方企勤 第二章 一元函数微分学 第16题

证 令 $f\left( x\right) = 1 - x + \frac{{x}^{2}}{2} - \frac{{x}^{3}}{3} + \cdots + {\left( -1\right) }^{n}\frac{{x}^{n}}{n}$ ,则当 $n$ 为奇数时,因为 $\mathop{\lim }\limits_{{x \rightarrow \pm \infty }}f\left( x\right) = \mp \infty$ ,所以当 $x > 0$ 充分大时 $f\left( x\right) \cdot f\left( {-x}\right) <$ 0,此时在(-x, x)内必有 $f\left( x\right) = 0$ 的实根. 又对 $\forall x \in$ $\left( {-\infty , + \infty }\right)$ ,有

$$
{f}^{\prime }\left( x\right) = - 1 + x - {x}^{2} + {x}^{3} + \cdots + {\left( -1\right) }^{n}{x}^{n - 1}
$$

$$
= \left\{ \begin{array}{ll} - \frac{1 + {x}^{n}}{1 + x}, & x \neq - 1, \\ - n, & x = - 1 \end{array}\right.
$$

$$
< 0\text{ . }
$$

由此可见, $f\left( x\right)$ 严格单调下降,从而当 $n$ 为奇数时, $f\left( x\right) = 0$ 恰有一实根.

当 $n$ 为偶数时,

$$
{f}^{\prime }\left( x\right) = - 1 + x - {x}^{2} + {x}^{3} + \cdots + {\left( -1\right) }^{n}{x}^{n - 1}
$$

$$
= \left( {-1 + x}\right) + \left( {-{x}^{2} + {x}^{3}}\right) + \cdots + \left( {-{x}^{n - 2} + {x}^{n - 1}}\right)
$$

$$
= \left( {x - 1}\right) + {x}^{2}\left( {x - 1}\right) + \cdots + {x}^{n - 2}\left( {x - 1}\right)
$$

$$
= \left( {x - 1}\right) \left( {1 + {x}^{2} + \cdots + {x}^{n - 2}}\right)
$$

$$
= \left( {x - 1}\right) \frac{1 - {x}^{n}}{1 - {x}^{2}}\left\{ \begin{array}{ll} < 0, & x < 1, \\ = 0, & x = 1, \\ > 0, & x > 1. \end{array}\right.
$$

由此可见点 $x = 1$ 是函数 $f\left( x\right)$ 在 $\left( {-\infty , + \infty }\right)$ 内的惟一极值点,并且是极小值点,从而在点 $x = 1$ 处达到函数 $f\left( x\right)$ 在 $\left( {-\infty , + \infty }\right)$ 内的最小值. 又

$$
f\left( 1\right) = \left( {1 - 1}\right) + \left( {\frac{1}{2} - \frac{1}{3}}\right) + \left( {\frac{1}{4} - \frac{1}{5}}\right) + \cdots
$$

$$
+ \left( {\frac{1}{n - 2} - \frac{1}{n - 1}}\right) + \frac{1}{n} > 0
$$

$$
\Rightarrow f\left( x\right) \geq f\left( 1\right) > 0\text{ , }
$$

于是,当 $n$ 为偶数时方程无实根.