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volume 6, issue 2, article 28, 2005.

Received 10 January, 2005;

accepted 03 February, 2005.

Communicated by:A. Lupa¸s

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Journal of Inequalities in Pure and Applied Mathematics

ON THE INEQUALITY OF P. TURÁN FOR LEGENDRE POLYNOMIALS

EUGEN CONSTANTINESCU

University of Sibiu Faculty of Sciences Department of Mathematics Str. I.Ra¸t nr. 7,

2400 Sibiu, Romania

EMail:egnconst68@yahoo.com

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On the Inequality of P. Turán for Legendre Polynomials

Eugen Constantinescu

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J. Ineq. Pure and Appl. Math. 6(2) Art. 28, 2005

http://jipam.vu.edu.au

Abstract Our aim is to prove the inequalities

1−x²

n(n+ 1)h_n≤ Pn(x) Pn+1(x)

Pn−1(x) P_n(x) ≤1−x²

2 , ∀x∈[−1,1], n= 1,2, . . . , wherehn:=Pn

k=11

k and(Pn)^∞_n=0are the Legendre polynomials . At the same time, it is shown that the sequence having as general term

n(n+ 1)P_n(x) P_n+1(x) Pn−1(x) Pn(x) is non-decreasing forx∈[−1,1].

2000 Mathematics Subject Classification:33C10, 26D20.

Key words: Orthogonal polynomials, Legendre polynomials, Turán Inequality, Posi- tivity.

1. Introduction

Let(P_n)^∞_n=0be the sequence of Legendre polynomials, that is

Pn(x) = 1

n!2ⁿ (x² −1)ⁿ(n)

=2F1

−n, n+ 1; 1;1−x 2

, where

2F₁(a, b;c;z) :=

∞

X

k=0

(a)_k(b)_k (c)k

· z^k k!, (a)_k:=a(a+ 1)· · ·(a+k−1), (a)₀ = 1.

Denote

∆_n(x) := P_n(x) P_n+1(x)

P_n−1(x) P_n(x) = [P_n(x)]²−Pn−1(x)P_n+1(x).

Note that P_n(1) = 1, P_n(−x) = (−1)ⁿP_n(−x), i.e. ∆_n(1) = ∆_n(−1) = 0.

For instance

∆₁(x) = 1−x²

2 , ∆₂(x) = 1−x⁴ 4 . Paul Turán [3] has proved the following interesting inequality

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Lemma 1.1 (A. Lupa¸s). Supposeϕ(x, t) := x² +t(1−x²)andP_n(x_k) = 0.

Then

(1.2) ∆n(x) = 1

πn(n+ 1) Z 1

−1

1−P_n(ϕ(x, t))

1−t · dt

√1−t² and

(1.3) ∆n(x) = 1−x² n(n+ 1)

n

X

k=1

P_n(x) x−x_k

2

(1−xxk) .

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2. Main Results

In this article our aim is to improve the Turán inequality (1.1).

Theorem 2.1. Ifx∈[−1,1],n ∈N, h_n :=Pn k=1

1 k,then

(2.1) 1−x²

n(n+ 1)h_n≤∆_n(x)≤ 1−x² 2 .

Proof. Let us denote T_k(t) = cos (k·arccost), γ₀ = _π¹, γ_k = _π² for k ≥ 1, and ϕ(x, t) = x² + t(1 −x²). According to addition formula for Legendre polynomials, we have

P_n(ϕ(x, t)) =π

n

X

k=0

(n−k)!

(n+k)!(1−x²)^k

P_n^(k)(x)²

γ_kT_k(t).

Ift= 1we find

1 = π

n

X

k=0

(n−k)!

(n+k)!(1−x²)^k

P_n^(k)(x)2

γ_k.

Therefore

1−P_n(ϕ(x, t)) 1−t = 2

n

X

k=1

(n−k)!

(n+k)!(1−x²)^k

P_n^(k)(x)2 1−T_k(t) 1−t

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This shows us that max

t∈[−1,1]

1−P_n(ϕ(x, t)) 1−t

= 1−P_n(ϕ(x, t)) 1−t

_t=1

= (1−x²)P_n⁰(1)

= n(n+ 1)

2 (1−x²).

Using the Lupa¸s identity (1.2) we obtain

∆_n(x)≤ 1−x²

2 , (n ≥1, x∈[−1,1]).

Taking into account the following well-known equalities

P_n(x) = 2n−1

n xPn−1(x)− n−1

n Pn−2(x), P₀(x) = 1, P₁(x) =x, (1−x²)P_n⁰(x) =n(P_n−1(x)−xP_n(x)) = (n+ 1) (xP_n(x)−P_n+1(x)), we obtain

k(k+1)∆k(x)−(k−1)k∆k−1(x) = (1−x²)

P_k⁰(x)Pk−1(x)−Pk(x)P_k−1⁰ (x) . The Christofell-Darboux formula for Legendre polynomials enables us to write

k(k+ 1)∆_k(x)−(k−1)k∆k−1(x) = 1−x² k

k−1

X

j=0

(2j+ 1) [P_j(x)]², k≥2.

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By summing fork ∈ {2,3, . . . , n}we give

n(n+ 1)∆_n(x) = (1−x²)h_n+ (1−x²)

n−1

X

k=1

1 k+ 1

k

X

j=1

(2j+ 1) [P_j(x)]²,

which implies∆_n(x)≥ ^(1−x_n(n+1)²^)hⁿ forx∈[−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(x)≥ n−1

n+ 1∆_n−1(x), x∈[−1,1], n≥2.

Proof. Let Π_m be the linear space of all polynomials, of degree≤ m, having real coefficients. Using a Lagrange-Hermite interpolation formula, every poly- nomialf fromΠ2n+1 withf(−1) = f(1) = 0may be written as

(2.2) f(x) = (1−x²)

n

X

k=1

P_n(x) P_n⁰(x_k)(x−x_k)

2

A_k(f;x),

where

A_k(f;x) = f(xk) + (x−xk)f⁰(xk) 1−x²_k .

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Let us observe that

Pn−1(x_k) = 1−x²_k

n P_n⁰(x_k), P_n+1(x_k) = −1−x²_k

n+ 1 P_n⁰(x_k), Pn−2(x_k) = 2n−1

n(n−1)x_k(1−x²_k)P_n⁰(x_k), (2.3)

P_n−1⁰ (x_k) = P_n+1⁰ (x_k) = x_kP_n⁰(x_k).

In (2.2) let us considerf ∈Π2n,where

f(x) =n(n+ 1)∆_n(x)−n(n−1)∆n−1(x).

From (2.3) we find

f(x_k) = (1−x²_k)²

n [P_n⁰(x_k)]², f⁰(x_k) = 0.

BecauseA_k(f;x) = ^1−x_n²^k [P_n⁰(x_k)]²,using (2.2) we give

f(x) = 1−x² n

n

X

k=1

P_n(x) x−x_k

2

(1−x²_k)≥0, x∈[−1,1].

Therefore

(n+ 1)∆_n(x)−(n−1)∆n−1(x)≥0 for x∈[−1,1].

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References

[1] A. LUPA ¸S, Advanced Problem 6517, Amer. Math. Monthly, (1986) p. 305;

(1988) p. 264.

[2] A. LUPA ¸S, 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, ˘Casopis pro pe˘stovani matematiky i fysky, 75 (1950) 113–122.