# Algebraic Polynomials with Random Coefficients with Binomial and Geometric Progressions

Authors: K. Farahmand; M. Sambandham
Journal: Journal of Applied Mathematics and Stochastic Analysis
 (2009)

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## Abstract

The expected number of real zeros of an algebraic polynomialao+a1x+a2x2+⋯+anxn with random coefficient
aj,j=0,1,2,…,n is known. The distribution of the
coefficients is often assumed to be identical albeit allowed to
have different classes of distributions. For the nonidentical case,
there has been much interest where the variance of the jth coefficient
is  var(aj)=(nj). It is shown that this class
of polynomials has significantly more zeros than the classical
algebraic polynomials with identical coefficients. However,
in the case of nonidentically distributed coefficients it is
analytically necessary to assume that the means
of coefficients are zero. In this work we
study a case when the moments of the coefficients have both
binomial and geometric progression elements. That is we assume E(aj)=(nj)μj+1 and var(aj)=(nj)σ2j. We show how the above expected number of real zeros is
dependent on values of σ2 and μ in various cases.


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## Body

## 1. Origin of Polynomials

Let(Ω,Pr,𝒜) be  a fixed  probability space, and for ω∈Ω let {aj(ω)}j=0n be a sequence of independent identically distributed
random variables defined on Ω.
There has been considerable work on obtaining the expected number of real zeros
of algebraic(1.1)Pn(x,ω)≡Pn(x)=∑j=0naj(ω)xj,and trigonometric ∑j=0naj(ω)cosjθ polynomials with random coefficients aj(ω)s. The study of the random algebraic
polynomials was initiated by Kac [1], and the recent works include [2, 3]. It is shown that under
general assumptions for the distribution of coefficients the expected number of
real zeros is asymptotic to (2/π)logn as n→∞.
For the case of random trigonometric polynomials, Dunnage [4] obtained the first result
which was later generalized by Wilkins Jr. [5, 6] and recently studied in [7, 8]. It is shown that, again for a
wide class of distributions for the coefficients, there are significantly
more real zeros in the case of trigonometric polynomial compared with the
algebraic case. The asymptotic value for the expected number of zeros for the
latter case is 2n/3.
Besides the comprehensive book of Bharucha-Reid and Sambandham [9], the earlier results of
general topics on random polynomials are reviewed in [10].Motivated by the interesting work of Edelman and
Kostlan [11], who,
among others, considered polynomials of the form ∑j=0naj(ω)(nj)1/2xj, [2, 12] obtained many characteristics, like the number of real
zeros or the number of maxima of these types of polynomials. This is
interesting as they showed that for this case of nonidentically distributed
coefficients the expected number of real zeros is O(n),
which is significantly more than the classical algebraic case but less than
that of trigonometric polynomials. Also in this direction of nonidentical
coefficients, a case in which the mean of coefficients aj(ω) increases with j is studied in [3, 13]. Now it would be interesting
to study a random polynomial formed by combining the above two distribution
laws. It is natural to ask, for instance, what would be the behavior of Pn(x) in (1.1) if for constants μ and σ the mean and variance of coefficients are E(aj(ω))=(nj)μj+1 and var(aj(ω))=(nj)σ2j.With the latter assumption of the distribution of the coefficients,
we first show that ifμ=0, the expected number of real zeros of Pn(x) denoted by ENn,P(0,∞)≡ENn(0,∞) is independent of σ.
The case of nonzero μ is studied in Theorem 1.2. The
analysis for the general case is complicated, and we only give the result for a
case that μ=σ2.
We prove the following theorem.Theorem 1.1.
Forμ=0 and σ2>0,
the expected number of real zeros of Pn(x) is independent of σ2.
That is(1.2)ENn(−∞,0)=ENn(0,∞)=n2.The analysis for the case ofμ≠0 would be complicated. Without loss of much
generality and certainly interest, we restrict ourselves to the case of μ=σ2. We
prove the following theorem.Theorem 1.2.
The expected number of real zeros
ofPn(x) for different values of μ satisfies(1.3)ENn(0,∞){=O(1),ifμ=σ>1,~(n2){1−arctan(2μ1−μ)},if0<μ=σ<1.For x negative and for every μ=σ2,(1.4)ENn(−∞,0)~n2.


## 2. Moments

In order to
obtain the expected number of real zeros we use a generalization of the well-known Kac-Rice formula initiated in [1, 14, 15]. To this end,
we need the following moments of Pn(x) and its dervative Pn′(x).
First, we assume the general assumptions on the means and the variances of
coefficients as stated above. That is, E(aj)=(nj)μj+1 and var(aj)=(nj)σ2j.
Since these coefficients are independent, it is easy to show(2.1)α=E(Pn(x))=μ∑j=0n(nj)(xμ)j=μ(1+μx)n,(2.2)β=E(Pn′(x))=μ2∑j=0nj(nj)(μx)j−1=nμ2(1+μx)n−1,(2.3)A2=var(Pn(x))=∑j=0n(nj)(σx)2j=(1+σ2x2)n,(2.4)B2=var(Pn′(x))=σ2∑j=0nj2(nj)(σx)2=nσ2(1+σ2x2)n−2(1+nσ2x2),and finally(2.5)C=cov(Pn(x),Pn′(x))=σ2x∑j=0nj(nj)(σx)2j−2=nσ2x(1+σ2x2)n−1. Then from (2.3)–(2.5) we can
obtain(2.6)Δ2=A2B2−C2=nσ2(1+σ2x2)2n−2.With the above notations, we can
now write the Kac-Rice for the expected number of real zeros of Pn(x) in the interval (a,b) as, see also [10, page 43],(2.7)ENn(a,b)=I1(a,b)+I2(a,b),where(2.8)I1(a,b)=∫abΔπA2exp(−α2B2+β2A2−2αβC2Δ2)dx,(2.9)I2(a,b)=∫ab2|βA2−Cα|πA3exp(−α22A2)erf(|βA2−Cα|2AΔ)dx, where as usual erf(x)=∫0xexp(−t2)dt.
Now we can progress and evaluate further the following terms required in the
Kac-Rice formulae (2.7)–(2.9). From (2.1)–(2.5) we can derive(2.10)α2B2+β2A2−2αβC=nμ2(1+μx)2n−2(1+σ2x2)n−2(nσ4x2+σ2+2μxσ2+μ2x2σ2+nμ2−2nμσ2x).This together with (2.6)
yields(2.11)α2B2+β2A2−2αβC2Δ2=μ2(1+μx)2n−2{n(σ2x−μ)2+σ2(1+μx)2}2σ2(1+σ2x2)n.


## 3. Proof of Theorems

First in the
case ofμ=0 from (2.7) and (2.3)–(2.6) by letting y=σx we can show(3.1)ENn(0,∞)=nπ∫0∞σ1+σ2x2dx=nπ∫0∞dy1+y2=n2.This proves Theorem 1.1. Now we
proceed with the more general case of μ≠0.
As explained above, in order to simplify the analysis we let μ=σ2.
This yields (2.11) to(3.2)α2B2+β2A2−2αβC2Δ2=fn(x,μ)gn(x,μ),where(3.3)gn(x,μ)=nμ2(x−1)2+μ(1+μx)2,and for all sufficiently large n,(3.4)fn(x,μ)=(1+μx)2n−2(1+μx2)n~(1+μx)2n(1+μx2)n=(1+μ2x2+2μx1+μx2)n.Now we assume x>0.
Then if we let μ>1,
since(3.5)1+μ2x2+2μx1+μx2>1,we can see that fn(x,μ)→∞ as n→∞.
Therefore the exponential term that appears in (2.8) tends to zero
exponentially fast. Hence the only contribution to ENn(0,∞) is from I2(0,∞).
In the following, we show that the latter is O(1).
To this end, we note that since from the definition for all x, erf(x)≤π/2,
then(3.6)I2(0,∞)<12π∫0∞βA2−CαA3exp(−α22A2)dx.Now we let u=α/A, and since (d/dx)(α/A)=(βA2−αC)/A3 from (3.6) we obtain(3.7)I2(0,∞)<12π∫0∞exp(−u22)du≤12.This completes the first part of
Theorem 1.2. On the other hand, if μ<1, the behavior of fn(x,μ) will depend on x.
That is for 0<x<2/(1−μ), fn(x,μ)→∞ as n→∞ and for x>2/(1−μ), fn(x,μ)→0 as n→∞.
Therefore the only contribution to ENn(0,∞) from I1 is in the interval (2/(1−μ),∞) as I1(0,2/(1−μ)) will tend to zero exponentially fast. Also for ν=μx,(3.8)I1(21−μ,∞)~nπ∫2/(1−μ)∞μ1+μx2dx=nπ∫2/(1−μ)∞dν1+ν2~(n2){1−arctan(2μ1−μ)}.The above argument for I2(0,∞) in (3.7) remains valid, and therefore we have
proof of the first part of Theorem 1.2.Forx<0 without loss of generality, we only consider
the case of μ>0 (since μ=σ2). For this case gn(x,μ) remains positive. However, for x2>ϵ/μ, where for a=1−loglogn10/logn we let ϵ=n−a, we have, (see also [10, page 31]),(3.9)(1+μx2)n>(1+ϵ)n{(1+n−a)na}n1−a=exp(n1−a)~n10.Hence(3.10)fn(x,μ)<(1+μx2)−n<n−10,which tends to zero very fast as n→∞.
Therefore the exponential term in I1 tends to be one, and hence(3.11)I1(ϵμ,∞)~nπ∫ϵ/μ∞dx1+x2~n2.Also in the interval (0,ϵ/μ),(3.12)I1(0,ϵμ)<∫0ϵ/μΔπA2dx<nμπ∫0ϵ/μdx1+μx2~n2πarctanϵ,which is small. This completes
the proof of Theorem 1.2.



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*Source: 725260-2009-04-07.xml*