###### tags: `數學問題` $\begin{align}\end{align}$ # 複數 令 $f(x)=\frac{1}{2}+\cos(x)+\cos(2x)+\cdots+\cos(10x)$。試證明 $$f(x)=\frac{\sin(\frac{21x}{2})}{2\sin(\frac{x}{2})}$$ --- ﹝法一﹞: \begin{align} f(x) & = \frac{1}{2}+\frac{2\sin(\frac{x}{2})\cos(x)+2\sin(\frac{x}{2})\cos(2x)+\cdots+2\sin(\frac{x}{2})\cos(10x)}{2\sin(\frac{x}{2})} \\ & =\frac{1}{2}+\frac{\sin\frac{3x}{2}-\sin\frac{x}{2}+\sin\frac{5x}{2}-\sin\frac{3x}{2}+\cdots+\sin\frac{21x}{2}-\sin\frac{19x}{2}}{2\sin(\frac{x}{2})} \\ & =\frac{1}{2}+\frac{\sin(\frac{21x}{2})-\sin(\frac{x}{2})}{2\sin(\frac{x}{2})}=\frac{\sin(\frac{21x}{2})}{2\sin(\frac{x}{2})} \end{align} ﹝法二﹞: 令 $z=\cos(x)+i\sin(x)$ $$\cos(kx)=\frac{z^k+\bar z^k}{2}$$ 所以 \begin{align} f(x) & = \frac{1}{2}+\frac{z+\bar z}{2}+\frac{z^2+\bar z^2}{2}+\cdots+\frac{z^{10}+\bar z^{10}}{2}\\ & =\frac{1}{2}\cdot\left(1+\frac{z(z^{10}-1)}{z-1}+\frac{\bar z(\bar z^{10}-1)}{\bar z -1}\right)\\ & =\frac{1}{2}\cdot\frac{2-z-\bar z+z^{11}-z+\bar z^{11}-\bar z}{(z-1)(\bar z-1)} \end{align} --- $f(x)=x^5+x^4+x^3+x^2+x+1$,$\omega$ 為方程式 $f(x)=0$ 的一根, 則 $(1-\omega)(1-\omega^2)(1-\omega^3)(1-\omega^4)(1-\omega^5)=$? --- $f(x)=(x+1)(x^4+x^2+1)$ 若 $\omega=-1$,則所求 $=0$ 若 $\omega$ 為虛根,則 $\omega^4+\omega^2+1=0 \Rightarrow \omega^2+\omega+1=0$ or $\omega^2-\omega+1=0$ $\omega^5=\pm\omega^2$ --- 令 $$w=\cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$$ 試求下列各式之值: (3) $(1+w^2)(1+w^4)(1+w^6)(1+w^8)(1+w^{10})(1+w^{12})$ --- 因為 $w$ 為方程式 $x^7=1$ 虛根,因此 $w^7=1$ 將所求次方超過 $7$ 的部份降下來,得到 \begin{align} & (1+w^2)(1+w^4)(1+w^6)(1+w^8)(1+w^{10})(1+w^{12})\\ = & (1+w^2)(1+w^4)(1+w^6)(1+w)(1+w^3)(1+w^5) \end{align} 又 $$x^6+x^5+x^4+x^3+x^2+x+1=(x-w)(x-w^2)(x-w^3)(x-w^4)(x-w^5)(x-w^6)$$ 令 $x=-1$ 代入上式,則 $$1=(1+w)(1+w^2)(1+w^3)(1+w^4)(1+w^5)(1+w^6)$$ 即為所求。 ﹝註﹞ 只要知道 $w$ 為方程式 $x^p=1$ ($p$ 為奇質數)的虛根,立刻寫出以下 $3$ 個性質: (1) $w^p=1$ $\rightarrow$ 用來降次方的,只要次方超過 $p$ 一律降下來; (2) $w^{p-1}+w^{p-2}+\cdots+w^2+w+1=0$ $\rightarrow$ 也就是次方連續 $p$ 個和為 $0$; (3) $x^{p-1}+x^{p-2}+\cdots+x^2+x+1=(x-w)(x-w^2)\cdots(x-w^{p-2})(x-w^{p-1})$ $\rightarrow$ 看看題目的所求像不像 $x$ 代多少進去?減號就代正的,加號就代負的。
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