Taylor Series of Multivariable Functions

高阶偏导数

二元函数的二阶偏导数有四种场景

$$\begin{array}{rl}\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial x}\right)& =\frac{{\partial }^{2}z}{\partial x^2}=f_{xx}(x,y)\\ \frac{\partial }{\partial y}\left(\frac{\partial z}{\partial x}\right)& =\frac{{\partial }^{2}z}{\partial x^2}y=f_{xy}(x,y)\\ \frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right)& =\frac{{\partial }^{2}z}{\partial y^2}x=f_{yx}(x,y)\\ \frac{\partial }{\partial y}\left(\frac{\partial z}{\partial y}\right)& =\frac{{\partial }^{2}z}{\partial y^2}=f_{yy}(x,y)\end{array}$$

类似地也可定义三阶偏导数如$\frac{{\partial }^{3}z}{\partial x^2\partial y}=\frac{\partial }{\partial y}\left(\frac{{\partial }^{2}z}{\partial x^2}\right)$

Tip

注意，一般来说$\frac{\partial z}{\partial x}y\ne \frac{\partial z}{\partial y}x$，但是若$f_{xy}(x,y),f_{yx}(x,y)$都在点$(x_0,y_0)$连续，则有$f_{xy}(x_0,y_0)=f_{yx}(x_0,y_0)$，这个结论对$n$元函数的混合偏导数也成立

在本节中，若无特殊说明，都认为分析的函数无数阶连续可导，从而混合偏导数和求导顺序无关

中值定理

Note

若区域$D$上任意两点的连线都含于$D$，则称$D$为凸区域

设二元函数$f$在凸开域$D\subset {\mathbb{R}}^2$上可微，则对任意两点$P(a,b),Q(a+h,b+k)\in D$，则存在实数$\theta \in (0,1)$使得

$$f(a+h,b+k)=f(a,b)+f_x(a+\theta h,b+\theta k)h+f_y(a+\theta h,b+\theta k)k$$

一般形式的推广

设凸域$D\subset {\mathbb{R}}^n$且$f:D\to \mathbb{R}$可微，则任给$a,b\in D$存在$\xi \in D$使得

$$f(b)-f(a)=Jf(\xi )\cdot (b-a)$$

其中$\xi =a+\theta (b-a),\theta \in (0,1)$，即$\xi$位于连接$a,b$的直线段上，而关于$Jf(\xi )$见Jacobian矩阵

Tip

这说明了如果一个多元函数可微，则它可以用线性函数逼近，利用这一点我 们可以做近似计算

此中值定理对向量值函数不成立

Taylor定理

记号约定

设${\alpha }_i\in {\mathbb{R}}^{+}(1\le i\le n)$，记$\mathbf{\alpha }=({\alpha }_1,\dots ,{\alpha }_n{)}^t$称为多重指标，记

$$\left|\mathbf{\alpha }\right|=\sum _{i=1}^n{\alpha }_i,\mathbf{\alpha }!=\prod _{i=1}^n{\alpha }_i!$$

如果$\mathbf{x}=(x_1\dots ,x_n{)}^t\in {\mathbb{R}}^n$，则记

$$x^{\mathbf{\alpha }}=x_1^{{\alpha }_1}\dots x_n^{{\alpha }_n}$$

和

$$D^{\mathbf{\alpha }}f({\mathbf{x}}^0)=\frac{{\partial }^{|\mathbf{\alpha }|}f}{{\partial }_{x_1}^{{\alpha }_1}\cdots {\partial }_{x_n}^{{\alpha }_n}}({\mathbf{x}}^0)$$

二元情况

若函数$f$在点$P_0(x_0,y_0)$的某邻域$U(P_0)$上有直到$n+1$阶的连续偏导数，则对$U(P_0)$上的任一点$(x_0+h,y_0+k)$都存在相应的$\theta \in (0,1)$，使得

$$\begin{array}{rl}f(x_0+h,y_0+k)& =f(x_0,y_0)+\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)f(x_0,y_0)\\ & +\frac{1}{2!}{\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{2}f(x_0,y_0)+\cdots \\ & +\frac{1}{n!}{\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{n}f(x_0,y_0)\\ & +\frac{1}{(n+1)!}{\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{n+1}f(x_0+\theta h,y_0+\theta k)\end{array}$$

其中

$${\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{m}f(x_0,y_0)=\sum _{i=0}^m{\mathrm{C}}_m^i\frac{{\partial }^m}{\partial x^i\partial y^{m-i}}f(x_0,y_0)h^i{k}^{m-i}$$

多元情况

设凸域$D\subset {\mathbb{R}}^n$，$f\in C^{m+1}(D),\mathbf{a}=(a_1,\dots ,a_n{)}^t\in D$. 则任给$\mathbf{x}\in D$存在$\theta \in (0,1)$使得

$$f(\mathbf{x})=\sum _{k=0}^m\sum _{|\mathbf{\alpha }|=k}\frac{D^{\mathbf{\alpha }}f(\mathbf{a})}{\mathbf{\alpha }!}\cdot (\mathbf{x}-\mathbf{a}{)}^{\mathbf{\alpha }}+\sum _{|\mathbf{\alpha }|=m+1}\frac{D^{\mathbf{\alpha }}f(\mathbf{a}+\theta (\mathbf{x}-\mathbf{a}))}{\mathbf{\alpha }!}\cdot (\mathbf{x}-\mathbf{a}{)}^{\mathbf{\alpha }}$$