Newton's Backward Interpolation explained with example. History of interpolation along with it's advantages and disadvantages. Applications of interpolation in computer sciences.
Newton's Backward Interpolation explained with example. History of interpolation along with it's advantages and disadvantages. Applications of interpolation in computer sciences.
These slides contain information about Euler method,Improved Euler and Runge-kutta's method.How these methods are helpful and applied to our questions are detailed discussed in the slides.
Here we focuses on Fixed-Point Iterative Technique for solving nonlinear Equations in Numerical Analysis. It is one of the opened-iterative techniques for finding roots called Fixed-Point of Non-linear Equations.
Computer Oriented Numerical Analysis
What is interpolation?
Many times, data is given only at discrete points such as .
So, how then does one find the value of y at any other value of x ?
Well, a continuous function f(x) may be used to represent the data values with f(x) passing through the points (Figure 1). Then one can find the value of y at any other value of x .
This is called interpolation
Newton’s Divided Difference Formula:
To illustrate this method, linear and quadratic interpolation is presented first.
Then, the general form of Newton’s divided difference polynomial method is presented.
Gauss jordan and Guass elimination methodMeet Nayak
This ppt is based on engineering maths.
the topis is Gauss jordan and gauss elimination method.
This ppt having one example of both method and having algorithm.
These slides contain information about Euler method,Improved Euler and Runge-kutta's method.How these methods are helpful and applied to our questions are detailed discussed in the slides.
Here we focuses on Fixed-Point Iterative Technique for solving nonlinear Equations in Numerical Analysis. It is one of the opened-iterative techniques for finding roots called Fixed-Point of Non-linear Equations.
Computer Oriented Numerical Analysis
What is interpolation?
Many times, data is given only at discrete points such as .
So, how then does one find the value of y at any other value of x ?
Well, a continuous function f(x) may be used to represent the data values with f(x) passing through the points (Figure 1). Then one can find the value of y at any other value of x .
This is called interpolation
Newton’s Divided Difference Formula:
To illustrate this method, linear and quadratic interpolation is presented first.
Then, the general form of Newton’s divided difference polynomial method is presented.
Gauss jordan and Guass elimination methodMeet Nayak
This ppt is based on engineering maths.
the topis is Gauss jordan and gauss elimination method.
This ppt having one example of both method and having algorithm.
Econometrics is a tough subject so is its homework and assignments in general. In case you are looking for econometrics homework help, you can rely on Economicshelpdesk. Our tutors are expert and we honour our client’s need as well as privacy.
To get a copy of the slides for free Email me at: japhethmuthama@gmail.com
You can also support my PhD studies by donating a 1 dollar to my PayPal.
PayPal ID is japhethmuthama@gmail.com
To get a copy of the slides for free Email me at: japhethmuthama@gmail.com
You can also support my PhD studies by donating a 1 dollar to my PayPal.
PayPal ID is japhethmuthama@gmail.com
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Digital Tools and AI for Teaching Learning and Research
Interpolation with Finite differences
1. Numerical Methods - Finite Differences
Dr. N. B. Vyas
Department of Mathematics,
Atmiya Institute of Tech. and Science,
Rajkot (Guj.)
niravbvyas@gmail.com
Dr. N. B. Vyas Numerical Methods - Finite Differences
2. Finite Differences
Forward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
Dr. N. B. Vyas Numerical Methods - Finite Differences
3. Finite Differences
Forward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The constant difference between two consecutive values of x is
called the interval of differences and is denoted by h.
Dr. N. B. Vyas Numerical Methods - Finite Differences
4. Finite Differences
Forward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The constant difference between two consecutive values of x is
called the interval of differences and is denoted by h.
The operator ∆ defined by
∆y0 = y1 − y0
Dr. N. B. Vyas Numerical Methods - Finite Differences
5. Finite Differences
Forward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The constant difference between two consecutive values of x is
called the interval of differences and is denoted by h.
The operator ∆ defined by
∆y0 = y1 − y0
∆y1 = y2 − y1
Dr. N. B. Vyas Numerical Methods - Finite Differences
6. Finite Differences
Forward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The constant difference between two consecutive values of x is
called the interval of differences and is denoted by h.
The operator ∆ defined by
∆y0 = y1 − y0
∆y1 = y2 − y1
.......
.......
∆yn−1 = yn − yn−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
7. Finite Differences
Forward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The constant difference between two consecutive values of x is
called the interval of differences and is denoted by h.
The operator ∆ defined by
∆y0 = y1 − y0
∆y1 = y2 − y1
.......
.......
∆yn−1 = yn − yn−1
is called Forward difference operator.
Dr. N. B. Vyas Numerical Methods - Finite Differences
8. Finite Differences
The first forward difference is ∆yn = yn+1 − yn
Dr. N. B. Vyas Numerical Methods - Finite Differences
9. Finite Differences
The first forward difference is ∆yn = yn+1 − yn
The second forward difference are defined as the difference of
the first differences.
∆2y0 = ∆(∆y0) = ∆(y1 − y0) = ∆y1 − ∆y0
= y2 − y1 − (y1 − y0) = y2 − 2y1 + y0
∆2y1 = ∆y2 − ∆y1
∆2yi = ∆yi+1 − ∆yi
In general nth forward difference of f is defined by
∆n
yi = ∆n−1
yi+1 − ∆n−1
yi
Dr. N. B. Vyas Numerical Methods - Finite Differences
10. Finite Differences
Forward Difference Table:
x y ∆y ∆2y ∆3y ∆4y ∆5y
x0 y0
∆y0
x1 y1 ∆2y0
∆y1 ∆3y0
x2 y2 ∆2y1 ∆4y0
∆y2 ∆3y1 ∆5y0
x3 y3 ∆2y2 ∆4y1
∆y3 ∆3y2
x4 y4 ∆2y3
∆y4
x5 y5
Dr. N. B. Vyas Numerical Methods - Finite Differences
11. Finite Differences
The operator ∆ satisfies the following properties:
1 ∆[f(x) ± g(x)] = ∆f(x) ± ∆g(x)
Dr. N. B. Vyas Numerical Methods - Finite Differences
12. Finite Differences
The operator ∆ satisfies the following properties:
1 ∆[f(x) ± g(x)] = ∆f(x) ± ∆g(x)
2 ∆[cf(x)] = c∆f(x)
Dr. N. B. Vyas Numerical Methods - Finite Differences
13. Finite Differences
The operator ∆ satisfies the following properties:
1 ∆[f(x) ± g(x)] = ∆f(x) ± ∆g(x)
2 ∆[cf(x)] = c∆f(x)
3 ∆m∆nf(x) = ∆m+nf(x), m, n are positive integers
4 Since ∆nyn is a constant, ∆n+1yn = 0 , ∆n+2yn = 0, . . .
i.e. (n + 1)th and higher differences are zero.
Dr. N. B. Vyas Numerical Methods - Finite Differences
14. Finite Differences
Backward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
Dr. N. B. Vyas Numerical Methods - Finite Differences
15. Finite Differences
Backward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The operator defined by
y1 = y1 − y0
Dr. N. B. Vyas Numerical Methods - Finite Differences
16. Finite Differences
Backward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The operator defined by
y1 = y1 − y0
y2 = y2 − y1
Dr. N. B. Vyas Numerical Methods - Finite Differences
17. Finite Differences
Backward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The operator defined by
y1 = y1 − y0
y2 = y2 − y1
.......
.......
yn = yn − yn−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
18. Finite Differences
Backward difference
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.
The operator defined by
y1 = y1 − y0
y2 = y2 − y1
.......
.......
yn = yn − yn−1
is called Backward difference operator.
Dr. N. B. Vyas Numerical Methods - Finite Differences
19. Finite Differences
The first backward difference is yn = yn − yn−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
20. Finite Differences
The first backward difference is yn = yn − yn−1
The second backward difference are obtain by the difference
of the first differences.
2y2 = ( y2) = (y2 − y1) = y2 − y1
= y2 − y1 − (y1 − y0) = y2 − 2y1 + y0
2y3 = y3 − y2
2yn = yn − yn−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
21. Finite Differences
The first backward difference is yn = yn − yn−1
The second backward difference are obtain by the difference
of the first differences.
2y2 = ( y2) = (y2 − y1) = y2 − y1
= y2 − y1 − (y1 − y0) = y2 − 2y1 + y0
2y3 = y3 − y2
2yn = yn − yn−1
In general nth backward difference of f is defined by
n
yi = n−1
yi − n−1
yi−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
22. Finite Differences
Backward Difference Table:
x y y 2y 3y 4y 5y
x0 y0
y1
x1 y1
2y2
y2
3y3
x2 y2
2y3
4y4
y3
3y4
5y5
x3 y3
2y4
4y5
y4
3y5
x4 y4
2y5
y5
x5 y5
Dr. N. B. Vyas Numerical Methods - Finite Differences
25. Finite Differences
Central difference
The operator δ defined by
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
26. Finite Differences
Central difference
The operator δ defined by
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1
is called Central difference operator.
Similarly, higher order central differences are defined as
δ2y1 = δy3
2
− δy1
2
Dr. N. B. Vyas Numerical Methods - Finite Differences
27. Finite Differences
Central difference
The operator δ defined by
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1
is called Central difference operator.
Similarly, higher order central differences are defined as
δ2y1 = δy3
2
− δy1
2
δ2y2 = δy5
2
− δy3
2
Dr. N. B. Vyas Numerical Methods - Finite Differences
28. Finite Differences
Central difference
The operator δ defined by
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1
is called Central difference operator.
Similarly, higher order central differences are defined as
δ2y1 = δy3
2
− δy1
2
δ2y2 = δy5
2
− δy3
2
.......
δ3y3
2
= δ2y2 − δ2y1
Dr. N. B. Vyas Numerical Methods - Finite Differences
29. Finite Differences
Central difference
The operator δ defined by
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1
is called Central difference operator.
Similarly, higher order central differences are defined as
δ2y1 = δy3
2
− δy1
2
δ2y2 = δy5
2
− δy3
2
.......
δ3y3
2
= δ2y2 − δ2y1
Dr. N. B. Vyas Numerical Methods - Finite Differences
30. Finite Differences
Central Difference Table:
x y δy δ2y δ3y δ4y δ5y
x0 y0
δy1
2
x1 y1 δ2y1
δy3
2
δ3y3
2
x2 y2 δ2y2 δ4y2
δy5
2
δ3y5
2
δ5y5
2
x3 y3 δ2y3 δ4y3
δy7
2
δ3y7
2
x4 y4 δ2y4
δy9
2
x5 y5
Dr. N. B. Vyas Numerical Methods - Finite Differences
31. Finite Differences
NOTE:
From all three difference tables, we can see that only the
notations changes not the differences.
y1 − y0 = ∆y0 = y1 = δy1
2
Dr. N. B. Vyas Numerical Methods - Finite Differences
32. Finite Differences
NOTE:
From all three difference tables, we can see that only the
notations changes not the differences.
y1 − y0 = ∆y0 = y1 = δy1
2
Alternative notations for the function y = f(x).
Dr. N. B. Vyas Numerical Methods - Finite Differences
33. Finite Differences
NOTE:
From all three difference tables, we can see that only the
notations changes not the differences.
y1 − y0 = ∆y0 = y1 = δy1
2
Alternative notations for the function y = f(x).
For two consecutive values of x differing by h.
Dr. N. B. Vyas Numerical Methods - Finite Differences
34. Finite Differences
NOTE:
From all three difference tables, we can see that only the
notations changes not the differences.
y1 − y0 = ∆y0 = y1 = δy1
2
Alternative notations for the function y = f(x).
For two consecutive values of x differing by h.
∆yx = yx+h − yx = f(x + h) − f(x)
Dr. N. B. Vyas Numerical Methods - Finite Differences
35. Finite Differences
NOTE:
From all three difference tables, we can see that only the
notations changes not the differences.
y1 − y0 = ∆y0 = y1 = δy1
2
Alternative notations for the function y = f(x).
For two consecutive values of x differing by h.
∆yx = yx+h − yx = f(x + h) − f(x)
yx = yx − yx−h = f(x) − f(x − h)
Dr. N. B. Vyas Numerical Methods - Finite Differences
36. Finite Differences
NOTE:
From all three difference tables, we can see that only the
notations changes not the differences.
y1 − y0 = ∆y0 = y1 = δy1
2
Alternative notations for the function y = f(x).
For two consecutive values of x differing by h.
∆yx = yx+h − yx = f(x + h) − f(x)
yx = yx − yx−h = f(x) − f(x − h)
δyx = yx+h
2
− yx−h
2
= f(x + h
2 ) − f(xh
2 )
Dr. N. B. Vyas Numerical Methods - Finite Differences
37. Finite Differences
Evaluate the following. The interval of difference being h.
1 ∆nex
Dr. N. B. Vyas Numerical Methods - Finite Differences
38. Finite Differences
Evaluate the following. The interval of difference being h.
1 ∆nex
2 ∆logf(x)
Dr. N. B. Vyas Numerical Methods - Finite Differences
39. Finite Differences
Evaluate the following. The interval of difference being h.
1 ∆nex
2 ∆logf(x)
3 ∆(tan−1x)
Dr. N. B. Vyas Numerical Methods - Finite Differences
40. Finite Differences
Evaluate the following. The interval of difference being h.
1 ∆nex
2 ∆logf(x)
3 ∆(tan−1x)
4 ∆2cos2x
Dr. N. B. Vyas Numerical Methods - Finite Differences
41. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
Dr. N. B. Vyas Numerical Methods - Finite Differences
42. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
Dr. N. B. Vyas Numerical Methods - Finite Differences
43. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Dr. N. B. Vyas Numerical Methods - Finite Differences
44. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
Dr. N. B. Vyas Numerical Methods - Finite Differences
45. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is defined as
Dr. N. B. Vyas Numerical Methods - Finite Differences
46. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is defined as
E−1f(x) = f(x − h);
Dr. N. B. Vyas Numerical Methods - Finite Differences
47. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is defined as
E−1f(x) = f(x − h);
E−nf(x) = f(x − nh);
Dr. N. B. Vyas Numerical Methods - Finite Differences
48. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is defined as
E−1f(x) = f(x − h);
E−nf(x) = f(x − nh);
If yx is the function f(x), then
Dr. N. B. Vyas Numerical Methods - Finite Differences
49. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is defined as
E−1f(x) = f(x − h);
E−nf(x) = f(x − nh);
If yx is the function f(x), then
Eyx = yx+h;
Dr. N. B. Vyas Numerical Methods - Finite Differences
50. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is defined as
E−1f(x) = f(x − h);
E−nf(x) = f(x − nh);
If yx is the function f(x), then
Eyx = yx+h;
E−1yx = yx−h;
Dr. N. B. Vyas Numerical Methods - Finite Differences
51. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is defined as
E−1f(x) = f(x − h);
E−nf(x) = f(x − nh);
If yx is the function f(x), then
Eyx = yx+h;
E−1yx = yx−h;
Enyx = yx+nh;
Dr. N. B. Vyas Numerical Methods - Finite Differences
52. Finite Differences
Other Difference Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is defined as
E−1f(x) = f(x − h);
E−nf(x) = f(x − nh);
If yx is the function f(x), then
Eyx = yx+h;
E−1yx = yx−h;
Enyx = yx+nh;
Dr. N. B. Vyas Numerical Methods - Finite Differences
53. Finite Differences
2. Averaging Operator µ:
It is defined as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
Dr. N. B. Vyas Numerical Methods - Finite Differences
54. Finite Differences
2. Averaging Operator µ:
It is defined as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
i.e. µyx =
1
2
yx+h
2
+ yx−h
2
;
Dr. N. B. Vyas Numerical Methods - Finite Differences
55. Finite Differences
2. Averaging Operator µ:
It is defined as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
i.e. µyx =
1
2
yx+h
2
+ yx−h
2
;
3. Differential Operator D:
Dr. N. B. Vyas Numerical Methods - Finite Differences
56. Finite Differences
2. Averaging Operator µ:
It is defined as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
i.e. µyx =
1
2
yx+h
2
+ yx−h
2
;
3. Differential Operator D:
It is defined as
Dr. N. B. Vyas Numerical Methods - Finite Differences
57. Finite Differences
2. Averaging Operator µ:
It is defined as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
i.e. µyx =
1
2
yx+h
2
+ yx−h
2
;
3. Differential Operator D:
It is defined as
Df(x) =
d
dx
f(x) = f (x);
Dr. N. B. Vyas Numerical Methods - Finite Differences
73. Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near the beginning a table,
the forward difference interpolation formula in used.
Dr. N. B. Vyas Numerical Methods - Finite Differences
74. Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near the beginning a table,
the forward difference interpolation formula in used.
Let yx = f(x) be a function which takes the values
yx0 , yx0+h, yx0+2h, . . . corresponding to the values
x0, x0 + h, x0 + 2h, . . . of x.
Dr. N. B. Vyas Numerical Methods - Finite Differences
75. Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near the beginning a table,
the forward difference interpolation formula in used.
Let yx = f(x) be a function which takes the values
yx0 , yx0+h, yx0+2h, . . . corresponding to the values
x0, x0 + h, x0 + 2h, . . . of x.
Suppose we want to evaluate yx when x = x0 + ph, where p is
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences
76. Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near the beginning a table,
the forward difference interpolation formula in used.
Let yx = f(x) be a function which takes the values
yx0 , yx0+h, yx0+2h, . . . corresponding to the values
x0, x0 + h, x0 + 2h, . . . of x.
Suppose we want to evaluate yx when x = x0 + ph, where p is
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences
77. Gregory - Newton Forward Interpolation Formula
Let it be yp. For any real number n, we have defined operator E
such that Enf(x) = f(x + nh).
∴ yx = yx0+ph = f (x0 + ph) = Epyx0 = (1 + ∆)p
y0
Dr. N. B. Vyas Numerical Methods - Finite Differences
78. Gregory - Newton Forward Interpolation Formula
Let it be yp. For any real number n, we have defined operator E
such that Enf(x) = f(x + nh).
∴ yx = yx0+ph = f (x0 + ph) = Epyx0 = (1 + ∆)p
y0
= 1 + p∆ +
p(p − 1)
2!
∆2 +
p(p − 1)(p − 2)
3!
∆3 + ... y0
Dr. N. B. Vyas Numerical Methods - Finite Differences
79. Gregory - Newton Forward Interpolation Formula
Let it be yp. For any real number n, we have defined operator E
such that Enf(x) = f(x + nh).
∴ yx = yx0+ph = f (x0 + ph) = Epyx0 = (1 + ∆)p
y0
= 1 + p∆ +
p(p − 1)
2!
∆2 +
p(p − 1)(p − 2)
3!
∆3 + ... y0
yx = y0 + p∆y0 +
p(p − 1)
2!
∆2
y0 +
p(p − 1)(p − 2)
3!
∆3
y0 + ...
is called Newton’s forward interpolation formula.
Dr. N. B. Vyas Numerical Methods - Finite Differences
80. Example
Ex. For the data construct the forward difference formula. Hence,
find f(0.5).
x -2 -1 0 1 2 3
f(x) 15 5 1 3 11 25
Dr. N. B. Vyas Numerical Methods - Finite Differences
81. Example
Ex. The population of the town in decennial census was as given
below estimate the population for the year 1895.
Year: 1891 1901 1911 1921 1931
Population(in thousand): 46 66 81 93 101
Dr. N. B. Vyas Numerical Methods - Finite Differences
82. Example
Ex. Estimate the value of production for the year 1984 using
Newton’s forward method for the following data:
Year: 1976 1978 1980 1982
Production: 20 27 38 50
Dr. N. B. Vyas Numerical Methods - Finite Differences
83. Gregory - Newton Backward Interpolation Formula
To estimate the value of a function near the end of a table, the
backward difference interpolation formula in used.
Dr. N. B. Vyas Numerical Methods - Finite Differences
84. Gregory - Newton Backward Interpolation Formula
To estimate the value of a function near the end of a table, the
backward difference interpolation formula in used.
Let yx = f(x) be a function which takes the values
yx0 , yx0+h, yx0+2h, . . . corresponding to the values
x0, x0 + h, x0 + 2h, . . . of x.
Dr. N. B. Vyas Numerical Methods - Finite Differences
85. Gregory - Newton Backward Interpolation Formula
To estimate the value of a function near the end of a table, the
backward difference interpolation formula in used.
Let yx = f(x) be a function which takes the values
yx0 , yx0+h, yx0+2h, . . . corresponding to the values
x0, x0 + h, x0 + 2h, . . . of x.
Suppose we want to evaluate yx when x = xn + ph, where p is
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences
86. Gregory - Newton Backward Interpolation Formula
Let it be yp. For any real number n, we have defined operator E
such that Enf(x) = f(x + nh).
∴ yx = yxn+ph = f (xn + ph) = Epyxn = (1 − )−p
yn
Dr. N. B. Vyas Numerical Methods - Finite Differences
87. Gregory - Newton Backward Interpolation Formula
Let it be yp. For any real number n, we have defined operator E
such that Enf(x) = f(x + nh).
∴ yx = yxn+ph = f (xn + ph) = Epyxn = (1 − )−p
yn
= 1 + p +
p(p + 1)
2!
2 +
p(p + 1)(p + 2)
3!
3 + ... yn
Dr. N. B. Vyas Numerical Methods - Finite Differences
88. Gregory - Newton Backward Interpolation Formula
Let it be yp. For any real number n, we have defined operator E
such that Enf(x) = f(x + nh).
∴ yx = yxn+ph = f (xn + ph) = Epyxn = (1 − )−p
yn
= 1 + p +
p(p + 1)
2!
2 +
p(p + 1)(p + 2)
3!
3 + ... yn
yx = yn + p yn +
p(p + 1)
2!
2
yn +
p(p + 1)(p + 2)
3!
3
yn + ...
is called Newton’s backward interpolation formula
Dr. N. B. Vyas Numerical Methods - Finite Differences
89. Example
Ex. Using Newtons backward difference interpolation, interpolate at
x = 1 from the following data.
x 0.1 0.2 0.3 0.4 0.5 0.6
f(x) 1.699 1.073 0.375 0.443 1.429 2.631
Dr. N. B. Vyas Numerical Methods - Finite Differences
90. Example
Ex. The table gives the distance in nautical miles of the visible
horizon for the given heights in feet above the earth’s surface.
Find the value of y when x=390 ft.
Height(x): 100 150 200 250 300 350 400
Distance(y): 10.63 13.03 15.04 16.81 18.42 19.90 21.47
Dr. N. B. Vyas Numerical Methods - Finite Differences
91. Stirling’s Interpolation Formula
To estimate the value of a function near the middle a table, the
central difference interpolation formula in used.
Let yx = f(x) be a functional relation between x and y.
If x takes the values x0 − 2h, x0 − h, x0, x0 + h, x0 + 2h, . . . and
the corresponding values of y are y−2, y−1, y0, y1, y2 . . . then we
can form a central difference table as follows:
Dr. N. B. Vyas Numerical Methods - Finite Differences
92. Stirling’s Interpolation Formula
x y
1st
difference
2nd
difference
3rd
difference
x0 − 2h y−2
∆y−2(= δy−3/2)
x0 − h y−1 ∆2y−2(= δ2y−1)
∆y−1(= δy−1/2) ∆3y−2(= δ3y−1/2)
x0 y0 ∆2y−1(= δ2y0) ∆
∆y0(= δy1/2) ∆3y−1(= δ3y1/2)
x0 + h y1 ∆2y0(= δ2y1)
∆y1(= δy3/2)
x0 + 2h y2
Dr. N. B. Vyas Numerical Methods - Finite Differences
94. Stirling’s Interpolation Formula
The Stirling’s formula in forward difference notation is
yp = y0 + p
∆y0 + ∆y−1
2
+
p2
2!
∆2y−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
95. Stirling’s Interpolation Formula
The Stirling’s formula in forward difference notation is
yp = y0 + p
∆y0 + ∆y−1
2
+
p2
2!
∆2y−1
+
p(p2 − 12)
3!
∆3y−1 + ∆3y−2
2
+
p2(p2 − 12)
4!
∆4y−2 + . . .
Dr. N. B. Vyas Numerical Methods - Finite Differences
97. Example
Ex. The function y is given in the table below:
Find y for x=0.0341
x: 0.01 0.02 0.03 0.04 0.05
y: 98.4342 48.4392 31.7775 23.4492 18.4542
Dr. N. B. Vyas Numerical Methods - Finite Differences
102. Example
Ex. Estimate the value of y(2.5) using Gauss’s forward formula given
that:
x: 1 2 3 4
y: 1 4 9 16
Dr. N. B. Vyas Numerical Methods - Finite Differences
103. Example
Ex. Interpolate by means of Gauss’s backward formula the
population for the year 1936 given the following table:
Year : 1901 1911 1921 1931 1941 1951
Population(in 1000s): 12 15 20 27 39 52
Dr. N. B. Vyas Numerical Methods - Finite Differences