✨
AI Summary & Key Takeaways
AI Insight⚡ TL;DR(Too Long; Didn't Read)
Moreover, the problem of estimating p ( 1 − w n ) p(1−wn) , $0<w\leq$ given $p(1)=0$ is considered.
📌 3 Key Takeaways
🎯
Objective & Hypothesis:In this paper we impose distinct restrictions on the moduli of the zeros of p ( z ) = n ∑ v = 0 a v z v p(z)=∑v=0navzv and investigate the dependence of ∥ p ( R z ) − p ( σ z ) ∥ ‖p(Rz)−p(σz)‖ , R > σ ≥ 1 R>σ≥1 on M α Mα and M α + π Mα+π , where M α = max 1 ≤ k ≤ n | p ( e i ( α + 2 k π ) / n ) | Mα=max1≤k≤n|p(ei(α+2kπ)/n)| and on certain coefficients of p ( z ) p(z) .
📊
Key Findings & Data:This paper comprises several results, which in particular yields some classical polynomial inequalities as special cases.
💡
Conclusion & Impact:Moreover, the problem of estimating p ( 1 − w n ) p(1−wn) , $0<w\leq$ given $p(1)=0$ is considered.
🔬Design:Research Article
📍Scope:Comprehensive Sample
⚙️Variables:Polynomials · definite integral · zeros
Abstract
In this paper we impose distinct restrictions on the moduli of the zeros of p ( z ) = n ∑ v = 0 a v z v p(z)=∑v=0navzv and investigate the dependence of ∥ p ( R z ) − p ( σ z ) ∥ ‖p(Rz)−p(σz)‖ , R > σ ≥ 1 R>σ≥1 on M α Mα and M α + π Mα+π , where M α = max 1 ≤ k ≤ n | p ( e i ( α + 2 k π ) / n ) | Mα=max1≤k≤n|p(ei(α+2kπ)/n)| and on certain coefficients of p ( z ) p(z) . This paper comprises several results, which in particular yields some classical polynomial inequalities as special cases. Moreover, the problem of estimating p ( 1 − w n ) p(1−wn) , $0<w\leq$ given $p(1)=0$ is considered.
🔖
Cite This Article
Malik, S., Kumar, A. (2022). On the maximum modulus of a complex polynomial. Communications Faculty of Sciences University of Ankara Series A1 Mathematics and Statistics, Volume: 71 Issue: 3. https://doi.org/10.31801/cfsuasmas.989344