The expected number of real zeros of the polynomial of the form a0+a1x+a2x2+⋯+anxn, where a0,a1,a2,…,an is a sequence of standard
Gaussian random variables, is known. For n large it is shown that this expected
number in (−∞,∞) is asymptotic to (2/π)logn. In this paper, we show that
this asymptotic value increases significantly to n+1 when we consider a
polynomial in the form a0(n0)1/2x/1+a1(n1)1/2x2/2+a2(n2)1/2x3/3+⋯+an(nn)1/2xn+1/n+1 instead. We give the motivation for our choice of
polynomial and also obtain some other characteristics for the polynomial, such
as the expected number of level crossings or maxima. We note, and present,
a small modification to the definition of our polynomial which improves our
result from the above asymptotic relation to the equality.

1. Introduction

The classical
random algebraic polynomial has previously been defined asT(x)≡Tn(x,ω)=∑j=0naj(ω)xj,where, for (Ω,𝒜,Pr) a fixed probability space, {aj(ω)}j=0n is a sequence of independent random variables
defined on Ω.
For n large, the expected number of real zeros of T(x),
in the interval (−∞,∞),
defined by EN0,T(−∞,∞),
is known to be asymptotic to (2/π)logn.
For this case the coefficients aj≡aj(ω) are assumed to be identical normal standard. This
asymptotic value was first obtained by the pioneer work of Kac [1] and was recently
significantly improved by Wilkins [2], who reduced the error term involved in this
asymptotic formula to O(1).
Since then, many other mathematical properties of T(x) have been studied and they are listed in
[3] and more recently
in [4].

The other class of random polynomials is introduced in
an interesting article of Edelman and Kostlan [5] in which the jth coefficients of T(x) in (1.1) have
nonidentical variance (nj).
It is interesting to note that in this case the expected number of zeros
significantly increased to n,
showing that the curve representing this type of polynomial oscillates
significantly more than the classical polynomial (1.1) with identical coefficients.
As it is the characteristic of (nj), j=0,1,2,…,n maximized at the middle term of j=[n/2], it is natural to conjecture that for other
classes of distributions with this property the polynomial will also oscillate
significantly more. This conjecture is examined in [6, 7]. This interesting and
unexpected property of the latter polynomial has its close relation to physics
reported by Ramponi [8], which together with its mathematical interest
motivated us to study the polynomialP(x)≡Pn(x,ω)=∑j=0naj(nj)1/2xj+1j+1.As we will see, because of the
presence of the binomial elements in (1.2), we can progress further than the
classical random polynomial defined in (1.1). However, even in this case the
calculation yields an asymptotic result rather than equality. With a small
change to the definition of the polynomial we show that the result improves. To
this end we defineQ(x)≡Qn(x,ω)=∑j=0naj(nj)1/2xj+1j+1+a*n+1,where a* is mutually independent of and has the same
distribution as {aj}j=0n.
We prove the following.

Theorem 1.1.

When the coefficients aj of P(x) are independent standard normal random
variables, then the expected number of real roots is asymptotic
toEN0,P(−∞,∞)~n+1.

Corollary 1.2.

With the same assumption as Theorem 1.1 for the
coefficients aj and a* one
has EN0,Q(−∞,∞)=n+1.

Also of interest is the expected
number of times that a curve representing the polynomial cuts a level K.
We assume K is any constant such that(i)K2≤enn2,(ii)1n2=o(K2),(iii)K2=o(enn2).For example, any absolute
constant K≠0 satisfies these conditions. Defining ENK,P as the expected number of real roots of P(x)=K,
we can generalize the above theorem to the following one.

Theorem 1.3.

When the
coefficients aj have the same distribution as in Theorem 1.1,
and K obeys the above conditions (i)–(iii), the
asymptotic estimate for the expected number of K-level crossings isENK,Q(−∞,∞)~ENK,P(−∞,∞)~n+1.

The other characteristic which also gives a good indication of the oscillatory
behavior of a random polynomial is the expected number of maxima or minima. We
denote this expected number by ENPM for polynomial P(x) given in (1.2) and, since the event of
tangency at the x-axis has probability zero, we note that this
is asymptotically the same as the expected number of real zeros of P′(x)=dP(x)/dx.
In the following theorem, we give the expected number of maxima of the
polynomial.

Theorem 1.4.

With the above assumptions on the
coefficients aj,
then the asymptotic estimate for the expected number of maxima of P(x) isENPM(−∞,∞)~n.

Corollary 1.5.

With the above assumptions for the
coefficients aj and a* one has ENQM(−∞,∞)~n.

2. Proof of Theorem <xref ref-type="statement" rid="thm1">1.1</xref>

We use a well-known Kac-Rice formula, [1, 9], in which it is proved thatEN0,P(a,b)=∫abΔπA2dx,where P′(x) represents the derivative with respect to x of P(x).
We denoteA2=var(P(x)),B2=var(P′(x)),C=cov(P(x),P′(x)),Δ2=A2B2−C2.Now, with our assumptions on the
distribution of the coefficients, it is easy to see thatA2=∑j=0n(nj)x2j+2j+1=(1+x2)n+1n+1−1n+1,B2=∑j=0n(nj)(j+1)x2j=(1+x2)n−1(1+x2+nx2),C=∑j=0n(nj)x2j+1=x(1+x2)n. We note that, for all
sufficiently large n and x bounded away from zero, from (2.3) we haveA2~(1+x2)n+1n+1.This together with (2.1), (2.4),
and (2.5) yieldsEN0(−∞,∞)~2π∫0ϵΔA2dx+2π∫ϵ∞n+11+x2dx,where ϵ>0, ϵ→0 as n→∞.
The second integral can be expressed as2n+1π{π2−arctanϵ}→n+1asn→∞.In the first integral, the
expression (Δ/A2) has a singularity at x=0:ΔA2=(n+1){(1+x2)2n−(1+x2)n−1(1+x2+nx2)}{(1+x2)n+1−1}2.Notice that the expression in
(2.9) is bounded from above:ΔA2<(n+1)(1−D)1+x2,whereD=1+nx2(1+x2)n−1−(1+x2)n{(1+x2)n−1}2=(n−1)(1+x2)n−2+(n−2)(1+x2)n−3+⋯+3(1+x2)2+2(1+x2)+1{(1+x2)n−1+(1+x2)n−2+⋯+(1+x2)+1}2. When x=0,
we haveD=n2−n2n2and therefore ΔA2<n+12n~n+12,which means that the integrand
in the first integral of (2.7) is bounded for every n.
When x>0,
it can easily be seen that1>D>∑j=0n−2(1+j)n2(1+x2)2n−2>0,and thereforeΔA2<n+11+x2.Hence, the first integral that
appears in (2.7) is bounded from above as follows:2π∫0ϵΔA2dx<2π∫0ϵn+11+x2dx=2(arctanϵ)n+1π=o(n+1)by the choice of ϵ.
Altogether, the value of the first integral in (2.7) is of a smaller order of
magnitude than the value of the second integral, and we have from (2.7)EN0(−∞,∞)~n+1which completes the proof of
Theorem 1.1.

In order to obtain the proof of Corollary 1.2, we note
that the above calculations remain valid for B2 and C.
However, for A2 we can obtain the exact value rather than the
asymptotic value. To this end, we can easily see thatAQ2=var(Q(x))=∑j=0n(nj)x2j+2j+1+1n+1=(1+x2)n+1n+1.Substituting this value instead
of (2.3) together with (2.4) and (2.5) in the Kac-Rice formula (2.1), we get a
much more straight forward expression than that in the above proof:EN0,Q(−∞,∞)=1π∫0∞n+11+x2dx=n+1.This gives the proof of
Corollary 1.2.

3. Level Crossings

To find the
expected number of K-level crossings, we use the following
extension to the Kac-Rice formula as it was used in [10]. It is shown that in the
case of normal standard distribution of the coefficientsENK(a,b)=I1(a,b)+I2(a,b)with I1(a,b)=∫abΔπA2exp(−B2K22Δ2)dx,I2(a,b)=∫ab2KCπA3exp(−K22A2)erf(−KC2AΔ)dx, where, as usual, erf(x)=∫0xexp(−t)dt≤π/2. Since changing x to −x leaves the distribution of the coefficients
unchanged, ENK(−∞,0)=ENK(0,∞).
Hence to what follows we are only concerned with x≥0.
Using (2.3)–(2.5) and (3.2) we obtainI1(−∞,∞)=2n+1π∫0∞11+x2exp(−K2(n+1)(1+x2+nx2)2(1+x2)n+1)dx.Using substitution x=tanθ in (3.4) we can see thatI1(−∞,∞)=J1(0,π2)=2n+1π∫0π/2exp(−K2(n+1)2(1+nsin2θ)cos2nθ)dθ,where the notation J1 emphasizes integration in θ.
In order to progress with the calculation of the integral appearing in (3.5),
we first assume θ>δ,
where δ=arccos(1−1/(nϵ)), where ϵ=1/{2log(nK)}.
This choice of ϵ is indeed possible by condition (i). Now since cosθ<(1−1/(nϵ)), we can show thatcos2nθ<(1−1nϵ)2n=((1−1nϵ)−nϵ)−2/ϵ~exp(−2ϵ)→0as n→∞.
Now we are in a position to evaluate the dominated term which appears in the
exponential term in (3.5). From (3.6), it is easy to see that for our choice of θK2n2cos2n−2θ<K2n2exp(−2ϵ)=K2n2exp(−4log(nK))=(Kn)−2→0,by condition (ii). Therefore,
for all sufficiently large n,
the argument of the exponential function in (3.5) is reduced to zero, and hence
the integrand is not a function
of θ and we can easily see by the bounded
convergence theorem and condition (iii) thatJ1(δ,π2)~n+1.Since the argument of the
exponential function appearing in (3.5) is always negative, it is straight
forward for our choice of δ and ϵ to see thatJ1(0,δ)<2n+1π∫0δdθ=2πn+1arccos(1−2log(nK)n)=o(n+1),by condition (iii). As I1(−∞,∞)=J1(0,δ)+J1(δ,π/2),
by (3.8) and (3.9) we see thatI1(−∞,∞)~n+1.Now we obtain an upper limit for I2 defined in (3.3). To this end, we let v=K/(2A).
Then we haveI2(−∞,∞)≤|K|2π∫−∞∞CA3exp(−K22A2)dx=2π∫0∞exp(−v2)dv≤2π.This together with (3.10) proves
that ENK,Q(−∞,∞)~n+1.
The theorem is proved for polynomial Q(x) given in (1.3).

Let us now prove the theorem for polynomial P(x) given in (1.2), that isENK,P(−∞,∞)~n+1.The proof in this case repeats
the proof for ENK,Q(−∞,∞) above, except that the equivalent of (3.4)
will be an asymptotic rather than an exact equality, and the derivation of the
equivalent of (3.9) is a little more involved, as shown below. Going back from
the new variable θ to the original variable x givesJ1(0,δ)=2π∫0tanδΔA2exp(−B2K22Δ2)dx<2π∫0tanδΔA2dx,where Δ/A2 is given by (2.9). Then by the same reasoning
as in the proof of Theorem 1.1,J1(0,δ)<2n+1πarctan(tanδ)=2n+1πδ=2n+1πarccos(1−2log(nK)n)=o(n+1),by condition (iii). This
completes the proof of Theorem 1.3.

4. Number of Maxima

In finding
the expected number of maxima of P(x),
we can find the expected number of zeros of its derivative P′(x).
To this end we first obtain the following characteristics needed in order to
apply them into the Kac-Rice formula (2.1),AM2=var(P′(x))=∑j=0n(nj)(j+1)x2j=(1+x2)n−1(1+x2+nx2),BM2=var((P′′(x))=∑j=0n(nj)j2(j+1)x2j−2=n(1+x2)n−3(2+4nx2+nx4+n2x4),CM=cov(P′(x),P′′(x))=∑j=0n(nj)j(j+1)x2j−1=nx(1+x2)n−2(2+x2+nx2). Hence from (4.1)–(4.3) we
obtainΔM2=AM2BM2−CM2=n(1+x2)2n−4[2+nx4+n2x4+2x2+2nx2].Now from (4.1) and (4.5) we
haveΔMAM2=n(2+nx4+n2x4+2x2+4nx2)(1+x2)(1+x2+nx2).As the value of x increases, the dominating terms in (4.5)
change. For accuracy therefore, the interval needs to be broken up. In this
case, the interval (0,∞) was divided into two subintervals. First,
choose ϵ<x<∞ such that ϵ=n−1/4,
thenΔMAM2~n1+x2.Substituting into the Kac-Rice formula (2.1) yieldsENPM(ϵ,∞)~1π∫ϵ∞n1+x2dx=n2.Now we choose 0<x<ϵ.
Since for n sufficiently large the term n2x4 is significantly larger than nx4 and also since for this range of x we can see 2x2<1, we can obtain an upper limit for (4.5) asΔMAM2<n(3+2n2x4+4nx2)1+nx2<n(3+6nx2+3n2x4)1+nx2=3n.Substituting this upper limit
into Kac-Rice formula, we can seeENPM(0,ϵ)=∫0ϵΔMπAM2dx<3nϵ=o(n1/4).This together with (4.7)
completes the proof of Theorem 1.4. To prove Corollary 1.5, it suffices to notice
that since Q′(x)=P′(x) and Q′′(x)=P′′(x),
all the arguments in the above proof apply to polynomial Q(x),
and we have therefore ENPM(a,b)=ENQM(a,b).

KacM.On the average number of real roots of a random algebraic equationWilkins,J. E.Jr.An asymptotic expansion for the expected number of real zeros of a random polynomialBharucha-ReidA. T.SambandhamM.FarahmandK.EdelmanA.KostlanE.How many zeros of a random polynomial are real?FarahmandK.NezakatiA.Algebraic polynomials with non-identical random coefficientsFarahmandK.SambandhamM.Real zeros of classes of random algebraic polynomialsRamponiA.A note on the complex roots of complex random polynomialsRiceS. O.Mathematical analysis of random noiseFarahmandK.On random algebraic polynomials