and, 一般来说,

$$g^{(r)} = \frac{d^r g}{dt^r} = \left( h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^r f \quad (r = 0, 1, 2, \dots, n)$$

Putting $t = 1$ 代入 的泰勒展开式中 $g$ 即得所需结果。.

同样的方法可以用来将结果推广到任意多个变量。. 对于函数 $f(\mathbf{x})$ , where $\mathbf{x} = (x_1, x_2, \dots, x_n)$ , 我们有

$$\begin{aligned} f(\mathbf{a} + \mathbf{h}) &= f(\mathbf{a}) + \sum_{i=1}^n h_i \frac{\partial f}{\partial x_i}(\mathbf{a}) + \frac{1}{2!} \left( \sum_{i=1}^n h_i \frac{\partial}{\partial x_i} \right)^2 f(\mathbf{a}) + \dots \ &\quad + \frac{1}{(m-1)!} \left( \sum_{i=1}^n h_i \frac{\partial}{\partial x_i} \right)^{m-1} f(\mathbf{a}) + \frac{1}{m!} \left( \sum_{i=1}^n h_i \frac{\partial}{\partial x_i} \right)^m f(\mathbf{a} + \theta \mathbf{h}) \end{aligned} \quad (9.36)$$

where $0 \leq \theta \leq 1$ , 只要所有偏导数都存在且连续.

通过令 $h = x - a$ and $k = y - b$ in (9.35), 我们有泰勒展开的如下另一种形式:

$$\begin{aligned} f(x, y) &= f(a, b) + \frac{1}{1!} \left[ (x - a) \frac{\partial}{\partial x} + (y - b) \frac{\partial}{\partial y} \right] f(a, b) \ &\quad + \frac{1}{2!} \left[ (x - a) \frac{\partial}{\partial x} + (y - b) \frac{\partial}{\partial y} \right]^2 f(a, b) \ &\quad + \dots \ &\quad + \frac{1}{n!} \left[ (x - a) \frac{\partial}{\partial x} + (y - b) \frac{\partial}{\partial y} \right]^n f(a + \theta(x - a), b + \theta(y - b)) \end{aligned} \quad (9.37)$$

即称为 泰勒展开 of $f(x, y)$ 关于该点 $(a, b)$ .

Example 9.39

求函数的泰勒级数 $f(x, y) = \sin xy$ 绕该点 $(1, \frac{1}{3}\pi)$ , 忽略次数项 3 及以上.

Solution From (9.37) 所求级数为

$$\begin{aligned} f(x, y) &= f(1, \frac{1}{3}\pi) + \frac{1}{1!} \left[ (x - 1) \frac{\partial}{\partial x} + (y - \frac{1}{3}\pi) \frac{\partial}{\partial y} \right] f(1, \frac{1}{3}\pi) \ &\quad + \frac{1}{2!} \left[ (x - 1) \frac{\partial}{\partial x} + (y - \frac{1}{3}\pi) \frac{\partial}{\partial y} \right]^2 f(1, \frac{1}{3}\pi) \dots \end{aligned}$$