多元函数的泰勒展开式公式推导

前言

最优化问题中总会遇到多元函数的泰勒展开公式，没有推导过总是感觉很抽象，参考如下链接，文章讲得很清晰。

多元函数的泰勒(Taylor)展开式

推导过程

• 一元函数在点 ${\mathcal{X}}_k$处的泰勒展开式为：
  $$f(x)=f\left(x_k\right)+\left(x-x_k\right)f^{\prime }\left(x_k\right)+\frac{1}{2!}{\left(x-x_k\right)}^2{f}^{\prime \prime }\left(x_k\right)+o^n$$
• 二元函数在点$\left(x_k,y_k\right)$处的泰勒展开式为：
  $$\begin{array}{c}f(x,y)=f\left(x_k,y_k\right)+\left(x-x_k\right)f_x^{\prime }\left(x_k,y_k\right)+\left(y-y_k\right)f_y^{\prime }\left(x_k,y_k\right)\\ +\frac{1}{2!}{\left(x-x_k\right)}^2{f}_{xx}^{\prime \prime }\left(x_k,y_k\right)+\frac{1}{2!}\left(x-x_k\right)\left(y-y_k\right)f_{xy}^{\prime \prime }\left(x_k,y_k\right)\\ +\frac{1}{2!}\left(x-x_k\right)\left(y-y_k\right)f_{yx}^{\prime \prime }\left(x_k,y_k\right)+\frac{1}{2!}{\left(y-y_k\right)}^2{f}_{yy}^{\prime \prime }\left(x_k,y_k\right)\\ +o^n\end{array}$$
  上公式化成矩阵形式如下，可以和后面的多元函数泰勒展开的矩阵形式相互对照理解。
  $$f(x,y)=f\left(x_k,y_k\right)+\left[f_x^{\prime }\left(x_k,y_k\right)f_y^{\prime }\left(x_k,y_k\right)\right]\left[\begin{array}{c}x-x_k\\ y-y_k\end{array}\right]$$
  $$+\frac{1}{2!}\left[x-x_{k}y-y_k\right]\left[\begin{array}{cc}{f}_{xx}^{\prime \prime }\left(x_k,y_k\right)& f_{xy}^{\prime \prime }\left(x_k,y_k\right)\\ f_{yx}^{\prime \prime }\left(x_k,y_k\right)& f_{yy}^{\prime \prime }\left(x_k,y_k\right)\end{array}\right]\left[\begin{array}{c}x-x_k\\ y-y_k\end{array}\right]$$
• 多元函数在$\left(x_k^1,x_k^2,\dots ,x_k^n\right)$ 处的泰勒展开公式：
  $$\begin{array}{c}f\left(x^1,x^2,\dots ,x^n\right)=f\left(x_k^1,x_k^2,\dots ,x_k^n\right)+{\sum }_{i=1}^n\left(x^i-x_k^i\right)f_{x^i}^{\prime }\left(x_k^1,x_k^2,\dots ,x_k^n\right)\\ +\frac{1}{2!}{\sum }_{i,j=1}^n\left(x^i-x_k^i\right)\left(x^j-x_k^j\right)f_{ij}^{\prime \prime }\left(x_k^1,x_k^2,\dots ,x_k^n\right)\\ +o^n\end{array}$$
• 多元函数泰勒展开式写成矩阵的形式：
  $$f(\mathbf{x})=f\left({\mathbf{x}}_k\right)+{\left[\nabla f\left({\mathbf{x}}_k\right)\right]}^T\left(\mathbf{x}-{\mathbf{x}}_k\right)+\frac{1}{2!}{\left[\mathbf{x}-{\mathbf{x}}_k\right]}^{T}H\left({\mathbf{x}}_k\right)\left[\mathbf{x}-{\mathbf{x}}_k\right]+o^n$$
  $$H\left({\mathbf{x}}_k\right)=\left[\begin{array}{cccc}\frac{{\partial }^{2}f\left(x_k\right)}{\partial x_1^2}& \frac{{\partial }^{2}f\left(x_k\right)}{\partial x_1\partial x_2}& \cdots & \frac{{\partial }^{2}f\left(x_k\right)}{\partial x_1\partial x_n}\\ \frac{{\partial }^{2}f\left(x_k\right)}{\partial x_2\partial x_1}& \frac{{\partial }^{2}f\left(x_k\right)}{\partial x_2^2}& \cdots & \frac{{\partial }^{2}f\left(x_k\right)}{\partial x_2\partial x_n}\\ ⋮& ⋮& \ddots & ⋮\\ \frac{{\partial }^{2}f\left(x_k\right)}{\partial x_n\partial x_1}& \frac{{\partial }^{2}f\left(x_k\right)}{\partial x_n\partial x_2}& \cdots & \frac{{\partial }^{2}f\left(x_k\right)}{\partial x_n^2}\end{array}\right]$$
• 当为二元时，
  $$\nabla f\left(x_k\right)=\left[\begin{array}{l}{f}_x^{\prime }\left(x_k,y_k\right)\\ f_y^{\prime }\left(x_k,y_k\right)\end{array}\right]$$

$$\left[x-x_k\right]=\left[\begin{array}{l}x-x_k\\ y-y_k\end{array}\right]$$
$$H\left(x_k\right)=\left[\begin{array}{ll}{f}_{xx}^{\prime \prime }\left(x_k,y_k\right)& f_{xy}^{\prime \prime }\left(x_k,y_k\right)\\ f_{yx}^{\prime \prime }\left(x_k,y_k\right)& f_{yy}^{\prime \prime }\left(x_k,y_k\right)\end{array}\right]$$