Correspondingly, three sharp perturbed trapezoid inequalities have been es- ... 2)(x − 3b−a. 2), if x ∈ (a+b. 2. ,b]. (3). Proof. It is not difficult to obtain from ...
SOOCHOW JOURNAL OF MATHEMATICS
Volume 33, No. 1, pp. 101-109, January 2007
A NOTE ON PERTURBED MIDPOINT INEQUALITIES BY ZHENG LIU
Abstract. Some sharp perturbed midpoint inequalities and related results are established. Applications in numerical integration are given.
1. Introduction In [2] we see that the remainder of the perturbed trapezoid inequality can be written in the kernel form Z b Z b b−a (b − a)2 ′ ′ f (x) dx − [f (a) + f (b)] + [f (b) − f (a)] = f (n) (x)pn (x) dx, 2 12 a a where 1 (b − a)2 , p2 (x) = − (x − a)(b − x) + 2 12 a+b 1 p3 (x) = (x − a)(x − )(b − x), 3! 2 and p4 (x) =
1 (x − a)2 (b − x)2 . 4!
Here we have given revised version for p2 (x) since the expression in [2] contained a misprint. Correspondingly, three sharp perturbed trapezoid inequalities have been esReceived September 8, 2005; revised June 15, 2006. AMS Subject Classification. 26D15. Key words. midpoint inequality, perturbed midpoint inequality, sharp bounds. 101
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ZHENG LIU
tablished as Z b b−a (b − a)2 ′ ′ [f (a) + f (b)] + [f (b) − f (a)] ≤ f (x) dx − 2 12 a Z b b−a (b − a)2 ′ f (x) dx − [f (a) + f (b)] + [f (b) − f ′ (a)] ≤ 2 12 a and Z b b−a (b − a)2 ′ f (x) dx − [f (a) + f (b)] + [f (b) − f ′ (a)] ≤ 2 12 a
Γ2 − γ2 √ (b − a)3 , 36 3 Γ3 − γ3 (b − a)4 , 384 M4 (b − a)5 , 720
where f : [a, b] → R is an nth-order differentiable mapping such that f (n) (n = 2, 3, 4) is integrable with Γn = supx∈(a,b) f (n) (x), γn = inf x∈(a,b) f (n) (x) (n = 2, 3) and M4 = supx∈(a,b) |f (4) (x)|.
The main purpose of this paper is to consider similar analysis on the per-
turbed midpoint inequalities. Some new sharp perturbed midpoint inequalities and related results are obtained. Applications in numerical integration are also given. 2. Main Result We first provide the following integral identities.
f (n) Z
b a
Lemma. Let f : [a, b] → R be an nth-order differentiable mapping such that (n = 2, 3) is integrable. Then we have the identity:
a+b (b − a)2 ′ f (x) dx − (b − a)f ( )− [f (b) − f ′ (a)] = 2 24
Z
b
qn (x)f (n) (x) dx, (1)
a
where q2 (x) := and q3 (x) :=
2 (x−a) − 2 (x−b)2 − 2
(b−a)2 24 , (b−a)2 24 ,
1 (x − a)( a+b − x)(x − 6 2 1 (b − x)(x − 6
a+b 2 )(x
−
if x ∈ [a, a+b 2 ], if x ∈ ( a+b 2 , b], 3a−b 2 ),
if x ∈ [a, a+b 2 ],
3b−a 2 ),
if x ∈ ( a+b 2 , b].
Proof. It is not difficult to obtain from integration by parts.
(2)
(3)
A NOTE ON PERTURBED MIDPOINT INEQUALITIES
103
Theorem 1. Let f : [a, b] → R be a twice differentiable mapping such that f ′′ is integrable with Γ2 = supx∈(a,b) f ′′ (x) and γ2 = inf x∈(a,b) f ′′ (x). Then we have Z b a+b (b − a)2 ′ )− [f (b) − f ′ (a)] f (x)dx − (b − a)f ( 2 24 a Γ2 − γ2 √ (b − a)3 , ≤ 36 3 Z b a+b (b − a)2 ′ f (x)dx − (b − a)f ( )− [f (b) − f ′ (a)] 2 24 a (b − a)2 ′ [f (b) − f ′ (a) − γ2 (b − a)], ≤ 12 and
Z b (b − a)2 ′ a+b )− [f (b) − f ′ (a)] f (x)dx − (b − a)f ( 2 24 a 2 (b − a) ≤ [Γ2 (b − a) − f ′ (b) + f ′ (a)], 12
where the constant
1√ 36 3
(4)
(5)
(6)
in the inequality (4) is the best one in the sense that it
cannot be replaced by a smaller one. Proof. By Lemma we see that Z
b a
a+b (b − a)2 ′ f (x) dx − (b − a)f ( )− [f (b) − f ′ (a)] = 2 24
and from (2) we get Z
a
b
Rb a
Z
b
q2 (x)f ′′ (x) dx,
a
q2 (x)dx = 0, it follows
(b − a)2 ′ a+b f (x) dx − (b − a)f ( )− [f (b) − f ′ (a)] = 2 24
Z
b a
q2 (x)[f ′′ (x) − C] dx, (7)
where C ∈ R is a constant. If we choose C =
γ2 +Γ2 2
in (7), then we have
Z b Γ −γ Z b a+b (b − a)2 ′ 2 2 ′ f (x) dx − (b − a)f ( )− [f (b) − f (a)] ≤ |q2 (x)| dx, 2 24 2 a a
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ZHENG LIU
and it is easy to calculate by substitution x = a + Z
b
a
b−a 2 t
that
a+b 2
Z (x − a)2 (b − a)2 (b − a)3 1 2 − |q2 (x)|dx = 2 dx = t − 2 24 8 a 0 Z 1 Z 1 (b − a)3 h √3 1 2 1 i (b−a)3 √ , = ( −t )dt+ (t2 − )dt = 8 3 3 18 3 √1 0 Z
1 dt 3
3
so the inequality (4) follows. If we choose C = γ2 in (7), then we have Z b Z b a+b (b−a)2 ′ ′ )− [f (b)−f (a)] ≤ max |q2 (x)| |f ′′ (x)−γ2 |dx, f (x)dx−(b−a)f ( 2 24 x∈[a,b] a a and the inequality (5) follows since maxx∈[a,b] |q2 (x)| =
(b−a)2 12 .
In a similar way we can prove that the inequality (6) holds. To explain the best constant 361√3 in the inequality (4), we can construct Rx Ry the function a ( a j(z) dz) dy to attain the equality in (4), where j(x) = γ2 for
a ≤ x < a+ x
