Mean number of real zeros of a random hyperbolic polynomial (Q1573704)
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scientific article; zbMATH DE number 1485599
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Mean number of real zeros of a random hyperbolic polynomial |
scientific article; zbMATH DE number 1485599 |
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Mean number of real zeros of a random hyperbolic polynomial (English)
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21 June 2001
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This paper provides an interesting result for the expected number of real zeros of a hyperbolic polynomial with random coefficients. For the integer \(n\geq 2\) suppose \(a_k(\omega)\), \(\omega\in\Omega\), \(k=0,1,\dots,n\), is a sequence of independent, normally distributed random variables defined on a probability space \((\Omega,{\mathcal A}, \text{Pr})\) with mean zero and variance one. Denote \(f_{np}(x,\omega) \equiv f(x)= \sum^n_{k=0} k^pa_k\text{cosh} kx\), where \(p\geq 0\) is an integer. Let \(\nu_{np}\) be the expected number of real zeros of \(f(x)\). It is shown that \(\nu_{np}= \pi^{-1}\log n+O (\sqrt{\log n})\). The fact that the leading term of \(\nu_{np}\) is independent of \(p\) is unexpected and interesting; for instance, for the random algebraic polynomial \(f(x)=\sum^n_{k=0} k^pa_kx^k\), \textit{M. Das} [J. Indian Math. Soc., n. Ser. 36, 53-63 (1972; Zbl 0293.60058)] showed that \(\nu_{np}\sim\pi^{-1}(1+ \sqrt {2p+1}) \log n\). Also in the case of the random trigonometric polynoial \(f(x)= \sum^n_{k=0} k^pa_k\cos kx\), \textit{M. Das} [Proc. Camb. Philos. Soc. 64, 721-729 (1968; Zbl 0169.48902)] and later the present author, in a series of papers, have shown that \(\nu_{np}=\{\sqrt {(2p+1)/(2p+3)} \}(2n+1)+ O(\sqrt n)\). For an updated development of the latter see the reviewer [``Topics in random polynomials'' (1998; Zbl 0949.60010)]. Given the above result of this paper it is therefore of great interest, as far as the dependence or independence of the principal term of \(\nu_{np}\) to \(p\) is concerned, to see which of the above behaviours the expected number of real zeros of random orthogonal polynomials will follow.
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