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 ${\displaystyle x_{1},\cdots ,x_{n}}$为变量，定义${\displaystyle k\geq 1}$且 ${\displaystyle p_{k}(x_{1},\cdots ,x_{n})}$为${\displaystyle k}$阶 幂和:

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

对于${\displaystyle k\geq 0}$ 定义 ${\displaystyle e_{k}(x_{1},\cdots ,x_{n})}$ 为 初等对称多项式，所以

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

那么牛顿恒等式可以表示为

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

对于所有的${\displaystyle n\geq 1}$ 以及 ${\displaystyle n\geq k\geq 1}$.

另外对于所有${\displaystyle k>n\geq 1}$.

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

我们可以带入前几个${\displaystyle k}$得到前几个式子

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

这些方程的形式和正确与否并不取决于变数的数量${\displaystyle n}$，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

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

在这里，LHS永远不会为零。这些等式允许以${\displaystyle p_{k}}$递归地表示${\displaystyle e_{i}}$

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

一般的，我们有

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

对于所有的 ${\displaystyle n\geq 1}$ 以及 ${\displaystyle n\geq k\geq 1}$。 另外对于所有${\displaystyle k>n\geq 1}$。 我们有

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

证明

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

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

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

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

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

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

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

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

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

参见

• 幂和对称多项式
• 初等对称多项式
• 对称多项式