泰勒级数是用无限项的连加式来表示一个函数。设$f(x)$在$x_0$处具有$n$阶导数，试找出一个关于$(x-x_0)$的$n$次多项式
$$p_n(x)=a_0+a_1(x-x_0)+a_2(x-x_0{)}^2+\cdots +a_n(x-x_0{)}^n$$

来近似表达$f(x)$，要求使得$p_n(x)$与$f(x)$之差是当 $x\to x_0$ 时比$(x-x_0{)}^n$高阶的无穷小。
设$p_n(x)$在$x_0$处的函数值及它的直到$n$阶导数在$x_0$处的值依次与$f(x_0),f^{\prime }(x_0),\cdots ,f^{(n)}(x_0)$相等，即满足
$$p_n(x_0)=f(x_0),p_n^{\prime }(x_0)=f^{\prime }(x_0),$$

$$p_n^{\prime \prime }(x_0)=f^{\prime \prime }(x_0),\cdots ,p_n^{(n)}(x_0)=f^{(n)}(x_0),$$
按这些等式来确定系数$a_0,a_1,a_2,\cdots ,a_n$。对$p_n(x)$求各阶导数，然后分别代入以上等式，得
$$a_0=f(x_0),1\cdot a_1=f^{\prime }(x_0),$$

$$2!a_2=f^{\prime \prime }(x_0),\cdots ,n!a_n=f^{(}n)(x_0)$$

可得
$$a_0=f(x_0),a_1=f^{\prime }(x_0),a_2=\frac{1}{2!}{f}^{\prime \prime }(x_0),\cdots ,a_n=\frac{1}{n!}{f}^{(n)}(x_0)$$

由此可得
$$p_n(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{f^{\prime \prime }(x_0)}{2!}(x-x_0{)}^2+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n$$

泰勒中值定理1

如果函数$f(x)$在$x_0$处具有$n$阶导数，那么存在$x_0$的一个领域，对于该邻域内的任一 $x$，有
$$f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{f^{\prime \prime }(x_0)}{2!}(x-x_0{)}^2+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n+R_n(x)$$

其中
$$R_n(x)=o((x-x_0{)}^n)$$

泰勒中值定理2

如果函数$f(x)$在$x_0$的某个邻域$U(x_0)$内具有$(n+1)$阶导数，那么对任一 $x\in U(x_0)$，有
$$f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{f^{\prime \prime }(x_0)}{2!}(x-x_0{)}^2+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n+R_n(x)$$

其中
$$R_n(x)=\frac{f^{(n+1)}(\zeta )}{(n+1)!}(x-x_0{)}^{n+1}$$

这里$\zeta$是$x_0$与$x$之间的某个值。

二元函数的泰勒级数

如果函数$f(x,y)$在$(x_0,y_0)$处具有$n$阶偏导数，那么存在$(x_0,y_0)$的一个领域，对于该邻域内的任一 $(x,y)$，有
$$\begin{array}{rl}f(x,y)=& f(x_0,y_0)+\\ & f_x^{\prime }(x_0,y_0)(x-x_0)+f_y^{\prime }(x_0,y_0)(y-y_0)+\\ & \frac{f_{xx}^{\prime \prime }(x_0,y_0)}{2!}(x-x_0{)}^2+\frac{f_{xy}^{\prime \prime }(x_0,y_0)}{2!}(x-x_0)(y-y_0)+\frac{f_{yx}^{\prime \prime }(x_0,y_0)}{2!}(x-x_0)(y-y_0)+\frac{f_{yy}^{\prime \prime }(x_0,y_0)}{2!}(y-y_0{)}^2\\ & \cdots +R_n(x,y)\end{array}$$

多元函数的泰勒级数

如果函数$f(x^n)$在$x_0^n$处具有$n$阶偏导数，其中$n=1,2,3,\cdots$，那么存在$x_0^n$的一个领域，对于该邻域内的任一 $x^n$，有
$$\begin{array}{rl}f(x^1,x^2,x^3,\cdots ,x^n)=& f(x_0^1,x_0^2,x_0^3,\cdots ,x_0^n)+\\ & \sum _{i=1}^n(x^i-x_0^i)f_{x^i}^{\prime }(x_0^1,x_0^2,x_0^3,\cdots ,x_0^n)+\\ & \frac{1}{2!}\sum _{i,j=1}^n(x^i-x_0^i)(x^j-x_0^j)f_{ij}^{\prime \prime }(x_0^1,x_0^2,x_0^3,\cdots ,x_0^n)+\\ & R_n(x^n)\end{array}$$

矢量函数的泰勒展开

对于矢量函数$\vec{F}(\vec{r})$，可将其看成三元函数$\vec{F}(x,y,z)$，根据上面的多元函数的泰勒级数，可以得到矢量函数的泰勒展开式为
$$\begin{array}{rl}\vec{F}(\vec{r})=& \vec{F}(x,y,z)\\ =& \vec{F}(x_0,y_0,z_0)+(x-x_0)\frac{\partial \vec{F}}{\partial x}+(y-y_0)\frac{\partial \vec{F}}{\partial y}+(z-z_0)\frac{\partial \vec{F}}{\partial z}+R_n(x,y,z)\\ =& \vec{F}({\vec{r}}_0)+[(\vec{r}-{\vec{r}}_0)\cdot \nabla ]\vec{F}+R_n(\vec{r})\end{array}$$

将泰勒级数写成矩阵形式

$$f(\mathbf{X})=f({\mathbf{X}}_0)+[\nabla f({\mathbf{X}}_0){]}^T+\frac{1}{2!}[\mathbf{X}-{\mathbf{X}}_k{]}^{T}H({\mathbf{X}}_0)(\mathbf{X}-{\mathbf{X}}_0)+R_n(\mathbf{X})$$

其中
$$H({\mathbf{X}}_0)=\left[\begin{array}{cccc}\frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_1^2}& \frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_1{x}_2}& \cdots & \frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_1{x}_n}\\ \frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_2{x}_1}& \frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_2^2}& \cdots & \frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_2{x}_n}\\ ⋮& ⋮& \ddots & ⋮\\ \frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_n{x}_1}& \frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_n{x}_2}& \cdots & \frac{{\partial }^{2}f({\mathbf{X}}_0)}{\partial x_n^2}\end{array}\right]$$