34306 無理有理一線牽

$Q_n=\displaystyle\Big (2\cos\frac{2\pi}{7} 2\cos\frac{4\pi}{7}\Big )^n +\Big (2\cos\frac{4\pi}{7} 2\cos\frac{6\pi}{7}\Big )^n +\Big (2\cos\frac{6\pi}{7} 2\cos\frac{2\pi}{7}\Big )^n$,

$Q_n=\displaystyle\frac{1}{2}[T_n^2-T_{2n}]$。

\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*}

$\vdots$

$\displaystyle\frac{x^{n+1}}{x-\alpha_m}=x^n+\sum_{i=1}^{n-1} \alpha_m^i x^{n-i}+\alpha_m^n+\frac{\alpha_m^{n+1}}{x-\alpha_m}$

$(a_n, a_{2n})\equiv (1, 1)$ 或 $(0, 0)$ $(\hskip -.25cm\mod 2)$

$(T_n, T_{2n})\equiv (1, 1)$ 或 $(0, 0)$ $(\hskip -.25cm\mod 2)$ $Q_n\in Z$ 。

$\equiv(1, 1, 0, 1, 0, 0, 1)$ $(\hskip -.25cm\mod 2)$

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 $m=k+1$, $a_{7k+8}=a_{7k+7}+a_{7k+5}\equiv 1+0 \equiv 1$ $(\hskip -.25cm\mod 2)$

$a_{7k+9}=a_{7k+8}+a_{7k+6}\equiv 1+0 \equiv 1$ $(\hskip -.25cm\mod 2)$

$a_{7k+10}=a_{7k+9}+a_{7k+7}\equiv 1+1 \equiv 0$ $(\hskip -.25cm\mod 2)$

$a_{7k+11}=a_{7k+10}+a_{7k+8}\equiv 0+1 \equiv 1$ $(\hskip -.25cm\mod 2)$

$a_{7k+12}=a_{7k+11}+a_{7k+9}\equiv 1+1 \equiv 0$ $(\hskip -.25cm\mod 2)$

$a_{7k+13}=a_{7k+12}+a_{7k+10}\equiv 0+0 \equiv 0$ $(\hskip -.25cm\mod 2)$

$a_{7k+14}=a_{7k+13}+a_{7k+11}\equiv 0+1 \equiv 1$ $(\hskip -.25cm\mod 2)$,

(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。

$f_n(x)\in Z[x]$, 且 $f_n(x)=x^3-T_nx^2+Q_nx-1$。

2. 令 $g_n(x)=\displaystyle\Big (x-\Big (2\cos\frac{2\pi}{7}\Big )^{2^n}\Big ) \Big (x-\Big (2\cos\frac{4\pi}{7}\Big )^{2^n}\Big ) \Big (x-\Big (2\cos\frac{6\pi}{7}\Big )^{2^n}\Big )$,

(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。

(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*}

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 $\displaystyle g_n(x)=x^n+2x^{n-1}-(n-2)x^{n-2}-2(n-2)x^{n-3} +\frac{1}{2}(n-3)(n-4)x^{n-4}+\frac{2}{2}(n-3)(n-4)x^{n-5}+\cdots+\Big ((-1)^{[\frac{n}{2}]} + (-1)^{[\frac{n+3}{2}]}\Big )$

(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

② 設 $n=k$ 成立, 即設 $f_k(x)$ 降羃前 6 項

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 $c_2=-1$,

(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

②設 $n=k$ 成立, 即設 $g_k(x)$ 降羃前 6 項

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 $c_2=0$,

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 $f_n(x)=\displaystyle\Big (x-\Big (2\cos\frac{\pi}{7}\Big )^n\Big ) \Big (x-\Big (2\cos\frac{3\pi}{7}\Big )^n\Big ) \Big (x-\Big (2\cos\frac{5\pi}{7}\Big )^n\Big )$,

(2)令 $g_n(x)=\displaystyle\Big (x-\Big (2\cos\frac{\pi}{7}\Big )^{2^n}\Big ) \Big (x-\Big (2\cos\frac{3\pi}{7}\Big )^{2^n}\Big ), \Big (x-\Big (2\cos\frac{5\pi}{7}\Big )^{2^n}\Big )$,

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 $g_n(x)\in Z[x]$, $g_{n+2}(x)=xg_{n+1}(x)-g_n(x)$,

### 參考文獻

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