例 4.1.1 （例4）以下是带$r$阶Lagrange余项的$r-1$阶Taylor公式：$$\begin{array}{}\text{(1)}& \begin{array}{rl}f({\mathit{x}}_0+\mathit{v})=f({\mathit{x}}_0)& +\sum _{k=1}^{r-1}\frac{1}{k!}\sum _{i_1,\cdots ,i_k=1}^n\frac{{\partial }^{k}f}{\partial x_{i_k}\cdots \partial x_{i_1}}({\mathit{x}}_0)v_{i_1}\cdots v_{i_k}\\ & +\frac{1}{r!}\sum _{i_1,\cdots ,i_r=1}^n\frac{{\partial }^{r}f}{\partial x_{i_r}\cdots \partial x_{i_1}}({\mathit{x}}_0+\theta \mathit{v})v_{i_1}\cdots v_{i_r}\end{array}\end{array}$$ 其中$\theta \in (0,1)$。具体来说，设$f:{\mathbb{R}}^2\to \mathbb{R}\in {\mathcal{C}}^2$，对于带2阶Lagrange余项的1阶Taylor公式，我们有$$\begin{array}{}\text{(2)}& \begin{array}{rl}f(x,y)& =f(x_0,y_0)+\frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)\\ & +\frac{1}{2}[\frac{{\partial }^{2}f}{\partial x^2}(x_1,y_1)(x-x_0{)}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}(x_1,y_1)(x-x_0)(y-y_0)+\frac{{\partial }^{2}f}{\partial y^2}(x_1,y_1)(y-y_0{)}^2]\end{array}\end{array}$$ 其中$x_1=x_0+\theta (x-x_0)$，$y_1=y_0+\theta (y-y_0)$，$\theta \in (0,1)$。所以本题需要计算到函数的所有二阶偏导数，以得到带Lagrange余项的1阶Taylor公式。

例 4.1.2 （例6）以下是带Peano余项的$r$阶Taylor公式：$$\begin{array}{}\text{(3)}& f({\mathit{x}}_0+\mathit{v})=\sum _{k=0}^r\frac{1}{k!}\sum _{i_1,\cdots ,i_k=1}^n\frac{{\partial }^{k}f}{\partial x_{i_k}\cdots \partial x_{i_1}}({\mathit{x}}_0)v_{i_1}\cdots v_{i_k}+o(\parallel \mathit{v}{\parallel }^r)\end{array}$$ 具体来说，设$f:{\mathbb{R}}^2\to \mathbb{R}\in {\mathcal{C}}^2$，对于带2阶Lagrange余项的1阶Taylor公式，我们有$$\begin{array}{}\text{(4)}& \begin{array}{rl}f(x,y)& =f(x_0,y_0)+\frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)\\ & +\frac{1}{2}[\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)(x-x_0{)}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}(x_0,y_0)(x-x_0)(y-y_0)+\frac{{\partial }^{2}f}{\partial y^2}(x_0,y_0)(y-y_0{)}^2]\\ & +o((x-x_0{)}^2+(y-y_0{)}^2)\end{array}\end{array}$$ 所以本题需要计算到函数的所有二阶偏导数，以得到带Peano余项的2阶Taylor公式。

例 4.1.3 （例10）计算可知$$\begin{array}{}\text{(7)}& \mathrm{\nabla }u=(\genfrac{}{}{0ex}{}{-(1+{\mathrm{e}}^y)\sin x}{{\mathrm{e}}^y(\cos x-1-y)})\end{array}$$ 解得驻点$$\begin{array}{}\text{(8)}& (\genfrac{}{}{0ex}{}{2k\pi }{0}),(\genfrac{}{}{0ex}{}{(2k+1)\pi }{-2}),k\in \mathbb{Z}\end{array}$$ 随后分情况讨论。

例 4.1.4 （例12）仍然需要注意参数的取值范围。本题的答案为：$$\begin{array}{}\text{(9)}& (1)=\pi \ln \frac{|a|+|b|}{2},(2)=\frac{\pi }{2}\mathrm{sgn}\lambda \ln (1+|\lambda |)\end{array}$$