复习一元函数泰勒展开

一元函数$f(x)$在点$x_0$展开：

$f(x) = \sum_{n=1} \frac{1}{n!} f^{(n)} (x-x_0)^n$

从二元到多元

二元函数

首先看二元函数的二阶展开，对于二元函数$f(x, y)$在点$(x_0, y_0)$处展开式为：

$\begin{align*} f(x, y) =& f(x_0, y_0) + f'_x(x_0,y_0)(x-x_0) + f'_y(x_0,y_0)(y-y_0) \\ & + \frac{1}{2!}[f''_{xx}(x_0, y_0)(x-x_0)^2 + f''_{xy}(x_0,y_0)(x-x_0)(y-y_0) \\ & + f''_{yx}(x_0,y_0)(x-x_0)(y-y_0) + f''_{yy}(x_0,y_0)(y-y_0)^2] + O(x^2) \end{align*}$

转换为矩阵形式为：

$\begin{align*} f(x, y) =& f(x_0, y_0) + f'_x(x_0,y_0)(x-x_0) + f'_y(x_0,y_0)(y-y_0) \\ & + \frac{1}{2!} [ \begin{pmatrix} x-x_0,y-y_0 \end{pmatrix} \begin{pmatrix} f''_{xx} & f''_{xy} \\ f''_{yx} & f''_{yy} \end{pmatrix} \begin{pmatrix} x-x_0 \\ y-y_0 \end{pmatrix} ] \end{align*}$

推广到m元函数

由上的公式可以推出函数$f(x1,x_2,\ldots \space,x_m)$在点$(x{k1}, x{k2}, \ldots \space ,x{k2})$处展开的公式为：

$\begin{align*} f(x_1,x_2,\ldots \space,x_m) =& f(x_{k1}, x_{k2}, \ldots \space ,x_{km}) \\ & + \sum_{i=1}^{m} f'_{x_{ki}}(x_{k1}, x_{k2}, \ldots \space ,x_{k2})(x_i - x_{ki}) \\ & + \frac{1}{2!} [ \begin{pmatrix} x-x_{k1}, x-x_{k2}, \ldots, x-x_{km} \end{pmatrix} \begin{pmatrix} f''_{x_1 x_1} & f''_{x_1 x_2} & \cdots & f''_{x_1 x_m} \\ f''_{x_2 x_1} & f''_{x_2 x_2} & \cdots & f''_{x_2 x_m} \\ \vdots & \vdots & & \vdots \\ f''_{x_m x_1} & f''_{x_m x_2} & & f''_{x_m x_m} \end{pmatrix} \begin{pmatrix} x-x_{k1} \\ x-x_{k2} \\ \vdots \\ x-x_{km} \end{pmatrix} ] \end{align*}$

若记

$\begin{align*} X &= (x_1, x_2, \cdots, x_m)\\ X_k &= (x_{k1}, x_{k2}, \cdots, x_{km}) \\ X-X_k &= (x_1-x_{k0}, x_2-x_{k2}, \cdots, x_m-x_{km}) \\ f'_X &= (f'_{x{k1}}, f'_{x_{k2}}, \cdots, f'_{x_{km}} ) \end{align*}$ $H(X) = \begin{pmatrix} f''_{x_1 x_1} & f''_{x_1 x_2} & \cdots & f''_{x_1 x_m} \\ f''_{x_2 x_1} & f''_{x_2 x_2} & \cdots & f''_{x_2 x_m} \\ \vdots & \vdots & & \vdots \\ f''_{x_m x_1} & f''_{x_m x_2} & & f''_{x_m x_m} \end{pmatrix}$

则上式可以写为:

$f(X) = f(X_k) + f'_X · (X-X_k)^T + \frac{1}{2!}(X-X_k)H(X)(X-X_k)^T$