新讲 第12章 多元微分学 第6题
📝 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
$$
{R}_{n + 1} = \mathop{\sum }\limits_{{\left| \beta \right| = n + 1}}\frac{n + 1}{\beta !}{\int }_{0}^{1}{\left( 1 - t\right) }^{n}{\partial }^{\beta }f\left( {a + {th}}\right) \mathrm{d}t{h}^{\beta }.
$$