Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 28, 2005
ON THE INEQUALITY OF P. TURÁN FOR LEGENDRE POLYNOMIALS
EUGEN CONSTANTINESCU
U NIVERSITY OF S IBIU
FACULTY OF S CIENCES
D EPARTMENT OF M ATHEMATICS
S TR . I.R AŢ NR . 7,
2400 S IBIU , ROMANIA
egnconst68@yahoo.com
Received 10 January, 2005; accepted 03 February, 2005
Communicated by A Lupaş
A BSTRACT. Our aim is to prove the inequalities
1 − x2
1 − x2
P (x)
Pn+1 (x)
hn ≤ n
≤
, ∀x ∈ [−1, 1], n = 1, 2, . . . ,
Pn−1 (x) Pn (x)
n(n + 1)
2
Pn
∞
where hn := k=1 k1 and (Pn )n=0 are the Legendre polynomials . At the same time, it is shown
that the sequence having as general term
n(n + 1)
Pn (x)
Pn+1 (x)
Pn−1 (x) Pn (x)
is non-decreasing for x ∈ [−1, 1].
Key words and phrases: Orthogonal polynomials, Legendre polynomials, Turán Inequality, Positivity.
2000 Mathematics Subject Classification. 33C10, 26D20.
1. I NTRODUCTION
Let (Pn )∞
n=0 be the sequence of Legendre polynomials, that is
1
1−x
2
n (n)
Pn (x) =
(x − 1)
= 2 F1 −n, n + 1; 1;
,
n!2n
2
where
∞
X
(a)k (b)k z k
F
(a,
b;
c;
z)
:=
· ,
2 1
(c)
k!
k
k=0
(a)k := a(a + 1) · · · (a + k − 1),
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
027-05
(a)0 = 1.
2
E UGEN C ONSTANTINESCU
Denote
∆n (x) :=
Pn (x)
Pn+1 (x)
= [Pn (x)]2 − Pn−1 (x)Pn+1 (x).
Pn−1 (x) Pn (x)
Note that Pn (1) = 1, Pn (−x) = (−1)n Pn (−x), i.e. ∆n (1) = ∆n (−1) = 0. For instance
1 − x2
1 − x4
, ∆2 (x) =
.
2
4
Paul Turán [3] has proved the following interesting inequality
∆1 (x) =
∀x ∈ (−1, 1),
∆n (x) > 0,
(1.1)
n ∈ {1, 2, . . . }.
In [1] – [2] are given the following remarkable representations of ∆n (x).
Lemma 1.1 (A. Lupaş). Suppose ϕ(x, t) := x2 + t(1 − x2 ) and Pn (xk ) = 0. Then
Z 1
1 − Pn (ϕ(x, t))
dt
1
(1.2)
∆n (x) =
·√
πn(n + 1) −1
1−t
1 − t2
and
2
n 1 − x2 X Pn (x)
(1.3)
∆n (x) =
(1 − xxk ) .
n(n + 1) k=1 x − xk
2. M AIN R ESULTS
In this article our aim is to improve the Turán inequality (1.1).
P
Theorem 2.1. If x ∈ [−1, 1], n ∈ N, hn := nk=1 k1 , then
1 − x2
1 − x2
hn ≤ ∆n (x) ≤
.
n(n + 1)
2
(2.1)
Proof. Let us denote Tk (t) = cos (k · arccos t), γ0 = π1 , γk = π2 for k ≥ 1, and ϕ(x, t) =
x2 + t(1 − x2 ). According to addition formula for Legendre polynomials, we have
Pn (ϕ(x, t)) = π
n
X
(n − k)!
k=0
If t = 1 we find
1=π
(n + k)!
n
X
(n − k)!
k=0
(n + k)!
2
(1 − x2 )k Pn(k) (x) γk Tk (t).
2
(1 − x2 )k Pn(k) (x) γk .
Therefore
n
X
2 1 − Tk (t)
1 − Pn (ϕ(x, t))
(n − k)!
=2
(1 − x2 )k Pn(k) (x)
1−t
(n + k)!
1−t
k=1
= 2π
n
X
(n − k)!
k=1
(n + k)!
k
(k) 2 X
(1 − x ) Pn (x)
(k − ν)γν Tν (t).
2 k
ν=0
This shows us that
max
t∈[−1,1]
1 − Pn (ϕ(x, t))
1−t
J. Inequal. Pure and Appl. Math., 6(2) Art. 28, 2005
1 − Pn (ϕ(x, t)) =
1−t
t=1
n(n + 1)
(1 − x2 ).
= (1 − x2 )Pn0 (1) =
2
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T URÁN I NEQUALITY
3
Using the Lupaş identity (1.2) we obtain
1 − x2
, (n ≥ 1, x ∈ [−1, 1]).
2
Taking into account the following well-known equalities
n−1
2n − 1
Pn (x) =
xPn−1 (x) −
Pn−2 (x), P0 (x) = 1, P1 (x) = x,
n
n
(1 − x2 )Pn0 (x) = n (Pn−1 (x) − xPn (x)) = (n + 1) (xPn (x) − Pn+1 (x)) ,
∆n (x) ≤
we obtain
0
k(k + 1)∆k (x) − (k − 1)k∆k−1 (x) = (1 − x2 ) Pk0 (x)Pk−1 (x) − Pk (x)Pk−1
(x) .
The Christofell-Darboux formula for Legendre polynomials enables us to write
k−1
1 − x2 X
k(k + 1)∆k (x) − (k − 1)k∆k−1 (x) =
(2j + 1) [Pj (x)]2 ,
k j=0
k ≥ 2.
By summing for k ∈ {2, 3, . . . , n} we give
n(n + 1)∆n (x) = (1 − x2 )hn + (1 − x2 )
n−1
X
k=1
which implies ∆n (x) ≥
(1−x2 )h
n
n(n+1)
k
1 X
(2j + 1) [Pj (x)]2 ,
k + 1 j=1
for x ∈ [−1, 1].
Another remark regarding ∆n (x) is the following :
Theorem 2.2. The sequence (n(n + 1)∆n (x))∞
n=1 , x ∈ [−1, 1], is non-decreasing, i.e.
n−1
∆n (x) ≥
∆n−1 (x), x ∈ [−1, 1], n ≥ 2.
n+1
Proof. Let Πm be the linear space of all polynomials, of degree ≤ m, having real coefficients. Using a Lagrange-Hermite interpolation formula, every polynomial f from Π2n+1 with
f (−1) = f (1) = 0 may be written as
2
n X
Pn (x)
2
(2.2)
f (x) = (1 − x )
Ak (f ; x),
0 (x )(x − x )
P
k
k
n
k=1
where
Ak (f ; x) =
f (xk ) + (x − xk )f 0 (xk )
.
1 − x2k
Let us observe that
1 − x2k 0
1 − x2k 0
Pn (xk ), Pn+1 (xk ) = −
P (xk ),
n
n+1 n
2n − 1
Pn−2 (xk ) =
xk (1 − x2k )Pn0 (xk ),
n(n − 1)
0
0
Pn−1 (xk ) = Pn+1
(xk ) = xk Pn0 (xk ).
Pn−1 (xk ) =
(2.3)
In (2.2) let us consider f ∈ Π2n , where
f (x) = n(n + 1)∆n (x) − n(n − 1)∆n−1 (x).
From (2.3) we find
f (xk ) =
(1 − x2k )2 0
2
[Pn (xk )] ,
n
J. Inequal. Pure and Appl. Math., 6(2) Art. 28, 2005
f 0 (xk ) = 0.
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4
E UGEN C ONSTANTINESCU
1−x2k
n
[Pn0 (xk )]2 , using (2.2) we give
2
n 1 − x2 X Pn (x)
(1 − x2k ) ≥ 0,
f (x) =
n k=1 x − xk
Because Ak (f ; x) =
x ∈ [−1, 1].
Therefore
(n + 1)∆n (x) − (n − 1)∆n−1 (x) ≥ 0 for
x ∈ [−1, 1].
R EFERENCES
[1] A. LUPAŞ, Advanced Problem 6517, Amer. Math. Monthly, (1986) p. 305; (1988) p. 264.
[2] A. LUPAŞ, On the inequality of P. Turán for ultraspherical polynomials, in Seminar on Numerical
and Statistical Calculus, University of Cluj-Napoca, Research Seminaries, Preprint Nr. 4 (1985)
82–87.
[3] P. TURÁN, On the zeros of the polynomials of Legendre, C̆asopis pro pes̆tovani matematiky i fysky,
75 (1950) 113–122.
J. Inequal. Pure and Appl. Math., 6(2) Art. 28, 2005
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