多元函数幂级数展开的 n 种常用等价形式

• 原文: https://zhuanlan.zhihu.com/p/62386078
• 发布日期: 2019-04-14
• 分类: 数学方法 / 物理数学工具箱

一元函数的幂级数展开

下面都是关于 $f\left(x\right)$ 在 $x=x_0$ 处展开的讨论.

初学时通常写作 f\left( x \right)=\sum\limits_{n=0}^{\infty }{\frac{f}^{\left( n \right)}\left( {x}_{0} \right)}{n!}{\left( x-{x}_{0} \right)}^{n} 其中 {f}^{\left( n \right)}\left( {x}_{0} \right)={\left. \frac{\text{d}^{n}f\left( \xi \right)}{\text{d}{\xi }^{n} \right|}_{\xi ={x}_{0}

物理学中常考察微小变动的 $n$ 阶近似,

也就会记: $\Delta x=x-x_0$, 这样就有:

f\left( {x}_{0}+\Delta x \right)=\sum\limits_{n=0}^{\infty }{\frac{f}^{\left( n \right)}\left( {x}_{0} \right)}{n!}\Delta {x}^{n}=\sum\limits_{n=0}^{\infty }{\frac{\Delta {x}^{n}{n!}{\left. \frac{\text{d}^{n}f\left( \xi \right)}{\text{d}{\xi }^{n} \right|}_{\xi ={x}_{0}=\sum\limits_{n=0}^{\infty }{\frac{\Delta {x}^{n}{n!}\frac{\text{d}^{n}f\left( {x}_{0} \right)}{\text{d}x_{0}^{n}.

其实最后那一下好像也没哪儿是这么写的, $x_0$究竟是不是个常数, 说到底还是看你自己怎么认为. 当然啦, 相对展开的过程而言 $x_0$ 肯定是个常数, 是个定点. 但是 {\left. \frac{d}^{n}f\left( \xi \right)}{d{\xi }^{n} \right|}_{\xi ={x}_{0} 这一小块仅仅只是为了求出这个函数的 $n$ 阶导函数, 然后考察导函数在 $x_0$ 处的取值罢了, 那么 $x_0$ 在这儿也可以看作变量, 最后求出来的结果还是一样的.

二元函数的幂级数展开

下面都是关于 $f\left(x_1,x_2\right)$ 在 $\left(x_1,x_2\right)=\left(a_1,a_2\right)$ 处展开的讨论.

与一元的情形相仿, 我们也记: $\Delta x_i=x_i-a_i,i\in \left\{1,2\right\}.$

简洁起见, 引入向量记号 $\vec{x}=\left(x_1,x_2\right),\vec{a}=\left(a_1,a_2\right)$, 这显然有 $\Delta \vec{x}=\vec{x}-\vec{a}.$

\begin{align} & f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\sum\limits_{i=1}^{2}{\Delta {x}_{i}{\left. \frac{\partial f\left( {\vec{\xi } \right)}{\partial {\xi }_{i} \right|}_{\vec{\xi }=\vec{a}+}\sum\limits_{i,j=1}^{2}{\frac{\Delta {x}_{i}\Delta {x}_{j}{2!}{\left. \frac{\partial }^{2}f\left( {\vec{\xi } \right)}{\partial {\xi }_{i}\partial {\xi }_{j} \right|}_{\vec{\xi }=\vec{a} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i,j,k=1}^{2}{\frac{\Delta {x}_{i}\Delta {x}_{j}\Delta {x}_{k}{3!}{\left. \frac{\partial }^{3}f\left( {\vec{\xi } \right)}{\partial {\xi }_{i}\partial {\xi }_{j}\partial {\xi }_{k} \right|}_{\vec{\xi }=\vec{a}+\cdot \cdot \cdot } \\ \end{align}

或者 f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\sum\limits_{i=1}^{2}{\Delta {x}_{i}\frac{\partial f\left( {\vec{a} \right)}{\partial {a}_{i}+}\sum\limits_{i,j=1}^{2}{\frac{\Delta {x}_{i}\Delta {x}_{j}{2!}\frac{\partial }^{2}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j}+\sum\limits_{i,j,k=1}^{2}{\frac{\Delta {x}_{i}\Delta {x}_{j}\Delta {x}_{k}{3!}\frac{\partial }^{3}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j}\partial {a}_{k}+\cdot \cdot \cdot }

回到最初的记录方法就是: \begin{align} & f\left( {\vec{x} \right)=f\left( {\vec{a} \right)+\sum\limits_{i=1}^{2}{\left( {x}_{i}-{a}_{i} \right)\frac{\partial f\left( {\vec{a} \right)}{\partial {a}_{i}+}\sum\limits_{i,j=1}^{2}{\frac{\left( {x}_{i}-{a}_{i} \right)\left( {x}_{j}-{a}_{j} \right)}{2!}\frac{\partial }^{2}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i,j,k=1}^{2}{\frac{\left( {x}_{i}-{a}_{i} \right)\left( {x}_{j}-{a}_{j} \right)\left( {x}_{k}-{a}_{k} \right)}{3!}\frac{\partial }^{3}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j}\partial {a}_{k}+\cdot \cdot \cdot } \\ \end{align}

三个记法没别的意思, 就是希望能熟练这种等价转换.

然后还有一个对不熟练求和记号的人而言比较直观的记法:

\begin{align} & f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2} \right)f\left( {\vec{a} \right)+\frac{1}{2!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2} \right)}^{2}f\left( {\vec{a} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{1}{3!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2} \right)}^{3}f\left( {\vec{a} \right)+\cdot \cdot \cdot \\ \end{align}

即 f\left( \vec{a}+\Delta \vec{x} \right)=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2} \right)}^{n}f\left( {\vec{a} \right)}.

甚至 f\left( \vec{a}+\Delta \vec{x} \right)=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left( \Delta \vec{x}\cdot \frac{\partial }{\partial \vec{a} \right)}^{n}f\left( {\vec{a} \right)}.

多元函数的幂级数展开

下面都是关于 $f\left(x_1,x_2,...,x_s\right)$ 在 $\left(a_1,a_2,...,a_s\right)$ 处展开的讨论.

虽然类比起来是简单的, 但我还是写出来给你看好了:

\begin{align} & f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\sum\limits_{i=1}^{s}{\Delta {x}_{i}{\left. \frac{\partial f\left( {\vec{\xi } \right)}{\partial {\xi }_{i} \right|}_{\vec{\xi }=\vec{a}+}\sum\limits_{i,j=1}^{s}{\frac{\Delta {x}_{i}\Delta {x}_{j}{2!}{\left. \frac{\partial }^{2}f\left( {\vec{\xi } \right)}{\partial {\xi }_{i}\partial {\xi }_{j} \right|}_{\vec{\xi }=\vec{a} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i,j,k=1}^{s}{\frac{\Delta {x}_{i}\Delta {x}_{j}\Delta {x}_{k}{3!}{\left. \frac{\partial }^{3}f\left( {\vec{\xi } \right)}{\partial {\xi }_{i}\partial {\xi }_{j}\partial {\xi }_{k} \right|}_{\vec{\xi }=\vec{a}+\cdot \cdot \cdot } \\ \end{align}

或者 f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\sum\limits_{i=1}^{s}{\Delta {x}_{i}\frac{\partial f\left( {\vec{a} \right)}{\partial {a}_{i}+}\sum\limits_{i,j=1}^{s}{\frac{\Delta {x}_{i}\Delta {x}_{j}{2!}\frac{\partial }^{2}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j}+\sum\limits_{i,j,k=1}^{s}{\frac{\Delta {x}_{i}\Delta {x}_{j}\Delta {x}_{k}{3!}\frac{\partial }^{3}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j}\partial {a}_{k}+\cdot \cdot \cdot }

回到最初的记录方法就是:

\begin{align} & f\left( {\vec{x} \right)=f\left( {\vec{a} \right)+\sum\limits_{i=1}^{s}{\left( {x}_{i}-{a}_{i} \right)\frac{\partial f\left( {\vec{a} \right)}{\partial {a}_{i}+}\sum\limits_{i,j=1}^{s}{\frac{\left( {x}_{i}-{a}_{i} \right)\left( {x}_{j}-{a}_{j} \right)}{2!}\frac{\partial }^{2}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i,j,k=1}^{s}{\frac{\left( {x}_{i}-{a}_{i} \right)\left( {x}_{j}-{a}_{j} \right)\left( {x}_{k}-{a}_{k} \right)}{3!}\frac{\partial }^{3}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j}\partial {a}_{k}+\cdot \cdot \cdot } \\ \end{align}

三个记法没别的意思, 就是希望你能熟练这种等价转换.

同样给出相对求和符号更直观的记法:

\begin{align} & f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2}+\cdot \cdot \cdot +\Delta {x}_{s}\frac{\partial }{\partial {a}_{s} \right)f\left( {\vec{a} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{1}{2!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2}+\cdot \cdot \cdot +\Delta {x}_{s}\frac{\partial }{\partial {a}_{s} \right)}^{2}f\left( {\vec{a} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{1}{3!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2}+\cdot \cdot \cdot +\Delta {x}_{s}\frac{\partial }{\partial {a}_{s} \right)}^{3}f\left( {\vec{a} \right)+\cdot \cdot \cdot \\ \end{align}

即 f\left( \vec{a}+\Delta \vec{x} \right)=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2}+\cdot \cdot \cdot +\Delta {x}_{s}\frac{\partial }{\partial {a}_{s} \right)}^{n}f\left( {\vec{a} \right)}

$★$特别地, 当$s=3$时:

\Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2}+\Delta {x}_{s}\frac{\partial }{\partial {a}_{3}=\Delta \vec{x}\cdot {\nabla }_{\vec{a} , 其中 {\nabla }_{\vec{a}=\left( \frac{\partial }{\partial {a}_{1},\frac{\partial }{\partial {a}_{2},\frac{\partial }{\partial {a}_{3} \right)

于是就可以写成

f\left( \vec{a}+\Delta \vec{x} \right)=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2}+\Delta {x}_{3}\frac{\partial }{\partial {a}_{3} \right)}^{n}f\left( {\vec{a} \right)}=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left( \Delta \vec{x}\cdot {\nabla }_{\vec{a} \right)}^{n}f\left( {\vec{a} \right)}

但一般也就展开到二阶: f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\Delta \vec{x}\cdot {\nabla }_{\vec{a}f\left( {\vec{a} \right)+\frac{1}{2!}{\left( \Delta \vec{x}\cdot {\nabla }_{\vec{a} \right)}^{2}f\left( {\vec{a} \right)+\cdot \cdot \cdot

举一个实际运用的例子,也就是电动力学中的电势多级展开:

将\ \varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\iiint_{V}'}{\frac{\rho \left( {\vec{x}'} \right)}{r}\text{d}{V}'} 中的 $\frac{1}{r}$ 展开是如何完成的呢?

前面得到的三元函数展开公式先写出来:

f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\Delta \vec{x}\cdot {\nabla }_{\vec{a}f\left( {\vec{a} \right)+\frac{1}{2!}{\left( \Delta \vec{x}\cdot {\nabla }_{\vec{a} \right)}^{2}f\left( {\vec{a} \right)+\cdot \cdot \cdot

其中的 {\vec{x} 相当于 \vec{x}-{\vec{x}' 或 $r$, 展开点 {\vec{a} 相当于 {\vec{x} 或 $R$, 而 $\Delta \vec{x}$ 相当于-{\vec{x}'.

再利用上述结论:

\begin{align} & \frac{1}{r}=\frac{1}{\left| \vec{x}-{\vec{x}' \right|}=f\left( \vec{x}-{\vec{x}' \right)=f\left( {\vec{x} \right)+\left( -{\vec{x}' \right)\cdot \nabla f\left( {\vec{x} \right)+\frac{1}{2}{\left( -\vec{x}\cdot \nabla \right)}^{2}f\left( {\vec{x} \right)\cdot \cdot \cdot \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{R}-{\vec{x}'\cdot \nabla \frac{1}{R}+\frac{1}{2}{\left( {x}'}_{1}\frac{\partial }{\partial {x}_{1}+{x}'}_{2}\frac{\partial }{\partial {x}_{2}+{x}'}_{3}\frac{\partial }{\partial {x}_{3} \right)}^{\text{2}\frac{\text{1}{R}+\cdot \cdot \cdot \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{R}-{\vec{x}'\cdot \nabla \frac{1}{R}+\frac{1}{2}\sum\limits_{i,j=1}^{3}{x}'}_{i}{x}'}_{j}\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}+}\cdot \cdot \cdot \\ \end{align}

其中 R=\left| {\vec{x} \right|, 且做多级展开时满足 $r\approx R=x\gg x^{\prime }.${\vec{x}'={x}'{\vec{e}_{x}+{y}'{\vec{e}_{y}+{z}'{\vec{e}_{z} 表源点, 而 {\vec{x} 表场点, 且记 \vec{r}=\vec{x}-{\vec{x}'. 此外 $\nabla$ 与 {\nabla }'} 也是完全不一样的东西,分别是对场坐标与源坐标求偏导.

于是乎:

\varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\iiint_{V}'}{\rho \left( {\vec{x}'} \right)\left[ \frac{1}{R}-{\vec{x}'\cdot \nabla \frac{1}{R}+\frac{1}{2}\sum\limits_{i,j=1}^{3}{x}'}_{i}{x}'}_{j}\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}+\cdot \cdot \cdot \right]d{V}'}. \varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\left[ \frac{1}{R}\iiint_{V}'}{\rho \left( {\vec{x}'} \right)d{V}'}-\nabla \frac{1}{R}\cdot \iiint_{V}'}{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'}+\frac{1}{6}\sum\limits_{i,j=1}^{3}{\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}\iiint_{V}'}{3{x}'}_{i}{x}'}_{j}\rho \left( {\vec{x}'} \right)d{V}'}+\cdot \cdot \cdot \right]. \varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\left[ \frac{1}{R}Q-\vec{P}\cdot \nabla \frac{1}{R}+\frac{1}{6}\sum\limits_{i,j=1}^{3}{D}_{ij}\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}+\cdot \cdot \cdot \right]. \varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\left[ \frac{1}{R}Q-\vec{P}\cdot \nabla \frac{1}{R}+\frac{1}{6}\overset{\rightrightarrows }{\mathop{D}\,:\nabla \nabla \frac{1}{R}+\cdot \cdot \cdot \right].

其中 \left\{ \begin{align} & Q=\iiint_{V}'}{\rho \left( {\vec{x}'} \right)d{V}'}, \\ & \vec{P}=\iiint_{V}'}{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'}, \\ & {D}_{ij}=\iiint_{V}'}{3{x}'}_{i}{x}'}_{j}\rho \left( {\vec{x}'} \right)d{V}'}, \\ & \overset{\rightrightarrows }{\mathop{D}\,=\iiint_{V}'}{3{\vec{x}'{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'}, \\ \end{align} \right. 且有 \overset{\rightrightarrows }{\mathop{D}\,:\nabla \nabla \frac{1}{R}=\sum\limits_{i,j=1}^{3}{D}_{ij}\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}.

如何理解 {\vec{x} 相当于 \vec{x}-{\vec{x}' 或 $r$, 展开点 {\vec{a} 相当于 {\vec{x} 或 $R$, 而 $\Delta \vec{x}$ 相当于-{\vec{x}'呢?

大致情况如图:

相对于 $\vec{r}$ 或 $\vec{R}$ 这个尺寸, 电荷分布区域 {V}'} 看起来与一个点无异. 值得注意的是, 原点必须取在区域 {V}'} 内这个展开才能够在低阶项就有足够高的近似程度, 因为零阶近似就是将电荷集中在原点的电势. 而区域 {V}'} 相对而言是如此之小以至于原点的选择在区域 {V}'} 内各处均是等价的.

所以当你要展开函数 f\left( {\vec{r} \right)=f\left( \vec{x}-{\vec{x}' \right)=\frac{1}{\left| \vec{x}-{\vec{x}' \right|}=\frac{1}{r} 时, 自然要认识到确定了整体特征的展开点就是 $\vec{r}=\vec{x}$ 这个点, 也就是 {x}'}=0 的点, 实际上还要认识到在计算 \varphi \left( {\vec{x} \right) 的积分 \frac{1}{4\pi {\varepsilon }_{0}\iiint_{V}'}{\frac{\rho \left( {\vec{x}'} \right)}{\left| \vec{x}-{\vec{x}' \right|}d{V}'} 里面 {\vec{x} 是个常数, 积分变量是 ${\vec{x}}^{\prime }$. 自然展开点的函数值就是 f\left( {\vec{x} \right)=\frac{1}{x}=\frac{1}{R} . 所以这个修正项 $\Delta \vec{x}$ 自然是由-{\vec{x}'来充当的.