牛頓恆等式

数学中，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对称多项式之間的關係。 牛顿在不知道阿爾伯特‧吉拉德（英语：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应用在许多數學领域，如伽罗瓦理论、不變量理論、群论、组合學，也被进一步应用於数学之外，如广义相对论。 令 x1, ..., xn 為變量, 定義 k ≥ 1 且 pk(x1, ..., xn) 為k階 冪和: p k ( x 1 , … , x n ) = ∑ i = 1 n x i k = x 1 k + ⋯ + x n k , {\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k ≥ 0 定義 ek(x1, ..., xn) 為 初等對稱多項式,所以 e 0 ( x 1 , … , x n ) = 1 , e 1 ( x 1 , … , x n ) = x 1 + x 2 + ⋯ + x n , e 2 ( x 1 , … , x n ) = ∑ 1 ≤ i < j ≤ n x i x j , e n ( x 1 , … , x n ) = x 1 x 2 ⋯ x n , e k ( x 1 , … , x n ) = 0 , for   k > n . {\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 k e k ( x 1 , … , x n ) = ∑ i = 1 k ( − 1 ) i − 1 e k − i ( x 1 , … , x n ) p i ( x 1 , … , x n ) , {\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n  ≥ 1 以及 n ≥k ≥ 1. 另外對於所有k > n ≥ 1. 0 = ∑ i = k − n k ( − 1 ) i − 1 e k − i ( x 1 , … , x n ) p i ( x 1 , … , x n ) , {\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k得到前幾個式子 e 1 ( x 1 , … , x n ) = p 1 ( x 1 , … , x n ) , 2 e 2 ( x 1 , … , x n ) = e 1 ( x 1 , … , x n ) p 1 ( x 1 , … , x n ) − p 2 ( x 1 , … , x n ) , 3 e 3 ( x 1 , … , x n ) = e 2 ( x 1 , … , x n ) p 1 ( x 1 , … , x n ) − e 1 ( x 1 , … , x n ) p 2 ( x 1 , … , x n ) + p 3 ( x 1 , … , x n ) . {\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 e 1 = p 1 , 2 e 2 = e 1 p 1 − p 2 = p 1 2 − p 2 , 3 e 3 = e 2 p 1 − e 1 p 2 + p 3 = 1 2 p 1 3 − 3 2 p 1 p 2 + p 3 , 4 e 4 = e 3 p 1 − e 2 p 2 + e 1 p 3 − p 4 = 1 6 p 1 4 − p 1 2 p 2 + 4 3 p 1 p 3 + 1 2 p 2 2 − p 4 , {\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pk遞歸地表示ei p 1 = e 1 , p 2 = e 1 p 1 − 2 e 2 = e 1 2 − 2 e 2 , p 3 = e 1 p 2 − e 2 p 1 + 3 e 3 = e 1 3 − 3 e 1 e 2 + 3 e 3 , p 4 = e 1 p 3 − e 2 p 2 + e 3 p 1 − 4 e 4 = e 1 4 − 4 e 1 2 e 2 + 4 e 1 e 3 + 2 e 2 2 − 4 e 4 ,     ⋮ {\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 p k ( x 1 , … , x n ) = ( − 1 ) k − 1 k e k ( x 1 , … , x n ) + ∑ i = 1 k − 1 ( − 1 ) k − 1 + i e k − i ( x 1 , … , x n ) p i ( x 1 , … , x n ) , {\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 n ≥ 1 以及 n ≥k ≥ 1。 另外對於所有k > n ≥ 1。 我們有 p k ( x 1 , … , x n ) = ∑ i = k − n k − 1 ( − 1 ) k − 1 + i e k − i ( x 1 , … , x n ) p i ( x 1 , … , x n ) , {\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}),} 設 f ( x ) = ( x − x 1 ) ( x − x 2 ) ⋯ ( x − x n ) = x n − σ 1 x n − 1 + ⋯ + ( − 1 ) n σ n {\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 k > n {\displaystyle k>n} 時,我們要證明的式子是 s k − σ 1 s k − 1 + σ 2 s k − 2 + ⋯ + ( − 1 ) n σ n s k − n = 0 ; {\displaystyle 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 − σ 1 x n − 1 + ⋯ + ( − 1 ) n σ n {\displaystyle f(x)=x^{n}-\sigma _{1}x^{n-1}+\cdots +(-1)^{n}\sigma _{n}} ,得 x k − n f ( x ) = x k − σ 1 x k − 1 + ⋯ + ( − 1 ) n σ n x k − n . {\displaystyle 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 ≤ i ≤ n ) , {\displaystyle f(x_{i})=0(1\leq i\leq n),} 求和得到 ∑ i = 1 n [ x i k − σ 1 x i k − 1 + ⋯ + ( − 1 ) n σ n x i k − n ] = 0 , {\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 s k − σ 1 s k − 1 + ⋯ + ( − 1 ) n σ n s k − n = 0. {\displaystyle s_{k}-\sigma _{1}s_{k-1}+\cdots +(-1)^{n}\sigma _{n}s_{k-n}=0.} 當 1 ≤ k ≤ n {\displaystyle 1\leq k\leq n} 時,我們要證明的式子是 s k − σ 1 s k − 1 + ⋯ + ( − 1 ) k − 1 σ k − 1 s 1 + ( − 1 ) k k σ k = 0. {\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 f ′ ( x ) = f ( x ) ∑ i = 1 n 1 x − x i = n x n − 1 + ⋯ + ( − 1 ) k ( n − k ) σ k x n − k − 1 + ⋯ {\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 f ( x ) ∑ i = 1 n ( x − 1 + x i x − 2 + x i 2 x − 3 + ⋯ ) = n x n − 1 + ⋯ + ( − 1 ) k ( n − k ) σ k x n − k − 1 + ⋯ {\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 ( x n − σ 1 x n − 1 + ⋯ + ( − 1 ) n σ n ) ( n x − 1 + s 1 x − 2 + s 2 x − 3 + ⋯ ) = n x n − 1 + ⋯ + ( − 1 ) k ( n − k ) σ k x n − k − 1 + ⋯ {\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 x n − k − 1 {\displaystyle x^{n-k-1}} 項系數,有 ( − 1 ) k σ k ⋅ n + ( − 1 ) k − 1 σ k − 1 s 1 + ( − 1 ) k − 2 σ k − 2 s 2 − ⋯ − σ 1 s k − 1 + s k = ( − 1 ) k ( n − k ) σ k , {\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. 冪和對稱多項式 初等對稱多項式 对称多项式