ome estimations for the Taylor’s remainder\(^{\bullet }\)
Hui Sun,\(^\ast \) Bo-Yong Long\(^{\S }\) Yu-Ming Chu\(^\S \)
April 30, 2013.
\(^{\ast }\) School of Mathematics and Computation Science, Hunan City University, Yiyang 413000, Hunan, China, e-mail: sunhui4624117@163.com.
\(^\S \)Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China, e-mail: lbymxy@163.com,chuyuming@hutc.zj.cn.
\(^\bullet \) The work of the first author has been supported by the Natural Science Foundation of the Department of Education of Hunan Province (Grant No. 13C127) and the work of the last author has been supported by the Natural Science Foundation of China (Grant No. 61374086).
In this paper, we establish several integral inequalities for the Taylor’s remainder by Grüss and Cheyshev inequalities.
MSC. 26D15.
Keywords. Taylor remainder, Grüss inequality, Cheyshev inequality.
1 Introduction
For two given integrable functions \(f\) and \(g\) on \([a,b]\), the Chebychev functional \(T(f,g)\) is defined by
\[ T(f,g)=\tfrac {1}{b-a}\int ^{b}_{a}f(x)g(x){\rm d}x-\tfrac {1}{b-a}\int ^{b}_{a}f(x){\rm d}x\cdot \tfrac {1}{b-a}\int ^{b}_{a}g(x){\rm d}x. \]
In 1935, Grüss
[
1
]
proved that
\begin{eqnarray} |T(f,g)|\leq \tfrac {1}{4}(M-m)(L-l)\end{eqnarray}
if
\[ m\leq f(x)\leq M,\qquad l\leq g(x)\leq L \]
for all \(x\in [a,b]\), where \(M, m, L\) and \(l\) are constants. Inequality (1.1) is called Grüss inequality.
The well-known Chebyshev inequality
[
2
]
can be stated as follows: if both \(f\) and \(g\) are increasing or decreasing, then
\[ T(f,g)\geq 0. \]
If one of the functions \(f\) and \(g\) is increasing and the other decreasing, then the above inequality is reversed.
In what follows \(n\) denotes a non-negative integer. We denote by \(R_{n,f}(x_{0},x)\) the \(n\)th Taylor remainder of the function \(f(x)\) with center \(x_{0}\), i.e.
\[ R_{n,f}(x_{0},x)=f(x)-\sum ^{n}_{k=0}\tfrac {(x-x_{0})^{k}}{k!}f^{(k)}(x_{0}). \]
The Taylor remainder has been the subject of intensive research
[
3
]
–
[
9
]
. In particular, many remarkable integral inequalities for the Taylor remainder can be found in the literature
[
5
]
–
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7
]
. The following Theorems \(A\) and \(B\) were proved by Gauchman in
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6
]
.
Theorem A
Let \(f(x)\) be a function defined on \([a,b]\) such that \(f(x)\in C^{n+1}[a,b]\) and \(m\leq f^{(n+1)}(x)\leq M\) for each \(x\in [a,b]\), where \(m\) and \(M\) are constants. Then
\begin{eqnarray*} \Big|\int ^{b}_{a}R_{n,f}(a,x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(a)}{(n+2)!}(b-a)^{n+1}\Big|\leq \tfrac {(b-a)^{n+2}}{4(n+1)!}(M-m) \end{eqnarray*}
and
\begin{eqnarray*} \Big|(-1)^{n+1}\int ^{b}_{a}R_{n,f}(b,x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(a)}{(n+2)!}(b-a)^{n+1}\Big|\leq \tfrac {(b-a)^{n+2}}{4(n+1)!}(M-m). \end{eqnarray*}
Theorem B
Let \(f(x)\) be a function defined on \([a,b]\) such that \(f(x)\in C^{n+1}[a,b]\). If \(f^{(n+1)}(x)\) is increasing on \([a,b]\), then
\begin{eqnarray*} & & -\tfrac {f^{(n+1)}(b)-f^{(n+1)}(a)}{4(n+1)!}(b-a)^{n+2}\leq \\ & & \leq \int ^{b}_{a}R_{n,f}(a,x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(a)}{(n+2)!}(b-a)^{n+1}\leq 0 \end{eqnarray*}
and
\begin{align*} 0& \leq (-1)^{n+1}\int ^{b}_{a}R_{n,f}(b,x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(a)}{(n+2)!}(b-a)^{n+1}\\ & \leq \tfrac {f^{(n+1)}(b)-f^{(n+1)}(a)}{4(n+1)!}(b-a)^{n+2}. \end{align*}
If \(f^{(n+1)}(x)\) is decreasing on \([a,b]\), then
\begin{align*} 0& \leq \int ^{b}_{a}R_{n,f}(a,x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(a)}{(n+2)!}(b-a)^{n+1}\leq \\ & \leq \tfrac {f^{(n+1)}(a)-f^{(n+1)}(b)}{4(n+1)!}(b-a)^{n+2} \end{align*}
and
\begin{eqnarray*} & & -\tfrac {f^{(n+1)}(a)-f^{(n+1)}(b)}{4(n+1)!}(b-a)^{n+2}\leq \\ & & \leq (-1)^{n+1}\int ^{b}_{a}R_{n,f}(b,x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(a)}{(n+2)!}(b-a)^{n+1}\leq 0. \end{eqnarray*}
It is the aim of this paper to establish several new inequalities for the Taylor remainder by Grüss and Cheyshev inequalities.
2 Main results
Lemma
2.1
Let \(f(x)\) be a function defined on \([a,b]\) and \(x_{0}\in (a,b)\). If \(f(x)\in C^{n+1}[a,b]\), then
\begin{eqnarray} \int ^{b}_{x_{0}}R_{n,f}(x_{0},x){\rm d}x=\int ^{b}_{x_{0}}\tfrac {(b-x)^{n+1}}{(n+1)!}f^{(n+1)}(x){\rm d}x \end{eqnarray}
and
\begin{eqnarray} \int ^{x_{0}}_{a}R_{n,f}(x_{0},x){\rm d}x=\int ^{x_{0}}_{a}\tfrac {(a-x)^{n+1}}{(n+1)!}f^{(n+1)}(x){\rm d}x. \end{eqnarray}
We only give the proof of (2.1) in detail, the similar argument leads to (2.2). It follows from the formula of integration by parts that
\begin{align*} & \qquad \int ^{b}_{x_{0}}\tfrac {(b-x)^{n+1}}{(n+1)!}f^{(n+1)}(x){\rm d}x=\\ & =\tfrac {(b-x)^{n+1}}{(n+1)!}f^{(n)}(x)|^{b}_{x_{0}}+\int ^{b}_{x_{0}}\tfrac {(b-x)^{n}}{n!}f^{(n)}(x){\rm d}x\\ & =-\tfrac {(b-x_{0})^{n+1}}{(n+1)!}f^{(n)}(x_{0})+\tfrac {(b-x)^{n}}{n!}f^{(n-1)}(x)|^{b}_{x_{0}}+\int ^{b}_{x_{0}}\tfrac {(b-x)^{n-1}}{(n-1)!}f^{(n-1)}(x){\rm d}x\\ & =-\tfrac {(b-x_{0})^{n+1}}{(n+1)!}f^{(n)}(x_{0})-\tfrac {(b-x_{0})^{n}}{n!}f^{(n-1)}(x_{0})-\cdots -(b-x_{0})f(x_{0})+\int ^{b}_{x_{0}}f(x){\rm d}x\\ & =\int ^{b}_{x_{0}}[f(x)-\sum ^{n}_{k=0}\tfrac {f^{(k)}(x_{0})}{k!}(x-x_{0})^{k}]{\rm d}x\\ & =\int ^{b}_{x_{0}}R_{n,f}(x_{0},x){\rm d}x. \end{align*}
Theorem
2.2
Let \(f(x)\in C^{n+1}[a,b]\), such that \(m_{1}\leq f^{(n+1)}(x)\leq M_{1}\) for \(x\in [a,x_{0}]\) and \(m_{2}\leq f^{(n+1)}(x)\leq M_{2}\) for \(x\in [x_{0},b]\), where \(m_{1}\), \(m_{2}\), \(M_{1}\) and \(M_{2}\) are constants. Then
\begin{eqnarray} & & \Big|\int ^{b}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\\ & & \quad -\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\Big|\nonumber \\ & & \leq \tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(M_{1}-m_{1})+\tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(M_{2}-m_{2}).\nonumber \end{eqnarray}
Let
\[ F(x)=f^{(n+1)}(x), \qquad G(x)=\tfrac {(b-x)^{n+1}}{(n+1)!}. \]
Then for any \(x\in [x_{0},b]\), we clearly see that
\[ m_{2}\leq F(x)\leq M_{2}, \qquad \quad 0\leq G(x)\leq \tfrac {(b-x_{0})^{n+1}}{(n+1)!}. \]
Making use of Grüss inequality (1.1) one has
\begin{align} & \quad \Big|\int ^{b}_{x_{0}}F(x)G(x){\rm d}x-\tfrac {1}{b-x_{0}}\int ^{b}_{x_{0}}F(x){\rm d}x\int ^{b}_{x_{0}}G(x){\rm d}x\Big|=\\ & =\Big|\int ^{b}_{x_{0}}\tfrac {(b-x)^{n+1}}{(n+1)!}f^{(n+1)}(x){\rm d}x-\tfrac {1}{b-x_{0}}\int ^{b}_{x_{0}}\tfrac {(b-x)^{n+1}}{(n+1)!}{\rm d}x\int ^{b}_{x_{0}}f^{(n+1)}(x){\rm d}x\Big|\nonumber \\ & \leq \tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(M_{2}-m_{2}).\nonumber \end{align}
Equation (2.1) and inequality (2.4) lead to the conclusion that
\begin{align} & \quad \Big|\int ^{b}_{x_{0}}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\Big|\leq \\ & \leq \tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(M_{2}-m_{2}).\nonumber \end{align}
Similarly, if \(x\in [a, x_{0}]\), then Grüss inequality (1.1) leads to
\begin{align} & \Big|\int ^{x_{0}}_{a}\tfrac {(a-x)^{n+1}}{(n+1)!}f^{(n+1)}(x){\rm d}x\\ & \quad -\tfrac {1}{x_{0}-a}\int ^{x_{0}}_{a}\tfrac {(a-x)^{n+1}}{(n+1)!}{\rm d}x\int ^{x_{0}}_{a}f^{(n+1)}(x){\rm d}x\Big|\leq \nonumber \\ & \leq \tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(M_{1}-m_{1}).\nonumber \end{align}
Equation (2.2) and inequality (2.6) imply that
\begin{align} & \quad \Big|\int ^{x_{0}}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\Big|\leq \\ & \leq \tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(M_{1}-m_{1}).\nonumber \end{align}
Therefore, inequality (2.3) follows form inequalities (2.5) and (2.7).
If take \(n=1\) in Theorem 2.2, then we have
Corollary
2.3
Let \(f(x)\in C^{2}[a,b]\) and \(m_{1}\leq f^{(2)}(x)\leq M_{1}\) for any \(x\in [a,x_{0}]\), \(m_{2}\leq f^{(2)}(x)\leq M_{2}\) for any \(x\in [x_{0},b]\), where \(m_{1}\), \(m_{2}\), \(M_{1}\) and \(M_{2}\) are constants. Then
\begin{align*} & \quad \Big|\int ^{b}_{a}f(x){\rm d}x-(b-a)f(x_{0})-\tfrac {f’(b)}{6}(b-x_{0})^{2}+\tfrac {f’(a)}{6}(a-x_{0})^{2}\nonumber \\ & \qquad -\tfrac {f’(x_{0})}{3}(a+b-2x_{0})(b-a)\Big|\leq \nonumber \\ & \leq \tfrac {(x_{0}-a)^{3}}{8}(M_{1}-m_{1})+\tfrac {(b-x_{0})^{3}}{8}(M_{2}-m_{2}). \end{align*}
In particular, if \(x_{0}=\tfrac {a+b}{2}\), then Corollary 2.3 becomes
Corollary
2.4
Let \(f(x)\in C^{2}[a,b]\) and \(m_{1}\leq f^{(2)}(x)\leq M_{1}\) for any \(x\in [a,\tfrac {a+b}{2}]\), \(m_{2}\leq f^{(2)}(x)\leq M_{2}\) for any \(x\in [\tfrac {a+b}{2},b]\), where \(m_{1}\), \(m_{2}\), \(M_{1}\) and \(M_{2}\) are constants. Then
\begin{align*} & \quad \Big|\tfrac {1}{b-a}\int ^{b}_{a}f(x){\rm d}x-f(\tfrac {a+b}{2})-\tfrac {1}{24}(b-a)(f’(b)-f’(a))\Big|\leq \nonumber \\ & \leq \tfrac {(b-a)^{2}}{64}(M_{1}-m_{1}+M_{2}-m_{2}). \end{align*}
If take \(n=0\) in Theorem 2.2, then we have
Corollary
2.5
Let \(f(x)\in C^{1}[a,b]\) and \(m_{1}\leq f'(x)\leq M_{1}\) for any \(x\in [a, x_{0}]\), \(m_{2}\leq f'(x)\leq M_{2}\) for any \(x\in [x_{0}, b]\), where \(m_{1}\), \(m_{2}\), \(M_{1}\) and \(M_{2}\) are constants. Then
\begin{eqnarray*} & & \Big|\int ^{b}_{a}f(x){\rm d}x-(b-a)f(x_{0})-\tfrac {f(b)-f(x_{0})}{2}(b-x_{0})-\tfrac {f(x_{0})-f(a)}{2}(a-x_{0})\Big|\leq \nonumber \\ & & \leq \tfrac {(x_{0}-a)^{2}}{4}(M_{1}-m_{1})+\tfrac {(b-x_{0})^{2}}{4}(M_{2}-m_{2}). \end{eqnarray*}
In particular, if \(x_{0}=\tfrac {a+b}{2}\), then Corollary 2.5 becomes
Corollary
2.6
Let \(f(x)\in C^{1}[a,b]\) and \(m_{1}\leq f'(x)\leq M_{1}\) for any \(x\in [a, \tfrac {a+b}{2}]\), \(m_{2}\leq f'(x)\leq M_{2}\) for any \(x\in [\tfrac {a+b}{2}, b]\), where \(m_{1}\), \(m_{2}\), \(M_{1}\) and \(M_{2}\) are constants. Then
\begin{align*} & \quad \Big|\tfrac {1}{b-a}\int ^{b}_{a}f(x){\rm d}x-\tfrac {1}{2}f(\tfrac {a+b}{2})-\tfrac {f(a)+f(b)}{4}\Big|\leq \nonumber \\ & \leq \tfrac {1}{16}(b-a)(M_{1}+M_{2}-m_{1}-m_{2}). \end{align*}
Theorem
2.7
Let \(f(x)\in C^{n+1}[a,b]\) and \(x_{0}\in [a,b]\), then the following statements are true:
(1) If \(n\) is an odd number and \(f^{(n+1)}(x)\) is increasing in \([a, b]\), then
\begin{eqnarray} & & \tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(f^{(n+1)}(x_{0})-f^{(n+1)}(a))\geq \\ & & \geq \int ^{b}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\nonumber \\ & & \quad -\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\nonumber \\ & & \geq -\tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(f^{(n+1)}(b)-f^{(n+1)}(x_{0}));\nonumber \end{eqnarray}
(2) If \(n\) is an odd number and \(f^{(n+1)}(x)\) is decreasing in \([a, b]\), then
\begin{eqnarray} & & -\tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(f^{(n+1)}(a)-f^{(n+1)}(x_{0}))\leq \\ & & \leq \int ^{b}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\nonumber \\ & & \quad -\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\nonumber \\ & & \leq \tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(f^{(n+1)}(x_{0})-f^{(n+1)}(b));\nonumber \end{eqnarray}
(3) If \(n\) is an even number and \(f^{(n+1)}(x)\) is increasing in \([a, b]\), then
\begin{eqnarray} & & -\tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(f^{(n+1)}(x_{0})-f^{(n+1)}(a))\\ & & -\tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(f^{(n+1)}(b)-f^{(n+1)}(x_{0}))\leq \nonumber \\ & & \leq \int ^{b}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\nonumber \\ & & \quad -\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\nonumber \\ & & \leq 0;\nonumber \end{eqnarray}
(4) If \(n\) is an even number and \(f^{(n+1)}(x)\) is decreasing in \([a, b]\), then
\begin{align} & \qquad \tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(f^{(n+1)}(a)-f^{(n+1)}(x_{0}))\\ & +\tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(f^{(n+1)}(x_{0})-f^{(n+1)}(b))\nonumber \\ & \geq \int ^{b}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\nonumber \\ & \quad -\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\nonumber \\ & \geq 0.\nonumber \end{align}
We divide the proof into two cases.
Case 1. \(x\in [x_{0},b]\). Let \(F(x)=f^{(n+1)}(x)\) and \(G(x)=\tfrac {(b-x)^{n+1}}{(n+1)!}\), then we clearly see that \(G(x)\) is decreasing in \([x_{0},b]\). We divide this case into two subcases.
Subcase 1.1. \(F(x)=f^{(n+1)}(x)\) is increasing in \([x_{0},b]\). It follows from the Chebyshev inequality that
\begin{eqnarray*} & & \int ^{b}_{x_{0}}\tfrac {(b-x)^{n+1}}{(n+1)!}f^{(n+1)}(x){\rm d}x-\tfrac {1}{b-x_{0}}\int ^{b}_{x_{0}}\tfrac {(b-x)^{n+1}}{(n+1)!}{\rm d}x\int ^{b}_{x_{0}}f^{(n+1)}(x){\rm d}x\leq 0. \end{eqnarray*}
Making use of equation (2.1) we get
\begin{eqnarray} & & \int ^{b}_{x_{0}}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\leq 0. \end{eqnarray}
From the monotonicity of \(F\) and \(G\) we have
\[ f^{(n+1)}(x_{0})\leq F(x)\leq f^{(n+1)}(b) \]
and
\[ 0\leq G(x)\leq \tfrac {(b-x_{0})^{n+1}}{(n+1)!} \]
for \(x\in [x_{0},b]\). Therefore, inequalities (2.5) and (2.12) lead to the conclusion that
\begin{eqnarray} & & \quad -\tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(f^{(n+1)}(b)-f^{(n+1)}(x_{0}))\leq \\ & & \leq \int ^{b}_{x_{0}}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\leq 0.\nonumber \end{eqnarray}
Subcase 1.2. \(F(x)=f^{(n+1)}(x)\) is decreasing in \([x_{0},b]\). The Chebyshev inequality implies that
\begin{eqnarray*} & & \int ^{b}_{x_{0}}\tfrac {(b-x)^{n+1}}{(n+1)!}f^{(n+1)}(x){\rm d}x-\tfrac {1}{b-x_{0}}\int ^{b}_{x_{0}}\tfrac {(b-x)^{n+1}}{(n+1)!}{\rm d}x\int ^{b}_{x_{0}}f^{(n+1)}(x){\rm d}x\geq 0. \end{eqnarray*}
Then equation (2.1) and inequality (2.5) lead to the conclusion that
\begin{eqnarray} & & \quad \tfrac {(b-x_{0})^{n+2}}{4(n+1)!}(f^{(n+1)}(x_{0})-f^{(n+1)}(b))\geq \\ & & \geq \int ^{b}_{x_{0}}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(b)-f^{(n)}(x_{0})}{(n+2)!}(b-x_{0})^{n+1}\geq 0.\nonumber \end{eqnarray}
Case 2. \(x\in [a, x_{0}]\). Let \(F(x)=f^{(n+1)}(x)\) and \(H(x)=\tfrac {(a-x)^{n+1}}{(n+1)!}\). We divide the discussion into four subcases.
Subcase 2.1. \(n\) is an odd number and \(F(x)=f^{(n+1)}(x)\) is increasing in \([a, x_{0}]\). Then \(H(x)=\tfrac {(a-x)^{n+1}}{(n+1)!}\) is increasing in \([a, x_{0}]\) and
\[ f^{(n+1)}(a)\leq F(x)\leq f^{(n+1)}(x_{0}) \]
for all \(x\in [a, x_{0}]\).
Making use of the Chebyshev inequality we get
\begin{eqnarray*} \int ^{x_{0}}_{a}\tfrac {(a-x)^{n+1}}{(n+1)!}f^{(n+1)}(x){\rm d}x-\tfrac {1}{x_{0}-a}\int ^{x_{0}}_{a}\tfrac {(a-x)^{n+1}}{(n+1)!}{\rm d}x\int ^{x_{0}}_{a}f^{(n+1)}(x){\rm d}x\geq 0. \end{eqnarray*}
Then equation (2.2) and inequality (2.7) imply that
\begin{eqnarray} & & \quad \tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(f^{(n+1)}(x_{0})-f^{(n+1)}(a))\geq \\ & & \geq \int ^{x_{0}}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\geq 0.\nonumber \end{eqnarray}
Subcase 2.2. \(n\) is an odd number and \(F(x)=f^{(n+1)}(x)\) is decreasing in \([a, x_{0}]\). Then \(H(x)\) is increasing in \([a, x_{0}]\). It follows from equation (2.2) and inequality (2.7) together with the Chebyshev inequality that
\begin{eqnarray} & & \quad -\tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(f^{(n+1)}(a)-f^{(n+1)}(x_{0}))\leq \\ & & \leq \int ^{x_{0}}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\leq 0.\nonumber \end{eqnarray}
Subcase 2.3. \(n\) is an even number and \(F(x)=f^{(n+1)}(x)\) is increasing in \([a, x_{0}]\). Then \(H(x)=\tfrac {(a-x)^{n+1}}{(n+1)!}\) is decreasing in \([a, x_{0}]\). Therefore, equation (2.2) and inequality (2.7) together with the Chebyshev inequality lead to the conclusion that
\begin{eqnarray} & & \quad -\tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(f^{(n+1)}(x_{0})-f^{(n+1)}(a))\leq \\ & & \leq \int ^{x_{0}}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\leq 0.\nonumber \end{eqnarray}
Subcase 2.4. \(n\) is an even number and \(F(x)=f^{(n+1)}(x)\) is decreasing in \([a, x_{0}]\). Then \(H(x)\) and \(F(x)\) have the same monotonicity in \([a, x_{0}]\). It follows from equation (2.2) and inequality (2.7) together with the Chebyshev inequality that
\begin{eqnarray} & & \quad \tfrac {(x_{0}-a)^{n+2}}{4(n+1)!}(f^{(n+1)}(a)-f^{(n+1)}(x_{0}))\geq \\ & & \geq \int ^{x_{0}}_{a}R_{n,f}(x_{0},x){\rm d}x-\tfrac {f^{(n)}(x_{0})-f^{(n)}(a)}{(n+2)!}(a-x_{0})^{n+1}\geq 0.\nonumber \end{eqnarray}
Therefore, inequality (2.8) follows from inequalities (2.13) and (2.15), inequality (2.9) follows from inequalities (2.14) and (2.16), inequality (2.10) follows from inequalities (2.13) and (2.17), and inequality (2.11) follows from inequalities (2.14) and (2.18).
If take \(n=0\) and \(x_{0}=\tfrac {a+b}{2}\) in Theorem 2.7, then we have
Corollary
2.8
Let \(f(x)\in C^{1}[a,b]\), then the following statements are true:
(1) If \(f'(x)\) is increasing in \([a, b]\), then
\begin{eqnarray} & & \quad -\tfrac {1}{16}(b-a)(f’(b)-f’(a))\leq \\ & & \leq \tfrac {1}{b-a}\int ^{b}_{a}f(x){\rm d}x-\tfrac {1}{2}f(\tfrac {a+b}{2})-\tfrac {f(a)+f(b)}{4}\leq 0.\nonumber \end{eqnarray}
(2) If \(f'(x)\) is decreasing in \([a, b]\), then
\begin{eqnarray} & & \quad \tfrac {1}{16}(b-a)(f’(a)-f’(b))\geq \\ & & \geq \tfrac {1}{b-a}\int ^{b}_{a}f(x){\rm d}x-\tfrac {1}{2}f(\tfrac {a+b}{2})-\tfrac {f(a)+f(b)}{4}\geq 0.\nonumber \end{eqnarray}
Let us recall the well known Hermite-Hadamard inequality:
\begin{eqnarray} f(\tfrac {a+b}{2})\leq (\geq )\tfrac {1}{b-a}\int ^{b}_{a}f(x){\rm d}x\leq (\geq )\tfrac {f(a)+f(b)}{2} \end{eqnarray}
if \(f(x)\) is convex (concave) in \([a,b]\).
Inequality (2.19) can be rewritten as
\begin{eqnarray} & & \quad \tfrac {1}{2}f(\tfrac {a+b}{2})+\tfrac {f(a)+f(b)}{4}-\tfrac {1}{16}(b-a)(f’(b)-f’(a))\leq \\ & & \leq \tfrac {1}{b-a}\int ^{b}_{a}f(x){\rm d}x\leq \tfrac {1}{2}f(\tfrac {a+b}{2})+\tfrac {f(a)+f(b)}{4}.\nonumber \end{eqnarray}
We clearly see that \(\tfrac {f(a)+f(b)}{2}\geq \tfrac {1}{2}f(\tfrac {a+b}{2})+\tfrac {f(a)+f(b)}{4}\) if \(f(x)\) is convex in \([a,b]\). Therefore, inequality (2.22) is an improvement of inequality (2.21) if \(f'(x)\) is decreasing in \([a, b]\).
Acknowledgement
The authors wish to thank the anonymous referees for their very careful reading of the manuscript and fruitful comments and suggestions.
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