📝 题目
例 6 中, 我们求得
$$ {\varphi }^{\left( k\right) }\left( t\right) = {\left( {h}_{1}\frac{\partial }{\partial {x}_{1}} + \cdots + {h}_{m}\frac{\partial }{\partial {x}_{m}}\right) }^{k}f\left( {a + {th}}\right) . $$
利用这计算结果, 就得到多元函数的泰勒公式
$$ f\left( {a + h}\right) = {T}_{n} + {R}_{n + 1}, $$
这里
$$ {T}_{n} = \mathop{\sum }\limits_{{p = 0}}^{n}\frac{1}{p!}{\left( \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}\frac{\partial }{\partial {x}_{i}}\right) }^{p}f\left( a\right) , $$
而余项 ${R}_{n + 1}$ 可以表示为
$$ {R}_{n + 1} = \frac{1}{\left( {n + 1}\right) !}{\left( \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}\frac{\partial }{\partial {x}_{i}}\right) }^{n + 1}f\left( {a + {\theta h}}\right) $$
$$ \left( {0 < \theta < 1}\right) \text{ , } $$
或者
$$ {R}_{n + 1} = \frac{1}{n!}{\int }_{0}^{1}{\left( 1 - t\right) }^{n}{\left( \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}\frac{\partial }{\partial {x}_{i}}\right) }^{n + 1}f\left( {a + {th}}\right) \mathrm{d}t. $$
定理 5 设函数 $f\left( x\right) = f\left( {{x}_{1},\cdots ,{x}_{m}}\right)$ 在点 $a = \left( {{a}_{1},\cdots ,{a}_{m}}\right)$ 邻近 $n$ 阶连续可微,则有
$$ f\left( {a + h}\right) = \mathop{\sum }\limits_{{p = 0}}^{n}\frac{1}{p!}{\left( \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}\frac{\partial }{\partial {x}_{i}}\right) }^{p}f\left( a\right) + o\left( {\parallel h{\parallel }^{n}}\right) . $$
这样的表示式被称为带小 $o$ 余项 (或佩亚诺型余项) 的泰勒公式.
💡 答案与解析
证明 根据定理 4 , 有这样的展式:
$$ f\left( {a + h}\right) = \mathop{\sum }\limits_{{p = 0}}^{{n - 1}}\frac{1}{p!}{\left( \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}\frac{\partial }{\partial {x}_{i}}\right) }^{p}f\left( a\right) + {R}_{n}, $$
$$ {R}_{n} = \frac{1}{n!}{\left( \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}\frac{\partial }{\partial {x}_{i}}\right) }^{n}f\left( {a + {\theta h}}\right) \;\left( {0 < \theta < 1}\right) . $$
由于各 $n$ 阶偏导数的连续性,对于
$$ {\alpha }_{1} + \cdots + {\alpha }_{m} = n,\;{\alpha }_{1},\cdots ,{\alpha }_{m}\text{ 为非负整数 } $$
应有
$$ \frac{{\partial }^{n}}{\partial {x}_{1}^{{a}_{1}}\cdots \partial {x}_{m}^{{a}_{m}}}f\left( {a + {\theta h}}\right) = \frac{{\partial }^{n}}{\partial {x}_{1}^{{a}_{1}}\cdots \partial {x}_{m}^{{a}_{m}}}f\left( a\right) + o\left( 1\right) . $$
又显然有
$$ {h}_{1}^{{a}_{1}}\cdots {h}_{m}^{{a}_{m}} = O\left( {\parallel h{\parallel }^{{a}_{1}}}\right) \cdots O\left( {\parallel h{\parallel }^{{a}_{m}}}\right) = O\left( {\parallel h{\parallel }^{n}}\right) . $$
所以
$$ {h}_{1}^{{\alpha }_{1}}\cdots {h}_{m}^{{\alpha }_{m}}\frac{{\partial }^{n}}{\partial {x}_{1}^{{\alpha }_{1}}\cdots \partial {x}_{m}^{{\alpha }_{m}}}f\left( {a + {\theta h}}\right) $$
$$ = {h}_{1}^{{a}_{1}}\cdots {h}_{m}^{{a}_{m}}\left( {\frac{{\partial }^{n}}{\partial {x}_{1}^{{a}_{1}}\cdots \partial {x}_{m}^{{a}_{m}}}f\left( a\right) + o\left( 1\right) }\right) $$
$$ = {h}_{1}^{{a}_{1}}\cdots {h}_{m}^{{a}_{m}}\frac{{\partial }^{n}}{\partial {x}_{1}^{{a}_{1}}\cdots \partial {x}_{m}^{{a}_{m}}}f\left( a\right) + o\left( {\parallel h{\parallel }^{n}}\right) . $$
由此得到
$$ {R}_{n} = \frac{1}{n!}{\left( \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}\frac{\partial }{\partial {x}_{i}}\right) }^{n}f\left( a\right) + o\left( {\parallel h{\parallel }^{n}}\right) . $$
采用重指标记号可以把多元函数的泰勒公式写成更紧凑的形式. 下面, 我们来介绍这种表示法.
设 ${\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{m}$ 是非负整数,我们把
$$ \alpha = \left( {{\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{m}}\right) $$
叫作以 ${\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{m}$ 为分量的一个重指标,并约定
$$ \left| \alpha \right| = {\alpha }_{1} + {\alpha }_{2} + \cdots + {\alpha }_{m}, $$
$$ \alpha ! = {\alpha }_{1}!{\alpha }_{2}!\cdots {\alpha }_{m}!. $$
对于 $h = \left( {{h}_{1},{h}_{2},\cdots ,{h}_{m}}\right) \in {\mathbb{R}}^{m}$ ,我们约定
$$ {h}^{a} = {h}_{1}^{{a}_{1}}{h}_{2}^{{a}_{2}}\cdots {h}_{m}^{{a}_{m}}. $$
我们还约定
$$ {\partial }^{a} = \frac{{\partial }^{\left| a\right| }}{\partial {x}_{1}^{{a}_{1}}\partial {x}_{2}^{{a}_{2}}\cdots \partial {x}_{m}^{{a}_{m}}}. $$
采用这些记号, 我们写出
$$ \frac{1}{p!}{\left( {h}_{1}\frac{\partial }{\partial {x}_{1}} + \cdots + {h}_{m}\frac{\partial }{\partial {x}_{m}}\right) }^{p} $$
$$ = \mathop{\sum }\limits_{{{a}_{1} + \cdots + {a}_{m} = p}}\frac{{h}_{1}^{{a}_{1}}\cdots {h}_{m}^{{a}_{m}}}{{\alpha }_{1}!\cdots {\alpha }_{m}!}\frac{{\partial }^{p}}{\partial {x}_{1}^{{a}_{1}}\cdots \partial {x}_{m}^{{a}_{m}}} $$
$$ = \mathop{\sum }\limits_{{\left| \alpha \right| = p}}\frac{{h}^{\alpha }}{\alpha !}{\partial }^{\alpha }. $$
于是,我们可以把 $m$ 元函数的泰勒公式写成更紧凑的形式
$$ f\left( {a + h}\right) = \mathop{\sum }\limits_{{p = 0}}^{n}\mathop{\sum }\limits_{{\left| \alpha \right| = p}}\frac{1}{\alpha !}{\partial }^{\alpha }f\left( a\right) {h}^{\alpha } + {R}_{n + 1} $$
$$ = \mathop{\sum }\limits_{{\left| a\right| = 0}}^{n}\frac{1}{\alpha !}{\partial }^{a}f\left( a\right) {h}^{a} + {R}_{n + 1} $$
$$ = \mathop{\sum }\limits_{{\left| a\right| \leq n}}\frac{1}{\alpha !}{\partial }^{a}f\left( a\right) {h}^{a} + {R}_{n + 1}. $$
余项 ${R}_{n + 1}$ 可以表示为
$$ {R}_{n + 1} = \mathop{\sum }\limits_{{\left| \beta \right| = n + 1}}\frac{1}{\beta !}{\partial }^{\beta }f\left( {a + {\theta h}}\right) {h}^{\beta } $$
$$ \left( {0 < \theta < 1}\right) \text{ , } $$
或者
$$ {R}_{n + 1} = \mathop{\sum }\limits_{{\left| \beta \right| = n + 1}}\frac{n + 1}{\beta !}{\int }_{0}^{1}{\left( 1 - t\right) }^{n}{\partial }^{\beta }f\left( {a + {th}}\right) \mathrm{d}t{h}^{\beta }. $$