多元函数的高阶微分公式 与 Taylor公式

设：
$m, n \in \mathbb N, m, n \ge 1,$
${\mathrm{D}}_j = \frac {\partial }{\partial x_j}$
${{\mathrm{D}}_j}^n = (\frac {\partial }{\partial x_j})^n = \frac {\partial^n }{\partial x_j^n}$
${{\mathrm{D}}_i}^m {{\mathrm{D}}_j}^n = (\frac {\partial }{\partial x_i})^m (\frac {\partial }{\partial x_j})^n= \frac {\partial^{m + n} }{\partial x_i^m \partial x_j^n}$

高阶微分公式

若： $y = f(\mathbf{x}): \mathbb R^n \rightarrow \mathbb R, k$ 阶可微， $n, k \in \mathbb N, n, k \ge 1,$
则：

$$\forall k \in \mathbb N, k \ge 1, {\mathrm{d} }^ky = \left [ \sum _{j = 1}^{n} \left (\mathrm{d} x_j {\mathrm{D}}_{j} \right ) \right ]^k y$$

证明：

$k = 1$ 时显然成立。
设$k$ 时成立，
由 $\left (\sum _{j = 1}^{n}x_j \right )^k = \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \prod_{j = 1}^{k} x_{\mathrm{n}_j}$ 可得：

${\mathrm{d} }^{k + 1}y = \mathrm{d} \left ( {\mathrm{d} }^ky \right ) = \mathrm{d} \left \{ \left [ \sum _{j = 1}^{n} \left (\mathrm{d} x_j {\mathrm{D}}_{j} \right ) \right ]^k y \right \} =$

$= \mathrm{d} \left [ \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \prod_{j = 1}^{k} \left (\mathrm{d} x_{n_j}{\mathrm{D}}_{{n_j}} \right ) y \right ]$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \left (\prod_{j = 1}^{k} \mathrm{d} x_{n_j} \right) \mathrm{d} \left (\prod_{j = 1}^{k} {\mathrm{D}}_{{n_j}} \right )y$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \left (\prod_{j = 1}^{k} \mathrm{d} x_{n_j} \right ) \left [ \sum_{1 \le \mathrm{n}_{k + 1} \le n} \mathrm{d} x_{\mathrm{n}_{k + 1}} {\mathrm{D}}_{{\mathrm{n}_{k + 1}}} \left (\prod_{j = 1}^{k} {\mathrm{D}}_{{n_j}} \right )y \right ]$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \left ( \prod_{j = 1}^{k} \mathrm{d} x_{n_j} \right ) \left ( \sum_{1 \le \mathrm{n}_{k + 1} \le n} \mathrm{d} x_{\mathrm{n}_{k + 1}} {\mathrm{D}}_{{\mathrm{n}_{k + 1}}} \prod_{j = 1}^{k} {\mathrm{D}}_{{n_j}} y \right )$

$= \sum _{1 \le \mathrm{n}_1, \cdots , \mathrm{n}_{k + 1} \le n }\left ( \prod_{j = 1}^{k} \mathrm{d} x_{n_j} \right ) \left ( \mathrm{d} x_{\mathrm{n}_{k + 1}} {\mathrm{D}}_{{\mathrm{n}_{k + 1}}} \prod_{j = 1}^{k} {\mathrm{D}}_{{n_j}} y \right )$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_{k + 1} \le n } \left (\prod_{j = 1}^{k + 1} \mathrm{d} x_{n_j} \right ) \left ( \prod_{j = 1}^{k + 1} {\mathrm{D}}_{{n_j}}y \right )$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_{k + 1} \le n } \prod_{j = 1}^{k + 1} \left (\mathrm{d} x_{n_j} {\mathrm{D}}_{{n_j}} \right )y$

$= \left [ \sum _{j = 1}^{n} \left (\mathrm{d} x_j {\mathrm{D}}_{j} \right ) \right ]^{k + 1} y$

Taylor公式

设 $f(x_1, \cdots, x _n)$ 在点 $\mathbf{x}$ 的邻域 $U$ 上有 $k + 1$ 阶连续偏导数，$k \in \mathbb N, k \ge 1,$ 则： $\forall \mathbf{x}' = \mathbf{x} + \Delta \mathbf{x} \in U, \exists \theta \in (0, 1),$
$\Delta y = \sum_{i = 1}^{k} \dfrac{1}{i!} \left [\sum _{j = 1}^{n} \left (\Delta x_j {\mathrm{D}}_{j} \right ) \right ]^i f(\mathbf{x}) + \dfrac{1}{(k + 1)!} \left [ \sum _{j = 1}^{n} \left (\Delta x_j {\mathrm{D}}_{j} \right ) \right ]^{k + 1} f(\mathbf{x} + \theta \Delta \mathbf{x})$

证明：

令 $\varphi (t) = f \left (\mathbf{x} + t \Delta \mathbf{x} \right ), t \in [0, 1],$
则由数学归纳法（证明在后面）可得：
$\forall i \in \mathbb N, i \ge 1,$ 若 $f(\mathbf{x})$ 在点 $(\mathbf{x} + t \Delta \mathbf{x})$ 有 $i$ 阶连续偏导数，则：
$\varphi ^{(i)} (t) = \left [\sum _{j = 1}^{n} \left (\Delta x_j {\mathrm{D}}_{j} \right ) \right ]^i f \left (\mathbf{x} + t \Delta \mathbf{x} \right ), \tag{*}$
于是 $\varphi (t)$ 在 $[0, 1]$ 有 $k$ 阶连续偏导数， 在 $(0, 1)$ 有 $k + 1$ 阶导数，
由 Taylor 公式， $\exists \theta \in (0, 1),$
$\Delta y = f(\mathbf{x} + \Delta \mathbf{x}) - f(\mathbf{x})$
$= \varphi (1) - \varphi (0) = \sum_{i = 1}^{k} \frac{ 1 }{i!} \cdot \varphi ^{(i)}(0) + \frac{ 1 }{(k + 1)!} \cdot \varphi ^{(k + 1)}(\theta)$
$\sum_{i = 1}^{k} \dfrac{1}{i!} \left [\sum _{j = 1}^{n} \left (\Delta x_j {\mathrm{D}}_{j} \right ) \right ]^i f(\mathbf{x}) + \dfrac{1}{(k + 1)!} \left [ \sum _{j = 1}^{n} \left (\Delta x_j {\mathrm{D}}_{j} \right ) \right ]^{k + 1} f(\mathbf{x} + \theta \Delta \mathbf{x})$

(*) 的证明：

由多元复合函数的求导法则， $i = 1$ 时显然成立。
设 $i = k$ 时成立，则 $i = k + 1$ 时：
由 $\left (\sum _{j = 1}^{n}x_j \right )^k = \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \prod_{j = 1}^{k} x_{\mathrm{n}_j}$ 可得：

$\varphi ^{(k + 1)} (t) = \dfrac {\mathrm{d}}{\mathrm{d} t} (\varphi ^{(k)} (t))$

$= \dfrac {\mathrm{d}}{\mathrm{d} t} \left [\sum _{j = 1}^{n} \left (\Delta x_j {\mathrm{D}}_{j} \right ) \right ]^k f \left (\mathbf{x} + t \Delta \mathbf{x} \right )$

$= \dfrac {\mathrm{d}}{\mathrm{d} t} \left [ \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \prod_{j = 1}^{k} (\Delta x_{n_j} {\mathrm{D}}_{n_j} ) f( \mathbf{x} + t \Delta \mathbf{x} ) \right ]$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \dfrac {\mathrm{d}}{\mathrm{d} t} \left [ \prod_{j = 1}^{k} (\Delta x_{n_j} {\mathrm{D}}_{n_j} ) f( \mathbf{x} + t \Delta \mathbf{x} ) \right ]$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \dfrac {\mathrm{d}}{\mathrm{d} t} \left [ \left ( \prod_{j = 1}^{k} \Delta x_{n_j} \right ) \left (\prod_{j = 1}^{k} {\mathrm{D}}_{n_j} \right ) f \left (\mathbf{x} + t \Delta \mathbf{x} \right ) \right ]$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \left ( \prod_{j = 1}^{k} \Delta x_{n_j} \right ) \dfrac {\mathrm{d}}{\mathrm{d} t} \left [ \prod_{j = 1}^{k} {\mathrm{D}}_{n_j} f \left (\mathbf{x} + t \Delta \mathbf{x} \right ) \right ]$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_k \le n } \left ( \prod_{j = 1}^{k} \Delta x_{n_j} \right ) \sum _{1 \le \mathrm{n}_{k + 1} \le n} \left( \Delta x_{n_{k + 1}} \mathrm{D}_{n_{k + 1}} \right ) \left [ \prod_{j = 1}^{k} {\mathrm{D}}_{n_j} f \left (\mathbf{x} + t \Delta \mathbf{x} \right ) \right ]$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_{k + 1} \le n } \left ( \prod_{j = 1}^{k} \Delta x_{n_j} \right ) \Delta x_{n_{k + 1}} \mathrm{D}_{n_{k + 1}} \prod_{j = 1}^{k} {\mathrm{D}}_{n_j} f \left (\mathbf{x} + t \Delta \mathbf{x} \right )$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_{k + 1} \le n } \left ( \prod_{j = 1}^{k + 1} \Delta x_{n_j} \right ) \left ( \prod_{j = 1}^{k + 1} {\mathrm{D}}_{n_j} \right ) f \left (\mathbf{x} + t \Delta \mathbf{x} \right )$

$= \sum _{1 \le \mathrm{n}_1, \cdots, \mathrm{n}_{k + 1} \le n } \prod_{j = 1}^{k + 1} \left (\Delta x_{n_j} {\mathrm{D}}_{n_j} \right ) f \left (\mathbf{x} + t \Delta \mathbf{x} \right )$

$= \left [\sum _{j = 1}^{n} \left (\Delta x_j {\mathrm{D}}_{j} \right ) \right ]^{k + 1} f \left (\mathbf{x} + t \Delta \mathbf{x} \right )$

推论

$\mathrm{d} f( \mathbf{x} ) \equiv 0 \Leftrightarrow f( \mathbf{x})$ 是常量。

证明：

1) $\Rightarrow$ : $\mathrm{d} f( \mathbf{x} ) \equiv 0 \Rightarrow \dfrac {\partial } {\partial x_j} f(\mathbf{x}) = 0, \forall j \in \mathbb N, 1 \le j \le n,$
由Taylor公式，
$\Delta y = \left [\sum _{j = 1}^{n} \left (\Delta x_j {\mathrm{D}}_{j} \right ) \right ] f(\mathbf{x} + \theta \Delta \mathbf{x})$
$= \sum _{j = 1}^{n} \Delta x_j \dfrac {\partial } {\partial x_j} f(\mathbf{x} + \theta \Delta \mathbf{x})$
$= 0$
$\Rightarrow f( \mathbf{x})$ 是常量。
2) $\Leftarrow$ : $f( \mathbf{x})$ 是常量 $\Rightarrow \mathrm{d} f( \mathbf{x} ) \equiv 0$