arXiv:1603.08055v1 [math.CA] 25 Mar 2016
AN IMPROVEMENT INEQUALITY OF SIMPSON’S TYPE FOR QUASI-CONVEX MAPPINGS WITH APPLICATIONS MOHAMMAD W. ALOMARI Abstract. In this paper, an inequality of Simpson type for quasi-convex mappings are proved. The constant in the classical Simpson’s inequality is improved. Furthermore, the obtained bounds can be (much) better than some recently obtained bounds. Application to Simpson’s quadrature rule is also given.
1. Introduction Suppose f : [a, b] → R, is fourth times continuously differentiable mapping on (a, b) and f (4) ∞ := sup f (4) (x) < ∞ . The following inequality x∈(a,b)
(1.1)
b Z 5
f (x) dx − (b − a) f (a) + 4f a + b + f (b) ≤ (b − a)
f (4) 6 2 2880 ∞ a
holds, and it is well known in the literature as Simpson’s inequality. It is well known that if the mapping f is neither four times differentiable nor is the fourth derivative f (4) bounded on (a, b), then we cannot apply the classical Simpson quadrature formula. In [11], Peˇcari´c and Varoˇsanec, obtained some inequalities of Simpson’s type for functions whose n-th derivative, n ∈ {0, 1, 2, 3} is of bounded variation, as follow: Theorem 1. Let n ∈ {0, 1, 2, 3}. Let f be a real function on [a, b] such that f (n) is function of bounded variation. Then b Z a+b (b − a) f (a) + 4f + f (b) (1.2) f (x) dx − 6 2 a
n+1
≤ Cn (b − a)
b _ f (n) , a
where, 1 1 1 1 , C1 = , C2 = , C3 = , 3 24 324 1152 f (n) is the total variation of f (n) on the interval [a, b]. C0 =
and Vab
Date: September 5, 2010. 2000 Mathematics Subject Classification. Primary 26D15, Secondary 26D10. Key words and phrases. Simpson’s inequality, quasi-convex function. 1
2
M.W. ALOMARI
Here to note that, the inequality (1.2) with n = 0, was proved in literature by Dragomir [6]. Also, Ghizzetti and Ossicini [10], proved that if f ′′′ is an absolutely Wb continuous mapping with total variation a (f ), then (1.2) with n = 3 holds. In recent years many authors were established several generalizations of the Simpson’s inequality for mappings of bounded variation and for Lipschitzian, monotonic, and absolutely continuous mappings via kernels, for refinements, counterparts, generalizations and several Simpson’s type inequalities see [4]–[17]. The notion of quasi-convex functions generalizes the notion of convex functions. More precisely, a function f : [a, b] → R, is said quasi-convex on [a, b] if f (λx + (1 − λ) y) ≤ sup {f (x) , f (y)} , for all x, y ∈ [a, b] and λ ∈ [0, 1]. Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex nor continuous. Example 1. The floor function floor (x) = ⌊x⌋, is the largest integer not greater than x, is an example of a monotonic increasing function which is quasi-convex but it is neither convex nor continuous. In the same time, one can note that the quasi-convex mappings may be not of bounded variation, i.e., there exist quasi-convex functions which are not of bounded variation. For example, consider the function f : [0, 2] → R, defined by x sin πx , x 6= 0 , f (x) = 0, x=0 therefore, f is quasi-convex but not of bounded variation on [0,2]. Therefore, we cannot apply the above inequalities. For recent inequalities concerning quasi-convex mappings see [1]–[5]. In this paper, we obtain an inequality of Simpson type via quasi-convex mapping. This approach allows us to investigate Simpson’s quadrature rule that have restrictions on the behavior of the integrand and thus to deal with larger classes of functions. In general, we show that our result is better than the classical inequality (1.1). 2. Inequalities of Simpson’s type for quasi-convex functions In order to prove our main results, we start with the following lemma: Lemma 1. Let f ′′′ : I ⊆ R → R be an absolutely continuous mapping on I ◦ , where a, b ∈ I with a < b. If f (4) ∈ L[a, b], then the following equality holds: (2.1)
Z
a
b
f (x) dx −
a+b (b − a) f (a) + 4f + f (b) 6 2 Z 1 5 = (b − a) p (t) f (4) (ta + (1 − t) b) dt, 0
SIMPSON’S INEQUALITIES
3
where, 1 3 24 t
p (t) =
1 24
Proof. We note that 5
I = (b − a)
2 3
t−
3
(t − 1)
t ∈ 0, 12
,
t−
1 3
, t∈
1 2, 1
.
1
Z
p (t) f (4) (ta + (1 − t) b) dt
0
=
(b − a) 24 +
5
Z
1/2
t3 t −
0
(b − a) 24
5
Z
1/2
2 3
f (4) (ta + (1 − t) b) dt 3
(t − 1)
t−
0
Integrating by parts, we get I=
1 3
f (4) (ta + (1 − t) b) dt.
1/2 1/2 ′′ f (3) (ta + (1 − t) b) 1 3 1 2 f (ta + (1 − t) b) − t− t − t 2 6 12 a−b (a − b) 0 0 1/2 1/2 f ′ (ta + (1 − t) b) f (ta + (1 − t) b) + 12 t2 − 16 t − t − 16 3 4 (a − b) (a − b) 0 0 1 Z 1/2 (3) f (ta + (1 − t) b) f (ta + (1 − t) b) 3 1 1 dt + 24 (t − 1) t − 3 + 4 a−b (a − b) 0 1/2 1 f ′′ (ta + (1 − t) b) 5 2 1 − 61 t3 − 12 t + 31 t − 12 2 (a − b) 1/2 1 1 ′ 1 2 5 1 f (ta + (1 − t) b) 5 f (ta + (1 − t) b) + 2t − 6t + 3 − t − 6 3 4 (a − b) (a − b) 1/2 1/2 Z 1 f (ta + (1 − t) b) + dt (a − b)4 1/2
1 3 24 t
2 3
Setting x = ta + (1 − t) b, and dx = (a − b)dt, gives Z b (b − a) a+b 5 f (x) dx − (b − a) · I = f (a) + 4f + f (b) , 6 2 a which gives the desired representation (2.1). Therefore, we may state our main result as follows: Theorem 2. Let f ′′′ : I ⊆ R → R be an absolutely mapping on I ◦ such (4)continuous (4) that f ∈ L[a, b], where a, b ∈ I with a < b. If f is quasi-convex on [a, b], then the following inequality holds: Z b (b − a) a+b (2.2) f (x) dx − f (a) + 4f + f (b) a 6 2 5 (4) a + b (4) (b − a) (4) (4) a + b sup f (a) , f ≤ + sup f , f (b) . 5760 2 2
4
M.W. ALOMARI
Proof. From Lemma 1, and since f is quasi-convex, we have Z b a+b (b − a) f (a) + 4f + f (b) f (x) dx − a 6 2 Z 1 = (b − a)5 p (t) f (4) (ta + (1 − t) b) dt ≤ (b − a)
Z
= (b − a)5
Z
5
0 1
0
|p (t)| f (4) (ta + (1 − t) b) dt
1/2
0 5
+ (b − a)
|p (t)| f (4) (ta + (1 − t) b) dt
Z
1
1/2
1/2
|p (t)| f (4) (ta + (1 − t) b) dt
· sup f (4) (b) , f (4) a + b dt 2 0 Z 1 (4) a + b (4) 1 5 3 1 + (b − a) 24 (t − 1) t − 3 · sup f , f (a) dt 2 1/2 5 (4) a + b (4) a + b (b − a) , f (b) , f sup f (4) (a) , f (4) + sup = 5760 2 2 5
≤ (b − a)
Z
1 3 24 t
t−
2 3
which completes the proof.
Corollary 1. Let f as in Theorem 2. (1) If f is decreasing, then we have Z b a+b (b − a) f (a) + 4f + f (b) (2.3) f (x) dx − a 6 2 5 (b − a) (4) (4) a + b ≤ (a) + f f . 5760 2
(2) If f is increasing, then we have Z b a+b (b − a) f (a) + 4f + f (b) (2.4) f (x) dx − a 6 2 5 (b − a) (4) a + b (4) ≤ f + (b) . f 5760 2
Example 2. Let f : [0, 1] → R be given by f (x) = exp(x2 ). Then f (0) = 1, f (1) = 2.718281828, f 21 = 1.284025417, f (4) (1) = 206.5894189, f (4) 21 =
R1 32.10063542, f (4) ∞ = 206.5894189 and 0 exp x2 dx = 1.462651746. Now, the error obtained by applying (1.1) is 0.07173243712. However, by applying (2.4) the error is 0.03586621856, which means that (2.4) is better than (1.1).
Corollary 2. Let f as in Theorem 2. If f (4) is exits, continuous and f (4) ∞ := sup f (4) (x) < ∞, then the inequality (2.2) reduced to (1.1).
x∈(a,b)
SIMPSON’S INEQUALITIES
5
Remark 1. 1- We note that the constants in (1.1) and (1.2) are improved. 2- The corresponding version of the inequality (2.2) for powers may be done by applying the H¨ older inequality and the power mean inequality. 3. Applications to Simpson’s Formula Let d be a division of the interval [a, b], i.e., d : a = x0 < x1 < ... < xn−1 < xn = b, hi = (xi+1 − xi )/2 and consider the Simpson’s formula (3.1)
S (f, d) =
n−1 X i=0
f (xi ) + 4f (xi + hi ) + f (xi+1 ) (xi+1 − xi ). 6
It is well known that if the mapping f (4) (x) exists on (a, b) and M = sup
f : [a, b] → R, is differentiable such that f (4) (x) < ∞, then
x∈(a,b)
(3.2)
I=
Zb
f (x) dx = S (f, d) + ES (f, d) ,
a
where the approximation error ES (f, d) of the integral I by the Simpson’s formula S (f, d) satisfies (3.3)
|ES (f, d)| ≤
n−1 M X 5 (xi+1 − xi ) . 2880 i=0
In the following we give a new estimation for the remainder term ES (f, d). Proposition 1. Let f ′′′ : I ⊆ R → R be an absolutely mapping on I ◦ (4)continuous (4) such that f ∈ L[a, b], where a, b ∈ I with a < b. If f is quasi-convex on [a, b], then in (3.2), for every division d of [a, b], the following holds: n−1 1 X xi + xi+1 5 (4) (4) |ES (f, d)| ≤ (xi+1 − xi ) sup f (xi ) , f 5760 i=0 2 xi + xi+1 (4) (4) + sup f , f (xi+1 ) . 2 Proof. Applying Theorem 2 on the subintervals [xi , xi+1 ], (i = 0, 1, ..., n − 1) of the division d, we get xi+1 Z x + x (x − x ) i i+1 i+1 i f (xi ) + 4f + f (xi+1 ) f (x) dx − 6 2 xi xi + xi+1 5 ≤ (xi+1 − xi ) sup f (4) (xi ) , f (4) 2 (4) xi + xi+1 (4) + sup f , f (xi+1 ) 2
6
M.W. ALOMARI
Summing over i from 0 to n − 1 and taking into account that f (4) is quasi-convex, we deduce that b Z n−1 X (4) (4) xi + xi+1 5 f (x) dx − S (f, d) ≤ 1 (xi+1 − xi ) sup f (xi ) , f 5760 2 i=0 a (4) xi + xi+1 (4) + sup f , f (xi+1 ) . 2 which completes the proof.
4. conclusion
For fourth times continuously differentiable mapping f on (a, b) and f (4) ∞ := sup f (4) (x) < ∞, the classical Simpson’s inequality holds. In this paper, we
x∈(a,b)
relax the conditions on Simpson’s inequality; namely, the proposed inequality (2.2) ◦ holds if f ′′′ : I ⊆ R → R is an absolutely continuous mapping on I such that ′′′ (4) f ∈ L[a, b] and f is quasi-convex on [a, b]. In general, the inequality (2.2) is better than the classical Simpson’s inequality (1.1) and the result(s) in Theorem 1, where we give an example shows that there exist a quasi-convex mapping which is not of bounded variation.
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[14] Y. Shi and Z. Liu, Some sharp Simpson type inequalities and applications, Appl. Math. E-Notes, 9 (2009), 205–215. [15] N.Ujevi´ c, Two sharp inequalities of Simpson type and applications, Georgian Math. J., 1 (11) (2004), 187–194. [16] N.Ujevi´ c, A generalization of the modified Simpson’s rule and error bounds, ANZIAM J., 47 (2005), E1–E13. [17] N. Ujevi´ c, New error bounds for the Simpson’s quadrature rule and applications, Comp. Math. Appl., 53 (2007), 64–72. Department of Mathematics, Faculty of Science and Information Technology, Irbid National University, 2600 Irbid 21110, Jordan E-mail address:
[email protected] 
