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[不等式] 实系数多项式的虚根的个数

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hbghlyj posted 2023-4-29 05:41 |Read mode
Last edited by hbghlyj 2023-4-29 09:00设 $a_1, a_2, \dots, a_n$ 是复数,令 $ \sigma _{k} $ 表示 $a_1, a_2, \dots, a_n$ 的 $k$ 阶基本对称多项式。定义基本对称均值
$$ S_{k}={\frac {\sigma _{k}}{\binom {n}{k}}} $$
若$S_k\in\mathbb R$,则多项式$P=\prod _{k=1}^{n}(x+a_{k})$为实系数多项式。
证明$P$的虚根的个数等于不等式$$S_{k-1}S_{k+1}\leq S_{k}^{2} $$不成立的个数。

来自Wikipedia牛頓不等式

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