方企勤 第二章 一元函数微分学 第22题
📝 题目
例 22 设 ${f}_{n}\left( x\right) = {x}^{n} + {x}^{n - 1} + \cdots + {x}^{2} + x$ . 求证:
(1)对任意自然数 $n > 1$ ,方程 ${f}_{n}\left( x\right) = 1$ 在 $\left( {\frac{1}{2},1}\right)$ 内只有一个根;
(2)设 ${x}_{n} \in \left( {\frac{1}{2},1}\right)$ 是 ${f}_{n}\left( x\right) = 1$ 的根,则 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \frac{1}{2}}$ .
💡 答案解析
证 (1) 因为 ${f}_{n}\left( 1\right) - 1 = n - 1 > 0$ ,以及
$$ {f}_{n}\left( \frac{1}{2}\right) - 1 = \frac{1}{2} + {\left( \frac{1}{2}\right) }^{2} + \cdots + {\left( \frac{1}{2}\right) }^{n} - 1 $$
$$ = \frac{\frac{1}{2} - {\left( \frac{1}{2}\right) }^{n + 1}}{1 - \frac{1}{2}} - 1 = - {\left( \frac{1}{2}\right) }^{n} < 0, \tag{6.25} $$
所以根据连续函数的中间值定理,存在 ${x}_{n} \in \left( {\frac{1}{2},1}\right)$ ,使得 ${f}_{n}\left( {x}_{n}\right) - 1$ $= 0$ . 又因为
$$ {f}_{n}^{\prime }\left( x\right) = 1 + {2x} + \cdots + n{x}^{n - 1} \geq 1 > 0\;\left( {\forall x \geq 0}\right) , $$
(6.26)
所以 ${f}_{n}\left( x\right)$ 严格单调增加,从而 ${f}_{n}\left( x\right) = 1$ 的根 ${x}_{n} \in \left( {\frac{1}{2},1}\right)$ 是惟一的.
(2)证法 1 根据微分中值定理, $\exists \xi \in \left( {\frac{1}{2},{x}_{n}}\right)$ ,使得
$$ {f}_{n}\left( {x}_{n}\right) - {f}_{n}\left( \frac{1}{2}\right) = {f}_{n}^{\prime }\left( \xi \right) \left( {{x}_{n} - \frac{1}{2}}\right) , $$
联合 (6.26) 及 (6.25) 式,我们有
$$ 0 \leq \left| {{x}_{n} - \frac{1}{2}}\right| \leq \left| {{f}_{n}\left( {x}_{n}\right) - {f}_{n}\left( \frac{1}{2}\right) }\right| = \left| {1 - \left\lbrack {1 - {\left( \frac{1}{2}\right) }^{n}}\right\rbrack }\right| = \frac{1}{{2}^{n}}. $$
于是根据极限的两边夹挤准则,即得 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \frac{1}{2}}$ .
证法 2 从考虑 ${x}_{n}$ 的单调性入手. 因为对 $\forall x > 0$ ,有 ${f}_{n}\left( x\right) <$ ${f}_{n + 1}\left( x\right)$ ,所以
$$ {f}_{n + 1}\left( {x}_{n + 1}\right) = 1 = {f}_{n}\left( {x}_{n}\right) < {f}_{n + 1}\left( {x}_{n}\right) . \tag{6.27} $$
又 ${f}_{n + 1}\left( x\right)$ 是严格单调增加的,从而由 (6.27) 式知 ${x}_{n + 1} < {x}_{n}$ ,也就是序列 ${x}_{n}$ 是单调下降的. 又 ${x}_{n} > \frac{1}{2}$ ,也就是序列 ${x}_{n}$ 是有下界的. 于是极限 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}}$ 是存在的,可设 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = a}$ . 注意到
$$ {x}_{n} \leq {x}_{2} < 1\left( {\forall n > 1}\right) \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}^{n + 1} = 0, $$
对等式
$$ 1 = {f}_{n}\left( {x}_{n}\right) = {x}_{n}^{n} + {x}_{n}^{n - 1} + \cdots + {x}_{n}^{2} + {x}_{n} = \frac{{x}_{n} - {x}_{n}^{n + 1}}{1 - {x}_{n}}, $$
两端令 $\displaystyle{n \rightarrow \infty}$ 取极限,得 $1 = \frac{a}{1 - a}$ ,即得 $a = \frac{1}{2}$ .
证法 3 注意到
$$ 1 = {f}_{n}\left( {x}_{n}\right) = {x}_{n}^{n} + {x}_{n}^{n - 1} + \cdots + {x}_{n}^{2} + {x}_{n} $$
$$ = \frac{{x}_{n} - {x}_{n}^{n + 1}}{1 - {x}_{n}} \Rightarrow {x}_{n} = \frac{1}{2} + \frac{{x}_{n}^{n + 1}}{2}, \tag{6.28} $$
以及 ${f}_{n}\left( x\right)$ 的严格单调增加性,
$$ \left. \begin{aligned} {f}_{n}\left( \frac{2}{3}\right) & = \frac{2}{3} + {\left( \frac{2}{3}\right) }^{2} + \cdots + {\left( \frac{2}{3}\right) }^{n} \\ & = \frac{\frac{2}{3} - {\left( \frac{2}{3}\right) }^{n + 1}}{1 - \frac{2}{3}} = 2 - {\left( \frac{2}{3}\right) }^{n} > 1 \\ {f}_{n}\left( {x}_{n}\right) & = 1 \end{aligned}\right\} \Rightarrow {x}_{n} < \frac{2}{3}. $$
(6.29)
联立 (6.28) 与 (6.29) 式, 解得
$$ \frac{1}{2} < {x}_{n} < \frac{1}{2} + {\left\lbrack \frac{2}{3}\right\rbrack }^{n + 1}\text{ 由夹挤准则 }\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \frac{1}{2}. $$