牛顿恒等式

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:

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

对于k ≥ 0 定义 e_k(x₁, ..., 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n  ≥ 1 以及 n ≥k ≥ 1.

另外对于所有k > n ≥ 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p_k递归地表示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 n ≥ 1 以及 n ≥k ≥ 1。 另外对于所有k > n ≥ 1。 我们有

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

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.

• 幂和对称多项式
• 初等对称多项式
• 对称多项式