牛顿恒等式

设

\[ P\left(x\right)=\sum_{k=0}^{n}a_{k}x^{k} \]

记 \(P\left(x\right)\) 的根为 \(\alpha_{k},k=1,2,\ldots,n\)，定义 \(S_{i}=\sum \alpha_{k}^{i}\)，并且令

\[ \begin{aligned} \sigma_1 &= \sum\alpha_{i} = -\dfrac{a_{n-1}}{a_{n}} \\ \sigma_2 &= \sum\alpha_{i}\alpha_{j} = \dfrac{a_{n-2}}{a_{n}} \\ \sigma_3 &= \sum\alpha_{i}\alpha_{j}\alpha_{k} = -\dfrac{a_{n-3}}{a_{n}} \\ &\vdots \\ \sigma_n &= \alpha_{1}\alpha_{2}\cdots\alpha_{n} = \left(-1\right)^{n}\dfrac{a_0}{a_{n}} \end{aligned} \]

则有

\[ \begin{aligned} S_{0} &= n \\ S_{1} &= 1\cdot\sigma_{1} \\ S_{2} &= \sigma_{1}S_{1}-2\cdot\sigma_{2} \\ S_{3} &= \sigma_{1}S_{2}-\sigma_{2}S_{1}+3\cdot\sigma_{3} \\ &\vdots \\ S_{n-1} &= \sigma_{1}S_{n-2}-\sigma_{2}S_{n-3}+\cdots+(-1)^{n-3}\sigma_{n-2}S_{1}+(-1)^{n-2}\cdot(n-1)\cdot\sigma_{n-1} \\ &\vdots \\ S_{m} &= \sum_{k=1}^{n}(-1)^{k+1}\sigma_{k}S_{m-k} \end{aligned} \]

这里 \(m\geqslant n\)。

更简洁地

\[ \boxed{ S_{m} = \begin{cases} n, & m=0 \\ \begin{aligned} (-1)^{m+1}\cdot\sigma_{m}\cdot m+\sum_{k=1}^{m-1}(-1)^{m-1}\sigma_{k}S_{m-k} \end{aligned}, & 1\leqslant m\lt n \\ \begin{aligned} \sum_{k=1}^{n}(-1)^{k+1}\sigma_{k}S_{m-k} \end{aligned}, & m\geqslant n \end{cases} } \]

当 \(n\) 为正无穷时，公式的第二行仍有意义。