牛頓恆等式

數學中，牛頓恆等式（英語：Newton's identities）描述了冪和對稱多項式和初等對稱多項式此兩種對稱多項式之間的關係。

牛頓在不知道阿爾伯特‧吉拉德（英語：Albert Girard）先前的成果下，於約1666年發現這些恆等式。這些恆等式目前已被應用在許多數學領域，如伽羅瓦理論、不變量理論、群論、組合學，也被進一步應用於數學之外，如廣義相對論。

數學陳述

對稱多項式

令 ${\displaystyle x_{1},\cdots ,x_{n}}$為變量，定義${\displaystyle k\geq 1}$且 ${\displaystyle p_{k}(x_{1},\cdots ,x_{n})}$為${\displaystyle k}$階 冪和:

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

對於${\displaystyle k\geq 0}$ 定義 ${\displaystyle e_{k}(x_{1},\cdots ,x_{n})}$ 為 初等對稱多項式，所以

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

那麼牛頓恆等式可以表示為

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

對於所有的${\displaystyle n\geq 1}$ 以及 ${\displaystyle n\geq k\geq 1}$.

另外對於所有${\displaystyle k>n\geq 1}$.

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

我們可以帶入前幾個${\displaystyle k}$得到前幾個式子

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

這些方程的形式和正確與否並不取決於變數的數量${\displaystyle n}$，這使得可以在對稱函數環中將它們稱為恆等式。在這個環之中我們有

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

在這裡，LHS永遠不會為零。這些等式允許以${\displaystyle p_{k}}$遞歸地表示${\displaystyle e_{i}}$

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

一般的，我們有

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

對於所有的 ${\displaystyle n\geq 1}$ 以及 ${\displaystyle n\geq k\geq 1}$。 另外對於所有${\displaystyle k>n\geq 1}$。 我們有

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

證明

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

當${\displaystyle k>n}$時，我們要證明的式子是${\displaystyle s_{k}-\sigma _{1}s_{k-1}+\sigma _{2}s_{k-2}+\cdots +(-1)^{n}\sigma _{n}s_{k-n}=0;}$

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

由於${\displaystyle f(x_{i})=0(1\leq i\leq n)}$，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${\displaystyle s_{k}-\sigma _{1}s_{k-1}+\cdots +(-1)^{n}\sigma _{n}s_{k-n}=0.}$

當${\displaystyle 1\leq k\leq n}$時，我們要證明的式子是${\displaystyle s_{k}-\sigma _{1}s_{k-1}+\cdots +(-1)^{k-1}\sigma _{k-1}s_{1}+(-1)^{k}k\sigma _{k}=0.}$

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

展開為形式冪級數，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${\displaystyle (x^{n}-\sigma _{1}x^{n-1}+\cdots +(-1)^{n}\sigma _{n})(nx^{-1}+s_{1}x^{-2}+s_{2}x^{-3}+\cdots )=nx^{n-1}+\cdots +(-1)^{k}(n-k)\sigma _{k}x^{n-k-1}+\cdots }$

對比兩邊的${\displaystyle x^{n-k-1}}$項系數，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.

參見

• 冪和對稱多項式
• 初等對稱多項式
• 對稱多項式