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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
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
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
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
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
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
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
Finite Differences
The first forward difference is ∆yn = yn+1 − yn
Dr. N. B. Vyas Numerical Methods - Finite Differences
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
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
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
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
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
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
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
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
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
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
Finite Differences
The first backward difference is yn = yn − yn−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
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
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
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
Finite Differences
Central difference
The operator δ defined by
δy1
2
= y1 − y0
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Central difference
The operator δ defined by
δy1
2
= y1 − y0
δy3
2
= y2 − y1
Dr. N. B. Vyas Numerical Methods - Finite Differences
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
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
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
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
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
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
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
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
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
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
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
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
Finite Differences
Evaluate the following. The interval of difference being h.
1 ∆nex
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Evaluate the following. The interval of difference being h.
1 ∆nex
2 ∆logf(x)
Dr. N. B. Vyas Numerical Methods - Finite Differences
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Finite Differences
Relation between the operators
1 ∆ = E − 1
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Relation between the operators
1 ∆ = E − 1
2 = 1 − E−1
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Relation between the operators
1 ∆ = E − 1
2 = 1 − E−1
3 δ = E
1
2 − E−1
2
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Relation between the operators
1 ∆ = E − 1
2 = 1 − E−1
3 δ = E
1
2 − E−1
2
4 µ =
1
2
{E
1
2 + E−1
2 }
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Relation between the operators
1 ∆ = E − 1
2 = 1 − E−1
3 δ = E
1
2 − E−1
2
4 µ =
1
2
{E
1
2 + E−1
2 }
5 ∆ = E = E = δE
1
2
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Relation between the operators
1 ∆ = E − 1
2 = 1 − E−1
3 δ = E
1
2 − E−1
2
4 µ =
1
2
{E
1
2 + E−1
2 }
5 ∆ = E = E = δE
1
2
6 E = ehD
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Relation between the operators
1 ∆ = E − 1
2 = 1 − E−1
3 δ = E
1
2 − E−1
2
4 µ =
1
2
{E
1
2 + E−1
2 }
5 ∆ = E = E = δE
1
2
6 E = ehD
7 (1 + ∆)(1 − ) = 1
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
Relation between the operators
1 ∆ = E − 1
2 = 1 − E−1
3 δ = E
1
2 − E−1
2
4 µ =
1
2
{E
1
2 + E−1
2 }
5 ∆ = E = E = δE
1
2
6 E = ehD
7 (1 + ∆)(1 − ) = 1
8 ∆ − = ∆ = δ2
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ
12 E−1
2 = µ −
1
2
δ
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ
12 E−1
2 = µ −
1
2
δ
13 ∆ =
1
2
δ2 + δ 1 +
δ2
4
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ
12 E−1
2 = µ −
1
2
δ
13 ∆ =
1
2
δ2 + δ 1 +
δ2
4
14 µδ =
1
2
∆E−1 +
1
2
∆
Dr. N. B. Vyas Numerical Methods - Finite Differences
Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ
12 E−1
2 = µ −
1
2
δ
13 ∆ =
1
2
δ2 + δ 1 +
δ2
4
14 µδ =
1
2
∆E−1 +
1
2
∆
Dr. N. B. Vyas Numerical Methods - Finite Differences
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Stirling’s Interpolation Formula
The Stirling’s formula in forward difference notation is
Dr. N. B. Vyas Numerical Methods - Finite Differences
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
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
Example
Ex.
Using Stirling’s formula find y35
x: 10 20 30 40 50
y: 600 512 439 346 243
Dr. N. B. Vyas Numerical Methods - Finite Differences
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
Central Difference
Gauss’s Forward interpolation formula:
Dr. N. B. Vyas Numerical Methods - Finite Differences
Central Difference
Gauss’s Forward interpolation formula:
Pn(x) = y0 + p∆y0 +
p(p − 1)
2!
∆2
y−1 +
(p + 1)p(p − 1)
3!
∆3
y−1+
(p + 1)p(p − 1)(p − 2)
4!
∆4
y−2 + ...
Dr. N. B. Vyas Numerical Methods - Finite Differences
Central Difference
Gauss’s Forward interpolation formula:
Pn(x) = y0 + p∆y0 +
p(p − 1)
2!
∆2
y−1 +
(p + 1)p(p − 1)
3!
∆3
y−1+
(p + 1)p(p − 1)(p − 2)
4!
∆4
y−2 + ...
Gauss’s Backward interpolation formula:
Dr. N. B. Vyas Numerical Methods - Finite Differences
Central Difference
Gauss’s Forward interpolation formula:
Pn(x) = y0 + p∆y0 +
p(p − 1)
2!
∆2
y−1 +
(p + 1)p(p − 1)
3!
∆3
y−1+
(p + 1)p(p − 1)(p − 2)
4!
∆4
y−2 + ...
Gauss’s Backward interpolation formula:
Pn(x) = y0 + p∆y−1 +
p(p + 1)
2!
∆2
y−1 +
(p + 1)p(p − 1)
3!
∆3
y−2+
(p + 2)(p + 1)p(p − 1)
4!
∆4
y−2 + ...
Dr. N. B. Vyas Numerical Methods - Finite Differences
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
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

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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
  • 23. Finite Differences Central difference The operator δ defined by δy1 2 = y1 − y0 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 24. Finite Differences Central difference The operator δ defined by δy1 2 = y1 − y0 δy3 2 = y2 − y1 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
  • 58. Finite Differences Relation between the operators 1 ∆ = E − 1 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 59. Finite Differences Relation between the operators 1 ∆ = E − 1 2 = 1 − E−1 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 60. Finite Differences Relation between the operators 1 ∆ = E − 1 2 = 1 − E−1 3 δ = E 1 2 − E−1 2 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 61. Finite Differences Relation between the operators 1 ∆ = E − 1 2 = 1 − E−1 3 δ = E 1 2 − E−1 2 4 µ = 1 2 {E 1 2 + E−1 2 } Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 62. Finite Differences Relation between the operators 1 ∆ = E − 1 2 = 1 − E−1 3 δ = E 1 2 − E−1 2 4 µ = 1 2 {E 1 2 + E−1 2 } 5 ∆ = E = E = δE 1 2 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 63. Finite Differences Relation between the operators 1 ∆ = E − 1 2 = 1 − E−1 3 δ = E 1 2 − E−1 2 4 µ = 1 2 {E 1 2 + E−1 2 } 5 ∆ = E = E = δE 1 2 6 E = ehD Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 64. Finite Differences Relation between the operators 1 ∆ = E − 1 2 = 1 − E−1 3 δ = E 1 2 − E−1 2 4 µ = 1 2 {E 1 2 + E−1 2 } 5 ∆ = E = E = δE 1 2 6 E = ehD 7 (1 + ∆)(1 − ) = 1 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 65. Finite Differences Relation between the operators 1 ∆ = E − 1 2 = 1 − E−1 3 δ = E 1 2 − E−1 2 4 µ = 1 2 {E 1 2 + E−1 2 } 5 ∆ = E = E = δE 1 2 6 E = ehD 7 (1 + ∆)(1 − ) = 1 8 ∆ − = ∆ = δ2 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 66. Finite Differences 9 1 + µ2δ2 = 1 + 1 2 δ2 2 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 67. Finite Differences 9 1 + µ2δ2 = 1 + 1 2 δ2 2 10 µ2 = 1 + 1 4 δ2 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 68. Finite Differences 9 1 + µ2δ2 = 1 + 1 2 δ2 2 10 µ2 = 1 + 1 4 δ2 11 E 1 2 = µ + 1 2 δ Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 69. Finite Differences 9 1 + µ2δ2 = 1 + 1 2 δ2 2 10 µ2 = 1 + 1 4 δ2 11 E 1 2 = µ + 1 2 δ 12 E−1 2 = µ − 1 2 δ Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 70. Finite Differences 9 1 + µ2δ2 = 1 + 1 2 δ2 2 10 µ2 = 1 + 1 4 δ2 11 E 1 2 = µ + 1 2 δ 12 E−1 2 = µ − 1 2 δ 13 ∆ = 1 2 δ2 + δ 1 + δ2 4 Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 71. Finite Differences 9 1 + µ2δ2 = 1 + 1 2 δ2 2 10 µ2 = 1 + 1 4 δ2 11 E 1 2 = µ + 1 2 δ 12 E−1 2 = µ − 1 2 δ 13 ∆ = 1 2 δ2 + δ 1 + δ2 4 14 µδ = 1 2 ∆E−1 + 1 2 ∆ Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 72. Finite Differences 9 1 + µ2δ2 = 1 + 1 2 δ2 2 10 µ2 = 1 + 1 4 δ2 11 E 1 2 = µ + 1 2 δ 12 E−1 2 = µ − 1 2 δ 13 ∆ = 1 2 δ2 + δ 1 + δ2 4 14 µδ = 1 2 ∆E−1 + 1 2 ∆ 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
  • 93. Stirling’s Interpolation Formula The Stirling’s formula in forward difference notation is 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
  • 96. Example Ex. Using Stirling’s formula find y35 x: 10 20 30 40 50 y: 600 512 439 346 243 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
  • 98. Central Difference Gauss’s Forward interpolation formula: Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 99. Central Difference Gauss’s Forward interpolation formula: Pn(x) = y0 + p∆y0 + p(p − 1) 2! ∆2 y−1 + (p + 1)p(p − 1) 3! ∆3 y−1+ (p + 1)p(p − 1)(p − 2) 4! ∆4 y−2 + ... Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 100. Central Difference Gauss’s Forward interpolation formula: Pn(x) = y0 + p∆y0 + p(p − 1) 2! ∆2 y−1 + (p + 1)p(p − 1) 3! ∆3 y−1+ (p + 1)p(p − 1)(p − 2) 4! ∆4 y−2 + ... Gauss’s Backward interpolation formula: Dr. N. B. Vyas Numerical Methods - Finite Differences
  • 101. Central Difference Gauss’s Forward interpolation formula: Pn(x) = y0 + p∆y0 + p(p − 1) 2! ∆2 y−1 + (p + 1)p(p − 1) 3! ∆3 y−1+ (p + 1)p(p − 1)(p − 2) 4! ∆4 y−2 + ... Gauss’s Backward interpolation formula: Pn(x) = y0 + p∆y−1 + p(p + 1) 2! ∆2 y−1 + (p + 1)p(p − 1) 3! ∆3 y−2+ (p + 2)(p + 1)p(p − 1) 4! ∆4 y−2 + ... 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