Personal Resilience in Project Management 2 - TV Edit 1a.pdf
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