x)$ 降羃前 6 項成立
當 $n=7$, $g_7(x)=x^7\!+\!2x^6\!-\!5x^5\!-\!10x^4\!+\!6x^3\!+\!12x^2\!-\!x\!-\!2$ $\Rightarrow g_7(x)$ 降羃前 6 項成立
②設 $n=k$ 成立, 即設 $g_k(x)$ 降羃前 6 項
為 $\displaystyle x^k+2x^{k-1}-(k\!-\!2)x^{k-2}-2(k\!-\!2)x^{k-3}+\frac{1}{2}(k\!-\!3)(k\!-\!4)x^{k-4}+\frac{2}{2}(k\!-\!3)(k\!-\!4)x^{k-5}$
設 $n=k+1$ 成立, 即設 $g_{k+1}(x)$ 降羃前 6 項
為 $\displaystyle x^{k+1}+2x^k-(k\!-\!1)x^{k-1}-2(k\!-\!1)x^{k-2}+\frac{1}{2}(k\!-\!2)(k\!-\!3)x^{k-3}+\frac{2}{2}(k\!-\!2)(k\!-\!3)x^{k-4}$
③當 $n=k+2$, $x\cdot g_{k+1}(x)-g_k(x)$ 降羃前『8』項 \begin{eqnarray*} \hbox{為} && x^{k+2} + 2x^{k+1} - (k-1)x^k - 2(k-1)x^{k-1} + \frac{1}{2}(k-2)(k-3)x^{k-2} \\ && + \frac{2}{2}(k-2)(k-3)x^{k-3} - \Big [x^k + 2x^{k-1} - (k-2)x^{k-2} - 2(k-2)x^{k-3} \\ && + \frac{1}{2}(k-3)(k-4)x^{k-4} + \frac{2}{2}(k-3)(k-4)x^{k-5}\Big ] \\ &=& x^{k+2} + 2x^{k+1} - kx^k - 2kx^{k-1} + \frac{1}{2}(k-1)(k-2)x^{k-2} \\ && + \frac{2}{2}(k-1)(k-2)x^{k-3} - \frac{1}{2}(k-3)(k-4)x^{k-4} \\ && - \frac{2}{2}(k-3)(k-4)x^{k-5} \end{eqnarray*} $\Rightarrow g_{k+2}(x)$ 降羃前 6 項成立, 由數學歸納法故得證。
(6). 設 $g_n(x)$ 常數項為 $c_n$, 由 $g_1(x)$ 與 $g_2(x)$, 知 $c_1=2$ 與 $c_2=0$,
由 $g_{n+2}(x)=x\cdot g_{n+1}(x)-g_n(x)$, 知 $c_n=-c_{n-2}$,
故 $c_n$ 為 $2, 0, -2, 0, \ldots\ldots$ 的週期數列, 相同於 $(-1)^{[\frac{n}{2}]}+(-1)^{[\frac{n+3}{2}]}$ 。
在此之前的分子都是偶數個 $\pi$, 若把分子改成奇數個 $\pi$, 由 \begin{eqnarray*} &&\Big (\cos\frac{\pi}{7}, \cos\frac{3\pi}{7}, \cos\frac{5\pi}{7}\Big ) =\Big (-\cos\frac{6\pi}{7}, -\cos\frac{4\pi}{7}, -\cos\frac{2\pi}{7}\Big ),\\ &&\Big \{2\cos\frac{(2k-1)\pi}{2n+1}\Big |1\le k\le n\Big \}= \Big \{-2\cos\frac{2k\pi}{2n+1}\Big |1\le k\le n\Big \}\\ {\hbox{及}} &&\Big \{2\cos\frac{(2k-1)\pi}{2n}\Big |1\le k\le n\Big \} =\Big \{-2\cos\frac{(2k-1)\pi}{2n}\Big |1\le k\le n\Big \}, \end{eqnarray*} 知其性質極為類似, 如下所列。
命題 9: 1. $\displaystyle\Big (x-2\cos\frac{\pi}{7}\Big ) \Big (x-2\cos\frac{3\pi}{7}\Big ) \Big (x-2\cos\frac{5\pi}{7}\Big )=x^3-x^2-2x+1$
2. 令 $K_n=\displaystyle\Big (2\cos\frac{\pi}{7}\Big )^n+\Big (2\cos\frac{3\pi}{7}\Big )^n+\Big (2\cos\frac{5\pi}{7}\Big )^n$ \begin{eqnarray*} \hbox{則} && (1) K_n = |T_n| \\[5pt] && (2) K_{n+3} = K_{n+2} + 2K_{n+1} - K_n \\[5pt] && (3) (K_{7m+1}, K_{7m+2}, \ldots, K_{7m+7}) \equiv (1, 1, 0, 1, 0, 0, 1) (\hskip -.4cm \mod 2) \hskip 2.5cm \\[5pt] && (4) (K_n, K_{2n}) \equiv (1, 1) \hbox{ 或 } (0, 0) (\hskip -.4cm \mod 2) \\[5pt] && (5) Q_n = \frac{1}{2}[T_n^2-T_{2n}] = \frac{1}{2}[K_n^2-K_{2n}] \\[5pt] && (6) (K_{7m+1}, K_{7m+2}, \ldots, K_{7m+7}) \equiv (1, 5, 4, 6, 2, 3) (\hskip -.4cm \mod 7) \end{eqnarray*}
3.(1)令 $f_n(x)=\displaystyle\Big (x-\Big (2\cos\frac{\pi}{7}\Big )^n\Big ) \Big (x-\Big (2\cos\frac{3\pi}{7}\Big )^n\Big ) \Big (x-\Big (2\cos\frac{5\pi}{7}\Big )^n\Big )$,
則 $f_n(x)\in Z[x]$, 且 $f_n(x)=x^3-|T_n|x^2+Q_nx-1$
(2)令 $g_n(x)=\displaystyle\Big (x-\Big (2\cos\frac{\pi}{7}\Big )^{2^n}\Big ) \Big (x-\Big (2\cos\frac{3\pi}{7}\Big )^{2^n}\Big ), \Big (x-\Big (2\cos\frac{5\pi}{7}\Big )^{2^n}\Big )$,
則 $g_n(x)\in Z[x]$, 且 $g_n(x)=x^3\!-\!r_nx^2\!+\!s_nx\!-\!1$, 式中 $r_{n+1}=r_n^2\!-\!2s_n$, $s_{n+1}=s_n^2\!-\!2r_n$,
當 $n\ge 2$ 時 $r_n$, $s_n$ 皆為 13 的倍數。
4. 令 $$f_n(x)=\displaystyle\Big (x\!-\!\Big (2\cos\frac{\pi}{7}2\cos\frac{3\pi}{7}\Big )^n\Big ) \Big (x\!-\!\Big (2\cos\frac{3\pi}{7}2\cos\frac{5\pi}{7}\Big )^n\Big ) \Big (x\!-\!\Big (2\cos\frac{5\pi}{7} 2\cos\frac{\pi}{7}\Big )^n\Big ),$$ 則 $f_n(x)\in Z[x]$, 且 $f_n(x)=x^3-Q_nx^2+|T_n|x-1$
5. (1)$\displaystyle\sum_{k=1}^n \cos\frac{(2k-1)\pi}{2n+1}=\frac{1}{2}$
(2)$ \displaystyle\sum_{k=1}^n \cos\frac{(2k-1)\pi}{2n}=0$
6. (1)令 $f_n(x)=\displaystyle\prod_{k=1}^n \Big (x-2\cos\frac{(2k-1)\pi}{2n+1}\Big )$, 則 $f_n(x)\in Z[x]$, $f_{n+2}(x)=xf_{n+1}(x)-$
$f_n(x)$, 且 $\displaystyle f_n(x)=x^n-x^{n-1}-(n-1)x^{n-2}+(n-2)x^{n-3}+\frac{1}{2}(n-2)(n-3)x^{n-4}$
$-\displaystyle\frac{1}{2}(n-3)(n-4)x^{n-5}+\cdots+ (-1)^{[\frac{n+1}{2}]}$
(2)令 $g_n(x)=\displaystyle\prod_{k=1}^n \Big (x-2\cos\frac{(2k-1)\pi}{2n}\Big )$, 則 $g_n(x)\in Z[x]$, $g_{n+2}(x)=xg_{n+1}(x)-g_n(x)$,
且 $g_n(x)=x^n-nx^{n-2}+\displaystyle\frac{1}{2}n(n-3)x^{n-4}+\cdots+\Big ((-1)^{[\frac{n}{2}]}+(-1)^{[\frac{n+1}{2}]}\Big )$
本文中的式子, 都經過撰寫 Mathematca 程式作預測與驗證, 不但省時與精確, 更能有效分析與歸納。
---本文作者任教於小港高中---