方企勤 第四章 级 数 第11题
📝 题目
例 11 设
$$ g\left( x\right) = \left\{ \begin{matrix} {x}^{2}\sin \frac{1}{x}, & x \neq 0, \\ 0, & x = 0, \end{matrix}\right. $$
令 $f\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{2}^{n}}g\left( {x - \frac{1}{n}}\right)$ . 求证:
(1) $f\left( x\right)$ 在 $\left( {\frac{1}{k + 1},\frac{1}{k - 1}}\right) \left( {k = 2,3,\cdots }\right)$ 上可导,且导数只在 $1/k$ 处不连续;
(2) $f\left( x\right)$ 在(0,1)上可导,且导数只在 $x = 1/k\left( {k = 2,3,\cdots }\right)$ 处不连续.
💡 答案解析
证 (1) 因为 $g\left( {x - \frac{1}{n}}\right) \in C\left( {0,1}\right)$ ,且
$$ \left| \frac{g\left( {x - \frac{1}{n}}\right) }{{2}^{n}}\right| \leq \frac{1}{{2}^{n}} $$
所以由连续性定理知 $f\left( x\right) \in C\left( {0,1}\right)$ . 又当 $n \neq k$ 时,
$$ {g}^{\prime }\left( {x - \frac{1}{n}}\right) = \left\{ \begin{array}{ll} 2\left( {x - \frac{1}{n}}\right) \sin \frac{1}{x - \frac{1}{n}} - \cos \frac{1}{x - \frac{1}{n}}, & x \neq \frac{1}{n}, \\ 0, & x = \frac{1}{n}, \end{array}\right. $$
因此 ${g}^{\prime }\left( {x - \frac{1}{n}}\right)$ 在 $\left( {\frac{1}{k + 1},\frac{1}{k - 1}}\right)$ 上连续,且 $\left| {{g}^{\prime }\left( {x - \frac{1}{n}}\right) }\right| \leq 3$ ,从而
$$ \mathop{\sum }\limits_{\substack{{n = 1} \\ {n \neq k} }}^{M}\frac{1}{{2}^{n}}{g}^{\prime }\left( {x - \frac{1}{n}}\right) $$
在 $\left( {\frac{1}{k + 1},\frac{1}{k - 1}}\right)$ 上一致收敛. 于是函数
$$ {f}_{k}\left( x\right) = \mathop{\sum }\limits_{\substack{{n = 1} \\ {n \neq k} }}^{\infty }\frac{1}{{2}^{n}}g\left( {x - \frac{1}{n}}\right) $$
在 $\left( {\frac{1}{k + 1},\frac{1}{k - 1}}\right)$ 上可导,且
$$ {f}_{k}^{\prime }\left( x\right) = \mathop{\sum }\limits_{\substack{{n = 1} \\ {n \neq k} }}^{\infty }\frac{1}{{2}^{n}}{g}^{\prime }\left( {x - \frac{1}{n}}\right) \in C\left( {\frac{1}{k + 1},\frac{1}{k - 1}}\right) . $$
又因为 $g\left( {x - \frac{1}{k}}\right)$ 在 $\left( {\frac{1}{k + 1},\frac{1}{k - 1}}\right)$ 上可导,导数在点 ${x}_{k} = \frac{1}{k}$ 处不连续,所以
$$ f\left( x\right) = \frac{1}{{2}^{k}}g\left( {x - \frac{1}{k}}\right) + {f}_{k}\left( x\right) $$
$$ = \frac{1}{{2}^{k}}g\left( {x - \frac{1}{k}}\right) + \mathop{\sum }\limits_{\substack{{n = 1} \\ {n \neq k} }}^{M}\frac{1}{{2}^{n}}g\left( {x - \frac{1}{n}}\right) $$
在 $\left( {\frac{1}{k + 1},\frac{1}{k - 1}}\right)$ 上可导,且导数只在点 ${x}_{k} = \frac{1}{k}$ 处不连续.
(2) 由 $\left( {0,1}\right) = \mathop{\bigcup }\limits_{{k = 2}}^{\infty }\left\lbrack {\frac{1}{k + 1},\frac{1}{k - 1}}\right)$ ,故由 (1) 知 $f\left( x\right)$ 在(0,1)上可导,且导数只在点 ${x}_{k} = \frac{1}{k}\left( {k = 2,3,\cdots }\right)$ 处不连续.