📝 题目
例 9 设 $f\left( x\right)$ 在 $\mathbf{R}$ 上二次可微,且 $\forall x \in \mathbf{R}$ ,有
$$ \left| {f\left( x\right) }\right| \leq {M}_{0},\;\left| {{f}^{\prime \prime }\left( x\right) }\right| \leq {M}_{2}. $$
(1)写出 $f\left( {x + h}\right)$ , $f\left( {x - h}\right)$ 关于 $h$ 的带拉格朗日余项的泰勒公式;
(2)求证:对 $\forall h > 0$ ,有 $\left| {{f}^{\prime }\left( x\right) }\right| \leq \frac{{M}_{0}}{h} + \frac{h}{2}{M}_{2}$ ；
(3)求证: $\left| {{f}^{\prime }\left( x\right) }\right| \leq \sqrt{2{M}_{0}{M}_{2}}$ .
💡 答案与解析
解(1)
$$ f\left( {x + h}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) h + \frac{{f}^{\prime \prime }\left( {x + {\theta }_{1}h}\right) }{2}{h}^{2}\;\left( {0 < {\theta }_{1} < 1}\right) ; $$
$$ f\left( {x - h}\right) = f\left( x\right) - {f}^{\prime }\left( x\right) h + \frac{{f}^{\prime \prime }\left( {x - {\theta }_{2}h}\right) }{2}{h}^{2}\;\left( {0 < {\theta }_{2} < 1}\right) . $$
(2)将第(1)小题得到的两个泰勒公式相减,得
$$ 2{f}^{\prime }\left( x\right) h = f\left( {x + h}\right) - f\left( {x - h}\right) + \frac{{f}^{\prime \prime }\left( {x - {\theta }_{2}h}\right) }{2}{h}^{2} - \frac{{f}^{\prime \prime }\left( {x + {\theta }_{1}h}\right) }{2}{h}^{2}. $$
由此,利用条件 $\left| {f\left( x\right) }\right| \leq {M}_{0},\left| {{f}^{\prime \prime }\left( x\right) }\right| \leq {M}_{2}$ ,即得
$$ \left| {{f}^{\prime }\left( x\right) }\right| \leq \frac{{M}_{0}}{h} + \frac{h}{2}{M}_{2}. \tag{5.7} $$
(3)设 $g\left( h\right) \frac{\text{ 定义 }}{}\frac{{M}_{0}}{h} + \frac{h}{2}{M}_{2}$ ,则有
$$ g\left( h\right) \geq 2\sqrt{\frac{{M}_{0}}{h} \cdot \frac{h}{2}{M}_{2}} = 2\sqrt{2{M}_{0}{M}_{2}}, $$
其中等号当 $\frac{{M}_{0}}{h} = \frac{h}{2}{M}_{2}$ 时,即当 $h = \sqrt{\frac{2{M}_{0}}{{M}_{2}}}$ 时成立. 将此 $h$ 值代入 (5.7)式,即得
$$ \left| {{f}^{\prime }\left( x\right) }\right| \leq \sqrt{\frac{{M}_{0}{M}_{2}}{2}} + \sqrt{\frac{{M}_{0}{M}_{2}}{2}} = \sqrt{2{M}_{0}{M}_{2}}. $$