泰勒级数整理

在学习变分法时发现了一些泰勒级数的trick,整理在这里。

n变量和的m次幂的展开式

n变量和的m次幂,及其展开式如下所示:

(1)(x1+x2+...+xn)m=i1=0ni2=0n...im=0n(xi1xi2...xim)(x_1+x_2+...+x_n)^m = \sum_{i_1=0}^n\sum_{i_2=0}^n...\sum_{i_m=0}^n(x_{i_1}x_{i_2}...x_{i_m}) \tag{1}

证明如下:(数学归纳法)

  1. m=1m=1

    (x1+x2+...+xn)1=i1=0nxi1(x_1+x_2+...+x_n)^1 = \sum_{i_1=0}^nx_{i_1}

    上式显然成立

  2. m=km=k

    设:

    (x1+x2+...+xn)m1=i1=0ni2=0n...im1=0n(xi1xi2...xim1)(x_1+x_2+...+x_n)^{m-1} = \sum_{i_1=0}^n\sum_{i_2=0}^n...\sum_{i_{m-1}=0}^n(x_{i_1}x_{i_2}...x_{i_{m-1}})

    成立,则:

    (x1+x2+...+xn)m=(x1+x2+...+xn)m1(x1+x2+...+xn)=[i1=0ni2=0n...im1=0n(xi1xi2...xim1)](x1+x2+...+xn)=j=1n[xji1=0ni2=0n...im1=0n(xi1xi2...xim1)]=i1=0ni2=0n...im=0n(xi1xi2...xim)\begin{aligned} (x_1+x_2+...+x_n)^m &= (x_1+x_2+...+x_n)^{m-1} (x_1+x_2+...+x_n) \\[2ex] &= [\sum_{i_1=0}^n\sum_{i_2=0}^n...\sum_{i_{m-1}=0}^n(x_{i_1}x_{i_2}...x_{i_{m-1}})](x_1+x_2+...+x_n) \\[2ex] &= \sum_{j=1}^n[x_j\sum_{i_1=0}^n\sum_{i_2=0}^n...\sum_{i_{m-1}=0}^n(x_{i_1}x_{i_2}...x_{i_{m-1}})] \\[2ex] &= \sum_{i_1=0}^n\sum_{i_2=0}^n...\sum_{i_m=0}^n(x_{i_1}x_{i_2}...x_{i_m}) \end{aligned}

证毕。

泰勒级数的另一种表达方法

多维函数的泰勒展开式如下所示:

T(x1,,xd)=n1=0nd=0(x1a1)n1(xdad)ndn1!nd!(n1++ndfx1n1xdnd)(a1,,ad)=f(a1,,ad)+j=1df(a1,,ad)xj(xjaj)+12!j=1dk=1d2f(a1,,ad)xjxk(xjaj)(xkak)++13!j=1dk=1dl=1d3f(a1,,ad)xjxkxl(xjaj)(xkak)(xlal)+\begin{aligned} T(x_1,\ldots,x_d) &= \sum_{n_1=0}^\infty \cdots \sum_{n_d = 0}^\infty \frac{(x_1-a_1)^{n_1}\cdots (x_d-a_d)^{n_d}}{n_1!\cdots n_d!}\,\left(\frac{\partial^{n_1 + \cdots + n_d}f}{\partial x_1^{n_1}\cdots \partial x_d^{n_d}}\right)(a_1,\ldots,a_d) \\ &= f(a_1, \ldots,a_d) + \sum_{j=1}^d \frac{\partial f(a_1, \ldots,a_d)}{\partial x_j} (x_j - a_j) \\[2ex] & \qquad \qquad + \frac{1}{2!} \sum_{j=1}^d \sum_{k=1}^d \frac{\partial^2 f(a_1, \ldots,a_d)}{\partial x_j \partial x_k} (x_j - a_j)(x_k - a_k) + \\ & \qquad \qquad + \frac{1}{3!} \sum_{j=1}^d\sum_{k=1}^d\sum_{l=1}^d \frac{\partial^3 f(a_1, \ldots,a_d)}{\partial x_j \partial x_k \partial x_l} (x_j - a_j)(x_k - a_k)(x_l - a_l) + \cdots \end{aligned}

以上来自:https://en.wikipedia.org/wiki/Taylor_series

由式(1)(1)可以改写泰勒级数为:

T(x1,,xd)=f(a1,,ad)+i=1df(a1,,ad)xi(xiai)+12!i=1dj=1d2f(a1,,ad)xixj(xiai)(xjaj)++13!i=1dj=1dk=1d3f(a1,,ad)xixjxk(xiai)(xjaj)(xkak)+=Δxi=xiaif+(i=1dxiΔxi)f+12!(i=1dxiΔxi)2f+13!(i=1dxiΔxi)3f+=k=01k!(i=1dxiΔxi)kf\begin{aligned} T(x_1,\ldots,x_d) &= f(a_1, \ldots,a_d) \\[2ex] & \qquad + \sum_{i=1}^d \frac{\partial f(a_1, \ldots,a_d)}{\partial x_i} (x_i - a_i) \\[2ex] & \qquad + \frac{1}{2!} \sum_{i=1}^d \sum_{j=1}^d \frac{\partial^2 f(a_1, \ldots,a_d)}{\partial x_i \partial x_j} (x_i - a_i)(x_j - a_j) + \\[2ex] & \qquad+ \frac{1}{3!} \sum_{i=1}^d\sum_{j=1}^d\sum_{k=1}^d \frac{\partial^3 f(a_1, \ldots,a_d)}{\partial x_i\partial x_j \partial x_k} (x_i - a_i)(x_j - a_j)(x_k - a_k) + \cdots \\[2ex] &\xlongequal{\Delta x_i=x_i-a_i} f+ (\sum_{i=1}^d \frac{\partial}{\partial x_i} \Delta x_i )f+\frac{1}{2!}(\sum_{i=1}^d \frac{\partial}{\partial x_i} \Delta x_i )^2f+\frac{1}{3!}(\sum_{i=1}^d \frac{\partial}{\partial x_i} \Delta x_i )^3f+\cdots\\[2ex] &= \sum_{k=0}^{\infty}\frac{1}{k!}(\sum_{i=1}^d \frac{\partial}{\partial x_i} \Delta x_i )^kf \end{aligned}

目前暂不讨论这种表达方式的严谨性,但是在变分法的推导中却应用了该表达方式,所以就推导下做个记录吧,以免以后会用到。

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