Abstract

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 𝑎0𝑛01/2𝑥/1+𝑎1𝑛11/2𝑥2/2+𝑎2𝑛21/2𝑥3/3++𝑎𝑛𝑛𝑛1/2𝑥𝑛+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)=𝑛𝑗=0𝑎𝑗(𝜔)𝑥𝑗,(1.1)where, for (Ω,𝒜,Pr) a fixed probability space, {𝑎𝑗(𝜔)}𝑛𝑗=0 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 𝑎𝑗𝑎𝑗(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)=𝑛𝑗=0𝑎𝑗𝑛𝑗1/2𝑥𝑗+1𝑗+1.(1.2)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)=𝑛𝑗=0𝑎𝑗𝑛𝑗1/2𝑥𝑗+1+𝑎𝑗+1𝑛+1,(1.3)where 𝑎 is mutually independent of and has the same distribution as {𝑎𝑗}𝑛𝑗=0. 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 to 𝐸𝑁0,𝑃(,)𝑛+1.(1.4)

Corollary 1.2. With the same assumption as Theorem 1.1 for the coefficients 𝑎𝑗 and 𝑎 one has 𝐸𝑁0,𝑄(,)=𝑛+1.(1.5)

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)𝐾2𝑒𝑛𝑛2,1(ii)𝑛2𝐾=𝑜2,(iii)𝐾2𝑒=𝑜𝑛𝑛2.(1.6)For example, any absolute constant 𝐾0 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 is 𝐸𝑁𝐾,𝑄(,)𝐸𝑁𝐾,𝑃(,)𝑛+1.(1.7)

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 EN𝑀𝑃 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) is 𝐸𝑁𝑀𝑃(,)𝑛.(1.8)

Corollary 1.5. With the above assumptions for the coefficients 𝑎𝑗 and 𝑎 one has 𝐸𝑁𝑀𝑄(,)𝑛.(1.9)

2. Proof of Theorem 1.1

We use a well-known Kac-Rice formula, [1, 9], in which it is proved thatEN0,𝑃(𝑎,𝑏)=𝑏𝑎Δ𝜋𝐴2𝑑𝑥,(2.1)where 𝑃(𝑥) represents the derivative with respect to 𝑥 of 𝑃(𝑥). We denote𝐴2=var𝑃(𝑥),𝐵2𝑃=var(𝑥),𝐶=cov𝑃(𝑥),𝑃(𝑥),Δ2=𝐴2𝐵2𝐶2.(2.2)Now, with our assumptions on the distribution of the coefficients, it is easy to see that𝐴2=𝑛𝑗=0𝑛𝑗𝑥2𝑗+2=𝑗+11+𝑥2𝑛+11𝑛+1,𝐵𝑛+1(2.3)2=𝑛𝑗=0𝑛𝑗(𝑗+1)𝑥2𝑗=1+𝑥2𝑛11+𝑥2+𝑛𝑥2,(2.4)𝐶=𝑛𝑗=0𝑛𝑗𝑥2𝑗+1=𝑥1+𝑥2𝑛.(2.5) We note that, for all sufficiently large 𝑛 and 𝑥 bounded away from zero, from (2.3) we have𝐴21+𝑥2𝑛+1𝑛+1.(2.6)This together with (2.1), (2.4), and (2.5) yieldsEN02(,)𝜋𝜖0Δ𝐴22𝑑𝑥+𝜋𝜖𝑛+11+𝑥2𝑑𝑥,(2.7)where 𝜖>0, 𝜖0 as 𝑛. The second integral can be expressed as2𝑛+1𝜋𝜋2arctan𝜖𝑛+1as𝑛.(2.8)In the first integral, the expression (Δ/𝐴2) has a singularity at 𝑥=0:Δ𝐴2=(𝑛+1)1+𝑥22𝑛1+𝑥2𝑛11+𝑥2+𝑛𝑥21+𝑥2𝑛+112.(2.9)Notice that the expression in (2.9) is bounded from above:Δ𝐴2<(𝑛+1)(1𝐷)1+𝑥2,(2.10)where𝐷=1+𝑛𝑥21+𝑥2𝑛11+𝑥2𝑛1+𝑥2𝑛12=(𝑛1)1+𝑥2𝑛2+(𝑛2)1+𝑥2𝑛3++31+𝑥22+21+𝑥2+11+𝑥2𝑛1+1+𝑥2𝑛2++1+𝑥2+12.(2.11) When 𝑥=0, we have𝑛𝐷=2𝑛2𝑛2(2.12)and therefore Δ𝐴2<𝑛+12𝑛𝑛+12,(2.13)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>𝑛2𝑗=0(1+𝑗)𝑛21+𝑥22𝑛2>0,(2.14)and thereforeΔ𝐴2<𝑛+11+𝑥2.(2.15)Hence, the first integral that appears in (2.7) is bounded from above as follows:2𝜋𝜖0Δ𝐴22𝑑𝑥<𝜋𝜖0𝑛+11+𝑥2𝑑𝑥=2(arctan𝜖)𝑛+1𝜋=𝑜𝑛+1(2.16)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+1(2.17)which 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 that𝐴2𝑄==var𝑄(𝑥)𝑛𝑗=0𝑛𝑗𝑥2𝑗+2+1𝑗+1=𝑛+11+𝑥2𝑛+1𝑛+1.(2.18)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,𝑄1(,)=𝜋0𝑛+11+𝑥2𝑑𝑥=𝑛+1.(2.19)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)(3.1)with 𝐼1(𝑎,𝑏)=𝑏𝑎Δ𝜋𝐴2𝐵exp2𝐾22Δ2𝐼𝑑𝑥,(3.2)2(𝑎,𝑏)=𝑏𝑎2𝐾𝐶𝜋𝐴3𝐾exp22𝐴2erf𝐾𝐶2𝐴Δ𝑑𝑥,(3.3) where, as usual, erf(𝑥)=𝑥0exp(𝑡)𝑑𝑡𝜋/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 𝑥0. Using (2.3)–(2.5) and (3.2) we obtain𝐼12(,)=𝑛+1𝜋011+𝑥2𝐾exp2(𝑛+1)1+𝑥2+𝑛𝑥221+𝑥2𝑛+1𝑑𝑥.(3.4)Using substitution 𝑥=tan𝜃 in (3.4) we can see that𝐼1(,)=𝐽1𝜋0,2=2𝑛+1𝜋0𝜋/2exp𝐾2(𝑛+1)21+𝑛sin2𝜃cos2𝑛𝜃𝑑𝜃,(3.5)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(11/(𝑛𝜖)), where 𝜖=1/{2log(𝑛𝐾)}. This choice of 𝜖 is indeed possible by condition (i). Now since cos𝜃<(11/(𝑛𝜖)), we can show thatcos2𝑛1𝜃<1𝑛𝜖2𝑛=11𝑛𝜖𝑛𝜖2/𝜖2exp𝜖0(3.6)as 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 𝜃𝐾2𝑛2cos2𝑛2𝜃<𝐾2𝑛22exp𝜖=𝐾2𝑛2exp4log(𝑛𝐾)=(𝐾𝑛)20,(3.7)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) that𝐽1𝜋𝛿,2𝑛+1.(3.8)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 that𝐽12(0,𝛿)<𝑛+1𝜋𝛿02𝑑𝜃=𝜋𝑛+1arccos12log(𝑛𝐾)𝑛=𝑜(𝑛+1),(3.9)by condition (iii). As 𝐼1(,)=𝐽1(0,𝛿)+𝐽1(𝛿,𝜋/2), by (3.8) and (3.9) we see that𝐼1(,)𝑛+1.(3.10)Now we obtain an upper limit for 𝐼2 defined in (3.3). To this end, we let 𝑣=𝐾/(2𝐴). Then we have𝐼2||𝐾||(,)2𝜋𝐶𝐴3𝐾exp22𝐴22𝑑𝑥=𝜋0exp𝑣22𝑑𝑣𝜋.(3.11)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.(3.12)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 gives𝐽12(0,𝛿)=𝜋0tan𝛿Δ𝐴2𝐵exp2𝐾22Δ22𝑑𝑥<𝜋0tan𝛿Δ𝐴2𝑑𝑥,(3.13)where Δ/𝐴2 is given by (2.9). Then by the same reasoning as in the proof of Theorem 1.1,𝐽12(0,𝛿)<𝑛+1𝜋2arctan(tan𝛿)=𝑛+1𝜋𝛿=2𝑛+1𝜋arccos12log(𝑛𝐾)𝑛=𝑜,𝑛+1(3.14)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),𝐴2𝑀𝑃=var=(𝑥)𝑛𝑗=0𝑛𝑗(𝑗+1)𝑥2𝑗=1+𝑥2𝑛11+𝑥2+𝑛𝑥2,𝐵(4.1)2𝑀=var(𝑃=(𝑥)𝑛𝑗=0𝑛𝑗𝑗2(𝑗+1)𝑥2𝑗2=𝑛1+𝑥2𝑛32+4𝑛𝑥2+𝑛𝑥4+𝑛2𝑥4,𝐶(4.2)𝑀𝑃=cov(𝑥),𝑃=(𝑥)𝑛𝑗=0𝑛𝑗𝑗(𝑗+1)𝑥2𝑗1=𝑛𝑥1+𝑥2𝑛22+𝑥2+𝑛𝑥2.(4.3) Hence from (4.1)–(4.3) we obtainΔ2𝑀=𝐴2𝑀𝐵2𝑀𝐶2𝑀=𝑛1+𝑥22𝑛42+𝑛𝑥4+𝑛2𝑥4+2𝑥2+2𝑛𝑥2.(4.4)Now from (4.1) and (4.5) we haveΔ𝑀𝐴2𝑀=𝑛2+𝑛𝑥4+𝑛2𝑥4+2𝑥2+4𝑛𝑥21+𝑥21+𝑥2+𝑛𝑥2.(4.5)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=𝑛1/4, thenΔ𝑀𝐴2𝑀𝑛1+𝑥2.(4.6)Substituting into the Kac-Rice formula (2.1) yieldsEN𝑀𝑃1(𝜖,)𝜋𝜖𝑛1+𝑥2𝑑𝑥=𝑛2.(4.7)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) asΔ𝑀𝐴2𝑀<𝑛3+2𝑛2𝑥4+4𝑛𝑥21+𝑛𝑥2<𝑛3+6𝑛𝑥2+3𝑛2𝑥41+𝑛𝑥2=3𝑛.(4.8)Substituting this upper limit into Kac-Rice formula, we can seeEN𝑀𝑃(0,𝜖)=𝜖0Δ𝑀𝜋𝐴2𝑀𝑑𝑥<𝑛3𝑛𝜖=𝑜1/4.(4.9)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 EN𝑀𝑃(𝑎,𝑏)=EN𝑀𝑄(𝑎,𝑏).