General Mathematics Vol. 15, No. 4 (2007), 101-105
Characterization theorem′s of Hermite
polynomials1
Ioan Ţincu
Abstract
The aim of this paper is to prove two equalities concerning the
roots of the Hermite polynomial. For the proof we used multiple
points Hermite interpolation.
2000 Mathematical Subject Classification: 33C45.
Key words: Interpolation, orthogonal polynomials, roots.
1
Introduction
³
´(n)
−x2
, n ∈ Z+ .
Let x ∈ (−∞, ∞) and Hn (x) = (−1) e
e
The following formulas are known:
n x2
(1.1)
y ′′ (x) − 2xy ′ (x) + 2ny(x) = 0 ,
x∈R,
y(x) = Hn (x)
(1.2) Hn+1 (x) − 2xHn (x) + 2nHn−1 (x) = 0 , n ∈ N ,
dHn (x)
(1.3)
= 2nHn−1 (x) for all n ∈ N \ {0} ,
dx
(1.4)
Hn (x)
= 2n .
x→∞
xn
lim
1
Received 19 December, 2007
Accepted for publication (in revised form) 29 December, 2007
101
n≥1
x∈R
102
Ioan Ţincu
2
Main results
Let the polynomials f (x) = Hn2 − Hn+1 (x)Hn−1 (x). From (1.4) we obtain
grad f = 2n − 1.
According to Hermite interpolation formula
(2.1)
f (x) = H2n−1 (x1 , x1 , x2 , x2 , ..., xn , xn ; f /x) =
n
X
ϕk (x)Ak (f ; x)
k=1
where
: x1 , x2 , ..., xn are the roots of Hn (x)
¸2
·
Hn (x)
: ϕk (x) =
(x − xk )Hn′ (xk )
·
¸
Hn′′ (xk )
′
: Ak (f ; x) = f (xk ) + (x − xk ) f (xk ) − ′
f (xk ) .
Hn (xk )
Further, we investigate Ak (f ; x). From (1.2) and (1.3) we obtain:
(2.2)
2
f (xk ) = 2nHn−1
(xk )
(2.3)
′
Hn+1
(xk ) = 0.
Using (1.1), (1.2), (2.2) and (2.3) we find
H ′ (xk )
f ′ (xk )
= n−1
= 2xk ,
f (xk )
Hn−1 (xk )
Hn′′ (xk )
= 2xk .
Hn′ (xk )
Therefore
½
· ′
¸¾
f (xk ) Hn′′ (xk )
2
Ak (f ; x) = f (xk ) 1 + (x − xk )
− ′
= 2nHn−1
(xk ).
f (xk )
Hn (xk )
We have
(2.4)
f (x) = 2n
n ·
X
k=1
Hn (x)
(x − xk )Hn′ (xk )
¸2
2
· Hn−1
(xk ).
Characterization theorem′ s of Hermite polynomials
103
From (2.4) we obtain Turán inequality
¯
¯
¯ Hn (x) Hn+1 (x)
2
f (x) = Hn (x) − Hn−1 (x)Hn+1 (x) = ¯¯
¯ Hn−1 (x) Hn (x)
Observe that
n ·
X
¯
¯
¯
¯ ≥ 0.
¯
¯
¸2
Hn (x)
2
− Hn−1 (x)Hn+1 (x) = 2n
Hn−1
(xk ),
′ (x )
(x
−
x
)H
k
k
n
k=1
¸2
n ·
X
Hn−1 (xk )
Hn−1 (x)Hn+1 (x)
, x 6= xk .
= 2n
1−
′
Hn2 (x)
(x
−
x
k )Hn (xk )
k=1
Hn2 (x)
Using (1.2) and (1.3) we calculate
·
¸
n
1 Hn′ (x)
Hn′ (x)
1 X
1
1−
·
,
2x − 2n
=
2n Hn (x)
Hn (x)
2n k=1 (x − xk )2
à n
!2
n
n
X 1
1 X
1
xX 1
=
+
,
1−
2
n k=1 x − xk
x
−
x
2n
(x
−
x
k
k)
k=1
k=1
n
n
X
X
1
1
xX 1
+
+
2
=
n k=1 x − xk k=1 (x − xk )2
(x
−
x
)(x
−
x
)
i
j
1≤i<j≤n
n
1 X
1
,
2n k=1 (x − xk )2
¶ n
µ
X
1
1
1 X
−
+
2
1+ 1−
2
2n k=1 (x − xk )
(x − xi )(x − xj )
1≤i<j≤n
n
xX 1
= 0.
−
n k=1 x − xk
1−
In conclusion, we proof the following theorem′ s:
Theorem 2.1. If {x1 , x2 , ..., xn } ⊂ R, xi 6= xj for i 6= j, i, j ∈ {1, 2, ..., n}
verifies
(2.5)
µ
¶ n
n
X
1 X
1
1
xX 1
1+ 1 −
−
=0
+2
2n k=1 (x − xk )2
(x
−
x
)(x
−
x
)
n
x
−
x
i
j
k
1≤i<j≤n
k=1
104
Ioan Ţincu
then Hn (xi ) = 0, i ∈ {1, 2, ..., n}.
Theorem 2.2. If {x1 , x2 , ..., xn } ⊂ R, xi 6= xj for i 6= j, i, j ∈ {1, 2, ..., n}
verifies
(2.6)
xj =
n
X
k=1
k6=j
1
, j = 1, 2, ..., n
xj − xk
then Hn (xi ) = 0, i ∈ {1, 2, ..., n}.
Proof. Let P (x) =
n
Y
(x − xk ). We obtain
k=1
n
X 1
P ′ (x)
1
−
=
,
P (x)
x − xj
x − xk
k=1
j ∈ {1, 2, ..., n},
k6=j
n
′
(x − xj )P (x) − P (x) X 1
=
,
(x − xj )P (x)
x − xk
k=1
k6=j
n
(x − xj )P ′ (x) − P (x) X
1
lim
=
,
x→xj
(x − xj )P (x)
xj − xk
k=1
k6=j
n
(2.7)
X
1
P ′′ (xj )
=
.
′
2P (xj )
xj − xk
k=1
k6=j
From (2.6) and (2.7) we have
(2.8)
2xj P ′ (xj ) − P ′′ (xj ) = 0 ,
j ∈ {1, 2, ..., n}.
Let h(x) = 2xP ′ (x) − P ′′ (x). From (2.8), we observe
h(xj ) = 0 ,
j = 1, 2, ..., n.
In conclusion exists cn ∈ R such that h(x) = cn P (x), then
(2.9)
P ′′ (x) − 2xP ′ (x) + cn P (x) = 0.
Characterization theorem′ s of Hermite polynomials
105
Because {H0 , H1 , ..., Hn } is base in Πn , exists ak ∈ R, k ∈ {0, 1, ..., n}
such that
n
X
ak Hk (x).
P (x) =
k=0
From (1.1) and (2.9) we obtain:
ak = 0 ,
k ∈ {0, 1, 2, ..., n − 1}
cn = 2n.
In conclusion, the polynomial P verifies following identity
(2.10)
P ′′ (x) − 2xP ′ (x) + 2nP (x) = 0.
Using (1.1) and (2.10) we obtain
P (x) = λn Hn (x) ,
λn ∈ R namely Hn (xi ) = 0 ,
i ∈ {1, 2, ..., n}.
References
[1] G. Gasper, On the extension of Turan′ s inequality to Jacobi polynomials, Duke Math. J. 38 (1971), 415-428.
[2] G. Szegö, Orthogonal Polynomials, Amer. Math. Soc. Providence, R.
I. 1985.
[3] I. Ţincu, Proofs of Turan′ s inequality, Mathematical Analysis and Approximation Theory, Burg Verlog, 2002.
”Lucian Blaga” University,
Faculty of Sciences,
Department of Mathematics,
I. Raţiu, No. 5-7,
2400 Sibiu, Romania
E-mail: tincuioan@yahoo.com