T035169178

American Journal of Engineering Research (AJER)

2014

American Journal of Engineering Research (AJER) e-ISSN : 2320-0847 p-ISSN : 2320-0936 Volume-03, Issue-05, pp-169-178 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 Mathematics, MITM, BPUT, BBSR, ODISHA, INDIA. Department of Mathematics, CET, BPUT, BBSR, ODISHA, INDIA.

2

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

Q n (Z) 

n

X j 0

j

z j, z  Z

which is a random Gaussian polynomial satisfying the normal distribution with mean zero and variance one i.e.

E ( X j)  0 and E( X j) 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

  log n as n   in the interval

approximated to 2

that the expected number of zeros

is

(- ,  ). 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

 

 2πlog n as n   .

same conditions is ENn 0,1 ~ 1

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.

I.

INTRODUCTION

Let X0, X1……Xn be a stationary Gaussian process satisfying

E ( X j)  0 and E( X j) 2  1 for j ≥ 0 as

sufficient conditions. Then the expected number of real zeros of the above polynomial Q n ( z ) 

n

X z j 0

j

j

is

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

Q n ( z)   X j z j j 0

(1)

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,

S f   n

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of its Fourier series development, and 0 ≤ m ≤

f   ≤

M≤



,

θ  - π, π

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American Journal of Engineering Research (AJER) where m and M are the lower and upper bounds of

f  

2014

. Then the expected number of real zeros of the above

 2πlog n as n   .

ENn0,1 ~ 1

polynomial is

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

Now we have partitioning the interval [0, 1] into three different sub intervals namely I n1, In2 & In3 the details of partitions are given bellow, Partition [0,1] into three sub intervals In1, In2, In3 given by

 

  1 1 I n  0, (1  ) log n    log log n  1 2 I n  (1  ), (1  ) n log n   log log n   3 I n  (1  ),1 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 estimate 2

EN n ( I n ) . 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

EN n (In )  EN n (I3 )   (log n) 1

i.e.

3

2.2.4 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 Let

T n( )

x  I n we derive upper and lower bounds for kn(x,x).

2

be a trigonometric polynomial of order n with real coefficients. Suppose we define

n  n  T n ( )   c v e iv ,   -  ,   and c v  R (2) and GT n , x   2   c v x v c 0 / 2  (3) v  -n  v 0 

Then

k T a

n

GT n , x   GT n , y  1  xy

, x, y  

r / (x, y) 

1 - x  1 - y  2

1

2

2

1

x,y  (0,1] Taking 0  d  r (x, y)   /

2

x,y  (0,1]

1  xy

We estimate kn(x,y) satisfying the conditions

0  d   and x, y 

2

In

We know from co-variance estimates

sup

r n ( x, y )  r / ( x, y )  0 as n  

2

x , y n r  ( x , y ) d 0

I

For x, y  I n  I n we have 1

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American Journal of Engineering Research (AJER)

f ( )

THEOREM -1 Suppose that



2014

can be uniformly approximated by the partial sums of its Fourier series

 G ( S n f , x)    0 for n  N 0 , x  0,1 Applying

development f ( )  A  ,     ,  & f (0)>0. If there is a constant α and an integer N0 such that co-variance estimates to find mean and variance of the

m E   a j X  j 0

m

a

series

j 0

j

X

we

j

have



 n v iv  n v iv  k n( x, y)     x e   y e  f( )d  v 0    v  0 

 1  xe  1  ye 



i 1

i 1

n 

Let T m ( ) 





We have by Cauchy’s residue theorem that

for m  0 and

Let k(x, y)  EQ(x)Q(y)   Limk n ( x, y )

4

f ( )d

2

m 2    2  A  aj j  j 0 

m

c

vm

   e 1  xe  1  ye  i

iv

1

i

1

v

e iv

(5)

d  2xv (1  xy)1





iv 1  xei  1  yei  d  2y v (1  xy) 1 and  e 1

1



b



h( x, y)   1  xe i

 1  ye  1

i 1

g ( ) d  C (6) Where C is a constant depending on B and b.

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

 

k n ( x, y )  k ( S n f , x, y )  J 0 a

b



J 1 k (S nf , x, y)   1  xe  i a

Where J 0  J 1 J 2 J 3 J 4 and

(7)

 1  ye  1

i  1

S nf  d

8

b

   1  xe  1  ye  b

J 2  k ( x, y ) 

i

1

i

1

f ( )d

9

b

(9)    1  xe  1  ye  b

J 3

1

i

i

1

( S n f ( )  f ( )d

10 J 4 k n ( x, y)  k ( x, y)

b

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



 S nf ( )



d  B  , for all n  0. & g ( ) S nf ( ) gives

J 1 C  

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

a (n)  sup S n f ( )  f ( )   b ,b 

By Cauchy’s inequality we have



J 3

2a(n) (1  x 2 )1/ 2 (1  y 2 )1/ 2

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k n( x, y)  k a ( S nf , x, y) Page 171

American Journal of Engineering Research (AJER)

2014

 

n 1 n 1   n 1   an   x  y  xy  C 1   1 / 2 1/ 2 2 2   1 x 1 y   For x, y  0,1 and where C is a constant depending upon b, A and B So w(n) (12) k n( x, y)  k a (S nf , x, y)  2 1/ 2 1  x  1  y 2 1/ 2 1 2 Where w(n)  0 as n   For x, y  n  n







I

Now G ( S n f , x ) 

I

n

1 2

  r vxv

(13)

v 1

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

G ( S n f , x)    0

x  In 2

when

, n  N 0 From Abel’s Theorem and Titchmarsh’s[4] conditions

 

we see that G(Snf,x) is uniformly convergent for x  0,1 and n

lim G(S 



x 1

f , x)  12   r v  f 0

(14)

v 1

f ( )  A  ,     ,   and f(0)>0 of Theorem1 and applying the uniform convergence of the series within a certain interval S n f ( )     ,   and S n f (0)  G ( S n f ,0) by adopting similar procedure as in the proof of Theorem -1 we see that J1=J2=0 PRROF OF THEOREM -2 Using the two conditions

holds for interval

x  I n  I n . For some µ we construct an interval   -  ,   and f( )    0 . Such an 1

exists



2

as

x  1  (log n)

f ( ) is continuous at θ=0 and 1 / 2



,1

f (0) >0. We consider the case when

and x=1 separately.



The following inequality holds for x, y  0,1

sup

1  xe

 i 1



 1 - cos 2 b



-1/2

, 2  b  0

0  b   

sup

1  xe i

1

 1 for

  b  2

We have

0  b   

n

sup  x e

 iv

v

for x  0,1 

0b    v  0

Substitution in kn(x,x) with 1  xe

 iv 2

n

replaced by

 xve

 iv 2

v 0 We got b

n

k n (x, x) -   x v e iv

2

f ( )d  C x 0,1

(15)

-b v 0

Substitute

f ( )  1 / 2 in (15)

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we have

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American Journal of Engineering Research (AJER) 2



n

-

 0

v  iv

 x e

2

n

b

v  iv

x e

d  

2014

d  C

(16)

 0

-b

Where C depends only on b and A. By simple calculation we have 

n

 x e v

- v  0

2  iv

b

n

d  2  x , x  0,1 (17) v 0



v

2  iv

-b v 0



1 / 2

n

 x e

2v

n

d ~ 2  x 2v , n  

(18)

v 0

,1 . and x  1  (log n) From our construction of [-b,b] we have b



2

n

b

 x v e iv f ( )d   

-b v 0

2

n

 x v e iv d (19)

-b v 0



So the desired result follows for x  1  (log n) When x=1 we have n

k n 1,1  n  1  2 n  j  1rj  2 n  1 n f 0

1 / 2



,1 .

(20)

j 1

Where σn f(θ) is the nth

Cesaro sum associated with f(θ). As f(θ) is continuous at θ=0 we have σ n 2

 m  m   m   iv  f(0)  f (0) as n   . Hence we have E   a j x j      a v e a v e iv  f ( )d       v 0  -  v  0  j 0 

given that f ( )  A  . Then by using Cauchy’s inequality we have 2

  m  E   a j x j   A  -  j 0 

m

a e v 0

2  iv

v

d

(21)

(2.2.35)

By simple calculation we have 

m

 a e

- v0

2 iv

v

m

d  2  a j

2

(22)

j 0

(2.2.36)





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

1 1 N n (a, b)       sgn Q n (a)Q n (b) 2 2 x> 0 1  x 0 Where sgn{x} =  0 - 1 x 0 

(23)

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

Define the event (24) U k Q n (x) has k more zeros in [a, b] For any interval [a,b]  [0,1] let

  (b  a)(1  b) 1 ,



  (n  1)(b  a), b  1 - (n  1)-1 ,1, (i) f ( ) is continuous at  =0 (ii)



b  0,1- (n  1)-1 ,

and

LEMMA -1 let

  230 and

f (0)  0

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

2014



f ( )  A  ,   -  ,  (iii) Then there is an integer N1 and an absolute constant C such that P(U k )  C 3k / 5

n  N1

k0

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

EN n (a, b)  EN n ([a, b])  C 6 / 5 , n N1 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:LEMMA -2 Let X0, X1……Xn be a set of n points satisfying

I

1 n

E ( X j)  0 and E( X j) 2  1 for j ≥ 0 .

(i)Let us consider a function f(θ ) which is uniformly approximated by the partial sums, Sn f(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





1





/ 2 log log n n  2

ENn( I n)  C 1

1/ 2



Then expected number of real zeros in the interval I n  0, (1 

f(θ ) .

bounds of



 1 ) is log n 

(25)

Where C is a finite constant PROOF. From the Kac-Rice [3] formula and using the postulates of Shankar [4] which states that b 1 1/ 2 1 / 2 2 2 EN n (a, b)  ( )  C n Bn 1  Rn dx (26)





a



B n  E (Q n ( x)) 2 , C n  E Q n ( x)





/





2

(27)

Rn  Cor Qn ( x),Qn ( x) d / Q n ( x) = Q n ( x ) Where dx /

(28)

Now we have to find an upper bound for We represent Bn=kn(x,x) by Noting that n

B n  2m  x 2 v

 

for x  1,0

v 0

n

x

n

2v

and using that

v 1

v

2

C n / B n with x In 1

f ( )  m  0,    ,   

n

v x  

From Lemma-2 we have C n  2A



2

v1

2 v1

 av eiv  summing 

x  0,1 

x 2 ( v 1)  2(1  x 2 ) 3

v 1

C n / B n  (2 A / m)(1  x 2( n 1) ) 1 1 - x 2

(29)

and applying Lemma-1 gives



-2

(30)

x 2( n1)  1, x  In for n  2 1

and

supx 1 x 2( n1)  0, x  In as In 1

n

1 - x   

2 n 1 1

We deduce that

is bounded

above by a constant. So

C n / B n  1 - x 

n  2 and x  In (31) 1

-2

Substituting the value of Cn/Bn and using the relation the interval

1

I

n

is





(1  R 2n)  1 gives the expected number of zeros in

ENn( I n)  C 1 / 2 / 2 log log n for n  2 (32)

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Hence the theorem is proved .

LEMMA-3 Let X0, X1……Xn be a set of n stationary points satisfying 0 Suppose that (i) there is an n 1   r jx j    0 2 j1

  ,  

interval

x  0,1

E ( X j)  0 and E( X j) 2  1 . for j ≥

and there is a constant α such that

n  N 0 , for some integer N0 there is a constant C such that

I   C2

  log logn for n  N 5   1 PROOF. To find an upper bound for Cn/Bn, x n and applying Lemma-1 and Lemma-2 we have EN n

1/ 2

1

n

I

n 1 G S n f , x     r j x j    0 for n  N 0 2 j 1

(33)

Let N n be any integer So the conditions of Theorem-2 are satisfied. From Theorem-2 and any such that 1 B n k n( x, x)  CK a (S nf , x, x), n  N4 & x  In (34)

C  0,1 we have an integer N4

Where C being any constant and C  (0,1) From LEMMA-2 we have

K a ( S n f , x, x )  2

G ( S n f , x) 1 x2

G(S nf , x)    0 for n  N 0 . So 2C

But

Bn 

, n  N 5  max (N0 ,N4 )

1 x2

(35)

In the same way as in Lemma- 2 we have

B n /C n  1 - x  , n  N 5 max (N0 ,N4 ) . Then the expected number of zeros in the interval -2

3

I

n

is

EN n

I    C2

1/ 2

1

n



LEMMA-4 Let

  log logn for n  N 5 (36) 

Hence the theorem is proved.

Z j, j  0 be a stationary sequence of uniformly mixing random variables with zero mean

and (i) E Z

2  j

 n (ii). E   Z   j 0

  for   0 2

  j    as n   

Then there exists a constant C such that 2 

n

EZ j 0

THEOREM- 2

j

 n  C E   Z  j 0

  j  

2 1 / 2

Let X0, X1……Xn be a set of n stationary real-valued uniformly mixing Gaussian process

E ( X j)  0 and E( X j) 2  1 for j ≥ 0 satisfying (i)    1 / 2 (k )   and (ii) f(θ ) >0 and also k 1 j satisfied by the sequence (-1) X j , j  0 . with



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

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American Journal of Engineering Research (AJER)

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 

We have

EN n I 1n  C log n1/2 log logn Where C is a constant.

PROOF: - Choose C such that

B0   X 0  c  and

P X 0  c  q  1 and Choose the events

B k   X 0  c......X k 1  c, X k  c and k  1,....n.

B   X 0  c...... X n  c 

Let 0  r  R .

Using the argument of Ibragimov and Maslova[2] We obtain. n

n

k 1

k 0

EN n  r , r    kP( B k )  nP( B)  (log R / r ) 1   H sdP Where

 (k ) i  H s log  sup (k!c) 1 Q n (R e )   0 ,  

Bk

(37)

Here Qn (k) (x) is the kth derivative of Qn (x) with respect to x.

We estimate P(Bk) and P(B) Define a sequence of random variables (Zk, k=0,…….n) by

1  q  Z k  -q 

Xk  c Xk  c

Now for k=1…..n we have

  k 1   Bk     Z j  kq  Z k  1  q and  j 0     B 0  ( Z 0  1  q)

We show that the conditions of Lemma-4 are satisfied by (Zk, k≥0) with   4 Clearly (Zk, k≥0) is a stationary sequence of uniformly random variables with mixing coefficient





 ( j ) . Using Ibragimov [2] and

2

1/ 2

( j)  

j 1

 n 1  E (38) gives   Z j  ~ n as n    j 0 

So condition (i) & (ii) is satisfied. Using, Markov’s inequality and Lemma-4 with

 n  P( B k )  P  Z j  kq / 2   C / k 3 for C    j 0    In the same way  n  P( B)  P  Z j  q(n  1) / 2   C / n 3 , C    j 0   

(40)



and

k  1 And

(39)

  4 gives

 kP( B k )  

k 1

And nP(B)  0, n  

We estimate the final term in using the method of Ibragimov and Maslova [2]



Bk

H sdP P( B k ) log wk  P( B k ) log T  C (1 i 0 ) exp( i 0 ) (41)

Where C is a constant, i0  ln(T ) and T>0 Then

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American Journal of Engineering Research (AJER)  n j ( j  1)....( j  k  1) R j  k  W k E   Xj  j k  k!c  

 1, k = 0  T   1 k , k  1 W k  C (1  R)  k 1 c 1 n

 k 0

Bk

 0

for

2014

Taking T to be the following function of k

using

Kac-rice

formula

and

noting

that

0  R  1 we have

C  n   H sdP  log   (k  1) P( B k )  1  R  k  0   n 1  (1   ) log k  D  log c k 1 k 1

n

 (1   ) (log k ) P( B k ) k 0

On Substituting r = 1- (logn)-1/2 and R=1-1/2 (logn)-1/2 in the above equation we get the expected number of real zeros in the interval

I n is C 1 / 2 / 2 log log n for n  2 where C and D are constants 1

(B) ESTIMATION OF NUMBER OF ZEROS IN THE INTERVAL :-

I  2 n



To find out the expected number of zeros in the interval number I n  (1  2



Let the expected number of zeros of the above interval is denoted by LEMMA (5) Suppose that 0  b   If (i) f ( ) can be uniformly approximated in

1 log n

 

), (1 

log log n  ) n 

EN n I 2n

[-b,b] by the partial sums of its Fourier series

development, S n f ( ) (ii) (iii) Then EN

f(0)>0

f ( )  A  ,   -  ,  

I  ~ (1 / 2 ) log n as n   2

n

n

(C) ESTIMATION OF NUMBER OF ZEROS IN THE INTERVAL ( LEMMA (6) If (i) f ( ) is continuous at (ii) f ( )  0

I

3 n

) :-

 0





(iii) f ( )  A  ,   -  ,  Then there is a constant and an integer N6 such that

EN n

I   C(log log n)

THEOREM -3

3

n

7/5

,

n N 6

Let X0, X1……Xn be a set of n points satisfying

Suppose that there is an interval [-b, b], 0<b≤ π, where

 

expected number of real zeros in the interval I n  (1  3

EN n

I   C(log log n) 3

n

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7/5

E ( X j)  0 and E( X j) 2  1 . For j ≥ 0.

f(θ ) is continuous at θ=0, θ  - π, π, Then

log log n  ),1 is n 

n N 6 Page 177

American Journal of Engineering Research (AJER)

2014

Where C is a finite constant PROOF:

Now we have to find an upper bound for



C n / B n  (2 A / m)(1  x 2( n1) ) 1 1 - x 2



-2

So

C n / B n with x  In using Kac-Rice formula 3

C n / B n  1 - x 

-2

n  2 and x  In 3

Substituting for Cn/Bn in the above equation and n taking the help of the inequality expected number of zeros in the interval

3

I

n

is EN

I  C(log log n) 3

n

n

7/5

(1  R 2n )  1 the

n N 6 .

Hence the theorem is proved.

REFERENCES [1] [2] [3] [4] [5] [6] [7]

Dunnage, J.E.A. The number of real zeros of a class of random algebraic polynomials (I), Proc. London Math. Soc. (3) 18 (1968), 439-460.

Ibragimov, I.A. and Maslova, N.B. The average number of real roots of random polynomials, Akad.

Nauk. SSSR. (199), (1971). 13-16.

Kac, M. Rice, S.O. On the average number of real roots of random algebraic equation (II), proc.

Londan. Math. Soc.(1949), 390-408.

Shenker, M. The mean number of real zeros of one class of random polynomials, Annals of Prob. 9 (1981), 510-512.

Titchmarsh, E.C. “Theory of Functions”, 2nd edn, The English language Book Society, Oxford University Press, 1939.

Yoshihara, G. (1979). The average number of real zeros of a random algebraic equation. Bull.Amer.

Math. Soc. 54, 125-134.

Zygmund, V. (1969) On the average number of real roots of a random algebraic efunctions. . Proc.

London Math. Soc. 11, 238-345.

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