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H¨older’s inequality; Power-mean inequality.

1

Introductions

To establish analytic inequalities, one of the most efficient way is the property of convexity of a dedicated function. Notedly, in the theory of higher transcenden-tal functions, there are many significant applications. We can use the integral inequalities in order to study qualitative and quantitative properties of integrals (see [6, 7, 9]). Thing continuing to bewilder us by indicating new inferences, new difficulties and also new open questions are a major mathematical outcome. The Hermite-Hadamard inequality: Let ϕ : I ⊆ R → R be a convex function and ı1, ı2∈ I with ı1< ı2. ϕ ı1+ ı2 2  ≤ 1 ı2− ı1 Z ı2 ı1 ϕ (κ) dκ ≤ ϕ (ı1) + ϕ (ı₂ 2)H (1) If ϕ is concave, this double inequality hold in the inverse way. See [1, 2, 5, 7] for details.

The Bullen inequality: 1 ı2− ı1 Z ı2 ı1 ϕ (κ) dκ ≤ 1 2  ϕ (ı1) + ϕ (ı2) 2 + ϕ  ı1+ ı2 2  , B (2)

provided that ϕ : [ı1, ı2] → R is a convex function on [ı1, ı2] (see for example

Lemma 1 [11]Let ϕ : I → R, I ⊂ R be a differentiable mapping on I◦, and ı1, ı2∈ I, ı1< ı2. If ϕ0 ∈ L ([ı1.ı2]) , t ∈ [0, 1] then Z 1 0 (1 − 2t)  ϕ0  tı1+ (1 − t)  ı1+ ı2 2  + ϕ0  t ı1+ ı2 2  + (1 − t) ı2  dt = 4 ı2− ı1  ϕ (ı1) + ϕ (ı2) 2 + ϕ  ı1+ ı2 2  − 2 ı2− ı1 Z ı2 ı1 ϕ (x) dx  . Here I◦ denotes the interior of I.

2

”Definition and Properties of Conformable

Frac-tional Derivative and Integral

The following definitions and theorems with respect to conformable fractional derivative and integral were referred in [12]-[17].

Definition 2 ( Conformable fractional derivative) Given a function ϕ : [0, ∞) → R. Then the ”conformable fractional derivative” of ϕ of order α is defined by

Dα(ϕ) (t) = lim ε→0

ϕ t + εt1−α
_{− ϕ (t)}

ε

for all t > 0, α ∈ (0, 1] . If ϕ is α−differentiable in some (0, ı1) , α > 0, limt→0+ϕ(α)(t)

exist, then define

ϕ(α)(0) = lim

t→0+ϕ

(α)_{(t) .}

We can write ϕ(α)_{(t) for D}

α(ϕ) (t) to denote the conformable fractional

deriva-tives of ϕ of order α.In addition, if the conformable fractional derivative of ϕ of order α exists, then we simply say ϕ is α−differentiable.

Theorem 3 Let α ∈ (0, 1] and ϕ, Λ be α−differentiable at a point t > 0. Then, 1) Dα(aϕ + bΛ) = aDα(ϕ) + bDα(Λ) , for all ı1, ı2∈ R,

2) Dα(λ) = 0, for all constant functions ϕ (t) = λ,

3) Dα(ϕΛ) = ϕDα(Λ) + ΛDα(ϕ) , 4) Dα _Λϕ
= Dα(ϕ)Λ+D_Λ2 α(Λ)ϕ, 5) If ϕ is differentiable, then Dα(ϕ) (t) = t1−α df dt(t) . Also, a) Dα(1) = 0 b) Dα(eat) = at1−αeat, ı1∈ R

c) Dα(sin (at)) = at1−αcos (at) , ı1∈ R

d) Dα(cos (at)) = −at1−αsin (at) , ı1∈ R

e) Dα 1_αtα
= 1 f) Dα sin t α α

= cos tα α
g) Dα cos t α α

= − sin tα α
h) Dα  etαα  = etαα.

Theorem 4 (Mean value theorem for conformable fractional differentiable func-tions). Let α ∈ (0, 1] and ϕ : [0, ∞) → R be a continuous on [ı1, ı2] and an

α−fractional differentiable mapping on (ı1, ı2) with 0 ≤ ı1 < ı2. Then, there

exist c ∈ (ı1, ı2) , such that

Dα(ϕ) (c) = ϕ (ı2) − ϕ (ı1) ı2 α − ı1 α .

Definition 5 (Conformable fractional integral). Let α ∈ (0, 1] and 0 ≤ ı1 <

ı2. A fucntion ϕ : [0, ∞) → R is α−fractional integrable on [ı1, ı2] if the

integral Z ı2 ı1 ϕ (x) dαx := Z ı2 ı1 ϕ (x) xα−1dx,

exists and is finite. All α−fractional integrable on [ı1, ı2] is indicated by

L1 α([ı1, ı2]) . Remark 6 Iı1 α (ϕ) (t) = I ı1 1 t α−1_ϕ
= Z t ı1 ϕ (x) x1−αdx,

where the integral is the usual Riemann improper integral and α ∈ (0, 1] . Theorem 7 Let ϕ : (ı1, ı2) → R be differentiable and α ∈ (0, 1] . Then, for all

t > ı1 we have

Iı1

αD ı1

α(ϕ) (t) = ϕ (t) − ϕ (ı1) .

Theorem 8 (Integration by parts). Let ϕ, Λ : [ı1, ı2] → R be two functions such

that ϕg is differentiable. Then, Z ı2 ı1 ϕ (x) Dı1 α(Λ) (x) dαx = ϕg|ı2 ı1− Z ı2 ı1 Λ (x) Dı1 α(ϕ) (x) dαx.

Theorem 9 Assume that ϕ : [ı1, ∞) → R such that ϕ(n)(t) is continuous and

α ∈ (n, n + 1] . Then, for all t > ı1 we have

Dı1

α (ϕ) (t) I ı1

α = ϕ (t) .

Theorem 10 Let α ∈ (0, 1] and ϕ : [ı1, ı2] → R be a continuous on [ı1, ı2] with

0 ≤ ı1< ı2. Then, |Iı1 α (ϕ) (x)| ≤ I ı1 α |ϕ| (x) . 287

Many studies in the literature on integral inequalities related to conformable fractional integration have been performed by many researchers. For more de-tails and properties concerning the conformable integral operators, we refer, for example,to the works [18]-[21]. In this paper, we establish the Bullen type inequalities for conformable fractional integral and we will investigate some in-tegral inequalities connected with Bullen-type inequalities for conformable frac-tional integral. The results presented here would provide generalizations of those given. In this study, some new Identity and Bullen type integral inequalities for differentiable functions are established, and are applied to produce some inequalities of special means.”

3

Bullen Type Inequalities for Conformable

Frac-tional Integral.

By using the following lemma, we will give some integral inequalities connected with Bullen-type inequalities for conformable fractional integral.

Lemma 11 Let α ∈ (0, 1] and ϕ : I ⊂ R+ _{→ R be an α−fractional}

differen-tiable function on (ı1, ı2) with 0 ≤ ı1 < ı2. If Dα(ϕ) be an α−fractional

inte-grable function on [ı1, ı2] , then the following identity for conformable fractional

integral holds: 1 4 Z 1 0 (1 − 2tα)  Dα(ϕ)  tαıα₁ + (1 − tα)ı α 1 + ıα2 2  (3) +Dα(ϕ)  tαı α 1 + ıα2 2 + (1 − t α_{) ı}α 2  dαt = α ıα 2− ı α 1 Z ı2 ı1 ϕ (xα) dαx − 1 2  ϕ (ıα 1) + ϕ (ıα2) 2 + ϕ  ıα 1+ ıα2 2 

Proof. Integrating by parts Z 1 0 (1 − 2tα) Dα(ϕ)  tαıα₁ + (1 − tα)ı α 1+ ıα2 2  dαt + Z 1 0 (1 − 2tα) Dα(ϕ)  tαı α 1 + ıα2 2 + (1 − t α_{) ı}α 2  dαt = (1 − 2tα) ϕ  tαıα1 + (1 − t α )ı α 1 + ıα2 2     1 0 +2α Z 1 0 ϕ  tαıα₁ + (1 − tα)ı α 1+ ı α 2 2  dαt + (1 − 2tα) ϕ  tαı α 1 + ıα2 2 + (1 − t α_{) ı}α 2     1 0 +2α Z 1 0 ϕ  tαı α 1 + ıα2 2 ı α 1 + (1 − t α_{) ı}α 2  dαt 288

= −ϕ (ıα1) − ϕ  ıα 1 + ıα2 2  + 4α ıα 1 − ıα2 Z ı1 ıα₁+ıα₂ 2 _α1 ϕ (x α ) dαx −ϕ (ıα 2) − ϕ  ıα 1 + ıα2 2  + 4α ıα 2 − ıα1 Z ı2 ıα₁+ıα₂ 2 _α1 ϕ (x α_{) d} αx = −  ϕ (ıα1) + ϕ (ıα2) + 2f  ıα 1+ ıα2 2  + 4α ıα 2 − ıα1 Z ı2 ı1 ϕ (xα) dαx.

Remark 12 If we choose α = 1 in the Lemma 11, then we obtain the Lemma 1.

Theorem 13 Let α ∈ (0, 1]and ϕ : I ⊂ R+ _{→ R be an α−fractional}

differen-tiable function on I◦and Dα(ϕ) be an α−fractional integrable function on I with

0 ≤ ı1< ı2. If |ϕ0| be a convex function on I, then the following inequality for

conformable fractional integral holds:

    α ıα 2 − ıα1 Z ı2 ı1 ϕ (xα) dαx − 1 2  ϕ (ıα 1) + ϕ (ıα2) 2 + ϕ  ıα 1 + ıα2 2     ≤ (ı α 2 − ıα1) 32 h (ıα1) α−1 |Dα(ϕ) (ıα1)| +2 ı α 1 + ı α 2 2 α−1    Dα(ϕ)  ıα 1+ ı α 2 2     + (ıα₂)α−1|Dα(ϕ) (ıα2)| # .

Proof. Since |ϕ0| is a convex function on I, by the using the properties Dα(ϕ ◦ Λ) (t) = ϕ0(Λ (t)) Dα(Λ (t)) and Dα(ϕ (t)) = t1−αϕ0(t) and by Lemma

11 and using the well known absolute value inequality, and we have     α ıα 2 − ıα1 Z ı2 ı1 ϕ (xα) dαx − 1 2  ϕ (ıα 1) + ϕ (ı2) 2 + ϕ  ıα 1 + ıα2 2     ≤ 1 4 Z 1 0 |1 − 2tα_|     Dα(ϕ)  tαıα1 + (1 − tα) ıα 1 + ıα2 2     dαt +1 4 Z 1 0 |1 − 2tα_|     Dα(ϕ)  tαı α 1 + ıα2 2 + (1 − t α_{) ı}α 2     dαt ≤ α (ı α 2 − ıα1) 8  (ıα₁)α−1|Dα(ϕ) (ı1)| Z 1 0 |1 − 2tα_{| t}α_d αt  +α (ı α 2 − ıα1) 8  ıα 1 + ıα2 2 α−1    Dα(ϕ)  ıα 1 + ıα2 2     Z 1 0 |1 − 2tα_{| (1 − t}α_{) d} αt ! +α (ı α 2 − ı α 1) 8  ıα 1 + ı α 2 2 α−1    Dα(ϕ)  ıα 1 + ı α 2 2     Z 1 0 |1 − 2tα_{| t}α_d αt ! 289

+α (ı α 2 − ıα1) 8  (ıα₂)α−1|Dα(ϕ) (ıα2)| Z 1 0 |1 − 2tα_{| (1 − t}α_{) d} αt  = (ı α 2− ıα1) 32 h (ıα₁)α−1|Dα(ϕ) (ıα1)| +2 ı α 1 + ıα2 2 α−1    Dα(ϕ)  ıα 1 + ıα2 2     + (ıα₂)α−1|Dα(ϕ) (ıα2)| # . where Z 1 0 |1 − 2tα_{| t}α_d αt = Z 1 0 |1 − 2tα_{| t}α_tα−1_dt = Z 1 21/α 0 (1 − 2tα) tαtα−1dt + Z 1 1 21/α (2tα− 1) tα_tα−1_dt = 1 4α and Z 1 0 |1 − 2tα_{| (1 − t}α_{) d} αt = Z 1 0 |1 − 2tα_{| (1 − t}α_{) t}α−1_dt = Z 1 21/α 0 (1 − 2tα) (1 − tα) tα−1dt + Z 1 1 21/α (2tα− 1) (1 − tα_{) t}α−1_dt = 1 4α.

Remark 14 If we choose α = 1 in Theorem (13), then we obtain the following inequality:     1 ı2− ı1 Z ı2 ı1 ϕ (x) dx − 1 2  ϕ (ı1) + ϕ (ı2) 2 + ϕ  ı1+ ı2 2     ≤ ı2− ı1 32  |ϕ0(ı1)| + 2     ϕ0 ı1+ ı2 2     + |ϕ0(ı2)|  .

Theorem 15 Let α ∈ (0, 1]and ϕ : I ⊂ R+ _{→ R be an α−fractional}

differen-tiable function on I◦and Dα(ϕ) be an α−fractional integrable function on I with

0 ≤ ı1< ı2 and p, q > 1, 1/p + 1/q = 1. If |ϕ0| q

be a convex function on I, then the following inequality for conformable fractional integral holds:

    α ıα 2 − ıα1 Z ı2 ı1 ϕ (xα) dαx − 1 2  ϕ (ıα 1) + ϕ (ıα2) 2 + ϕ  ıα 1 + ıα2 2     290

≤ α (ı α 2 − ıα1) 8  1 α (1 + p) 1p 1 2α 1q ×   (ı qα 1 ) α−1 (Dα(ϕ) (ıα1)) q + ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 q! 1 q + ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 q + (ıqα₂ )α−1(Dα(ϕ) (ıα2)) q1q # .

Proof. Since |ϕ0|q is a convex function on I, by the using the properties Dα(ϕ ◦ Λ) (t) = ϕ0(Λ (t)) Dα(Λ (t)) and Dα(ϕ (t)) = t1−αϕ0(t) and by Lemma

11 and using the well known H¨older inequality, we have     α ı2− ı1 Z ı2 ı1 ϕ (xα) dαx − 1 2  ϕ (ı1) + ϕ (ı2) 2 + ϕ  ı1+ ı2 2     ≤ 1 4 Z 1 0 (1 − 2tα)pdαt p1Z 1 0  Dα(ϕ)  tαıα₁ + (1 − tα)ı α 1 + ıα2 2 q dαt 1q +1 4 Z 1 0 (1 − 2tα)pdαt 1pZ 1 0  Dα(ϕ)  tαı α 1+ ıα2 2 + (1 − t α_{) ı}α 2 q dαt 1q ≤ α (ı α 2 − ıα1) 8  1 α (1 + p) 1_p (ıqα₁ )α−1(Dα(ϕ) (ıα1)) qZ 1 0 tαdαt + ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 qZ 1 0 (1 − tα) dαt !1_q +α (ı α 2 − ıα1) 8  1 α (1 + p) 1p _{ ı}α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 qZ 1 0 tαdαt + (ıqα₂ )α−1(Dα(ϕ) (ıα2)) qZ 1 0 (1 − tα) dαt  1 q = α (ı α 2 − ı α 1) 8  ₁ α (1 + p) 1p ₁ 2α 1q ×   (ı qα 1 ) α−1 (Dα(ϕ) (ıα1)) q + ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 q! 1 q + ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 q + (ıqα₂ )α−1(Dα(ϕ) (ıα2)) q1q # . where Z 1 0 (1 − 2tα)pdαt = Z 1 0 (1 − 2tα)ptα−1dt 291

= Z 1 21/α 0 (1 − 2tα)ptα−1dt + Z 1 1 21/α (2tα− 1)ptα−1dt = 1 α (1 + p) and Z 1 0 tαdαt = Z 1 0 tαtα−1dt = 1 2α Z 1 0 1 − tαdαt = Z 1 0 (1 − tα) tα−1dt = 1 2α

Corollary 16 If we choose α = 1 in Theorem 15, then we obtain the following inequality.     1 ı2− ı1 Z ı2 ı1 ϕ (x) dx −1 2  ϕ (ı1) + ϕ (ı2) 2 + ϕ  ı1+ ı2 2     ≤ ı2− ı1 8  ₁ 1 + p 1p_{ 1} 2 1q × "_ (ϕ0(ı1)) q +  ϕ0 ı1+ ı2 2 q1q +  ϕ0 ı1+ ı2 2 q + (ϕ0(ı2)) q 1 q# .

Remark 17 If we choose p = q = 2 in Corollary 16, then we obtain the follow-ing inequality.     1 ı2− ı1 Z ı2 ı1 ϕ (x) dx − 1 2  ϕ (ı1) + ϕ (ı2) 2 + ϕ  ı1+ ı2 2     ≤ ı2− ı1 8 1 61/2 ×   (ϕ 0_(ı 1)) 2 +  ϕ0 ı1+ ı2 2 2! 1 2 +  ϕ0 ı1+ ı2 2 2 + (ϕ0(ı2)) 2 !12 .

Theorem 18 Let α ∈ (0, 1]and ϕ : I ⊂ R+ _{→ R be an α−fractional}

differen-tiable function on I◦and Dα(ϕ) be an α−fractional integrable function on I with

0 ≤ ı1 < ı2 and q ≥ 1. If |ϕ0|q be a convex function on I, then the following

inequality for conformable fractional integral holds:     α ıα 2 − ıα1 Z ı2 ı1 ϕ (xα) dαx − 1 2  ϕ (ıα 1) + ϕ (ıα2) 2 + ϕ  ıα 1 + ıα2 2     ≤ α (ı α 2 − ıα1) 8  ₁ 2α 1−1q ₁ 4α 1q × 292

(ıqα₁ )α−1(Dα(ϕ) (ıα1)) q + ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 q! 1 q +  ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 q + (ıqα₂ )α−1(Dα(ϕ) (ıα2)) q !1q . Proof. Since |ϕ0|q is a convex function on I, by the using the properties Dα(ϕ ◦ Λ) (t) = ϕ0(Λ (t)) Dα(Λ (t)) and Dα(ϕ (t)) = t1−αϕ0(t) and assume

that q ≥ 1, by Lemma 11 and using the well known power-mean inequality, we have     α ıα 2 − ıα1 Z ı2 ı1 ϕ (xα) dαx − 1 2  ϕ (ıα 1) + ϕ (ıα2) 2 + ϕ  ıα 1 + ıα2 2     ≤ α (ı α 2 − ıα1) 8 Z 1 0 (1 − 2tα) dαt 1−1q ×  (ıqα₁ )α−1(Dα(ϕ) (ıα1)) qZ 1 0 (1 − 2tα) tαdαt + ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 qZ 1 0 (1 − 2tα) (1 − tα) dαt !1q +α (ı α 2 − ı α 1) 8 Z 1 0 (1 − 2tα) dαt 1− 1 q ×  ıα 1+ ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 qZ 1 0 (1 − 2tα) tαdαt + (ıqα₂ )α−1(Dα(ϕ) (ı2)) qZ 1 0 (1 − 2tα) (1 − tα) dαt 1q = α (ı α 2 − ıα1) 8  ₁ 2α 1−1q ₁ 4α 1q × (ıqα₁ )α−1(Dα(ϕ) (ıα1)) q + ı α 1 + ı α 2 2 q(α−1) Dα(ϕ)  ıα 1 + ı α 2 2 q! 1 q +  ı α 1 + ıα2 2 q(α−1) Dα(ϕ)  ıα 1 + ıα2 2 q + (ıqα₂ )α−1(Dα(ϕ) (ıα2ı2)) q !q1 . where Z 1 0 (1 − 2tα) dαt = Z 1 21/α 0 (1 − 2tα) tα−1dt + Z 1 1 21/α (2tα− 1) tα−1_{dt =} 1 2α and Z 1 0 (1 − 2tα) tαdαt 293

= Z 1 21/α 0 (1 − 2tα) tαtα−1dt + Z 1 1 21/α (2tα− 1) tα_tα−1_{dt =} 1 4α, Z 1 0 (1 − 2tα) (1 − tα) dαt = Z 1 21/α 0 (1 − 2tα) (1 − tα) tα−1dt + Z 1 1 21/α (2tα− 1) (1 − tα_{) t}α−1_{dt =} 1 4α.

Corollary 19 If we choose α = 1 in Theorem 18, then we obtain the following inequality.     1 ı2− ı1 Z ı2 ı1 ϕ (x) dx − 1 2  ϕ (ı1) + ϕ (ı2) 2 + ϕ  ı1+ ı2 2     ≤ ı2− ı1 8  1 2 1+1q × "  (ϕ0(ı1)) q +  ϕ0 ı1+ ı2 2 q1q +  ϕ0 ı1+ ı2 2 q + (ϕ0(ı2)) q 1 q# .

Remark 20 If we choose q = 1 in Corollary 19, then we obtain Remark 14.

4

Applications

be the arithmetic mean, generalized logarithmic mean for ı1, ı2> 0 respectively.

Proposition 21 Let s ∈ (0, 1] , ı1, ı2> 0, then

    Ln_n(ıα₁, ıα₂) −1 2[A (ı nα 1 , ı nα 2 ) + A n_(ıα 1, ı α 2)]     ≤ n (ı α 2 − ıα1) 16 h A(ıα1) n−1 , (ıα2) n−1 + An−1(ıα1, ıα2) i .

Proof. The claim follows from Theorem 13 applied to convex function ϕ (κ) = κn where n ∈ N.

Example 22 If we take α = 1, n = 2 in Proposition 21, then we can obtain inequality following  L2₂(ı1, ı2) − AA (ı1, ı2) , A2(ı1, ı2)   ≤ (ı2− ı1) 4 A (ı1, ı2) .

Proposition 23 Let α ∈ (0, 1] , ı1, ı2> 0, p, q > 1, then

    Ln_n(ıα₁, ıα₂) −1 2[A (ı nα 1 , ı nα 2 ) + A n_(ıα 1, ı α 2)]     ≤ nα (ı α 2 − ı α 1) 8  ₁ α (1 + p) 1p ₁ 2α 1q ×    (ıα₁)n−1 q +  ı α 1 + ıα2 2 n−1!q! 1 q +  ı α 1 + ıα2 2 n−1!q +(ıα₂)n−1 q1q # .

Proof. The claim follows from Theorem 15 applied to convex ϕ (κ) = κn_where

n ∈ N.

Example 24 If we take α = 1, p = q = n = 2 in Proposition 23, then we can obtain the following inequality

    L22(ı1, ı2) − 1 2A ı 2 1, ı22
+ A2(ı1, ı2)      ≤ ı2− ı1 4 ×   ı 2 1+  ı1+ ı2 2 2! 1 2 +  ı1+ ı2 2 2 + ı2₂ !  1 2 .

Proposition 25 Let α ∈ (0, 1] , ı1, ı2> 0, q ≥ 1, then

    Ln_n(ıα₁, ıα₂) −1 2[A (ı nα 1 , ı nα 2 ) + A n_(ıα 1, ı α 2)]     ≤ α (ı α 2 − ıα1) 8  ₁ 2α 1−1q  ₁ 4α 1q ×    (ıα₁)n−1 q +  ı α 1 + ıα2 2 n−1!q! 1 q +  ı α 1 + ıα2 2 n−1!q +(ıα2) n−1q1q # . 295

Proof. The claim follows from Theorem 18 applied to convex ϕ (κ) = κnwhere n ∈ N.

Example 26 If we take α = 1, q = 2, n = 2 in Proposition 25, then we can obtain inequality following

    L22(ı1, ı2) − 1 2A ı 2 1, ı 2 2
+ A 2 (ı1, ı2)      ≤ (ı2− ı1) 8  1 8 12 × ı2₁+ ı1+ ı2 2 2! 1 2 +  ı1+ ı2 2 2 + ı2₂ !12 . 296

297

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