 Open Access
 Authors : Muhammad Amir Ashraf, Dr. Muhammad Zafar Iqbal, Kiran Shahzadi, Ayesha Kiran, Balawal Mehmood
 Paper ID : IJERTV13IS060055
 Volume & Issue : Volume 13, Issue 06 (June 2024)
 Published (First Online): 13062024
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
on Novel Approximation of (P,Q) – Trapezoidal and Midpoint Inequalities and Their Implementation in Postquantum Calculus
Muhammad Amir Ashraf, Dr. Muhammad Zafar Iqbal, Kiran Shahzadi, Ayesha Kiran, Balawal Mehmood
Department of Mathematics and Statistics
University of Agriculture Faisalabad, 38000, Faisalabad, Punjab, Pakistan
ABSTRACT:
In this work, we will show several improvements of Hermite Hadamard inequalities for convex functions using the concept of postquantum calculus. We will also demonstrate how the recent discoveries relate to the previous findings. Furthermore, we will present some applications of recently discovered inequalities along with several mathematical illustrations.
KEYWORDS: (p, q)HermiteHadamard Inequality; (p, q) derivative; (p, q)integral; postquantum calculus; convex function.
INTRODUCTION:
The field of qanalysis, which Euler introduced in response to the increasing need for mathematical models that may explain quantum computing, has been the subject of several contemporary investigations. Due to its versatility, Q calculus has become a key tool in the domains of quantum theory, mechanics, relativity, and number theory and also in combinatorics, fundamental hypergeometric functions, and orthogonal polynomials[1], [2].
Early scholars were the first to recognize the importance of convexity in a number of areas, therefore the idea of convexity has a long history. Overall, qcalculus is a powerful tool that has found applications in a wide range of fields. Its ability to describe noncommuting operators and its development of qanalogues of mathematical tools have made it a useful tool in quantum mechanics, computer science, and statistics, among other fields. As research into quantum computing continues to grow, it is likely that qcalculus will play an increasingly important role in the development of new algorithms and computational techniques[3].
One of the benefits of postquantum calculus is that it can be used to address issues that qcalculus cannot. In specific quantum systems, for instance, the behavior of a system may be affected by both the location and momentum of a particle. These systems may be modelled using the (p, q) calculus.[4]
Post quantum calculus is an extension of qcalculus, which is a generalization of the traditional calculus that allows for calculations with qnumbers.
Qnumbers are a mathematical construct that generalizes the concept of integers and can be used to represent quantum systems[5]. Qcalculus has been applied to problems in physics, finance, and computer science, among other fields. Inequalities are crucial for addressing mathematical issues, but they also have useful uses in a variety of realworld contexts, including data analysis, function optimization, and setting conditions for certain outcomes. They are a crucial subject in any mathematics curriculum because they provide a basis for more complex mathematical ideas and concepts. A mathematical function that meets the convexity property is said to be a convex function[6]. In plain English, a function is said to be convex if the line segment joining any two points on its graph sits wholly above the graph. In other words, the function seems "bowlshaped" or "curves upward" when viewed from above.
If function () is convex on interval , if any two points 1 and 2 in and any where 0 <<1
[1 + (1 )2] (1) + (1 )(2).The HermiteHadamard inequality is an important finding in mathematical analysis that shows a strong correlation among the average values of convex functions. This discrepancy was first studied in the late 19th century and is named after Charles Hermite and Jacques Hadamard[7][8]. By history, construction and uses of the HermiteHadamard inequality emphasizing its importance across a range of mathematical specialties and related disciplines. The Hermite Hadamard inequality of (p, q) Calculus is that:
Let u: [1, 2] R be a convex differentiable function on [1, 2] and 0 < q < p 1. Then we have
1 + 2
1 2+(1)1
(1) + (2)
(
)
+ (
) ()1 ,
+
. (4)
2 1 1
Similarly, the utilizing the right (p, q)integral, demonstrated HermiteHadamard inequality for convex mappings, then
1 + 2
( )
+
1
(2 1)
2
()2 ,
1+(1)2
(1) + (2)
+
. (5)
The main objectives of research is that the use (p, q)calculus approaches, establish various extensions of HermiteHadamard for q and (p, q) function. The establish connection between the new result and the previous finding results.
PRELIMINARIES
In this section we discuss some definition of quantum calculus and also implement on it.
Definition no 1: The qnumber, which is represented by the quantum number notation [], is defined for any positive integer n.
1
[] =1
= 1 + + 2 + 3 + + 1, (2.1)
Where the degree of the polynomial is n1 in relation to the deformation component q, which can take on either real or complex
numbers.
Definition no 2: The (p, q) numbers is introduced by
[], =, (2.2)
Where [], is symmetric then [], = [], and if = 1then the (p, q) numbers are reduced into qnumbers and if = 1 then reduced back into ordinary numbers n[9].
It means that the (p, q) numbers of Post Quantum calculus are performing same role as qnumbers of Quantum calculus.
Definition no 3:[10], [11] Let u:[1, 2] be continuous. The left qderivative of u at x [1, 2] is
() = () ( + (1 )1) ,
(2.3)
1
If 1 = 0 1 () = ()then (1) become
(1 )( 1) 1
() ()
This equation is Similar to q Jackson derivative formula. And the right qderivative is
() =
(1 ) , 0 . (2.4)
2 () = ( + (1 )1) () ,
(2.5)
(1 )(2 ) 2
Definition no 4: [12] Let u:[1, 2] and it is continuous on[1, 2]. The qintegral of u at x [1, 2]is
2
()2 = (1 )(2 1) (2 + (1 )1).
(2.6)
If 1=0 then equation (2) becomes
1
=
2
()2 = (1 )(2) (2). (2.7)
0
The equation (4) is similar to qJackson Integral. And the right qintegral is
=
2
()2 = (1 )(2 1) (1 + (1 )2). (2.8)
1
=
Definition no 5: [13] Let : [1, 2 ] is continuous function, and (, )1 derivative of at [1, 2]is define as:
With 0 < < 1.
1
,
() = ( + (1 )1) ( + (1 )1)
( )( 1)
(2.9),
Definition no 6: Similarly, (, )2 derivative of at [1, 2] is define as:
2 ,
() = ( + (1 )2) ( + (1 )2)
( )( 2)
(2.10)
Definition no 7: Let : [1, 2 ] is continuous function, and (, )1 integral of at [1, 2]is define as:
()1 , = ( )( 1) + 1 (
+1 + (1
+1) 1), (2.11)
With 0 < < 1.
1
=
Definition no 8: [14] Similarly, (, )2 Integral of at [1, 2]is define as:
2
With 0 < < 1.
()2 ( )(2 )
=
+ 1
( + 1
+ (1
+ 1
) 2) ,
(2.12)
Lemma 1: If a function : [1, 2] is Convex 1<2, then the following inequality hold for (, )2 Integral:
(
Where 0 < < 1
1 + 2
)
[2],1
(2 1)
2
()2 ,
1+(1)2
(1) + (2) [2],
, (2.13)
Lemma 2: If a function : [1, 2] is Convex 1< 2, then following inequality satisfied for (, )1 Integral:
(
Where 0 < < 1
1 + 2 [2],
) (
1 1+(1)2
)
()1 ,
2 1 1
(1) + (2) [2],
, (2.14)

New Trapezoidal type inequalities for (p, q)integrals
1
Lemma 3: If u: [1, 2] is p, qdifferentiable function that way[ ,] and [2 ,] are continuous and integrable
on [1, 2], then
(1) + (2) 1
2+(1)1
2
{ () , + ()2 ,}
2
2 1 1
2(2 1)
1
1
1+(1)2
1
= [ ,(2 + (1 )1), + 2 ,(1 + (1 )2 ), ]
Proof:
2 0 1 0
1
1 = 1 ,(2 + (1 )1),
0
1
= (
)( ) (2 + (1 )1) (2 + (1 )1),
2 1
0
+1
+1
= (
) [ +1 ( 2 + (1 ) 1) +1 (+1 2 + (1 +1) 1)]
2 1 =0
=0
1
1 +1
+1
+1
= (
) [ (
2 + (1 ) 1) +1 ( +1 2 + (1 +1) 1)]
2 1
1
=0
=0
1 +1
+1
+1 1
= (
) [
( 2 + (1
) 1) +1 (
+1 2 + (1
+1) 1) + (2)]
2 1
=0
=0
1 1
1
= (
) [(
) (
2 + (1 ) 1) + (2)]
2 1
1
=0
1
2+(1)1
= (
) [(
) ()1 , + (2)]
2 1 2
1 1
1
2+(1)1
(2 1)1 = [( ) ()1 , + (2)] (1)
Similarly,
2 1 1
1
2 = 2 ,(1 + (1 )2),
0
1
= (
)( ) (1 + (1 )2) (1 + (1 )2),
2 1
0
+1
+1
= (
) [ +1 (+1 1 + (1 +1) 2) +1 (
1 + (1 ) 2)]
2 1 =0
=0
1
+1
+1
+1
1
= (
) [ +1 (+1 1 + (1 +1) 2) (
1 + (1 ) 2)]
2 1 =0
=0
1
+1
+ 1
+1 1
1
= (
) [
+1 (
+ 1 2 + (1
+1) 1) +
(1)
(
1 + (1
) 2)]
2 1
=0
=0
1 1
1
= (
) [(
) (
2 + (1
) 1) +
(2)]
2 1
1
=0
1
2
=
(2 1)
[(2 1)
()2 , + (1)]
1(1)2
1 2+(1)1
(2 1)2 = [( ) ()1 , + (1)] (2)
By equation 1 and 2 we get
2 1 1
(1) + (2) 1
2+(1)1
2
{ () , + ()2 ,}
2
2 1 1
2(2 1)
1
1
1+(1)2
1
= [ ,(2 + (1 )1), + 2 ,(1 + (1 )2),]
Hence proved.
2 0 1 0
1
Theorem 1: If   and 2 are convex, then the following inequality holds according to Lemma (3) hypotheses:
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2 2(2 1)
1
1
1+(1)2
(2 1)
([3], [2],)( ,(1) + 2 ,(2))
[ ,(2) + 2 ,(1) + 1 ]2[3], 1
[2],Proof: we know by lemma 3 and Taking mode both sides
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2
2 1 1
2(2 1)
1
1
1+(1)2
1
= [ ,(2 + (1 )1), +  2 ,(1 + (1 )2),]
2 0 1 0
1
By using convexity of  , and 2 ,
2 1 1 1
[ { ,(2) + (1 ) ,(1),} + {2 ,(1) + (1 )2 ,(2)},]
2 0 1 1 0
2 1 1 1 1
= [ ,(2) 2, +  ,(1) (1 ), + 2 ,(1) 2,
2 1 0 1 0 0
1
+ 2 ,(2) (1 ),]
0
2 1
= [
( ) + 
( ) [2],[3], + 2
( )
+ 2
[2], [3],( ) ]
2 1
,
2 [3],
1
,
1 [2],[3],
,
1 [3],
,
2 [2],[3],
2 1
= [{
( ) + 2
( )}
+ {
( ) + 2
([2], [3],)
( )}
2 1
, 2
,
1 [3],
1
, 2
, 1
[2],
[3]]
,
Then, we get
(2 1)
([3], [2],)( ,(1) + 2 ,(2))
[ ,(2) + 2 ,(1) + 1 ]Hence proved.
2[3], 1
[2],1
Theorem 2: If, ,,2 , 1and it is convex function, then inequality holds according to Lemma (3)
hypotheses:
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2 2(2 1)
1
1
1+(1)2
(2 1)
1

, ,(2) + ([3], [2],) ,(2)
[( 1
2[2],
1 )

,
1

,2 ,(1) + ([3], [2],)2 ,(1)
+ ( ) ]

,

Proof: We know by lemma 3 and theorem 1 give
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2
2 1 1
2(2 1)
1
1
1+(1)2
1
= [ ,(2 + (1 )1), +  2 ,(1 + (1 )2),]
2 0 1 0
By using power mean inequality
2 1 1
11 1
1
2
[( ,)
0
( 1 ,(2 + (1 )1),)
0
1
1
11 1
+ ( ,)
0
( 2 ,(1 + (1 )2),) ]
0
1
By the convexity of  , and 2 ,, we have
2 1 1
11 1
1
2
[( ,)
0
( {1 ,(2) + (1 )1 ,(1)},)
0
1
1
11 1
+ ( ,)
0
( {2 ,(1) + (1 )2 ,(2)},) ]
0
By taking integration we get
(2 1)
1

, ,(2) + ([3], [2],) ,(2)
[( 1
2[2],
1 )

,
1

,2 ,(1) + ([3], [2],)2 ,(1)
+ ( ) ]

,

Hence proved.
1
Theorem 3: If  2  > 1and it is convex, then inequality holds according to Lemma (3) hypotheses:
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2 2(2 1)
1
1
1+(1)2
(2 1)
=
2
1
1
( )
[ + 1],1
 ,(2) + ( + 1) ,(1)
[( 1 1 ) [2],1
2 ,(1) + ( + 1)2 ,(2)
+ ( ) ] [2],
Where 1 + 1 = 1
Proof: The lemma 1 and theorem 1 give
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2
2 1 1
2(2 1)
1
1
1+(1)2
1
= [ ,(2 + (1 )1), + 2 ,(1 + (1 )2),]
2 0 1 0
2 1 1 1
[  ,(2 + (1 )1), + 2 ,(1 + (1 )2),]
2 0 1 0
By Holder inequality and lemma 1, we get
2 1
1
1
[( ())1
1
(  (2 + (1 )1))
2 0 0 1
1 1
1 1
+ ( ())
0
( 2 (1 + (1 )2)) ]
0
1
By the convexity of  , and 2 ,, we have
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2 2(2 1)
1
1
1+(1)2
2 1
1
1
[( (), )1
1
( { ,(2) + (1 ) ,(1)},)
2 0 0 1 1
1 1
1 1
+ ( (),)
0
( {2 ,(1) + (1 )2 ,(2)},) ]
0
By taking integration we get
(2 1)
=
2
1
1
( )
[ + 1],1
 ,(2) + ( + 1) ,(1)
[( 1 1 ) [2],1
2 ,(1) + ( + 1)2 ,(2)
+ ( ) ] [2],
Thus, the proof is completed.


New Midpoint type inequalities for (p, q)integrals
1
Lemma 4: If u: [[1, 2] R is p, qdifferentiable function that way[ ,] and [2 ,] are continuous and integrable
on[1, 2], then
2 + 1
1 2+(1)1
2
(
) { () , + ()2 ,}
2
2(2 1)
1
1
1
1+(1)2
1
= 2 1 2
(
+ (1 ) )
+ ( 1)
(
+ (1 ) )
2
[ 1 , 2 0
1 , 1
2
1
, 2
1 ,
1
2
2
+
(
+ (1 ) )
1
+ (1 )2
(
+ (1 ) )
]
, 1
0
2 , 1
2
, 1
2 ,
Proof: It is simple to demonstrate by following the technique described in Lemma 3
1
Theorem 4: If  , and 2 ,are convex, then the following inequality holds according to Lemma (4) hypotheses:
2 + 1
1 2+(1)1
2
 (
) [ () , + ()2 ,]
2 2(2 1)
1
1
1+(1)2
2 1
42 32 3 32
2(23 23 + 22 + 22 + 32
2 [1 ,(2)
4([3],
[2],) + 1 ,(1)
8([3]
,
[2],)
+ 2 ,(1)
42 32 3 32
4([3],[2],)
+ 2 ,(2)
2(23 23 + 22 + 22 + 32
]
8([3],[2],)
Proof: It is simple to demonstrate by following the technique described in theorem 1
1
Theorem 5: If ,, 2 , 1and it is convex function, then the inequality holds according to Lemma (4)
hypotheses:
2 + 1
1 2+(1)1
2
 (
) [ () , + ()2 ,]
2 2(2 1)
11
1
1
1+(1)2
1
2
2 1
[3], +
2 4[2]
[( )
,
Ã— (1 ,(2)
8[3],
+ 1 ,(1)
8([3],
)
[2],)2
11
6 [2],
1
2
5 2 2
4[2]
+ ( )
,
Ã— (1 ,(2)
11
8([3],
[2],) + 1 ,(1)
)
8[3],
1
2
[3], +
4[2]
+ ( )
,
Ã— ( 2 ,(1)
8[3],
+  2 ,(2)
)
8([4], + [2],)
2
11
6 [2],
1
2
5 2 2
4[2]
+ ( )
,
Ã— ( 2 ,(1)
8([3],[2],)
+  2 ,(2)
) ]
8[3],
Proof: It is simple to demonstrate by following the technique described in theorem 2.
Theorem 6: If1 , 2 , > 1and it is convex function, then the inequality holds by using Lemma (4) hypotheses:
2 + 1
1 2+(1)1
2
 (
) [ () , + ()2 ,]
2 2(2 1)
1
1
1+(1)2
2 1
1
1
1
1 1 + 2
2 [ (2+1[ + 1] )
Ã— (1 (2)
4[2]
,
+ 1 (1) 4[2]
)
,
1
1
1
+ (  1)
2
Ã— (1 (2)
3
4[2],
+ 1 (1)
1
6 1
)
4[2],
1
1
1
1
1 + 2
+ (2+1[ + 1] )
Ã— ( 2 (1) 4[2]
,
+  2 (2) 4[2]
)
,
1
1
1
+ ( 1 )
2
Ã— (2 (1)
3
4[2],
+ 2 (2)
1
6 1
) ]
4[2],
Where 1 + 1 = 1
Proof: It is simple to demonstrate by following the technique described in theorem 3

Examples of (P, Q)HermiteHadamard inequality
This section Discuss the effectiveness of the newly build inequalities by giving examples and discuss these it useful like basic HermiteHadamard inequality.
Example 1: let = 2 + 2 is convex function on [0, 1] with = = 1 then the L.H.S of inequality is that
2
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2 2(2 1)
1
1
1+(1)2
2 + 3
= 
2
1
2(1 0)
1
2
{ (2
0
1
1
2
+ 2)1 , + (
2
+ 2)2
,}
The right side of inequality is
(2 1)
= 0.63
([3], [2],)( ,(1) + 2 ,(2))
[ ,(2) + 2 ,(1) + 1 ]
2[3], 1
[2],So, its clear
= 0.74
Hence proved.
0.63 < 0.74
Example 2: let = 2 + 2 is convex function on [0, 1] with p=q=1 = 2 then L.H.S of inequality is following
2
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2 2(2 1)
1
1
1+(1)2
2 + 3
= 
2
1
2(1 0)
1
2
{ (2
0
1
1
2
+ 2)1 , + (
2
+ 2)2
,}
The R.H.S of inequality is become
= 0.63
(2 1)
2[2],
1
[2], ,(2) + ([3], [2],) ,(2) [( 1 1 ) [3],1
[2],2 ,(1) + ([3], [2],)2 ,(1)+ ( ) ] [3],
So, its clear
= 0.81
Hence proved.
0.63 < 0.81
Example 3: let = 2 + 2 is convex function on [0, 1] with p=q=1 = = 2 then the L.H.S of the inequality is
2
(1) + (2) 1
2+(1)1
2
 { () , + ()2 ,}
2 2(2 1)
1
1
1+(1)2
2 + 3
= 
2
1
2(1 0)
1
2
{ (2
0
1
1
2
+ 2)1 , + (
2
+ 2)2
,}
The R.H.S of inequality is become
= 0.63
(2 1)
=
2
1
1
( )
[ + 1],1
 ,(2) + ( + 1) ,(1)
[( 1 1 ) [2],1
2 ,(1) + ( + 1)2 ,(2)
+ ( ) ] [2],
Where 1 + 1 =
So, its clear
= 0.94
Hence proved.
0.63 < 0.94
Example 4: let = 2 + 2 is convex function on [0, 1] with = = 1 then the L.H.S of inequality is that
2
2 + 1
1 2+(1)1
2
 (
) [ () , + ()2 ,]
2 2(2 1)
1
1
1+(1)2
5 1
= 
1
2( 2
)
1
+ (2 + 2)2
}
0
2 2(1 0) {
+ 2 1
, 1
2
,
The right side of inequality is
2 1
42 32 3 32
= 0.27
2(23 23 + 22 + 22 + 32
2 [1 ,(2)
4([3],
[2],) + 1 ,(1)
8([3]
,
[2],)
+ 2 ,(1)
42 32 3 32
4([3],[2],)
+ 2 ,(2)
2(23 23 + 22 + 22 + 32
]
8([3],[2],)
So, its clear
= 0.44
Hence proved.
0.27 < 0.44
Example 5: let = 2 + 2 is convex function on [0, 1] with p=q=1 = 2 then L.H.S of inequality is following
2
2 + 1
1 2+(1)1
2
 (
) [ () , + ()2 ,]
2 2(2 1)
5 1
1
1
1
1+(1)2
1
= 
2( 2
)
+ (2 + 2)2 }
0
2 2(1 0) {
+ 2 1
, 1
2
,
The R.H.S of inequality is become
2 1
11
= 0.27
1
2
[3], +
2 4[2]
[( )
,
Ã— (1 ,(2)
8[3],
+ 1 ,(1)
8([3],
)
[2],)2
11
6 [2],
1
2
5 2 2
4[2]
+ ( )
,
Ã— (1 ,(2)
11
8([3],
[2],) + 1 ,(1)
)
8[3],
1
2
[3], +
4[2]
+ ( )
,
Ã— ( 2 ,(1)
8[3],
+  2 ,(2)
)
8([4], + [2],)
2
11
6 [2],
1
2
5 2 2
4[2]
+ ( )
,
Ã— ( 2 ,(1)
8([3],[2],)
+  2 ,(2)
) ]
8[3],
So, its clear
= 0.48
Hence proved.
0.27 < 0.48
Example 6: let = 2 + 2 is convex function on [0, 1] with p=q=1 = = 2 then the L.H.S of the inequality is
2
2 + 1
1 2+(1)1
2
 (
) [ () , + ()2 ,]
2 2(2 1)
5 1
1
1
1
1+(1)2
1
= 
2( 2
)
+ (2 + 2)2 }
0
2 2(1 0) {
+ 2 1
, 1
2
,
The R.H.S of inequality is become
= 0.27
2 1
1
1
1
1 1 + 2
2 [ (2+1[ + 1] )
Ã— (1 (2)
4[2]
,
+ 1 (1) 4[2]
)
,
1
1
1
+ (  1)
2
Ã— (1 (2)
3
4[2],
+ 1 (1)
1
6 1
)
4[2],
1
1
1
1
1 + 2
+ (2+1[ + 1] )
Ã— ( 2 (1) 4[2]
,
+  2 (2) 4[2]
)
,
1
1
1
+ ( 1 )
2
Ã— (2 (1)
3
4[2],
+ 2 (2)
1
6 1
) ]
4[2],
Where 1 + 1 =
So, its clear
= 0.56
Hence proved.
0.27 < 0.56

CONCLUSIONS
In the (p, q)calculus framework, we prove new versions of the trapezoidal and midpoint inequalities for differentiable convex functions. Additionally, we use famous HÃ¶lder and power mean inequality for (p, q)differentiable functions to construct (p, q)midpoint and (p, q)type trapezoidal inequalities. These novel results have applications in determining certain error boundaries for the trapezoidal and midpoint principles in p, qintegration formulae, which are crucial in numerical analysis. The possibility that post quantum coordinated convex mappings might lead to new inequality formulations by mathematicians working in this area is an interesting one.
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