📝 题目
例 5 设 ${f}_{n}\left( x\right) = \cos x + {\cos }^{2}x + \cdots + {\cos }^{n}x$ . 求证:
(1)对任意自然数 $n$ ,方程 ${f}_{n}\left( x\right) = 1$ 在 $\lbrack 0,\pi /3)$ 内有且仅有一个根;
(2)设 ${x}_{n} \in \left\lbrack {0,\frac{\pi }{3}}\right)$ 是 ${f}_{n}\left( x\right) = 1$ 的根,则 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \frac{\pi }{3}}$ .
💡 答案与解析
证 (1) 当 $n = 1$ 时, ${x}_{1} = 0$ 适合 ${f}_{1}\left( {x}_{1}\right) = \cos 0 = 1$ ; 当 $n > 1$ 时, ${f}_{n}\left( 0\right) = n > 1$ ,而 ${f}_{n}\left( \frac{\pi }{3}\right) = 1 - \frac{1}{{2}^{n}} < 1$ ,根据连续函数的中间值定理, $\exists {x}_{n} \in \left( {0,\frac{\pi }{3}}\right)$ ,使得 ${f}_{n}\left( {x}_{n}\right) = 1$ . 又
$$ {f}_{n}^{\prime }\left( x\right) = - \sin x\left( {1 + 2\cos x + \cdots + n{\cos }^{n - 1}x}\right) < 0,\;x \in \left( {0,\pi /3}\right) . $$
这意味着 ${f}_{n}\left( x\right)$ 在 $\left( {0,\pi /3}\right)$ 内严格单调下降,从而上述 ${x}_{n}$ 惟一,即方程 ${f}_{n}\left( x\right) = 1$ 在 $\lbrack 0,\pi /3)$ 内有且仅有一个根. 到此 (1) 证毕.
关于 (2) 下面给出几种证法.
证法 1 因为
$$ {f}_{n}\left( \frac{\pi }{4}\right) = \cos \frac{\pi }{4} + {\cos }^{2}\frac{\pi }{4} + \cdots + {\cos }^{n}\frac{\pi }{4} $$
$$ = \left( {\sqrt{2} + 1}\right) \times \left( {1 - {\left( \frac{1}{\sqrt{2}}\right) }^{n}}\right) , $$
所以 $\mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( \frac{\pi }{4}\right) = \sqrt{2} + 1 > 1$ . 由极限定义, $\exists N$ ,使得
$$ {f}_{n}\left( \frac{\pi }{4}\right) > 1\;\left( {\forall n \geq N}\right) . $$
又 ${f}_{n}\left( x\right)$ 对 $x$ 严格单调下降, ${f}_{n}\left( {x}_{n}\right) = 1$ ,故有 ${x}_{n} > \pi /4\left( {\forall n \geq N}\right)$ . 进一步用微分中值定理,对 $\forall n \geq N,\exists {\xi }_{n} \in \left( {{x}_{n},\pi /3}\right)$ (此时 ${\xi }_{n} > {x}_{n} >$ $\pi /4)$ ,使得
$$ \left| {{f}_{n}\left( {x}_{n}\right) - {f}_{n}\left( \frac{\pi }{3}\right) }\right| = \left| {{f}_{n}^{\prime }\left( {\xi }_{n}\right) }\right| \left| {{x}_{n} - \frac{\pi }{3}}\right| $$
$$ \geq \left| {\sin {\xi }_{n}}\right| \left| {{x}_{n} - \frac{\pi }{3}}\right| \geq \frac{1}{\sqrt{2}}\left| {{x}_{n} - \frac{\pi }{3}}\right| , $$
从而 $0 \leq \left| {{x}_{n} - \frac{\pi }{3}}\right| \leq \sqrt{2}\left| {{f}_{n}\left( {x}_{n}\right) - {f}_{n}\left( \frac{\pi }{3}\right) }\right| = \frac{\sqrt{2}}{{2}^{n}}$ . 于是根据极限两边夹挤原理,推出 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \frac{\pi }{3}}$ .
证法 2 考虑辅助函数 $f\left( x\right) \frac{\operatorname{定义}\cos x}{1 - \cos x}\left( {0 < x \leq \frac{\pi }{3}}\right)$ . 因
$$ {f}^{\prime }\left( x\right) = - \frac{\sin x}{{\left( 1 - \cos x\right) }^{2}} < 0\;\left( {\forall 0 < x < \frac{\pi }{3}}\right) , $$
所以 $f\left( x\right)$ 在 $(0,\pi /3\rbrack$ 内严格单调下降. 因此, $\forall 0 < \varepsilon < \pi /3$ ,
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( {\frac{\pi }{3} - \varepsilon }\right) = f\left( {\frac{\pi }{3} - \varepsilon }\right) > f\left( \frac{\pi }{3}\right) = 1, $$
从而 $\exists N\left( \varepsilon \right)$ ,使得 $\forall n \geq N\left( \varepsilon \right)$ 时,有 ${f}_{n}\left( {\frac{\pi }{3} - \varepsilon }\right) > 1$ . 又 ${f}_{n}\left( x\right)$ 对 $x$ 严格单调下降, ${f}_{n}\left( {x}_{n}\right) = 1$ ,故有
$$ \frac{\pi }{3} - \varepsilon < {x}_{n} < \frac{\pi }{3}\;\left( {\forall n \geq N\left( \varepsilon \right) }\right) , $$
也就有 $\left| {{x}_{n} - \pi /3}\right| < \varepsilon \left( {\forall n \geq N\left( \varepsilon \right) }\right)$ ,即得 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \pi /3}$ .
证法 3 先证明 ${x}_{n}$ 是单调增加序列. 事实上,
$$ {f}_{n + 1}\left( x\right) > {f}_{n}\left( x\right) \;\left( {x \in \lbrack 0,\pi /3}\right) ). $$
由 ${f}_{n + 1}\left( {x}_{n + 1}\right) = 1 = {f}_{n}\left( {x}_{n}\right) < {f}_{n + 1}\left( {x}_{n}\right)$ ,及 ${f}_{n + 1}\left( x\right)$ 对 $x$ 在 $\lbrack 0,\pi /3)$ 内严格单调下降,即得 ${x}_{n} < {x}_{n + 1}\left( {n = 1,2,\cdots }\right)$ . 由 (1) 知 ${x}_{n} < \pi /3$ ,于是 $\left\{ {x}_{n}\right\}$ 是单调增加的有上界序列,从而极限 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}}$ 存在. 设 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = a}$ . 下面证明 $a = \pi /3$ .
由 ${x}_{n} < {x}_{n + 1}$ ,推出 $\cos {x}_{n} > \cos {x}_{n + 1}$ ,因此
$$ \cos {x}_{n} < \cos {x}_{2} < 1\;\left( {\forall n > 2}\right) , $$
进而有 $0 < {\cos }^{n + 1}{x}_{n} < {\cos }^{n + 1}{x}_{2}$ . 根据两边夹挤原理,可知 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{\cos }^{n + 1}{x}_{n} =}$ 0. 于是由
$$ 1 = {f}_{n}\left( {x}_{n}\right) = \frac{\cos {x}_{n} - {\cos }^{n + 1}{x}_{n}}{1 - \cos {x}_{n}} $$
两边令 $\displaystyle{n \rightarrow + \infty}$ 取极限,即得 $1 = \frac{\cos a}{1 - \cos a}$ ,从而 $a = \frac{\pi }{3}$ .
证法 4 因为 ${f}_{n}\left( x\right)$ 对 $n$ 单调增加,根据微分中值定理,
$$ \exists \xi \in \left( {\min \left( {{x}_{n},{x}_{n + 1}}\right) ,\max \left( {{x}_{n},{x}_{n + 1}}\right) }\right) \text{ , } $$
使得
$$ {f}_{n + 1}^{\prime }\left( \xi \right) \left( {{x}_{n + 1} - {x}_{n}}\right) = {f}_{n + 1}\left( {x}_{n + 1}\right) - {f}_{n + 1}\left( {x}_{n}\right) $$
$$ = 1 - {f}_{n + 1}\left( {x}_{n}\right) < 1 - {f}_{n}\left( {x}_{n}\right) = 0. $$
又 ${f}_{n + 1}^{\prime }\left( \xi \right) < 0$ ,即得 ${x}_{n + 1} - {x}_{n} > 0$ ,这说明 ${x}_{n} \leq {x}_{n + 1},{x}_{n}$ 单调上升且有界. 设 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = a}$ ,则 $a > {x}_{n} > 0$ . 因为 ${f}_{n}\left( x\right)$ 的极限函数
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = \frac{\cos x}{1 - \cos x}\;\left( {x \in \left\lbrack {{x}_{2},\frac{\pi }{3}}\right\rbrack }\right) $$
在 $\left\lbrack {{x}_{2},\frac{\pi }{3}}\right\rbrack$ 上连续,且 ${f}_{n + 1}\left( x\right) > {f}_{n}\left( x\right)$ . 根据狄尼定理,我们有
$$ {f}_{n}\left( x\right) \overset{-\text{ 致 }}{ \rightarrow }\frac{\cos x}{1 - \cos x}\;\left( {n \rightarrow \infty ,x \in \left\lbrack {{x}_{2},\frac{\pi }{3}}\right\rbrack }\right) , $$
从而 $1 \equiv {f}_{n}\left( {x}_{n}\right) \rightarrow \frac{\cos a}{1 - \cos a}\left( {n \rightarrow \infty }\right)$ ,即有 $\cos a = \frac{1}{2}$ ,或 $a = \frac{\pi }{3}$ .