International Journal of Mathematics Research.
ISSN 0976-5840 Volume 5, Number 5 (2013), pp. 441-447
© International Research Publication House
http://www.irphouse.com
Expected Number of Real Zeros of Random
Ultraspherical Polynomial
L. Sahoo and M.K. Mahanti
Gopabandhu Science College, Athgarh, Cuttack
E-mail [email protected]
College of Basic Science and Humanities,
Department of Mathematics, OUAT, Bhubaneswar
E-mail [email protected]
Abstract
Let y0 (w) , y1 (w) , …, yn (w) be a sequence of independent random variables
with mathematical expectation zero and variance one. We show that the
polynomial ∑
k (w) ψk (t) , where ψk (t) is the normalized orthogonal
ultraspherical polynomial, has n/√3 real zeros for large values of n.
2000 Mathematics Subject Classification. Primary 60H99;Secondary 26c99.
Key words and phrases. Random polynomial, expected number of zeros,
Kac-Rice Formula.
1.Introduction
Let fn (t) = ∑
k (w) k (t) be a random polynomial, where y0 (w) , y1 (w) , … yn
(w) is a sequence of mutually independent, normally distributed random variables
with mathematical expectation zero and variance one ; ( 0 (t) , 1 (t) , …) is a
sequence of real valued polynomials (functions) and (b0, b1, …) a sequence of real
constants. Kac [4 ] showed that, when bk=1 and k (t) =tk, the expected number of
times the random polynomial fn (t) crosses the t- axis is asymptotic to (1 / π ) log n .
When b0=0, bk=1 for k≠0 and k (t) = cosk (cos
) , J.E.A. Dunnage [2] estimated
that the expected number of crossings of fn (t) is asymptotic to 2n/√3 in the interval (1, 1) .
It is interesting to observe that while tk, s are a set of functions monotonic in (-∞,
0]and [0, ∞) , cosk (cos
) for each k, oscillates k times between -1 and 1. The fact
that the average number of zeros of y=0 when k (t) = cosk (cos
) is proportional
442
L. Sahoo and M.K. Mahanti
to the number of individual oscillations of k (t) about the t-axis, draws attention to
the question as to how far the oscillatory nature of
( ) decisively affects the zeros
of y=o. Although the answer remains still inconclusive, we attempt to show that for
large n, the above equation may be expected to have c.n, (c>0) number of real roots
when k (t) happens to be the ultraspherical classical orthogonal Gegebauer
polynomial. In other words the oscillatory property of n (t) is also shared by ∑
k
(w) bk k (t) . It was Das[1] who first initiated the work on Random Orthogonal
polynomial. Taking bk k (t) as normalized orthogonal Legendre Polynomial, he
showed that expected number of real zeros of the polynomial is asymptotic to n/√3) .
Das’s polynomial can be considered as a special case of our polynomial f (t).
Now, k (t) is associated with a weight function u (t) = (1-t2) -1/2, t>1/2
corresponding to the interval (-1, 1) over which the integral of u (t) k2 (t) is a positive
number hk .We take bk=hk1/2.Then the integral of ψk2 (t) =bk2 k2 (t) over the given
interval is unity, so that each of the terms of the polynomial ∑
k (w) ψk (t)
=∑
k (w) bk k (t) has same weightage in the same sense.
Thus, in what follows, we find the average number of zeros of the equation
f (t) = ∑
(1.1)
k (w) ψk (t) =0.
Let ENn (α, β) denote the expected number of real zeros of fn (t) in (α, β) . We
prove the following theorem
THEOREM: If {yk (w) }nk=0 is a sequence of mutually independent, normally
distributed random variables with mean zero and variance unity;. Then
EN n (−∞, ∞) ~
n
.
3
2. FORMULA FOR ENn (α, β)
Following the procedure of Kac [4 ], we obtain
( )
ENn (f;a, b) =
where
Xn (t) =Xn=∑
Yn (t) =Yn=∑
Zn (t) =Zn=∑
( )
( )
[
[
[
( )
/
(2.1)
(t) ]2
(t) ][ψ'k (t) ]
(t) ]2
provided that XnZn-Yn >0 which holds good by Cauchy's inequality.
is the coefficient of
(t) .The famous
Let us put µk=rnhn-1rn+1-1 where
Christofell-Darboux formula ([3], p.159) of the theory of orthogonal functions reads
as follows :
( ) ( )
( )
( )
∑
(µ)
( ) =μ
(2.2)
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Expected Number of Real Zeros of Random Ultraspherical Polynomial
Putting µ=t+γ, expanding both the sides of (3.1.3) by Taylor’s series and equating
the coefficients of γ we obtain that
-1
2
'
'
∑ [ k (t) ]2= ∑
k [ k (t) ] =µn[ n+1 (t) n (t) - n+1 (t) n (t) ], (2.3)
∑ [ ( )][
(t) ] = ∑
[ k
(t)
(t) ] = [
∑
[
k
((t)
(t)
(t) -
(t) ]= [
n+1
(t)
(t)
n
(t) ],
(t) -
Differentiating (2.4) , we get
∑
[
(t) + (t)
(t) ]= [
(t)
(t) (t)
(t) ].
Therefore,
∑
[
( )
(t) =∑
[
(t) ]2= [
(t) ]+ [
Thus
ENn (f, a, b) =
(t)
n
(t) -
n+1
(t)
(t)
(t)
n+1
(2.4)
(t)
t) ]
(t) +
(t)
(t) -
(t) (t) ].
(2.5)
(t) dt,
(2.6)
where
g2n (t) =
( )
( )
( )
( )
,
(t)
(t) (t)
Rn (t) =
Vn (t) =1/2[
(t)
(t) (t)
( ) ].
(t) , Un (t) =
(t)
(t) (t)
( ) ] and Wn (t) =1/6[
(t)
(t)
(t) ,
(t) -
3.Proof of the theorem
For the sake of convenience for proving the theorem, we break up the interval (-1, 1)
into three sub intervals namely (i) (-1+Є, 1+Є) , (ii) (-1, -1+Є) , and (iii) (1-Є, 1) . We
choose Є=n-1/ (4+δ) .
In the section 3.1, we find out the average number of zeros in the first interval. In
the section 3.2 We prove that the number of zeros in the other two intervals are
negligible in comparison to those in the interval (i) .
EXPECTED NUMBER OF ZEROS IN THE INTERVAL (-1+Є, 1-Є)
From [17, ], We have
(t) = (2λ+1) t (t) -n (n+2λ) n (t)
(1-t2)
Thus
2 (1-t2) Vn (t) =n (n+2λ) Rn (t) - (2n+1+2λ)
(1-t2) Un (t) = (2λ+1) t Rn (t) - (2n+1+2λ)
(t)
(t)
n+1
n+1
n
(t) ,
(t) .
(3.1)
(3.2)
(3.3)
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L. Sahoo and M.K. Mahanti
Differentiating (3.1) we have
-2t (t) + (1-t2)
(t) = (2λ+1)
(t) + (2λ+1) t
Therefore it is easy to derive that
(
)
6 (1-t2) Wn (t) =(2 + 3) (
Rn (t)
)
( + 2 )]Rn (t)
( )
(t) ( + 1)
+[(2 + 1)
(
)
+ (
)
From [17, p . 279], we have
(1
)
( ) ) = (2 + )
Therefore
(1
) ( )
=( + ) ( )
( )
(
)
(
( )
)
(2 + 1)
( )
( )
(t) -n (n+2λ)
(t) .
( )
( ) .
(3.4)
( )
(2 + 1 + )
( )
( ).
(3.5)
Gegenbauer polynomials can be considered as a special case of the Jacobi
( , )
Polynomial
(t) , where = =
(see Szego[5 ] page 80].For large n, We
shall use the asymptotic estimate of n (t) as
(t) ~
n
(1
) /
(
)
(1 + )
+
(
) ,
(3.6)
=
where χ= (nѳ+λѳ-λπ/2) and t=cos . (We have taken
=
in asymptotic
( , )
(Szego[ 5] page 190)
estimate of
Hence from (3.5) and (3.6) we have
(t) =
(
(1
)
)
(1 + )
(1
) +
ѳ
It is easy derive from (3.6) that
( )
( )
( )
=o
( )
,
( )
=o
and
( )
( )
( )
=0
(
)
.
Therefore, from (3.2) and (3.4) we have
( )
(
= (
( )
( )
(
=
)
(
[
)(
(
)
( )
)
-
(
)
( )
( )
)
(
) R ( ))
+0
(
)
(
.
Thus,
( )
=
[ (
)
(
) (
)
( )
=
(
)
+0
(
)
.
Also, we can derive from (3.3) that
+0
(
)
)(
) (
)
( )]
)(
(
)
)
,
( )
( )
(
)(
)
( )
]
=-
445
Expected Number of Real Zeros of Random Ultraspherical Polynomial
( )
( )
=0 ( (
) +0
)
(
.
)
So
( )
( )=
=
( )
( )
1+0(
) /
√ (
( )
( )
)
)
(
/
>2
For the range (-1+ , 1- ) , we notice that 1(
=
ô)
, as previously specified . Thus (1
=
(
) =0
ô)
(
ô)
(
ô)
, where
.
Therefore,
( )=
1+0(
(
Thus from (2.6)
( ; 1+ , 1
=
) .
)
√ (
)
1+0(
)
) =
√
1+0(
)
as sin (1 ) ) ~ /2 .
Number of zeros in (-1, -1+Є) , and (1-Є, 1) .
Here we show that in the ranges (1- , 1) and (-1, -1+ ) the number of zeros of (1 .
1) is negligibly small in comparison to those in ( 1 + Є, 1 Є).
Let
( ) ( ),
(3.7)
f (z) =f ( ( ), ) =∑
where ( ) denotes the random vector ( ( ) , ( ) , …………., ( ) ) .Now f
( ) (1) is a random variable with mean zero and variance
( ( ), 1) =∑
(1)
(1) 0and hence has the distribution function
=∑
dv.
√
Now
Let = max
)=p(∏
p(
(| ( )| > ))
=1
/
) =
p (|f (1) |
(3.8)
/
(| ( )|). Then
)=∏
| ( )|
>1
/
dv
/
( |
( >
( )|
).
)
<
=(∏
.
(1
(3.9)
446
L. Sahoo and M.K. Mahanti
Let
= max
(
(
)
)
( )
(
|
1 + 2Є
|.For the Gegenbauer polynomials,
for λ>1/2[see 17, p.281].Hence
)
/
=
<α1
/
=
where
is a
constant.From the integral representation of Gegenbauer polynomial [3], we have
( )=
!
(
( )
)
/(
Remembering that =
(1 + 2
exp 2
)
(
) <
)
(1
( +
)
and
)
.
, we have from above representation
![ ( ) ]
, where
) (
(1 + 2 )
<
(1 + 2 ) ) <
are constants involving λ only .
Hence
/
( )
=
/
<
where A is a constant .Also
(1 + 2
) =∑
)|
1 ∑
=n
exp 2
( )
,
(1 + 2
(3.10)
∑
)
|
.Hence from (3 . 9) , it follows that P (
( )|
(1 + 2
(1 + 2
)
)
1
.
This together with (3 . 10) , gives
1+2
P
where
exp (2
)
1
,
(3 . 11)
+ 5/2.
=
So from (3 . 8) and (3 . 11) , we obtain
(
P
)
( )
An exp (2n
+ 2n )
>1
>1- .
(3 . 12)
Let n ( ) denote the number of zeros of f ( ( ), ) =0inside the circle |
.It is easy to see that the number of zeros of (2 .1) inside the interval 1not exceed n ( ) .
By Jensen's theorem, (Tetchmarsh[6]) we have
(
( )
2
)
( )
1|
1 does
2
+
,
for f (1)
0, except for a set of measure at most 2/n, as is evident from (3 .12) .
Thus, we obtain that the number of zeros of (1 .1) in (1- , 1) is at most 0
with probability at least 1
(1.1) in (-1, -1+ ) , so that
,
.Identical result is obtainable for the number of zeros of
Expected Number of Real Zeros of Random Ultraspherical Polynomial
(f;-1, -1+ ) =0
447
.
The above derivation together with the estimate of
( ; 1+ , 1
), proves
that
( ; 1, 1) =
√
+0
.
Reference
[1] Das M., Real zeros of a random sum of orthogonal polynomials, Proc. Amer.
Math. Soc. 27 (1971) , 147{153.
[2] Dunnage, J.E.A .The number of Real Zeros of Random Trigonometric
Polynomial, Proc. London Math. Soc. (3) 16 (1966)
[3] Erdélyi, A “Higher Transcendental Functions”, Vol-II, Mc. Graw Hill Book
Company, New York, 1953
[4] Kac M.On the average number of real roots of a random algebraic equation,
Bull. Amer. Math. Soc. 49 (1943) , 314-320. MR0007812
[5] Szeg o G. “Orthogonal Polynomials”, Amer. Math. Soc., Rhode Island,
1939.53-54
[6] Titchmarsh, E.C.” Theory of functions”, 2nd Edition, The English Language
Book Society, Oxford University Press
448
L. Sahoo and M.K. Mahanti