EXPECTED NUMBER OF REAL ZEROS OF A CLASS OF RANDOM HYPERBOLIC POLYNOMIAL

Let y1(ω), y2(ω), . . . , yn(ω) be independent and normally distributed random variables with mean zero and variance one. For large values of n, it is proved that the the expected number of times the random hyperbolic polynomial y1(ω)sinht+ y2(ω)sinh2t+ · · ·+ yn(ω)sinhnt crosses the line y=K is (1/π) log n +O(1) as long as 0 ≤ K ≤ √n. AMS Subject Classification: 60H99, 26C99

Let EN n,K (α, β) be the expected number of times f n (t) crosses the line t = K.
Received: September 14, 2013 c 2014 Academic Publications, Ltd. url: www.acadpubl.eu§ Correspondence author A lot of information about the behavior of a polynomial can be obtained by determining EN n,K (α, β) in different subintevals (α, β) of R.To calculate the expected number of crossings of different type of random polynomial the famous Kac-Rice formula(see [1], [2] and [4]) is a widely used technique.Applying the formula, Das [3] calculated the expected number of axis crossings of a random polynomial having hyperbolic elements, namely in a neighbourhood that does not contain zero.It is to be noted that Das's approach for calculation of expected number of real zeros of g n (t) cannot be emulated for our random polynomial f n (t) in a neighbourhood of the origin.We still do not have a satisfactory estimate of EN n,0 (−∞, ∞).Later using the Kac-Rice formula, Farahmand and Hannigan [5] have calculated EN n,K (−∞, ∞).
In their estimation they have tried to get over the obstacle one may face near zero by including an upper bound of expected number of level crossings near zero in their result.This move succeeds in providing the asymptotic estimate of EN n,K (−∞, ∞) as (1/π) log n, but brings about a rather unwelcome large error term of the order √ log n.Clearly,by reducing the error term, a more accurate picture of random polynomial can be obtained.Thus our objective in this paper is twofold.Firstly we calculate EN n,0 (−∞, ∞) and then we show that the error terem in the asyymptotic estimate of EN n,K (−∞, ∞) is infact O(1).We prove the following two theorems.
Theorem 1.If the coefficients of f n (t) are normally distributed independent random variables with mean zero and variance one, then for large values of n: EN where 0.7731.. < C 2 < 0.806772.
It is to be observed that if we calculate EN n,0 (−∞, ∞) from EN n,K (−∞, ∞), which has been found out in [5], by replacing K by zero, then we obtain a large error term √ log n.We apply a little calculus to overcome this obstacle in the proof of the Theorem 1.

Proof of the Theorem 1
Let us consider an interval (α, β) which does not include zero.The Kac-Rice formula for expected number of zeros of f n (t) in this interval is where: For the ease of computation involved, we shall calculate the integral of (1) after obtaining the dominant terms in X n , Y n and Z n in some suitably chosen subintervals of (-∞,∞).Let us consider the interval (2m −1 log m, ∞) first.In this interval we find that n s sinh s t(sinh nt) −1 , where s is a finite positive number, is a monotonically decreasing function of t and tends to zero for sufficiently large values of n.With the help of this information, we can obtain that and Thus by (1) we have Our next aim is to calculate EN n,0 (1/2 n , 2m −1 log m).To this end, we write g m (t) in the following manner where τ = mt and η (τ ) = 1/ sinh t − 1/t.
Then X n ,Y n and Z n can be written as where We note that η (τ ) has an absolutely and uniformly convergent power series representation in (0, π) [6, page 35].Hence η (τ ) , η ′ (τ ) ,and η ′′ (τ ) are bounded in (0,1).As a consequence a i , b i and c i are bounded in (0,1) and we obtain the following relations in 1/2 n , 2m and 8τ (sinh τ It is to be noted that the o(1) terms mentioned in ( 7) where 2 )/(3a 2 1 ).We now calculate EN n,0 (3/m, 2m −1 log m) using (10).The following inequality, which can be shown to be valid for τ ≥ 3 if a little algebra is applied, helps to find the bounds of of δ n (τ ).Thus Using ( 11), ( 10) and ( 1) we can obtain with the help of standard results of integration and numeric integration that where where −0.114472 < L < −0.097636.Since δ n (τ ) is bounded for t ∈ (1/2 n , 3/m), it is also possible to calculate EN n (1/2 n , 3/m) with the help of (10) using numeric integration.Thus We next calculate EN n,0 (−1/2 n , 1/2 n ) .The formula (1) cannot be applied in this interval.However, an upper bound of will be sufficient for our purpose.To this end, we shall first calculate the expected number of real zeros of f ′ n (t) which we shall denote as Using Kac-Rice formula for f ′ n (t), we note that where We note that f n (t) has at least one zero in (−1/2 n , 1/2 n ), since f n (t) vanishes at t = 0. Since (15) is true, by Roll's theorem, f n (t) can have at most one zero in (−1/2 n , 1/2 n ).Thus, we conclude that Finally, we observe from (1) that the expected number of zeros of (α, β) and (−β, −α) are equal.Therefore, the expected number of zeros in (−∞, −1/2 n ), is same as those in (1/2 n , ∞).The proof of the Theorem 1 now follows from ( 6), ( 12), ( 13) and (16).