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~
12π 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)
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 ~ 12π 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)
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 11 1 2 (1 )1
eiv xe i yei d xv xy
And 1 11 1 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
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
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 01 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)
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
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
1isbounded 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