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<h1>ome estimations for the Taylor’s remainder\(^{\bullet }\)</h1>
<p class="authors">
<span class="author">Hui Sun,\(^\ast \) Bo-Yong Long\(^{\S }\) Yu-Ming Chu\(^\S \)</span>
</p>
<p class="date">April 30, 2013.</p>
</div>
<p>\(^{\ast }\) School of Mathematics and Computation Science, Hunan City University, Yiyang 413000, Hunan, China, e-mail: <span class="tt">sunhui4624117@163.com</span>. </p>
<p>\(^\S \)Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China, e-mail: <span class="tt">lbymxy@163.com,chuyuming@hutc.zj.cn</span>. </p>
<p>\(^\bullet \) 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). </p>

<div class="abstract"><p> In this paper, we establish several integral inequalities for the Taylor’s remainder by Grüss and Cheyshev inequalities. </p>
<p><b class="bf">MSC.</b> 26D15. </p>
<p><b class="bf">Keywords.</b> Taylor remainder, Grüss inequality, Cheyshev inequality. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>

<p>For two given integrable functions \(f\) and \(g\) on \([a,b]\), the Chebychev functional \(T(f,g)\) is defined by </p>
<div class="displaymath" id="a0000000003">
  \[ T(f,g)=\tfrac {1}{b-a}\int ^{b}_{a}f(x)g(x){\rm d}x-\tfrac {1}{b-a}\int ^{b}_{a}f(x){\rm d}x\cdot \tfrac {1}{b-a}\int ^{b}_{a}g(x){\rm d}x. \]
</div>
<p>In 1935, Grüss <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span> proved that </p>
<div class="displaymath" id="a0000000004">
  \begin{eqnarray} |T(f,g)|\leq \tfrac {1}{4}(M-m)(L-l)\end{eqnarray}
</div>
<p> if </p>
<div class="displaymath" id="a0000000005">
  \[ m\leq f(x)\leq M,\qquad l\leq g(x)\leq L \]
</div>
<p> for all \(x\in [a,b]\), where \(M, m, L\) and \(l\) are constants. Inequality (1.1) is called Grüss inequality. </p>
<p>The well-known Chebyshev inequality <span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> can be stated as follows: if both \(f\) and \(g\) are increasing or decreasing, then </p>
<div class="displaymath" id="a0000000006">
  \[ T(f,g)\geq 0. \]
</div>
<p>If one of the functions \(f\) and \(g\) is increasing and the other decreasing, then the above inequality is reversed. </p>
<p>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. </p>
<div class="displaymath" id="a0000000007">
  \[ R_{n,f}(x_{0},x)=f(x)-\sum ^{n}_{k=0}\tfrac {(x-x_{0})^{k}}{k!}f^{(k)}(x_{0}). \]
</div>
<p>The Taylor remainder has been the subject of intensive research <span class="cite">
	[
	<a href="#3" >3</a>
	]
</span>–<span class="cite">
	[
	<a href="#9" >9</a>
	]
</span>. In particular, many remarkable integral inequalities for the Taylor remainder can be found in the literature <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span>–<span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>. The following Theorems \(A\) and \(B\) were proved by Gauchman in <span class="cite">
	[
	<a href="#6" >6</a>
	]
</span>. </p>
<p><div class="theorema_thmwrapper " id="a0000000008">
  <div class="theorema_thmheading">
    <span class="theorema_thmcaption">
    Theorem A
    </span>
  </div>
  <div class="theorema_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000009">
  \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*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000010">
  \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*}
</div>

  </div>
</div> </p>
<p><div class="theoremb_thmwrapper " id="a0000000011">
  <div class="theoremb_thmheading">
    <span class="theoremb_thmcaption">
    Theorem B
    </span>
  </div>
  <div class="theoremb_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000012">
  \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*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000013">
  \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*}
</div>
<p>If \(f^{(n+1)}(x)\) is decreasing on \([a,b]\), then </p>
<div class="displaymath" id="a0000000014">
  \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*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000015">
  \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*}
</div>

  </div>
</div> </p>
<p>It is the aim of this paper to establish several new inequalities for the Taylor remainder by Grüss and Cheyshev inequalities. </p>
<h1 id="a0000000016">2 Main results</h1>

<p><div class="lemma_thmwrapper " id="a0000000017">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.1</span>
  </div>
  <div class="lemma_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000018">
  \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}
</div>
<p> and </p>
<div class="displaymath" id="a0000000019">
  \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}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000020">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>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 </p>
<div class="displaymath" id="a0000000021">
  \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*}
</div>

<p><div class="theorem_thmwrapper " id="a0000000022">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000023">
  \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}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Let </p>
<div class="displaymath" id="a0000000025">
  \[ F(x)=f^{(n+1)}(x), \qquad G(x)=\tfrac {(b-x)^{n+1}}{(n+1)!}. \]
</div>
<p> Then for any \(x\in [x_{0},b]\), we clearly see that </p>
<div class="displaymath" id="a0000000026">
  \[ m_{2}\leq F(x)\leq M_{2}, \qquad \quad 0\leq G(x)\leq \tfrac {(b-x_{0})^{n+1}}{(n+1)!}. \]
</div>
<p> Making use of Grüss inequality (1.1) one has </p>
<div class="displaymath" id="a0000000027">
  \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}
</div>
<p>Equation (2.1) and inequality (2.4) lead to the conclusion that </p>
<div class="displaymath" id="a0000000028">
  \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}
</div>
<p>Similarly, if \(x\in [a, x_{0}]\), then Grüss inequality (1.1) leads to </p>
<div class="displaymath" id="a0000000029">
  \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}
</div>
<p>Equation (2.2) and inequality (2.6) imply that </p>
<div class="displaymath" id="a0000000030">
  \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}
</div>
<p>Therefore, inequality (2.3) follows form inequalities (2.5) and (2.7). </p>
<p>If take \(n=1\) in Theorem 2.2, then we have </p>
<p><div class="corollary_thmwrapper " id="a0000000031">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.3</span>
  </div>
  <div class="corollary_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000032">
  \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*}
</div>

  </div>
</div> </p>
<p>In particular, if \(x_{0}=\tfrac {a+b}{2}\), then Corollary 2.3 becomes </p>
<p><div class="corollary_thmwrapper " id="a0000000033">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.4</span>
  </div>
  <div class="corollary_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000034">
  \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*}
</div>

  </div>
</div> </p>
<p>If take \(n=0\) in Theorem 2.2, then we have </p>
<p><div class="corollary_thmwrapper " id="a0000000035">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.5</span>
  </div>
  <div class="corollary_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000036">
  \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*}
</div>

  </div>
</div> </p>
<p>In particular, if \(x_{0}=\tfrac {a+b}{2}\), then Corollary 2.5 becomes </p>
<p><div class="corollary_thmwrapper " id="a0000000037">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.6</span>
  </div>
  <div class="corollary_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000038">
  \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*}
</div>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="a0000000039">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.7</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f(x)\in C^{n+1}[a,b]\) and \(x_{0}\in [a,b]\), then the following statements are true: </p>
<p>(1)&#8195;If \(n\) is an odd number and \(f^{(n+1)}(x)\) is increasing in \([a, b]\), then </p>
<div class="displaymath" id="a0000000040">
  \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}
</div>
<p>(2)&#8195;If \(n\) is an odd number and \(f^{(n+1)}(x)\) is decreasing in \([a, b]\), then </p>
<div class="displaymath" id="a0000000041">
  \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}
</div>
<p>(3)&#8195;If \(n\) is an even number and \(f^{(n+1)}(x)\) is increasing in \([a, b]\), then </p>
<div class="displaymath" id="a0000000042">
  \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}
</div>
<p>(4)&#8195;If \(n\) is an even number and \(f^{(n+1)}(x)\) is decreasing in \([a, b]\), then </p>
<div class="displaymath" id="a0000000043">
  \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}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000044">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We divide the proof into two cases. </p>
<p><i class="it">Case 1.</i> \(x\in [x_{0},b]\).&#8195;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. </p>
<p><i class="it">Subcase 1.1.</i> \(F(x)=f^{(n+1)}(x)\) is increasing in \([x_{0},b]\). It follows from the Chebyshev inequality that </p>
<div class="displaymath" id="a0000000045">
  \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*}
</div>
<p> Making use of equation (2.1) we get </p>
<div class="displaymath" id="a0000000046">
  \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}
</div>
<p>From the monotonicity of \(F\) and \(G\) we have </p>
<div class="displaymath" id="a0000000047">
  \[ f^{(n+1)}(x_{0})\leq F(x)\leq f^{(n+1)}(b) \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000048">
  \[ 0\leq G(x)\leq \tfrac {(b-x_{0})^{n+1}}{(n+1)!} \]
</div>
<p> for \(x\in [x_{0},b]\). Therefore, inequalities (2.5) and (2.12) lead to the conclusion that </p>
<div class="displaymath" id="a0000000049">
  \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}
</div>
<p><i class="it">Subcase 1.2.</i> \(F(x)=f^{(n+1)}(x)\) is decreasing in \([x_{0},b]\). The Chebyshev inequality implies that </p>
<div class="displaymath" id="a0000000050">
  \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*}
</div>
<p> Then equation (2.1) and inequality (2.5) lead to the conclusion that </p>
<div class="displaymath" id="a0000000051">
  \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}
</div>
<p><i class="it">Case 2.</i>&#8195;\(x\in [a, x_{0}]\).&#8195;Let \(F(x)=f^{(n+1)}(x)\) and \(H(x)=\tfrac {(a-x)^{n+1}}{(n+1)!}\). We divide the discussion into four subcases. </p>
<p><i class="it">Subcase 2.1.</i>&#8195;\(n\) 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 </p>
<div class="displaymath" id="a0000000052">
  \[ f^{(n+1)}(a)\leq F(x)\leq f^{(n+1)}(x_{0}) \]
</div>
<p> for all \(x\in [a, x_{0}]\). </p>
<p>Making use of the Chebyshev inequality we get </p>
<div class="displaymath" id="a0000000053">
  \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*}
</div>
<p> Then equation (2.2) and inequality (2.7) imply that </p>
<div class="displaymath" id="a0000000054">
  \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}
</div>
<p><i class="it">Subcase 2.2.</i> \(n\) 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 </p>
<div class="displaymath" id="a0000000055">
  \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}
</div>
<p><i class="it">Subcase 2.3.</i> \(n\) 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 </p>
<div class="displaymath" id="a0000000056">
  \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}
</div>
<p><i class="it">Subcase 2.4.</i> \(n\) 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 </p>
<div class="displaymath" id="a0000000057">
  \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}
</div>
<p>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). </p>
<p>If take \(n=0\) and \(x_{0}=\tfrac {a+b}{2}\) in Theorem 2.7, then we have </p>
<p><div class="corollary_thmwrapper " id="a0000000058">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.8</span>
  </div>
  <div class="corollary_thmcontent">
  <p>Let \(f(x)\in C^{1}[a,b]\), then the following statements are true: </p>
<p>(1)&#8195;If \(f'(x)\) is increasing in \([a, b]\), then </p>
<div class="displaymath" id="a0000000059">
  \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}
</div>
<p>(2)&#8195;If \(f'(x)\) is decreasing in \([a, b]\), then </p>
<div class="displaymath" id="a0000000060">
  \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}
</div>

  </div>
</div> </p>
<p>Let us recall the well known Hermite-Hadamard inequality: </p>
<div class="displaymath" id="a0000000061">
  \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}
</div>
<p> if \(f(x)\) is convex (concave) in \([a,b]\). </p>
<p>Inequality (2.19) can be rewritten as </p>
<div class="displaymath" id="a0000000062">
  \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}
</div>
<p>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 \([a, b]\). </p>
<p><div class="acknowledgement_thmwrapper " id="a0000000063">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgement
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>The authors wish to thank the anonymous referees for their very careful reading of the manuscript and fruitful comments and suggestions. </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="1">1</a></dt>
  <dd><p><i class="sc">G. Grüss</i>, <i class="it">Über das Maximum des absoluten Betrages von \(\tfrac {1}{b-a}\int ^{b}_{a}f(x)g(x){\rm d}x-\tfrac {1}{(b-a)^{2}}\int ^{b}_{a}f(x){\rm d}x\int ^{b}_{a}g(x){\rm d}x\)</i>, Math. Z., <b class="bf">39</b>(1) (1935), pp. 215-226. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><i class="sc">D. S. Mitrinović, J. E. Pec̆arić</i> and <i class="sc">A. M. Fink</i>, <i class="it">Classical and New Inequalities in Analysis</i>, Kluwer Academic Publishers Group, Dordrecht, 1993. </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><i class="sc">S. S. Dragomir</i>, <i class="it">New estimation of the remainder in Taylor’s formula using Grüss’ type inequalities and applications</i>, Math. Inequal. Appl., <b class="bf">2</b>(2)(1999), pp.&#160; 183–193. </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">G. A. Anastassiou</i> and <i class="sc">S. S. Dragomir</i>, <i class="it">On some estimates of the remainder in Taylor’s Formula</i>, J. Math. Anal. Appl., <b class="bf">263</b>(1)(2001), pp.&#160; 246–263. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">H. Gauchman</i>, <i class="it">Some integral inequalities involving Taylor’s remainder I</i>, JIPAM. J. Inequal. Pure Appl. Math., <b class="bf">3</b>(2), Article 26, 9 pages, 2002. </p>
</dd>
  <dt><a name="6">6</a></dt>
  <dd><p><i class="sc">H. Gauchman</i>, <i class="it">Some integral inequalities involving Taylor’s remainder II</i>, JIPAM. J. Inequal. Pure Appl. Math., <b class="bf">4</b>(1), Article 1, 5 pages, 2003. </p>
</dd>
  <dt><a name="7">7</a></dt>
  <dd><p><i class="sc">L. Bougoffa</i>, <i class="it">Some estimations for the integral Taylor’s remainder</i>, JIPAM. J. Inequal. Pure Appl. Math., <b class="bf">4</b>(5), Article 84, 4 pages, 2003. </p>
</dd>
  <dt><a name="8">8</a></dt>
  <dd><p><i class="sc">D.-Y. Hwang</i>, <i class="it">Improvements of some integral inequalities involving Taylor’s remainder</i>, J. Appl. Math. Comput., <b class="bf">16</b>(1-2)(2004), pp.&#160; 151–163. </p>
</dd>
  <dt><a name="9">9</a></dt>
  <dd><p><i class="sc">E. Talvila</i>, <i class="it">Estimates of the remainder in Taylor’s theorem using the Henstock-Kurzweil integral</i>, Czechoslovak Math. J., <b class="bf">55</b>(4) (2005), pp.&#160; 933–940. </p>
</dd>
</dl>


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