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arXiv 1802.02708math.NTmath.RA
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关于多项式非实根的几点注记

Some remarks on the non-real roots of polynomials

Shuichi Otake, Tony Shaska

摘要:

令$f \in { \mathbb R} ( t) [x]$由$ f(t, x) = x^n + t \cdot g(x) $和$β_1 < \dots < β_m$给出,$ f(t, x) = x^n + t \cdot g(x) $和$β_1 < \dots < β_m$是$f(t, x)$关于$x$的判别式$Δ_{(f, x)} (t)$的不同实根。令$γ$为$g(x)=\sum_{k=0}^s t_{s-k} x^{s-k}$的实根的个数。对于任何$ξ> | β_m |$,如果$n-s$是奇数,则$f(ξ, x)$的实根的数目是$γ+1$,并且如果$n-s$是偶数,则$f(ξ, x)$的实根的数目分别是$γ$,$γ+2$,如果$t_s>0$或$t_s < 0$。上述结果的一个特例是构造一族在$\mathbb Q$上可约的全复多项式。

英文摘要:

Let $f \in { \mathbb R} ( t) [x]$ be given by $ f(t, x) = x^n + t \cdot g(x) $ and $β_1 < \dots < β_m$ the distinct real roots of the discriminant $Δ_{(f, x)} (t)$ of $f(t, x)$ with respect to $x$. Let $γ$ be the number of real roots of $g(x)=\sum_{k=0}^s t_{s-k} x^{s-k}$. For any $ξ> | β_m |$, if $n-s$ is odd then the number of real roots of $f(ξ, x)$ is $γ+1$, and if $n-s$ is even then the number of real roots of $f(ξ, x)$ is $γ$, $γ+2$ if $t_s>0$ or $t_s < 0$ respectively. A special case of the above result is constructing a family of totally complex polynomials which are reducible over $\mathbb Q$.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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