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w w w . a j e r . o r g Page 169 American Journal of Engineering Research (AJER)

e-ISSN: 2320-0847 p-ISSN : 2320-0936 Volume-03, Issue-05, pp-169-179 www.ajer.org Research Paper Open Access

The Expected Number of Real Zeros of Random Polynomial A. K. Mansingh1, P.K.Mishra2

1 Department of Basic Science and Humanities, MITM, BPUT, BBSR, ODISHA, INDIA.

2 Department of Basic Science and Humanities, CET, BPUT, BBSR, ODISHA, INDIA.

Abstract:- In this paper we have estimated the number of real zeros of

Z z

, z X (Z)

Q

n

0 j

j j

n which is a random Gaussian polynomial satisfying the normal distribution with mean zero and variance one i.e. E(X j)0 and E(Xj)2 1 for j ≥ 0. Much research works has been done on the same polynomial with different co-efficients satisfying the above condition and found that the expected number of zeros is approximated to

 2 logn as nin the interval (-,). Our present work is to estimate the number of zeros in the interval [0, 1] and found that the expected number of real zeros of the above polynomial under same conditions is ENn 0,1~

 

1 logn asn . Our result gives better approximation as compared to results given by Yoshihara [6]

Keywords: - Independent identically distributed random variables, random algebraic polynomial, random algebraic equation, real roots.

INTRODUCTION

Let X0, X1……Xn be a stationary Gaussian process satisfying E(X j)0 and E(Xj)21 for j ≥ 0 as sufficient conditions. Then the expected number of real zeros of the above

polynomial

n

j j j

n z X z

Q

0

)

( is

2

1 ( log n ) as n tends to infinity. The expected number of zeros of a random trigonometric polynomial has been studied by Dunnage[1] and estimated the number of zeros in the interval [0,1] which is approximated to (1/2)logn where

n but in the same problem we consider a polynomial which is piece wise continuous and differentiable in the same interval and used normal distribution with mean zero and variance one and applied Gaussian process. Ibragimov and Maslova [2] had worked with same mean and variance under different conditions and had got the expected number of zeros in the same interval is approximated to the result of Dunnage [1].

Let us consider the Gaussian polynomial

n

j

j j

n z X z

Q

0

)

( (1)

(2)

w w w . a j e r . o r g Page 170 Which is a piece-wise continuous and differentiable polynomial within a closed interval [0, 1]

and satisfying same conditions. Suppose that there is an interval [-b,b] , 0<b≤ π, where

 

f is uniformly approximated by the partial sums, Snf  of its Fourier series

development, and 0 ≤ m ≤ f  ≤ M ≤ , θ-π,π where m and M are the lower and upper bounds of f  . Then the expected number of real zeros of the above polynomial is

 

 0,1 ~ 1 logn

ENn as n.

Let the number of zeros of the above polynomial is denoted by ENn. .The main aim of our work is to estimateENn  0,1 .

Now we have partitioning the interval [0, 1] into three different sub intervals namely In1, In2

& In3

the details of partitions are given bellow,



 

1 log ), 1 log

(

log ) 1 log

( ), log 1 1

(

) log 1 1

( , 0

3 2 1

n I n

n n n

I

n I

n n n

For n≥3,

Let Nn (a,b) be the number of sign changes of a piece-wise linear approximation to Qn(x).

First we estimateENn(In2). Then at last section we have approximated ENn(In1

) and ENn(In3

) and found that the expected number of zeros of both the intervals are equivalent to σ(logn) as n tends to infinity

i.e. ENn(In1)ENn(I33) (logn)

COVARIANCE ESTIMATES:

Let kn(x,y)=E{Qn(x)Qn(y)} and rn(x,y )=Cor {Qn(x) , Qn(y)}. We estimate kn(x,y) and rn(x,y) for x,y satisfying certain conditions. For xIn2 we derive upper and lower bounds for kn(x,x).

Let Tn() be a trigonometric polynomial of order n with real coefficients. Suppose we define

- , and c R

, e c )

( v

n

-n v

iv

v

Tn (2)

2 / x c 2

, 0

n

0 v

v

v c

x T

G n (3)

Then

  x,y (0,1]

1 y - 1 x - y) 1 (x, r

(0,1]

y x, 1

, , ,

,

12 12

2 2

/

xy xy

y T G x T y G x

Tn n n

ka

Taking 0dr/(x, y)

(3)

w w w . a j e r . o r g Page 171 We estimate kn(x,y) satisfying the conditions

d

0 And x, y In

2

We know from co-variance estimates

have w e

y x, For

n as 0 )

, ( )

, (

2 1

/

0 ) , ( ,

sup

2

I In n

n

d y x r

y x

y x r y x r In

THEOREM -1 Suppose that f() can be uniformly approximated by the partial sums of its Fourier series development f() A,,& f (0)>0. If there is a constant α and an integer N0 such that G(Snf,x)0for nN0 , x  0,1 Applying co-variance estimates to find mean and variance of the series j

m

j jX

a

0

we have

0 m for 2

0 2 2

0





m

j j

m

j

jX A aj

a

E and

f( )d

) , (

0

 0

n

v iv v n

v

iv v

n x y x e y e

k

Q(x)Q(y) ( , )

E y) k(x,

et k x y

L n

Limn

 

ye f d

xe i 1 i ( )

1 1 1  4

m

m v

iv ve

c

) ( T

Let m (5)

We have by Cauchy’s residue theorem that 1  111 2 (1 )1

eiv xe i yei d xv xy

And 1  111 2 (1 )1

e iv xe i yei d yv xy

xe   ye g d C

y x h

b

b

i

i

1 () 1

) ,

( 1 1 (6)

Where C is a constant depending on B and b.

Taking m=n in equation (5) we have m=n and Tn(θ)=Snf(θ), where Snf(θ) is the nth partial sum of the Fourier series development of f(θ). Now for x, y 0,1 we have

) 0

, , ( ) ,

(x y S f x y J

kn ka n (7)

Where J0 J1J2J3J4 and

(4)

w w w . a j e r . o r g Page 172

     

b

b

n i i

n

a S f x y xe ye S f d

k

J1 ( , , ) 1 11 1  8

 

b

b

i

i ye f d

xe y

x k

J2 ( , ) 1 11 1 ()  9

 

b

b

n i

i ye S f f d

xe

J3 1 11 1( () ()  10

) , ( ) ,

4 k (x y k x y

J n (11)

From the conditions of Zygmund [7] we found that there is a constant B not depending on n such that

0.

n all for B

d f

Sn

, )

(

& g()Snf() gives J1C Where C depends on b and B & g(θ)= f(θ) gives

Let

( ) ( )

sup )

(

,

S f f

n

a n

b b

By Cauchy’s inequality we have

2 / 1 2 2

/ 1

3 (1 2) (1 )

) ( 2

y x

n J a

) , , ( )

,

(x y k S f x y

kn a n

 

 

   

x y

xy xn yn n

n C a

2 2 1

1 1

2 / 2 1

/ 1

1 1 1

For x,y 0,1 and where C is a constant depending upon b, A and B So

   1 (1)

) , , ( ) ,

( 1/2

2 2 / 21

y x

n y w

x f S k y x

kn a n

(12)

Where w(n)0 as n For x,yI1n In 2

Now

n

v v v

nf x r x

S G

1 2

) 1

,

( (13)

We show that there is a constant α>0 and an integer N0 such that

0 2

n ,

x 0

) ,

(S f x when I n N

G n From Abel’s Theorem and Titchmarsh[4]

conditions we see that G(Snf ,x) is uniformly convergent for x 0,1 and

 

n

v v x

f r x

f S G

1 2 1

1

0 )

,

lim ( (14)

PRROF OF THEOREM -2 Using the two conditions f()A,, and f(0)>0 of Theorem-1 and applying the uniform convergence of the series within a certain interval

) ,

(

f

Sn and Snf(0)G(Snf,0) by adopting similar procedure as in the proof of Theorem -1 we see that J1=J2=0 holds for xI1nIn2. For some µ we construct an

(5)

w w w . a j e r . o r g Page 173 interval-,and f() 0. Such an interval exists as f()is continuous at θ=0 and

) 0 (

f >0. We consider the case when x

1(logn)1/2,1

and x=1 separately.

The following inequality holds for x,y0,1

1-cos b

, 2 0

1 1 2 -1/2

0

sup

b xe i

b

2 for

1

1 1

0

sup

b xe i

b

We have

0,1

x for

0 0

sup

n iv

v v

b

e x

Substitution in kn(x,x) with 1xeiv 2 replacedby

2

0

n iv

v

ve x

We got

0,1

x C ) ( -

x) (x,

k 2

b

b -

n

0 v

n   xveiv f d (15)

Substitute f () 1/2 in (15) we have C n

0 n

0 -b

2

-

2

 

 xve iv d b xve iv d

(16)

Where C depends only on b and A.

By simple calculation we have

 0,1 x , 2

n

0 v

2 2

- n

0 v

 

iv v

ve d x

x

(17)

 

~2 , n

n

0 v

2 2

b -

n

0 v

v b

iv

ve d x

x (18)

and x

1(logn)1/2,1

.

From our construction of [-b,b] we have

f d x e d

e x

b

iv v b

iv v

2

b -

n

0 v 2

b -

n

0 v

)

(  

  (19)

So the desired result follows forx1(logn)1/2,1.

When x=1 we have

 1,1 1 2 1 2  1  0

1

f n r j n

n n

n

j n j

k

(20)

(6)

w w w . a j e r . o r g Page 174 Where σn f(θ) is the nth Cesaro sum associated with f(θ). As f(θ) is continuous at θ=0 we have σn f(0) f(0) as n. Hence we

have

a e f d

e a x

a E

v iv v v

iv v j

j ( )

-

m

0 m

0 m 2

0

j





given that f()A. Then by using Cauchy’s inequality we have

d

e a A x

a

E j j   v iv





- m 2

0 v m 2

0 j

(21)

By simple calculation we have

  m

j iv j

ve d a

a

0 2

- m 2

0 v

2

(22)

GENERAL APPROXIMATION FORMULA : - For x a,b we approximate Qn(x) by a process which linearly interpolates between Qn(a) and Qn(b). It is convenient to count the sign changes of this process in [a,b] by

Q (a)Q (b)

2 sgn 1 2 ) 1 ,

( n n

b a

Nn (23)

Where sgn{x} =

0 x 1 -

0 x 0

0

>

x 1

The results of this section depend on an upper bound for the number of zeros of Qn(x) in an interval [a,b]. Define the event

b]

[a, in zeros more k has ) (x Q

Uk n (24)

For any interval [a,b] [0,1] let

 

1-(n 1) ,1,

b ), )(

1 (

and , 1) (n - 0,1 b , ) 1 )(

(

1 -

-1 1

a b n

b a b

LEMMA -1 let 230 and

(i) f() is continuous at =0 (ii) f(0)0

(iii) f()A,-,

Then there is an integer N1 and an absolute constant C such that 0

k N n )

(Uk C 3k/5 1

P

COROLLARY-1 Under the conditions of Lemma-1 we have

N n , ])

, ([

) ,

(a b EN a b C6/5 1

ENn n

Where C is an absolute constant.

Now we have to estimate the number of zeros in the three sub intervals.

(A). ESTIMATION OF NUMBER OF ZEROS IN THE INTERVAL:- I1n

(7)

w w w . a j e r . o r g Page 175 LEMMA -2 Let X0, X1……Xn be a set of n points satisfying E(X j)0 and E(Xj)21 for j ≥ 0 .

(i)Let us consider a function f(θ)which is uniformly approximated by the partial sums, Snf(θ)in the interval [-π,π]

(ii) In the interval 0 ≤ m ≤ f(θ) ≤ M ≤ and θ-π,π where m and M are the lower and upper bounds off(θ). Then expected number of real zeros in the interval

)

log 1 1 ( ,

1 0

In n is ENn(I1n)C1/2/2loglogn n2 (25)

Where C is a finite constant

PROOF: - From the Kac-Rice [3] formula and using the postulates of Shankar [4] which states that

 a b dx

EN C B Rn

b

a n n n

2 2 2 / 1 2 /

1 1

1) ( ) ,

( (26)

( )

E C , )) (

(Q x 2 n Q / x 2

E

Bn n n (27)

Q (x),Q /(x)

Cor

Rn n n (28)

Where ( )

dx

d Qn x =Qn/(x)

Now we have to find an upper bound for Cn/Bn with xI1n

We represent Bn=kn(x,x) by Noting that f()m0,, and applying Lemma-1 gives 2

0

2

n

v v

n m x

B for x 1,0 From Lemma-2 we have

 

n 1

1 2

2 2 v

iv v v

n A v x ae

C summing

n

1 2

v

x v and using that 2( 1) 2(1 2) 3

1

2

n v x v x

v

0,1

x (29)

 1-x

) 1

)(

/ 2 (

/ n 2(n1) 1 2 -2

n B A m x

C (30)

2 n for x

,

1 I1n

) 1 (

2 n

x

x2(n 1) I1nas n

x 0,x

sup I

and 1

n

We deduce that

1-x2 n1

1is

bounded above by a constant. So

 1-x -2 n 2and x I1n

/ n

n B

C (31)

Substituting the value of Cn/Bn and using the relation (1 R2n) 1 gives the expected number of zeros in the interval I1n is ENn(I1n)C1/2/2loglogn for n2 (32) Hence the theorem is proved.

LEMMA-3 Let X0, X1……Xn be a set of n stationary points satisfying E(Xj)0 and 1

) (Xj 2

E . for j ≥ 0 Suppose that (i) there is an interval , and there is a constant

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