Pseudo-boolean functions and the multiplicity of the zeros of polynomials

被引:0
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作者
Tamás Erdélyi
机构
[1] Texas A&M University,Department of Mathematics
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关键词
Boolean Function; Absolute Constant; Acta Arith; Chebyshev Space; Bernstein Type Inequality;
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摘要
A highlight of this paper states that there is an absolute constant c1 > 0 such that every polynomial P of the form P(z) = Σj=0najzj, aj ∈ ℂ with \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\left| {a_0 } \right| = 1, \left| {a_j } \right| \leqslant M^{ - 1} \left( {\begin{array}{*{20}c} n \\ j \\ \end{array} } \right), j = 1,2, \ldots ,n,$$\end{document} for some 2 ≤ M ≤ en has at most \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n - \left\lfloor {{c_1}\sqrt {n\log M} } \right\rfloor $$\end{document} zeros at 1. This is compared with some earlier similar results reviewed in the introduction and closely related to some interesting Diophantine problems. Our most important tool is an essentially sharp result due to Coppersmith and Rivlin asserting that if Fn = {1, 2, …, n}, there exists an absolute constant c > 0 such that \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\left| {P(0)} \right| \leqslant \exp (cL)\mathop {\max }\limits_{x \in {F_n}} \left| {P(x)} \right|$$\end{document} for every polynomial P of degree at most \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$m \leqslant \sqrt {nL/16} $$\end{document} with 1 ≤ L < 16n. A new proof of this inequality is included in our discussion.
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页码:91 / 108
页数:17
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