5th National Conference on Role of Engineers in Nation Building, 3rd and 4th March, 2017
INTRODUCTION OF YOGI’S THREE POINTS AVERAGE RULE OF NUMERICAL INTEGRATION AND ITS COMPARATIVE STUDY Yogendra Vishwakarma
Dr. Ajazul Haque
Jayesh Jain
R. Prajapati
Assistant Professor Viva Institute Of Technology Virar (E),Mumbai(India)
Assistant Professor Viva Institute Of Technology Virar (E),Mumbai(India)
Assistant Professor Viva Institute Of Technology Virar (E),Mumbai(India)
Assistant Professor Viva Institute Of Technology Virar (E),Mumbai(India)
[email protected]
[email protected]
[email protected]
ramashankar.p28@gmail. com.
ABSTRACT The present paper is an effort to explore a new technique for solving the problem of numerical integration. The numerical integration problem plays a very important role in numerical analysis and it has a wide range of application in various domains of science and technology. This is a marvelous mathematical method for analyzing and interpreting the result of scientific and technical problems.
Keywords Numerical integration, approximating, infimum, supremum, integral and quadrature.
while the lower Riemann sum of the function f(x) with respect to the partition Ƥ is defined as L (f, Ƥ)= ∑𝑛𝑘=1 𝑚𝑘 𝛿𝑘 Therefore, the upper integral of the function f(x) on closed interval [a, b] is defined as U( f)= inf U(f, Ƥ) and the lower integral of the function f(x) on closed interval [a, b] is defined as L(f)=Sup L(f, Ƥ),
1. INTRODUCTION Numerical integration is a very useful mathematical method to get an approximate solution which correspond to a very closed value with the exact value of the problem. In this study, we discuss the various methods of numerical integration and we introduce a new method for solving the problem of numerical integration which is giving more precise result compared of to other method (Mid-point rule). There are many reasons as of why such approximations can be useful. Not every function can be analytically integrated, even if a closed integration formula exists, it may still not be the most efficient way of calculating the integral. In addition, it can happen that we need to integrate an unknown function, in which only some samples of the function are known. Suppose y=f(x) is a continuous and bounded function defined on a closed interval [a, b].Let Ƥ={a=x0,x1,x2,........,xn=b} be a partition of the a closed interval [a, b].For each k, we define Mk(f )=
sup
f(x)
𝑥𝜖[xk−1,xk]
and mk(f )=
inf
𝑥𝜖[xk−1,xk]
f(x)
Let 𝛿xk= xk - xk-1 be the strip of width say h. The upper Riemann sum of the function f(x) with respect to the partition Ƥ is defined as
where both infimum and supremum are taken over all possible partition, Ƥ,of the closed interval [a, b]. If the upper and lower integral of f(x) are equal, i.e., U(f) = L(f) to each other and their common value is denoted by 𝑏
∫𝑎 𝑓(𝑥)dx and is referred to as the Riemann integral of f(x) a 𝑏
simpler approach for approximating the value of ∫𝑎 𝑓(𝑥)dx would be to compute the product of the value of the function at one of the end points of the interval by the length of the interval. In case we chose the end point where the function is evaluated 𝑏
to be x=a, we obtain ∫𝑎 𝑓(𝑥)dx=(b-a)f(a). This approximation is called the rectangular method. Numerical integration formula also referred to as integration rules or quadrature, and hence we can referred above integral as rectangular rule or rectangular quadrature .The points x0,x1,x2,......xn that are used in the quadrature formula as called quadrature points .A variation on the rectangular rule is the midpoint rule .Similarly to the rectangular rule, we 𝑏
approximate the value of the integral ∫𝑎 𝑓(𝑥)dx by multiplying the length of the interval by the value of the function at one point only this time we replace the value of the function at the 1
centre point (a+b), i.e., 2
𝑏
U(f, Ƥ)=
∑𝑛𝑘=1 𝑀𝑘
𝛿𝑘 ,
∫𝑎 𝑓(𝑥)dx ≅(b-a) f (
𝑎−𝑏 2
),
5th National Conference on Role of Engineers in Nation Building, 3rd and 4th March, 2017
the midpoint quadrature is a more accurate quadrature than the rectangular rule.
Therefore the total area of the region on [a, b] is approximately equal to the sum of area of these sub-rectangles
1.1 Three Points Average Rule Let y = f(x) be continuous real valued integrable function which is defined on [a, b]. Dividing the interval [a, b] into n even number of sub-interval so that width of each sub interval in same (say h) and is given by
∴ ∫𝑎 𝑓(𝑥) dx = 2h f (
𝛿x=h=
𝑏
f(
𝑥𝑛−2 +𝑥𝑛−1 +𝑥𝑛
3
𝑏
f(
) +2h 𝑓(
𝑥2 +𝑥3 +𝑥4 3
) +............+ 2h
).
3
∴ ∫𝑎 𝑓(𝑥) dx = 2h[ f (
𝑥𝑛 −𝑥0 𝑛
𝑥0 +𝑥1 +𝑥2
𝑥𝑛−2 +𝑥𝑛−1 +𝑥𝑛 3
𝑥0 +𝑥1 +𝑥2
) + 𝑓(
3
𝑥2 +𝑥3 +𝑥4 3
) +............+
…………………… (*)
)
This method is called Yogi’s three points average rule or simply three points average rule of numerical integration. [*This rule is purely introduced by the first author and this formula has never seen in anywhere in the entire life of study. If anybody anywhere find it exists, please reject the same by showing the proof and the author has no objection. This is a part of the author’s study and any suggestion and improvement of this result will be always most welcome.] 2 2𝑥
Consider the integral∫1
dx, the exact value of this
𝑥 2 +4
integral is given by 2 2𝑥
∫1
8
dx =[log(𝑥 2 + 4)]12 = log(8) –log(5) =log( ) = log(1.6)
𝑥 2 +1
5
= 0.4700. Let us divide the interval [1, 2] into 4 equal parts here a=1,b=2, 𝑏−𝑎 2−1
and h=
Figure (a) Let Ƥ={a=x0,x1,x2.....xn=b} be the corresponding partition of [a, b] so, in general kth interval is given by [xk-1, xk]. Now, definite interval is defined as a limit of Riemann sum so any Rsum could be used as an approximate to the interval [a, b]. Now, let us consider first two sub interval, i.e. [x0, x1] and [x1, x2], so width of sub interval is h= x1-x0 and h=x2-x1 adding both we get 2h=x2- x0.
𝑛
=
1
= = 0.25
4
4
x0=a=1 x1=x0+h = 1+0.25 =1.25 x2= x1+h = 1.25+0.25 =1.50 x3=x2+h = 1.50+0.25 =1.75 x4=x3+h = 1.75+0.25 =2.0
The height of sub - rectangles would be the centre point of extreme (we can also say that average height) which is defined as f(
𝑥0 +𝑥1 +𝑥2 3
𝑥0 +𝑥1 +𝑥2 1+1.25+1.50
𝑥0 +𝑥1 +𝑥2 3
0
2 2𝑥
𝑥
𝑥2 +𝑥3 +𝑥4 3
=
3
∫1
)
Similarly, area of 2nd two sub-rectangle is A24 = ∫𝑥 4 𝑓(𝑥) dx = 2h f (
3
=1.25
𝑥2 +𝑥3 +𝑥4 1.50+1.75+2.0 𝑥
∴Area of first two sub-rectangle is A02 = ∫𝑥 2 𝑓(𝑥) dx = 2h f (
=
3
).
3
dx =2h [f(
𝑥 2 +4
=1.75
𝑥0 +𝑥1 +𝑥2 3
)+ f(
𝑥2 +𝑥3 +𝑥4 3
)]
=2×2.5[f(1.25) + f(1.75)]
2
)
=0.50[0.4494+0.4956]
⋮
=0.50×0.945
& so on the area of last two sub rectangle is given by
=0.4725
𝑥
𝐴(𝑛−2) n = ∫𝑥 𝑛 𝑓(𝑥) dx = 2h f ( 𝑛−2
𝑥𝑛−2 +𝑥𝑛−1 +𝑥𝑛 3
)
Let us solve the above problem by the methods mid-point rule, Trapezoidal rule and Simpson’s rules.
5th National Conference on Role of Engineers in Nation Building, 3rd and 4th March, 2017
The approximate values of these rules with absolute error can be tabulated as follows:
1.2 Mid-Point Rule x0=1, x1 =1.25, x2 =1.50, x3 =1.75, x4 =2.0 𝑥0 +𝑥4 2 𝑥1 +𝑥2 2 𝑥2 +𝑥3 2 𝑥3 +𝑥4 2
1+1.25
=
2
=
2.25 2
1.25+1.50
=
1.50+1.75
=
2 1.75+2
=
=
2
2 2𝑥 ∫1 𝑥 2+4dx
Approxim ate value
Absolute Error
Three points average rule
0.4725
0.0025
Mid-point rule
0.4706
0.0006
Trapezoidal rule
0.4688
0.0012
Simpson’s 1/3rd rule
0.4681
0.0019
Simpson’s 3/8th rule
0.4389
0.0311
Exact value
0.4700
0.0000
1.8282
0.0004
Mid-point rule
1.8236
0.0042
Trapezoidal rule
1.8276
0.0002
Simpson’s 1/3rd rule
1.8278
0.0000
Simpson’s 3/8th rule
1.8278
0.0000
Exact value
1.8278
0.0000
Function
Method
= 1.275
2
=
=1.125
3.25 2
3.75 2
= h[f (
= 1.875
𝑥0 +𝑥4 2
= 1.625
) + f(
𝑥1 +𝑥2 2
) + f(
𝑥2 +𝑥3 2
) + f(
𝑥3 +𝑥4 2
)]
=0.25 [f(1.125) + f(1.275) + f(1.625) + f(1.875)] =0.25[0.4273 + 0.4668 + 0.4894 + 0.4990]
2 2𝑥
∫1
dx,
𝑥 2 +1
= 0.4706 1.3 Trapezoidal Rule
n= 4
x0=1, x1 =1.25, x2 =1.50, x3 =1.75, x4 =2.0 𝑦0 =0.4, 𝑦1 = 0.4494, 𝑦2 = 0.48, 𝑦3 =0.4956, 𝑦4 =0.5 2 2𝑥
∫1
ℎ
dx = [(𝑦0 + 𝑦4 ) + 2(𝑦1 + 𝑦2 + 𝑦3)]
𝑥 2 +1
2
=
0.25 2
Three points average rule
[(0.4 + 0.5) + 2(0.4494 + 0.48 + 0.4956)]
=0.4688. 1.4 Simpson’s 1/3rd Rule x0=1, x1 =1.25, x2 =1.50, x3 =1.75, x4 =2.0 𝑦0 =0.4, 𝑦1 = 0.4494, 𝑦2 = 0.48, 𝑦3 =0.4956, 𝑦4 =0.5 2 2𝑥
∫1
ℎ
dx = [(𝑦0 + 𝑦4 ) + 4(𝑦1 + 𝑦3) + 2 𝑦2 ]
𝑥 2 +1
3
=
0.25 3
5.2
∫4.0 log 𝑒 (𝑥)dx ,
[(0.4 + 0.5) + 4(0.4494 + 0.4956) +2(0.48 )]
=0.4681. n=6 1.5 Simpson’s 3/8 th Rule x0=1, x1 =1.25, x2 =1.50, x3 =1.75, x4 =2.0 𝑦0 =0.4, 𝑦1 = 0.4494, 𝑦2 = 0.48, 𝑦3=0.4956, 𝑦4 =0.5 2 2𝑥
∫1
3ℎ
dx =
𝑥 2 +1
=
8
[(𝑦0 + 𝑦4 ) + 3(𝑦1 + 𝑦2) + 2 𝑦3 ]
3×0.25 8
[(0.4 + 0.5) + 3(0.4494 + 0.48) +2(0.4956 )]
= 0.4389.
CONCLUSION From the above result one can conclude that that three points average rule gives better accurate result than mid-point rule but closer/higher to the exact value of integral and this technique of calculation is faster way of analyzing the result. Sometimes
5th National Conference on Role of Engineers in Nation Building, 3rd and 4th March, 2017
the value of integral varies from problem to problem and this rule is applicable if only if n is even. If n is large it will produce more accurate result.
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Brian Bradie, Sixth Impression, Pearson Prentice Hall. [11] Applied Numerical Analysis, Seventh Edition, Curtis F. Gerald, Patrick O. Wheateley, Pearson. [12] Introductory Methods of Numerical Analysis by S.S. Sastry, Prentice Hall of India.
