Journal of Inequalities in Pure and
Applied Mathematics
ON THE INEQUALITY OF P. TURÁN FOR LEGENDRE
POLYNOMIALS
volume 6, issue 2, article 28,
2005.
EUGEN CONSTANTINESCU
University of Sibiu
Faculty of Sciences
Department of Mathematics
Str. I.Raţ nr. 7,
2400 Sibiu, Romania
EMail: egnconst68@yahoo.com
Received 10 January, 2005;
accepted 03 February, 2005.
Communicated by: A. Lupaş
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
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
P
where hn := nk=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)
On the Inequality of P. Turán for
Legendre Polynomials
Pn+1 (x)
Eugen Constantinescu
Pn−1 (x) Pn (x)
Title Page
is non-decreasing for x ∈ [−1, 1].
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2000 Mathematics Subject Classification: 33C10, 26D20.
Key words: Orthogonal polynomials, Legendre polynomials, Turán Inequality, Positivity.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
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J. Ineq. Pure and Appl. Math. 6(2) Art. 28, 2005
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1.
Introduction
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
· ,
2 F1 (a, b; c; z) :=
(c)k
k!
k=0
(a)k := a(a + 1) · · · (a + k − 1),
(a)0 = 1.
Denote
Eugen Constantinescu
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∆n (x) :=
Pn (x)
Pn+1 (x)
Pn−1 (x) Pn (x)
= [Pn (x)]2 − Pn−1 (x)Pn+1 (x).
Note that Pn (1) = 1, Pn (−x) = (−1)n Pn (−x), i.e. ∆n (1) = ∆n (−1) = 0.
For instance
1 − x4
1 − x2
∆1 (x) =
, ∆2 (x) =
.
2
4
Paul Turán [3] has proved the following interesting inequality
(1.1)
On the Inequality of P. Turán for
Legendre Polynomials
∆n (x) > 0,
∀x ∈ (−1, 1),
n ∈ {1, 2, . . . }.
In [1] – [2] are given the following remarkable representations of ∆n (x).
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Lemma 1.1 (A. Lupaş). Suppose ϕ(x, t) := x2 + t(1 − x2 ) and Pn (xk ) = 0.
Then
Z 1
1
1 − Pn (ϕ(x, t))
dt
(1.2)
∆n (x) =
·√
πn(n + 1) −1
1−t
1 − t2
and
n
(1.3)
1 − x2 X
∆n (x) =
n(n + 1) k=1
Pn (x)
x − xk
2
(1 − xxk ) .
On the Inequality of P. Turán for
Legendre Polynomials
Eugen Constantinescu
Title Page
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2.
Main Results
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
(n + k)!
2
(1 − x2 )k Pn(k) (x) γk Tk (t).
On the Inequality of P. Turán for
Legendre Polynomials
Eugen Constantinescu
Title Page
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If t = 1 we find
1=π
n
X
(n − k)!
k=0
(n + k)!
2
(1 − x ) Pn(k) (x) γk .
2 k
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n
X
(n − k)!
2 1 − Tk (t)
1 − Pn (ϕ(x, t))
=2
(1 − x2 )k Pn(k) (x)
1−t
(n + k)!
1−t
k=1
= 2π
n
X
k=1
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k
X
2
(n − k)!
(1 − x2 )k Pn(k) (x)
(k − ν)γν Tν (t).
(n + k)!
ν=0
J. Ineq. Pure and Appl. Math. 6(2) Art. 28, 2005
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This shows us that
max
t∈[−1,1]
1 − Pn (ϕ(x, t))
1−t
1 − Pn (ϕ(x, t)) =
1−t
t=1
= (1 − x2 )Pn0 (1)
n(n + 1)
(1 − x2 ).
=
2
Using the Lupaş identity (1.2) we obtain
∆n (x) ≤
1 − x2
,
2
On the Inequality of P. Turán for
Legendre Polynomials
(n ≥ 1,
x ∈ [−1, 1]).
Eugen Constantinescu
Taking into account the following well-known equalities
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2n − 1
n−1
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)) ,
Pn (x) =
we obtain
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0
k(k+1)∆k (x)−(k−1)k∆k−1 (x) = (1−x ) Pk0 (x)Pk−1 (x) − Pk (x)Pk−1
(x) .
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The Christofell-Darboux formula for Legendre polynomials enables us to write
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2
k−1
1 − x2 X
(2j + 1) [Pj (x)]2 ,
k(k + 1)∆k (x) − (k − 1)k∆k−1 (x) =
k j=0
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k ≥ 2.
J. Ineq. Pure and Appl. Math. 6(2) Art. 28, 2005
http://jipam.vu.edu.au
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 )hn
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−1 (x), x ∈ [−1, 1], n ≥ 2.
∆n (x) ≥
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
(2.2)
f (x) = (1 − x2 )
n X
k=1
where
Pn (x)
0
Pn (xk )(x −
2
xk )
Ak (f ; x),
f (xk ) + (x − xk )f 0 (xk )
Ak (f ; x) =
.
1 − x2k
On the Inequality of P. Turán for
Legendre Polynomials
Eugen Constantinescu
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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
On the Inequality of P. Turán for
Legendre Polynomials
f (x) = n(n + 1)∆n (x) − n(n − 1)∆n−1 (x).
Eugen Constantinescu
From (2.3) we find
(1 − x2k )2 0
2
f (xk ) =
[Pn (xk )] ,
n
Because Ak (f ; x) =
1−x2k
n
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0
f (xk ) = 0.
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[Pn0 (xk )]2 , using (2.2) we give
2
n 1 − x2 X Pn (x)
f (x) =
(1 − x2k ) ≥ 0,
n k=1 x − xk
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(n + 1)∆n (x) − (n − 1)∆n−1 (x) ≥ 0 for
x ∈ [−1, 1].
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References
[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.
On the Inequality of P. Turán for
Legendre Polynomials
Eugen Constantinescu
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