New Inequalities of Hermite-Hadamard Type for Functions whose Derivatives Absolute Values are Quasi-Convex (original) (raw)
NEW INEQUALITIES OF HERMITE-HADAMARD TYPE FOR FUNCTIONS WHOSE DERIVATIVES ABSOLUTE VALUES ARE QUASI-CONVEX
ÇETİN YILDIZ ⋅\cdot , AHMET OCAK AKDEMİR ⋅\cdot, AND MERVE AVCI ∙\bullet
Abstract. In this paper we establish some estimates of the right hand side of a Hermite-Hadamard type inequality in which some quasi-convex functions are involved.
1. INTRODUCTION
Let f:I⊂R→Rf: I \subset \mathbb{R} \rightarrow \mathbb{R} be a convex function defined on the interval II of real numbers and a,b ∈I\in I, with a<ba<b. The following inequality, known as the Hermite-Hadamard inequality for convex functions, holds:
f(a+b2)≤1b−a∫abf(x)dx≤f(a)+f(b)2f\left(\frac{a+b}{2}\right) \leq \frac{1}{b-a} \int_{a}^{b} f(x) d x \leq \frac{f(a)+f(b)}{2}
In recent years many authors have established several inequalities connected to Hermite-Hadamard’s inequality. For recent results, refinements, counterparts, generalizations and new Hermite-Hadamard type inequalities see [2],[4] and [5].
We recall that the notion of quasi-convex functions generalizes the notion of convex functions. More precisely, a function f:[a,b]→Rf:[a, b] \rightarrow \mathbb{R} is said to be quasiconvex on [a,b][a, b] if
f(λx+(1−λ)y)≤max{f(x),f(y)}f(\lambda x+(1-\lambda) y) \leq \max \{f(x), f(y)\}
for any x,y∈[a,b]x, y \in[a, b] and λ∈[0,1]\lambda \in[0,1]. Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex (see[3]).
Recently, D.A. Ion [3] established two inequalities for functions whose first derivatives in absolute value are quasi-convex. Namely, he obtained the following results:
Theorem 1. Let f:Io⊂R→Rf: I^{o} \subset \mathbb{R} \rightarrow \mathbb{R} be a differentiable mapping on Io,a,b∈IoI^{o}, a, b \in I^{o} with a<ba<b. If ∣f′∣\left|f^{\prime}\right| is quasi-convex on [a,b][a, b], then the following inequality holds:
∣f(a)+f(b)2−1b−a∫abf(u)du∣≤b−a4{max∣f′(a)∣,∣f′(b)∣}\left|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \leq \frac{b-a}{4}\left\{\max \left|f^{\prime}(a)\right|,\left|f^{\prime}(b)\right|\right\}
Theorem 2. Let f:Io⊂R→Rf: I^{o} \subset \mathbb{R} \rightarrow \mathbb{R} be a differentiable mapping on Io,a,b∈IoI^{o}, a, b \in I^{o} with a<ba<b. If ∣f′∣pb−1\left|f^{\prime}\right|^{\frac{p}{b-1}} is quasi-convex on [a,b][a, b], then the following inequality holds:
∣f(a)+f(b)2−1b−a∫abf(u)du∣≤b−a2(p+1)pp(max{∣f′(a)∣pp−1,∣f′(b)∣pp−1})p−1p\left|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \leq \frac{b-a}{2(p+1)^{\frac{p}{p}}}\left(\max \left\{\left|f^{\prime}(a)\right|^{\frac{p}{p-1}},\left|f^{\prime}(b)\right|^{\frac{p}{p-1}}\right\}\right)^{\frac{p-1}{p}}
- 2000 Mathematics Subject Classification. Mathematics Subject Classification. 26A51, 26D10. Key words and phrases. quasi-convex functions, hölder inequality, power mean inequality. eorresponding author. ↩︎
In [1], Alomari et al. obtained the following results.
Theorem 3. Let f:Io⊂[0,∞)→Rf: I^{o} \subset[0, \infty) \rightarrow \mathbb{R} be a differentiable mapping on IoI^{o} such that f′∈L[a,b]f^{\prime} \in L[a, b], where a,b∈Ioa, b \in I^{o} with a<ba<b. If ∣f′∣\left|f^{\prime}\right| is quasi-convex on [a,b][a, b], then the following inequality holds:
∣f(a)+f(b)2−1b−a∫abf(u)du∣≤b−a8[max{∣f′(a+b2)∣,∣f′(a)∣}+max{∣f′(a+b2)∣,∣f′(b)∣}]\begin{aligned} \left|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \leq & \frac{b-a}{8}\left[\max \left\{\left|f^{\prime}\left(\frac{a+b}{2}\right)\right|,\left|f^{\prime}(a)\right|\right\}\right. \\ & \left.+\max \left\{\left|f^{\prime}\left(\frac{a+b}{2}\right)\right|,\left|f^{\prime}(b)\right|\right\}\right] \end{aligned}
Theorem 4. Let f:Io⊂[0,∞)→Rf: I^{o} \subset[0, \infty) \rightarrow \mathbb{R} be a differentiable mapping on IoI^{o} such that f′∈L[a,b]f^{\prime} \in L[a, b], where a,b∈Ioa, b \in I^{o} with a<ba<b. If ∣f′∣pp−1\left|f^{\prime}\right|^{\frac{p}{p-1}} is quasi-convex on [a,b][a, b], for p>1p>1 then the following inequality holds:
∣f(a)+f(b)2−1b−a∫abf(u)du∣≤b−a4(1p+1)1p[(max{∣f′(a+b2)∣pp−1,∣f′(a)∣pp−1})p−1p+(max{∣f′(a+b2)∣pp−1,∣f′(b)∣pp−1})p−1p]\begin{aligned} & \left|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \\ \leq & \frac{b-a}{4}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left[\left(\max \left\{\left|f^{\prime}\left(\frac{a+b}{2}\right)\right|^{\frac{p}{p-1}},\left|f^{\prime}(a)\right|^{\frac{p}{p-1}}\right\}\right)^{\frac{p-1}{p}}\right. \\ & \left.+\left(\max \left\{\left|f^{\prime}\left(\frac{a+b}{2}\right)\right|^{\frac{p}{p-1}},\left|f^{\prime}(b)\right|^{\frac{p}{p-1}}\right\}\right)^{\frac{p-1}{p}}\right] \end{aligned}
Theorem 5. Let f:Io⊂R→Rf: I^{o} \subset \mathbb{R} \rightarrow \mathbb{R} be a differentiable mapping on Io,a,b∈IoI^{o}, a, b \in I^{o} with a<ba<b. If ∣f′∣q\left|f^{\prime}\right|^{q} is quasi-convex on [a,b],q≥1[a, b], q \geq 1, then the following inequality holds:
∣f(a)+f(b)2−1b−a∫abf(u)du∣≤b−a8[(max{∣f′(a+b2)∣q,∣f′(a)∣q})1q+(max{∣f′(a+b2)∣q,∣f′(b)∣q})1q]\begin{aligned} & \left|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \\ \leq & \frac{b-a}{8}\left[\left(\max \left\{\left|f^{\prime}\left(\frac{a+b}{2}\right)\right|^{q},\left|f^{\prime}(a)\right|^{q}\right\}\right)^{\frac{1}{q}}\right. \\ & \left.+\left(\max \left\{\left|f^{\prime}\left(\frac{a+b}{2}\right)\right|^{q},\left|f^{\prime}(b)\right|^{q}\right\}\right)^{\frac{1}{q}}\right] \end{aligned}
The main purpose of this study is to generalize the Theorem 3, Theorem 4 and Theorem 5 for quasi-convex functions using the new Lemma.
2. HERMITE-HADAMARD TYPE INEQUALITIES
Lemma 1. Let f:I⊂R→Rf: I \subset \mathbb{R} \rightarrow \mathbb{R} be a differentiable mapping on IoI^{o} where a,b∈Ia, b \in I with a<ba<b. If f′∈L[a,b]f^{\prime} \in L[a, b], then the following equality holds:
(b−x)f(b)+(x−a)f(a)b−a−1b−a∫abf(u)du=(x−a)2b−a∫01(t−1)f′(tx+(1−t)a)dt+(b−x)2b−a∫01(1−t)f′(tx+(1−t)b)dt\begin{aligned} \frac{(b-x) f(b)+(x-a) f(a)}{b-a}-\frac{1}{b-a} \int_{a}^{b} f(u) d u= & \frac{(x-a)^{2}}{b-a} \int_{0}^{1}(t-1) f^{\prime}(t x+(1-t) a) d t \\ & +\frac{(b-x)^{2}}{b-a} \int_{0}^{1}(1-t) f^{\prime}(t x+(1-t) b) d t \end{aligned}
Theorem 6. Let f:Io⊂R→Rf: I^{o} \subset \mathbb{R} \rightarrow \mathbb{R} be a differentiable mapping on Io,a,b∈IoI^{o}, a, b \in I^{o} with a<ba<b. If ∣f′∣\left|f^{\prime}\right| is quasi-convex on [a,b][a, b], then the following inequality holds:
∣(b−x)f(b)+(x−a)f(a)b−a−1b−a∫abf(u)du∣≤(x−a)22(b−a)max{∣f′(x)∣,∣f′(a)∣}+(b−x)22(b−a)max{∣f′(x)∣,∣f′(b)∣}\begin{aligned} & \left|\frac{(b-x) f(b)+(x-a) f(a)}{b-a}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \leq \frac{(x-a)^{2}}{2(b-a)} \max \left\{\left|f^{\prime}(x)\right|,\left|f^{\prime}(a)\right|\right\} \\ & +\frac{(b-x)^{2}}{2(b-a)} \max \left\{\left|f^{\prime}(x)\right|,\left|f^{\prime}(b)\right|\right\} \end{aligned}
Proof. From Lemma 1, we have
∣(b−x)f(b)+(x−a)f(a)b−a−1b−a∫abf(u)du∣≤(x−a)2b−a∫01(1−t)∣f′(tx+(1−t)a)∣dt+(b−x)2b−a∫01(1−t)∣f′(tx+(1−t)b)∣dt≤(x−a)2b−a∫01(1−t)max{∣f′(x)∣,∣f′(a)∣}dt+(b−x)2b−a∫01(1−t)max{∣f′(x)∣,∣f′(b)∣}dt=(x−a)2b−amax{∣f′(x)∣,∣f′(a)∣}∫01(1−t)dt+(b−x)2b−amax{∣f′(x)∣,∣f′(b)∣}∫01(1−t)dt=(x−a)22(b−a)max{∣f′(x)∣,∣f′(a)∣}+(b−x)22(b−a)max{∣f′(x)∣,∣f′(b)∣}\begin{aligned} & \left|\frac{(b-x) f(b)+(x-a) f(a)}{b-a}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \\ \leq & \frac{(x-a)^{2}}{b-a} \int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) a)\right| d t \\ & +\frac{(b-x)^{2}}{b-a} \int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) b)\right| d t \\ \leq & \frac{(x-a)^{2}}{b-a} \int_{0}^{1}(1-t) \max \left\{\left|f^{\prime}(x)\right|,\left|f^{\prime}(a)\right|\right\} d t \\ & +\frac{(b-x)^{2}}{b-a} \int_{0}^{1}(1-t) \max \left\{\left|f^{\prime}(x)\right|,\left|f^{\prime}(b)\right|\right\} d t \\ = & \frac{(x-a)^{2}}{b-a} \max \left\{\left|f^{\prime}(x)\right|,\left|f^{\prime}(a)\right|\right\} \int_{0}^{1}(1-t) d t \\ & +\frac{(b-x)^{2}}{b-a} \max \left\{\left|f^{\prime}(x)\right|,\left|f^{\prime}(b)\right|\right\} \int_{0}^{1}(1-t) d t \\ = & \frac{(x-a)^{2}}{2(b-a)} \max \left\{\left|f^{\prime}(x)\right|,\left|f^{\prime}(a)\right|\right\} \\ & +\frac{(b-x)^{2}}{2(b-a)} \max \left\{\left|f^{\prime}(x)\right|,\left|f^{\prime}(b)\right|\right\} \end{aligned}
which completes the proof.
Remark 1. In Theorem 6,if we choose x=a+b2x=\frac{a+b}{2}, we obtain (1.1) inequality.
Theorem 7. Let f:Io⊂R→Rf: I^{o} \subset \mathbb{R} \rightarrow \mathbb{R} be a differentiable mapping on Io,a,b∈IoI^{o}, a, b \in I^{o} with a<ba<b. If ∣f′∣pp−1\left|f^{\prime}\right|^{\frac{p}{p-1}} is quasi-convex on [a,b],p>1[a, b], p>1, then the following inequality holds:
∣(b−x)f(b)+(x−a)f(a)b−a−1b−a∫abf(u)du∣≤(x−a)2b−a(1p+1)1p(max{∣f′(x)∣pp−1,∣f′(a)∣pp−1})p−1p+(b−x)2b−a(1p+1)1p(max{∣f′(x)∣pp−1,∣f′(b)∣pp−1})p−1p\begin{aligned} & \left|\frac{(b-x) f(b)+(x-a) f(a)}{b-a}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \\ \leq & \frac{(x-a)^{2}}{b-a}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left(\max \left\{\left|f^{\prime}(x)\right|^{\frac{p}{p-1}},\left|f^{\prime}(a)\right|^{\frac{p}{p-1}}\right\}\right)^{\frac{p-1}{p}} \\ & +\frac{(b-x)^{2}}{b-a}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left(\max \left\{\left|f^{\prime}(x)\right|^{\frac{p}{p-1}},\left|f^{\prime}(b)\right|^{\frac{p}{p-1}}\right\}\right)^{\frac{p-1}{p}} \end{aligned}
where q=p/(p−1)q=p /(p-1).
Proof. From Lemma 1 and using well known Hölder inequality, we have
∣(b−x)f(b)+(x−a)f(a)b−a−1b−a∫abf(u)du∣≤(x−a)2b−a∫01(1−t)∣f′(tx+(1−t)a)∣dt+(b−x)2b−a∫01(1−t)∣f′(tx+(1−t)b)∣dt≤(x−a)2b−a(∫01(1−t)pdt)1p(∫01∣f′(tx+(1−t)a)∣pp−1dt)p−1p+(b−x)2b−a(∫01(1−t)pdt)1p(∫01∣f′(tx+(1−t)b)∣pp−1dt)p−1p≤(x−a)2b−a(∫01(1−t)pdt)1p(∫01max{∣f′(x)∣pp−1,∣f′(a)∣pp−1}dt)p−1p+(b−x)2b−a(∫01(1−t)pdt)1p(∫01max{∣f′(x)∣pp−1,∣f′(b)∣pp−1}dt)p−1p=(x−a)2b−a(1p+1)1p(max{∣f′(x)∣pp−1,∣f′(a)∣pp−1})p−1p+(b−x)2b−a(1p+1)1p(max{∣f′(x)∣pp−1,∣f′(b)∣pp−1})p−1p\begin{aligned} & \left|\frac{(b-x) f(b)+(x-a) f(a)}{b-a}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \\ \leq & \frac{(x-a)^{2}}{b-a} \int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) a)\right| d t \\ & +\frac{(b-x)^{2}}{b-a} \int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) b)\right| d t \\ \leq & \frac{(x-a)^{2}}{b-a}\left(\int_{0}^{1}(1-t)^{p} d t\right)^{\frac{1}{p}}\left(\int_{0}^{1}\left|f^{\prime}(t x+(1-t) a)\right|^{\frac{p}{p-1}} d t\right)^{\frac{p-1}{p}} \\ & +\frac{(b-x)^{2}}{b-a}\left(\int_{0}^{1}(1-t)^{p} d t\right)^{\frac{1}{p}}\left(\int_{0}^{1}\left|f^{\prime}(t x+(1-t) b)\right|^{\frac{p}{p-1}} d t\right)^{\frac{p-1}{p}} \\ \leq & \frac{(x-a)^{2}}{b-a}\left(\int_{0}^{1}(1-t)^{p} d t\right)^{\frac{1}{p}}\left(\int_{0}^{1} \max \left\{\left|f^{\prime}(x)\right|^{\frac{p}{p-1}},\left|f^{\prime}(a)\right|^{\frac{p}{p-1}}\right\} d t\right)^{\frac{p-1}{p}} \\ & +\frac{(b-x)^{2}}{b-a}\left(\int_{0}^{1}(1-t)^{p} d t\right)^{\frac{1}{p}}\left(\int_{0}^{1} \max \left\{\left|f^{\prime}(x)\right|^{\frac{p}{p-1}},\left|f^{\prime}(b)\right|^{\frac{p}{p-1}}\right\} d t\right)^{\frac{p-1}{p}} \\ = & \frac{(x-a)^{2}}{b-a}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left(\max \left\{\left|f^{\prime}(x)\right|^{\frac{p}{p-1}},\left|f^{\prime}(a)\right|^{\frac{p}{p-1}}\right\}\right)^{\frac{p-1}{p}} \\ & +\frac{(b-x)^{2}}{b-a}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left(\max \left\{\left|f^{\prime}(x)\right|^{\frac{p}{p-1}},\left|f^{\prime}(b)\right|^{\frac{p}{p-1}}\right\}\right)^{\frac{p-1}{p}} \end{aligned}
where 1p+1q=1\frac{1}{p}+\frac{1}{q}=1, which completes the proof.
Remark 2. In Theorem 7, if we choose x=a+b2x=\frac{a+b}{2}, we obtain (1.2) inequality.
Theorem 8. Let f:Io⊂R→Rf: I^{o} \subset \mathbb{R} \rightarrow \mathbb{R} be a differentiable mapping on Io,a,b∈IoI^{o}, a, b \in I^{o} with a<ba<b. If ∣f′∣q\left|f^{\prime}\right|^{q} is quasi-convex on [a,b],q≥1[a, b], q \geq 1, then the following inequality holds:
∣(b−x)f(b)+(x−a)f(a)b−a−1b−a∫abf(u)du∣≤(x−a)22(b−a)(max{∣f′(x)∣q,∣f′(a)∣q})1q+(b−x)22(b−a)(max{∣f′(x)∣q,∣f′(b)∣q})1q\begin{aligned} \left|\frac{(b-x) f(b)+(x-a) f(a)}{b-a}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \leq & \frac{(x-a)^{2}}{2(b-a)}\left(\max \left\{\left|f^{\prime}(x)\right|^{q},\left|f^{\prime}(a)\right|^{q}\right\}\right)^{\frac{1}{q}} \\ & +\frac{(b-x)^{2}}{2(b-a)}\left(\max \left\{\left|f^{\prime}(x)\right|^{q},\left|f^{\prime}(b)\right|^{q}\right\}\right)^{\frac{1}{q}} \end{aligned}
Proof. From Lemma 1 and using the well known power mean inequality, we have
∣(b−x)f(b)+(x−a)f(a)b−a−1b−a∫abf(u)du∣≤(x−a)2b−a∫01(1−t)∣f′(tx+(1−t)a)∣dt+(b−x)2b−a∫01(1−t)∣f′(tx+(1−t)b)∣dt≤(x−a)2b−a(∫01(1−t)dt)1−1a(∫01(1−t)∣f′(tx+(1−t)a)∣qdt)1q+(b−x)2b−a(∫01(1−t)dt)1−1a(∫01(1−t)∣f′(tx+(1−t)b)∣qdt)1q\begin{aligned} & \left|\frac{(b-x) f(b)+(x-a) f(a)}{b-a}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \\ \leq & \frac{(x-a)^{2}}{b-a} \int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) a)\right| d t \\ & +\frac{(b-x)^{2}}{b-a} \int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) b)\right| d t \\ \leq & \frac{(x-a)^{2}}{b-a}\left(\int_{0}^{1}(1-t) d t\right)^{1-\frac{1}{a}}\left(\int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) a)\right|^{q} d t\right)^{\frac{1}{q}} \\ & +\frac{(b-x)^{2}}{b-a}\left(\int_{0}^{1}(1-t) d t\right)^{1-\frac{1}{a}}\left(\int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) b)\right|^{q} d t\right)^{\frac{1}{q}} \end{aligned}
Since ∣f′∣q\left|f^{\prime}\right|^{q} is quasi-convex we have
∫01(1−t)∣f′(tx+(1−t)a)∣qdt≤12max{∣f′(x)∣q,∣f′(a)∣q}\int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) a)\right|^{q} d t \leq \frac{1}{2} \max \left\{\left|f^{\prime}(x)\right|^{q},\left|f^{\prime}(a)\right|^{q}\right\}
and
∫01(1−t)∣f′(tx+(1−t)b)∣qdt≤12max{∣f′(x)∣q,∣f′(b)∣q}\int_{0}^{1}(1-t)\left|f^{\prime}(t x+(1-t) b)\right|^{q} d t \leq \frac{1}{2} \max \left\{\left|f^{\prime}(x)\right|^{q},\left|f^{\prime}(b)\right|^{q}\right\}
Therefore, we have
∣(b−x)f(b)+(x−a)f(a)b−a−1b−a∫abf(u)du∣≤(x−a)22(b−a)(max{∣f′(x)∣q,∣f′(a)∣q})1q+(b−x)22(b−a)(max{∣f′(x)∣q,∣f′(b)∣q})1q\begin{aligned} \left|\frac{(b-x) f(b)+(x-a) f(a)}{b-a}-\frac{1}{b-a} \int_{a}^{b} f(u) d u\right| \leq & \frac{(x-a)^{2}}{2(b-a)}\left(\max \left\{\left|f^{\prime}(x)\right|^{q},\left|f^{\prime}(a)\right|^{q}\right\}\right)^{\frac{1}{q}} \\ & +\frac{(b-x)^{2}}{2(b-a)}\left(\max \left\{\left|f^{\prime}(x)\right|^{q},\left|f^{\prime}(b)\right|^{q}\right\}\right)^{\frac{1}{q}} \end{aligned}
Remark 3. In Theorem 8, if we choose x=a+b2x=\frac{a+b}{2}, we obtain (1.3) inequality.
REFERENCES
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- ⋅\cdot ATATÜRK UNIVERSITY, K.K. EDUCATION FACULTY, DEPARTMENT OF MATHEMATICS, 25240, CAMPUS, ERZURUM, TURKEY
E-mail address: yildizcetiindyahoo.com ↩︎
- GRADUATE SCHOOL OF NATURAL AND APPLIED SCIENCES, AĞRI İBRAHİM ÇEÇEN UNIVERSITY, AĞRI, TURKEY
E-mail address: ahmetakdemir@ağri.edu.tr