General Mathematics Vol. 16, No. 4 (2008), 41–46
About Fejér s sum
Eugen Constantinescu
Abstract
In this paper we will show an improvement of Fejér inequality,
n
sin kφ
k
k=1
n
sin kφ
k=1
Pk (x) =
k
≥
≥
16π(1 − Pn (cos φ))
or
(n + 1) sin φ
π(1 − cos nφ)
, where
2
(n + 1) sin φ
2n−5
n
d
1
(x2 − 1)n is Legendre polynomial
2 k! dx
k
2000 Mathematics Subject Classification: 41A55, 41A80, 65D32
1
Main results
Using the quadrature formulas of Bouzitat, we present an improvement of
Fejér inequality:
n
sin kφ
k=1
k
> 0, ∀φ ∈ (0, π), n ∈ N∗ .
41
About Fejér s sum
42
For (α, β) ∈ (−1, +∞) × (−1, +∞) we define the functional I (α,β) : Π →
R. It is known that
1
f (x)(1 − x)α (1 + x)β dx, f ∈ Πn
I (α,β) (f ) =
−1
where Πn is the set of all polynomials of degree less or equal to n. The
following result is well-known.
Theorem 1. Let P ∈ Πn with degree [P ] = n and P ≥ 0 on [−1, 1]. If
n + 1 then
m= n
,
d
=
2
2
I (α,β) (P ) ≥ 2α+β+1 B(α + 1, β + 1)
Γ(m + 1)(β + 1)d
(α + 2)m (α + β + 2)d
We note with Pk the Legendre polynomial, and with Uk the Cebyshev
second species polynomial. We have
n
d
1
Pk (x) = k
(x2 − 1)n ,
2 k! dx
Uk (x) =
sin((k + 1) arccos x)
√
.
(k + 1) 1 − x2
It is well-known (see [9]) that |Pn (x)| ≤ 1, x ∈ (−1, 1) and |Pn (±)| = 1.
Theorem 2. For all x ∈ (−1, 1) and n ∈ N the inequality
n
Uk (x) ≥
k=0
16π 1 − Pn+1 (x)
·
holds.
n+2
1 − x2
Proof. For x ∈ (−1, 1). A. Lupaş [6] established the identity
n
sin(k arccos x)
k=1
k
√
=
1−x
2
1
−1
dy
1 − Pn (y)
·√
.
1−y
x−y
About Fejér s sum
43
x−1
1
For y(t, x) := x +
2 t + 2 , we have
√
1
n
2
dt
1 − Pn+1 (y(t, x))
.
Uk (x) =
Hn (t, x) √
, Hn (t, x) :=
2(1 + x)
1 − y(t, x)
1−t
k=0
−1
We observe that the Hn (·, x) is a polynomial of effective degree n, and,
1 − Pn+1 (x)
additionally Hn (·, x) ≥ 0 on [−1, 1]. But Hn (1, x) =
1 − x . We have
n
Uk (x) ≥ μn ·
k=0
1 − Pn+1 (x)
1 − x2
where denoting m = [n/2], d = [(n + 1)/2] = n − m we have
22n+3 m!(m + 1)!(d + 1)!d!
=
μn :=
(2m + 2)!(2d + 2)!
8πΓ(m + 1)Γ(d + 1)
.
3
3
Γ m+
Γ d+
2
2
From the convexity of log Γ : (0, ∞) → (0, ∞) we have the next evalua
1
Γ x+
1
1
2
≤√ .
≤
Γ(x + 1)
x
1
x+
2
tions
which implies:
So
8π
(m + 1)(d + 1)
≤ μn ≤ $
8π
16π
, μn ≥ n + 2 ,
1
1
d+
m+
2
2
n
%
1 − Pn+1 (x)
·
Uk (x) ≥ n16π
and thus, the assertion is proved.
+
2
1 − x2
k=0
Corollary 1. For all φ ∈ [0, π] the following inequality is true:
(1)
n
sin kφ
k=1
k
≥
16π(1 − Pn (cos φ))
.
(n + 1) sin φ
About Fejér s sum
44
For proving our main result we will need the following lemmas:
Lemma 1. (T. Koorwinder [1]).
following development:
Let gk,n
&α,β '
a,b
be the coefficients of the
α, β
(a,b)
=
gk,n
Rk (x).
a, b
k=0
& '
If a ≤ b, α + β ≥ a + b ’si β − α ≤ b − a, then gn,k α,β
≥ 0, 0 ≤ k ≤ n.
a,b
Rn(α,β) (x)
n
Lemma 2. If α ≥ a ≥ − 12 , and
Tn(α) (x) :=
(α)
(α + 1)n
(1 − Rn(α,α) (x)),
(n + 2α + 1)n
(a)
then Tn (x) ≥ Tn (x) for all x ∈ [−1, 1].
Proof. Taking into account the hypergeometric form of the Jacobi polynomials, through identification of xn coefficients we deduce:
α, β
(a + 1)n (n + α + β + 1)n
.
=
(2)
gn,n
a, b
(α + 1)n (n + a + b + 1)n
(a,b)
On the other hand, for x = 1 and taking into account that Rk (1) = 1 we
n
& '
%
deduce that
gk,n α,β
= 1. We consider a = b ≥ − 21 and α = β. For
a,b
k=0
α ≥ a ≥ − 21 and x ∈ [−1, 1] we have
n
α, α (a,a)
(α,α)
1 − Rk (x)
gk,n
1 − Rn (x) =
a, a
k=0
'
α, α &
≥ gn,n
1 − Rn(a,a) (x)
a, a
(α)
(a)
and Tn (x) ≥ Tn (x), ∀x ∈ [−1, 1]. So, if we choose α = 0 and a ∈
[−1/2, 0] for x ∈ [−1, 1] we have
1 − Pn (x) ≥
(a + 1)n (n + 1)n
(1 − Rn(a,a) (x))
(n + 2a + 1)n n!
About Fejér s sum
45
and
1 − Pn (x) ≥
1
2
2n−1 (1
− Tn (x)).
Corollary 2. For all φ ∈ [0, π] the next inequality is true
n
sin kφ
k=1
k
≥
π(1 − cos nφ)
2
(n + 1) sin φ
2n−5
References
[1] R. Askey, Jacobi polynomials, I: New proofs of Koornwinder,s Laplace
type integral representation and Bateman,s bilinear sum, Math. Res.
Center, University of Wisconsin, MRC Technical Summary Report 1216,
(1972).
[2] R. Askey, Ortyhogonal Polynomials and Special Functions, Regional
Conf. Lect. Appl. Math., vol. 21, SIAM, Philadelphia, PA, 1975.
[3] R. Askey and G. Gasper, Positive Jacobi polynomial sums, (II), Amer.
J. Math., 98 (1976), 709-737.
[4] R. Askey and G. Gasper, Inequalities for polynomials, in The Bieberbach Conjecture, Proc. of the Symposium on the Ocassion of the Proof,
mathematical Surveys and Monographs, 21, Amer. Mathematical Society, 1986, 7-32
[5] A. Ghizzetti, A. Ossicini, Quadrature Formulae, Birkhäuser Verlag,
ISNM, vol. 13, Basel, Stuttgart, 1970.
About Fejér s sum
46
[6] A. Lupaş, Advanced Problem 6585, Amer. Math. Monthly (1988), p. 880;
(1990) p. 859-860.
[7] A. Lupaş, An identity for ultraspherical polynomials, Revue d, Analyse
numerique et de Théorie de l, approximation, tome 24, no 1-2, 1995,
181-185.
[8] G. Szegö, Orthogonal Polynomials, Amer. Math. Soc. Coloq. Publication, vol. 23, 4th ed., Amer. Math. Soc., Providence, R. I., 1975.
”Lucian Blaga” University of Sibiu
Faculty of Sciences
Department of Mathematics
Str. Dr. I. Ratiu, no. 5–7
550012 Sibiu - Romania
E-mail address: egnconst68@yahoo.com