牛顿恒等式

数学中，牛顿恒等式（英语：Newton's identities）描述了幂和对称多项式和初等对称多项式此两种对称多项式之间的关系。

牛顿在不知道阿尔伯特‧吉拉德（英语：Albert Girard）先前的成果下，于约1666年发现这些恒等式。这些恒等式目前已被应用在许多数学领域，如伽罗瓦理论、不变量理论、群论、组合学，也被进一步应用于数学之外，如广义相对论。

数学陈述

对称多项式

令 x₁, ..., x_n 为变量, 定义 k ≥ 1 且 p_k(x₁, ..., x_n) 为k阶幂和:

p_{k}(x_{1},\ldots ,x_{n})=\sum \nolimits _{i=1}^{n}x_{i}^{k}=x_{1}^{k}+\cdots +x_{n}^{k},

对于k ≥ 0 定义 e_k(x₁, ..., x_n) 为初等对称多项式,所以

{\begin{aligned}e_{0}(x_{1},\ldots ,x_{n})&=1,\\e_{1}(x_{1},\ldots ,x_{n})&=x_{1}+x_{2}+\cdots +x_{n},\\e_{2}(x_{1},\ldots ,x_{n})&=\textstyle \sum _{1\leq i<j\leq n}x_{i}x_{j},\\e_{n}(x_{1},\ldots ,x_{n})&=x_{1}x_{2}\cdots x_{n},\\e_{k}(x_{1},\ldots ,x_{n})&=0,\quad {\text{for}}\ k>n.\\\end{aligned}}

那么牛顿恒等式可以表示为

ke_{k}(x_{1},\ldots ,x_{n})=\sum _{i=1}^{k}(-1)^{i-1}e_{k-i}(x_{1},\ldots ,x_{n})p_{i}(x_{1},\ldots ,x_{n}),

对于所有的n ≥ 1 以及 n ≥k ≥ 1.

另外对于所有k > n ≥ 1.

0=\sum _{i=k-n}^{k}(-1)^{i-1}e_{k-i}(x_{1},\ldots ,x_{n})p_{i}(x_{1},\ldots ,x_{n}),

我们可以带入前几个k得到前几个式子

{\begin{aligned}e_{1}(x_{1},\ldots ,x_{n})&=p_{1}(x_{1},\ldots ,x_{n}),\\2e_{2}(x_{1},\ldots ,x_{n})&=e_{1}(x_{1},\ldots ,x_{n})p_{1}(x_{1},\ldots ,x_{n})-p_{2}(x_{1},\ldots ,x_{n}),\\3e_{3}(x_{1},\ldots ,x_{n})&=e_{2}(x_{1},\ldots ,x_{n})p_{1}(x_{1},\ldots ,x_{n})-e_{1}(x_{1},\ldots ,x_{n})p_{2}(x_{1},\ldots ,x_{n})+p_{3}(x_{1},\ldots ,x_{n}).\\\end{aligned}}

这些方程的形式和正确与否并不取决于变数的数量n,这使得可以在对称函数环中将它们称为恒等式。在这个环之中我们有

{\begin{aligned}e_{1}&=p_{1},\\2e_{2}&=e_{1}p_{1}-p_{2}=p_{1}^{2}-p_{2},\\3e_{3}&=e_{2}p_{1}-e_{1}p_{2}+p_{3}={\tfrac {1}{2}}p_{1}^{3}-{\tfrac {3}{2}}p_{1}p_{2}+p_{3},\\4e_{4}&=e_{3}p_{1}-e_{2}p_{2}+e_{1}p_{3}-p_{4}={\tfrac {1}{6}}p_{1}^{4}-p_{1}^{2}p_{2}+{\tfrac {4}{3}}p_{1}p_{3}+{\tfrac {1}{2}}p_{2}^{2}-p_{4},\\\end{aligned}}

在这里，LHS永远不会为零。这些等式允许以p_k递归地表示e_i

{\begin{aligned}p_{1}&=e_{1},\\p_{2}&=e_{1}p_{1}-2e_{2}=e_{1}^{2}-2e_{2},\\p_{3}&=e_{1}p_{2}-e_{2}p_{1}+3e_{3}=e_{1}^{3}-3e_{1}e_{2}+3e_{3},\\p_{4}&=e_{1}p_{3}-e_{2}p_{2}+e_{3}p_{1}-4e_{4}=e_{1}^{4}-4e_{1}^{2}e_{2}+4e_{1}e_{3}+2e_{2}^{2}-4e_{4},\\&{}\ \ \vdots \end{aligned}}

一般的,我们有

p_{k}(x_{1},\ldots ,x_{n})=(-1)^{k-1}ke_{k}(x_{1},\ldots ,x_{n})+\sum _{i=1}^{k-1}(-1)^{k-1+i}e_{k-i}(x_{1},\ldots ,x_{n})p_{i}(x_{1},\ldots ,x_{n}),

对于所有的 n ≥ 1 以及 n ≥k ≥ 1。另外对于所有k > n ≥ 1。我们有

p_{k}(x_{1},\ldots ,x_{n})=\sum _{i=k-n}^{k-1}(-1)^{k-1+i}e_{k-i}(x_{1},\ldots ,x_{n})p_{i}(x_{1},\ldots ,x_{n}),

证明

设 $f(x)=(x-x_{1})(x-x_{2})\cdots (x-x_{n})=x^{n}-\sigma _{1}x^{n-1}+\cdots +(-1)^{n}\sigma _{n}$ .

当 $k>n$ 时,我们要证明的式子是 $s_{k}-\sigma _{1}s_{k-1}+\sigma _{2}s_{k-2}+\cdots +(-1)^{n}\sigma _{n}s_{k-n}=0;$

由 $f(x)=x^{n}-\sigma _{1}x^{n-1}+\cdots +(-1)^{n}\sigma _{n}$ ,得 $x^{k-n}f(x)=x^{k}-\sigma _{1}x^{k-1}+\cdots +(-1)^{n}\sigma _{n}x^{k-n}.$

由于 $f(x_{i})=0(1\leq i\leq n),$ 求和得到 $\sum _{i=1}^{n}[x_{i}^{k}-\sigma _{1}x_{i}^{k-1}+\cdots +(-1)^{n}\sigma _{n}x_{i}^{k-n}]=0,$ 故 $s_{k}-\sigma _{1}s_{k-1}+\cdots +(-1)^{n}\sigma _{n}s_{k-n}=0.$

当 $1\leq k\leq n$ 时,我们要证明的式子是 $s_{k}-\sigma _{1}s_{k-1}+\cdots +(-1)^{k-1}\sigma _{k-1}s_{1}+(-1)^{k}k\sigma _{k}=0.$

注意到 $f'(x)=f(x)\sum _{i=1}^{n}{\frac {1}{x-x_{i}}}=nx^{n-1}+\cdots +(-1)^{k}(n-k)\sigma _{k}x^{n-k-1}+\cdots$

展开为形式幂级数,得 $f(x)\sum _{i=1}^{n}(x^{-1}+x_{i}x^{-2}+x_{i}^{2}x^{-3}+\cdots )=nx^{n-1}+\cdots +(-1)^{k}(n-k)\sigma _{k}x^{n-k-1}+\cdots$

即 $(x^{n}-\sigma _{1}x^{n-1}+\cdots +(-1)^{n}\sigma _{n})(nx^{-1}+s_{1}x^{-2}+s_{2}x^{-3}+\cdots )=nx^{n-1}+\cdots +(-1)^{k}(n-k)\sigma _{k}x^{n-k-1}+\cdots$

对比两边的 $x^{n-k-1}$ 项系数,有 $(-1)^{k}\sigma _{k}\cdot n+(-1)^{k-1}\sigma _{k-1}s_{1}+(-1)^{k-2}\sigma _{k-2}s_{2}-\cdots -\sigma _{1}s_{k-1}+s_{k}=(-1)^{k}(n-k)\sigma _{k},$ 即得.

参见