GSB 3204: Statistics & Numerical Analysis
Yasin K. Kowa
Email:kowayasin@gmail.com
April 30, 2019
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Introduction
Definition
Numerical Methods are methods for solving problems numerically
(in terms of number) on computer or calculator or by hand.
Numerical methods emphasize the implementation of algorithms.
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Introduction
Definition
Numerical Methods are methods for solving problems numerically
(in terms of number) on computer or calculator or by hand.
Numerical methods emphasize the implementation of algorithms.
Definition
Numerical methods provide systematic methods of solving
problems in numerical form. Normally starts from an initial data,
using high precision digital computers, steps in algorithms and
finally obtaining the results.
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GSB 3204: Statistics & Numerical Analysis
Introduction
Definition
The numerical methods give approximate results or solution, that
is they have errors.
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GSB 3204: Statistics & Numerical Analysis
Introduction
Definition
The numerical methods give approximate results or solution, that
is they have errors.
Definition
That is True Value=approximate+error
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GSB 3204: Statistics & Numerical Analysis
Error
Error
Error refers to the difference between the true value and the
estimated (approximated) value
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GSB 3204: Statistics & Numerical Analysis
Error
Error
Error refers to the difference between the true value and the
estimated (approximated) value
Error
Error=Actual value-Estimated value
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GSB 3204: Statistics & Numerical Analysis
Error
Error
Approximate values or numbers are those that represent the
numbers to a certain degree of accuracy.
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GSB 3204: Statistics & Numerical Analysis
Error
Error
Approximate values or numbers are those that represent the
numbers to a certain degree of accuracy.
Error
But error is different from mistakes (blunders), the last one are
the deviations due to human factors such as misreading a number,
misreading from a scale or using faulty instrument, misprinting,
misfeeding e.t.c
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GSB 3204: Statistics & Numerical Analysis
Error
Sources of Error
When a computational procedure is involved in solving a scientific
-mathematical problem, errors often will be involved in the process
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GSB 3204: Statistics & Numerical Analysis
Error
Sources of Error
When a computational procedure is involved in solving a scientific
-mathematical problem, errors often will be involved in the process
Sources of Error
(a) Error due to method itself
Sometimes numerical method by virtue of what they are,
introduce some error in the result. This error is called
Discretization error
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GSB 3204: Statistics & Numerical Analysis
Error
Sources of Error
(b) Truncation Error
This occurs due to the failure to do the problem in a finite
steps, we are forced to terminate our calculations somewhere
or carry a few terms.
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GSB 3204: Statistics & Numerical Analysis
Error
Sources of Error
(b) Truncation Error
This occurs due to the failure to do the problem in a finite
steps, we are forced to terminate our calculations somewhere
or carry a few terms.
Sources of Error
For instance Taylor’s series,
f 0 (a)
f 00 (a)
f (x ) = f (a) +
(x − a) +
(x − a)2 + . . .
1!
2!
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GSB 3204: Statistics & Numerical Analysis
Error
Sources of Error
(c) Round-off Error
Is the error that a rises due to the fact that when we have
large number of digits and it will be necessary to cut them to
usable number of figures. This process is called rounding off
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GSB 3204: Statistics & Numerical Analysis
Error
Sources of Error
(c) Round-off Error
Is the error that a rises due to the fact that when we have
large number of digits and it will be necessary to cut them to
usable number of figures. This process is called rounding off
Sources of Error
(d) Propagated Error
This is an error in the succeeding steps of the process due to
the occurrence of earlier error.
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GSB 3204: Statistics & Numerical Analysis
Error
Sources of Error
(e) Modeling Error
Mathematical modeling is a process when mathematical
equations are used to represent a physical systems. This
modeling introduces errors and are called modeling error
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GSB 3204: Statistics & Numerical Analysis
Error
Accuracy
Accuracy refers to how closely a computed or measured value
agrees with the true value.
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GSB 3204: Statistics & Numerical Analysis
Error
Accuracy
Accuracy refers to how closely a computed or measured value
agrees with the true value.
Precision
Precision refers to how closely individual computed or measured
values agree with each other.
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GSB 3204: Statistics & Numerical Analysis
Error
Calculation of Errors
Absolute Error =⇒ ex = |x − x ∗ | where
x − represents a true value of a quantity
x ∗ − represent approximate value
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GSB 3204: Statistics & Numerical Analysis
Error
Calculation of Errors
Absolute Error =⇒ ex = |x − x ∗ | where
x − represents a true value of a quantity
x ∗ − represent approximate value
Calculation of Errors
When
(i) x ∗ > x , the number x ∗ is said to be an approximation with
excess
(ii) x ∗ < x , the number x ∗ is said to be an approximation with
deficit
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GSB 3204: Statistics & Numerical Analysis
Error
Calculation of Errors
For instance√
2 ,an approximation with deficit
x ∗ = 1.4 ≈ √
∗
x = 2.24 ≈ 5, an approximation with excess
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GSB 3204: Statistics & Numerical Analysis
Error
Calculation of Errors
For instance√
2 ,an approximation with deficit
x ∗ = 1.4 ≈ √
∗
x = 2.24 ≈ 5, an approximation with excess
Calculation of Errors
From True Value=Approximate+error, we can write as
x = x ∗ ± ex
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Relative Error
The relative error is the ratio of the absolute error to actual figure,
that is
|x − x ∗ |
ex
rx =
=
x
x
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Relative Error
The relative error is the ratio of the absolute error to actual figure,
that is
|x − x ∗ |
ex
rx =
=
x
x
Relative Error
Where rx is the relative error, the relative error is expressed as
percentage.That is
ex
rx =
× 100%
x
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Example-1
Suppose that you have the task of measuring the length of bridge
and rivet and come up with 9999 cm and 9 cm respectively. If the
true values are 10000 cm and 10 cm respectively. Compute
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Example-1
Suppose that you have the task of measuring the length of bridge
and rivet and come up with 9999 cm and 9 cm respectively. If the
true values are 10000 cm and 10 cm respectively. Compute
Example-1
(a) the true error
(b) relative error in %
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Solution-1
(a) Case I
Given x = 10000cm
x ∗ = 9999cm
ex = |10000 − 9999|cm=1cm
Case II
x=10cm
x ∗ = 9cm
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Solution-1
(a) Case I
Given x = 10000cm
x ∗ = 9999cm
ex = |10000 − 9999|cm=1cm
Case II
x=10cm
x ∗ = 9cm
Solution-1
ex = |10 − 9| cm = |9 − 10| cm = 1cm
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Solution-1
(b) Case I
ex
1 cm
rx =
× 100% =
× 100% = 0.01%
x
10000 cm
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Solution-1
(b) Case I
ex
1 cm
rx =
× 100% =
× 100% = 0.01%
x
10000 cm
Solution-1
Case II
ex
1cm
rx =
× 100% =
= 10%
x
10cm
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Exercise-1
Which of the following is more accurate estimate?
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Exercise-1
Which of the following is more accurate estimate?
Exercise-1
√
0.84 × 45
(a) 10 as an estimate for
or
0.6
(b) 150cm2 as the area of circle of diameter 14cm
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Exercise-2
1
Three approximate values of the number are given as 0.30, 0.33
3
and 0.34. Which of these values is the best approximate?
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Theorem
In addition and subtraction on error bound for the results, is given
as the sum of error bounds for the terms.
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Theorem
In addition and subtraction on error bound for the results, is given
as the sum of error bounds for the terms.
Proof
For addition
ex +y = |(x + y ) − (x ∗ + y ∗ )| = |x − x ∗ + y − y ∗ |
By triangular rule
ex +y ≤ |x − x ∗ | + |y − y ∗ | =⇒ ex +y ≤ ex + ey
For subtraction
ex −y = |(x −y )−(x ∗ −y ∗ )| = |(x −x ∗ )−(y −y ∗ )| ≤ |x −x ∗ |+|y −y ∗ |
ex −y = ex + ey
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Theorem
In multiplication and division, a bound for the relative error of the
results is given by the sum of the bounds for the relative errors of
the given terms or numbers.
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Theorem
In multiplication and division, a bound for the relative error of the
results is given by the sum of the bounds for the relative errors of
the given terms or numbers.
Proof
For multiplication we what to prove that rx .y = rx + ry
exy
|xy − x ∗ y ∗ |
|xy − (x − ex )(y − ey )|
rx .y =
=
=
xy
xy
xy
|xy − xy + xey + yex + ex ey |
|xey + yex + ex ey |
rxy =
=
xy
xy
Neglecting the small product we have
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Proof
rxy =
ey
ex
xey + yex
=
+
xy
y
x
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Proof
rxy =
ey
ex
xey + yex
=
+
xy
y
x
Proof
rxy ≈ rx + ry
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Proof
For division, we want to prove r yx ≈ rx + ry
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GSB 3204: Statistics & Numerical Analysis
Calculation of Errors
Proof
For division, we want to prove r yx ≈ rx + ry
Proof
r yx =
x
y
−
x
y
x∗
y∗
=
x
y
−
x −ex
y −ey
x
y
=
x (y − ey ) − y (x − ex ) y
y (y − e y )
x
r yx =
xy − xey − yx + yex y
−xey + yex y
=
y (y − ey )
x
y (y − ey ) x
ry =
ye x − xe y
(y − ey )−1
y
x
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Introduction
Consider a function y = f (x ) defined on (a, b), x and y are the
independent and dependent variables respectively. If the points
x0 , x1 , . . . , xn are taken at equidistance.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Introduction
Consider a function y = f (x ) defined on (a, b), x and y are the
independent and dependent variables respectively. If the points
x0 , x1 , . . . , xn are taken at equidistance.
Introduction
That is xi = x0 + ih, i = 0, 1, 2, . . . , n then the value of y when
x = xi is denoted as yi , where yi = f (xi )
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Introduction
Then y1 − y0 , y2 − y1 , y3 − y2 , . . . yn − yn−1 are called the
differences of y . Denoting these differences by
4y0 , 4y1 , . . . , 4yn−1 respectively, we have
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Introduction
Then y1 − y0 , y2 − y1 , y3 − y2 , . . . yn − yn−1 are called the
differences of y . Denoting these differences by
4y0 , 4y1 , . . . , 4yn−1 respectively, we have
Introduction
4y0 = y1 −y0 , 4y1 = y2 −y1 , 4y2 = y3 −y2 · · ·4yn−1 = yn −yn−1
Where 4 is the forward difference operator and
4y0 = y1 −y0 , 4y1 = y2 −y1 , 4y2 = y3 −y2 · · ·4yn−1 = yn −yn−1
are called first forward differences
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Introduction
The differences of the first forward differences are called second
forward differences and are denoted by
42 y0 , 42 y1 , . . . , 42 yn−1
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Introduction
The differences of the first forward differences are called second
forward differences and are denoted by
42 y0 , 42 y1 , . . . , 42 yn−1
Introduction
Similarly we can define the third, fourth, . . .
We will consider three types of finite differences namely
Forward Finite differences
Backward Finite differences
Central Finite differences
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Forward Finite Differences
The forward finite differences or simply differences operator is
denoted by 4 and is defined as
4f (x ) = f (x + h) − f (x ), we can write in terms of y as
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Forward Finite Differences
The forward finite differences or simply differences operator is
denoted by 4 and is defined as
4f (x ) = f (x + h) − f (x ), we can write in terms of y as
Forward Finite Differences
4yi = yi+1 − yi , i = 0, 1, 2, . . . , n − 1.
The differences of the first differences are called the second
differences and they are denoted by 42 y0 , 42 y1 , . . . , 42 yn−1
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Forward Finite Differences
Hence
42 y0 = 4y1 − 4y0 = (y2 − y1 ) − (y1 − y0 ) = y2 − 2y1 + y0
42 y1 = 4y2 − 4y1 = (y3 − y2 ) − (y2 − y1 ) = y3 − 2y2 + y1
..
.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Forward Finite Differences
Hence
42 y0 = 4y1 − 4y0 = (y2 − y1 ) − (y1 − y0 ) = y2 − 2y1 + y0
42 y1 = 4y2 − 4y1 = (y3 − y2 ) − (y2 − y1 ) = y3 − 2y2 + y1
..
.
Forward Finite Differences
But also
43 y0 = 42 y1 − 42 y0 = (y3 − 2y2 + y1 ) − (y2 − 2y1 + y0 ) that is
43 y0 = y3 − 3y2 + 3y1 − y0
43 y1 = y4 − 3y3 + 3y2 − y1
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Forward Finite Differences
Generally
4n+1 yi = 4n yi+1 − 4n yi , n = 0, 1, 2, . . . and i = 0, 1, 2, . . .
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Forward Finite Differences Table
The difference table is the standard format for displaying
differences
x
x0
y
y0
x1
y1
4y
42 y
43 y
44 y
4y0
42 y0
43 y0
4y1
x2
42 y1
y2
42 y2
x3
y3
x4
y4
44 y0
43 y1
42 y2
4y3
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Forward Finite Differences Table
The forward difference table is also called as a diagonal difference
table, the y0 term in the table above is called the leading term.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Forward Finite Differences Table
The forward difference table is also called as a diagonal difference
table, the y0 term in the table above is called the leading term.
Forward Finite Differences Table
The differences 4y0 , 42 y0 , 43 y0 . . . are called the leading
differences. In the difference table x is called argument and y as
function or entry.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Backward Finite Differences
The backward difference operator is denoted by 5 and is defined
as 5f (x ) = f (x ) − f (x − h) or
5yi = yi − yi−1 , i = n, n − 1, . . . , 1
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Backward Finite Differences
The backward difference operator is denoted by 5 and is defined
as 5f (x ) = f (x ) − f (x − h) or
5yi = yi − yi−1 , i = n, n − 1, . . . , 1
Backward Finite Differences
5y1 = y1 − y0 , 5y2 = y2 − y1 , 5y3 = y3 − y2 · · · 5 yn = yn − yn−1
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Backward Finite Differences
The second differences are denoted by
52 y2 , 52 y3 , 52 y4 , · · · 52 yn
52 y2 = 5(5y − 2) = 5(y2 − y1 ) = 5y2 − 5y1 =
(y2 − y1 ) − (y1 − y0 ) = y2 − 2y1 + y0
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Backward Finite Differences
The second differences are denoted by
52 y2 , 52 y3 , 52 y4 , · · · 52 yn
52 y2 = 5(5y − 2) = 5(y2 − y1 ) = 5y2 − 5y1 =
(y2 − y1 ) − (y1 − y0 ) = y2 − 2y1 + y0
Backward Finite Differences
52 y3 = y3 − 2y2 + y1
52 y4 = y4 − 2y3 + y2
Generally 5k yi = 5k−1 yi − 5k−1 yi−1 , i=n, n-1, . . . ,k
where 50 yi = yi , 51 yi = 5yi
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Backward Finite Differences Table
x
x0
y
y0
5y
52 y
V53 y
54 y
55 y
5y1
x1
52 y2
y1
53 y3
5y2
x2
52 y3
y2
5y3
x3
54 y4
53 y
4
52 y4
y3
5y4
x4
y4
The Backward Difference is also known as horizontal differences,
and its table is as follows
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Backward Finite Differences Table
52 y
53 y
x
x0
y
y0
5y
x1
y1
5y1
x2
y2
5y2
52 y2
x3
y3
5y3
52 y3
53 y3
x4
y4
5y4
52 y4
53 y4
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54 y
54 y4
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Finite Differences
The central difference operator is σ and is defined as
σf (x ) = f (x + h2 ) − f (x − h2 ) where h is the interval of differences.
In terms of y , the first central difference is written as
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Finite Differences
The central difference operator is σ and is defined as
σf (x ) = f (x + h2 ) − f (x − h2 ) where h is the interval of differences.
In terms of y , the first central difference is written as
Central Finite Differences
σyi = yi+ 1 − yi− 1
2
2
where yi+ 1 = f (x + h2 ) and yi− 1 = f (x − h2 )
2
2
That is
σy 1 = y1 − y0
2
σy 3 = y2 − y1
2
..
.
σyn− 1 = yn − yn−1
2
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Finite Differences
The second central differences are given by
σ 2 yi = σyi+ 1 − σyi− 1
2
2
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Finite Differences
The second central differences are given by
σ 2 yi = σyi+ 1 − σyi− 1
2
2
Central Finite Differences
σ 2 yi = σyi+ 1 − σy1− 1 = (yi+1 − yi ) − (yi − yi−1 )
2
2
σ 2 yi = yi+1 − 2yi + yi−1
Generally σ n yi = σ n−1 yi+ 1 − σ n−1 yi− 1
2
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2
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Finite Differences Table
x
x0
y
y0
σy
σ2y
σ3y
σ4y
σ5y
σ6y
σy 1
2
x1
y1
σ 2 y1
σ3y 3
σy 3
2
x2
x3
y2
σy 5
σ3y 5
2
2
y3
σ 2 y3
σ3y
σy 7
2
x4
y4
σ 2 y4
σ3y
σy 9
2
x5
2
σ 2 y2
y5
σ2y
7
2
σ 4 y2
σ5y 5
2
σ 4 y3
σ5y
σ 4 y4
σ 6 y3
7
2
9
2
5
σy 11
2
x6
y6
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example-3
(a) Construct the forward difference table and the horizontal tab;e
for the following data
x
y
1
4
2
6
3
9
4
12
5
17
Example-3
(b) Construct a forward difference table for the following data
x
y
0
0
10
0.174
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20
0.347
30
0.518
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example-3
(c) Obtain the backward differences for the function f (x ) = x 3
from x = 1 to 1.05 to three decimal places take h = 0.01
Example-4
Calculate differences up to the fourth order for the function y = x 3
from x = 1 to x = 6 take h = 1
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Solution -4
x
1
y
1
2
8
3
27
4
64
5
125
6
216
4y
42 y
43 y
44 y
7
12
19
6
18
37
0
6
24
61
0
6
30
91
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Solution -4
From the table above we see that third differences are constant (6)
and the fourth differences as zero.
Solution -4
Thus the finite difference of order n of an algebraic polynomial of
degree n is constant.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example- 5
Find the missing values in the following table
x
y
45
3
50
-
55
2
60
-
65
-2.4
Example- 6
Assuming that the following values of y belongs to polynomial to
degree 4, compute the next three values
x
y
0
1
1
-1
2
1
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3
-1
4
1
5
-
6
-
7
-
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Error Propagation in a Difference Table
Let y0 , y1 , . . . , yn be the true values of a function and suppose
the value y2 to be affected with an error ξ.
Error Propagation in a Difference Table
Thus its value is y2 + ξ. Then the successive differences of the y
are as shown below:
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Error Propagation in a Difference Table
x
x0
y
y0
x1
y1
x2
y2 + ξ
x3
y3
x4
y4
42 y
4y
4y0
42 y0 + ξ
4y1 + ξ
42 y1 − 2ξ
4y2 − ξ
42 y2 + ξ
4y3
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Error Propagation in a Difference Table
From the table above, error have the following characteristics
(i) The effect of the error increases with the order of the
differences
(ii) The errors in any column are binomial coefficients with
alternating signs
Error Propagation in a Difference Table
(iii) The algebraic sum of the errors in any difference column is
zero
(iv) The error influences a triangular portion of the difference table
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example -6
The following table gives the values of a polynomial of degree five.
It is given that f (4) is in error. Correct the value of f (4)
Example -6
x
y
1
0.975
2
-0.6083
3
-3.5250
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4
-5.5250
5
-6.3583
6
4.2250
GSB 3204: Statistics & Numerical Analysis
7
36.4750
Finite Differences
Example -7
The following table gives the values of a polynomial of degree five.
It is given that f (3) is in error. Correct the error
Example -7
x
y
0
1
1
2
2
33
3
254
4
1054
5
3126
6
7777
ans. ξ = −10, f (3) = 244
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Properties of the Operator 4
(i) If c is a constant then 4c = 0
(ii) 4 is distributive
4(f (x ) ± g(x )) = 4f (x ) ± 4g(x )
(iii) If c is constant then
4(cf (x )) = c 4 f (x )
Properties of the Operator 4
(iv) If m and n are positive integers then
4m 4n f (x ) = 4m+n f (x )
(v) 4(f (x ).g(x )) = f (x ) 4 g(x ) + g(x ) 4 f (x )
f (x )
g(x ) 4 f (x ) − f (x ) 4 g(x )
(v) 4
=
g(x )
g(x )g(x + h)
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Other Difference Operators
(a) Shift Operator E
The shift operator is defined as Ef (x ) = f (x + h), in terms of
y as Eyi = yi+1
E 2 f (x ) = E (Ef (x )) = E (f (x + h)) = f (x + 2h)
Generally E n f (x ) = f (x + nh) or E n yi = yi+nh
Other Difference Operators
The inverse shift operator E −1 is defined as E −1 f (x ) = f (x − h)
E −2 f (x ) = f (x − 2h)
Generally E −n f (x ) = f (x − nh)
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Other Difference Operators
(b) Average Operator
µ is defined as
µf (x ) = 12 f (x + h2 ) + f (x − h2 )
µyi =
1
2
yi+ 1 + yi− 1
2
2
Other Difference Operators
(c) Differential Operator, D
The differential operator is usually denoted by D where
d
d
, Df (x ) =
f (x ) = f 0 (x )
D=
dx
dx
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Relation between Difference Operators
Operator
4
5
σ
µ
Definition
4f (x ) = f (x + h) − f (x )
5f (x ) = f (x ) − f (x − h)
σf (x ) = f (x + h2 ) − f (x − h2 )
µf (x ) =
1
2
f (x +
h
2
+ f (x − h2 )
Relation
(E − 1)f (x )
(1 − E −1 )f (x )
1
1
(E 2 − E − 2 )f (x )
1
1 1
E 2 + E − 2 f (x )
2
Relation between Difference Operators
To link different operators with differential operator D we consider
the Taylor’s formula or series
h2 f 00 (x )
+ ...
f (x + h) = f (x ) + hf 0 (x ) +
2!!
h2 D 2
Ef (x ) = 1 + hD +
+ . . . f (x )
2!
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example- 8
Show that 4 log f (x ) = log 1 +
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4f (x )
f (x )
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example- 8
Show that 4 log f (x ) = log 1 +
4f (x )
f (x )
Example-9
Evaluate
42
E
!
x3
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Exercise
Prove that (1 + 4)(1 − 4) = 1
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Exercise
Prove that (1 + 4)(1 − 4) = 1
Exercise
Prove that 45 = 4 − 5
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example-9
Estimate the missing term in the following table by using shift
operator
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example-9
Estimate the missing term in the following table by using shift
operator
Exercise
x
y
0
4
1
3
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2
4
3
-
4
12
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Representing a Polynomial Using Factorial Notation
A polynomial of degree n can be expressed as factorial polynomial
of the same degree.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Representing a Polynomial Using Factorial Notation
A polynomial of degree n can be expressed as factorial polynomial
of the same degree.
Representing a Polynomial Using Factorial Notation
Let f(x) be a polynomial of degree n which is expressed in factorial
notation and let
f (x ) = a0 + a1 x + a2 x 2 + · · · + an x n , where a0 , a1 , a2 , . . . an are
constants and an 6= 0
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Representing a Polynomial Using Factorial Notation
4f (x ) = 4(a0 + a1 x + a2 x 2 + · · · + an x n , where a0 , a1 , a2 , . . . an )
From 4f (x ) = f (x + h) − f (x ) then
4f (x ) = a1 h + 2a2 hx + · · · + nan hx n−1
42 f (x ) = 4(a1 h + 2a2 hx + · · · + nan hx n−1 )
42 f (x ) = 2a2 h2 · · · + h2 n(n − 1)an x n−2
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Representing a Polynomial Using Factorial Notation
4f (x ) = 4(a0 + a1 x + a2 x 2 + · · · + an x n , where a0 , a1 , a2 , . . . an )
From 4f (x ) = f (x + h) − f (x ) then
4f (x ) = a1 h + 2a2 hx + · · · + nan hx n−1
42 f (x ) = 4(a1 h + 2a2 hx + · · · + nan hx n−1 )
42 f (x ) = 2a2 h2 · · · + h2 n(n − 1)an x n−2
Representing a Polynomial Using Factorial Notation
4f (x ) is a polynomial of degree n − 1 where the coefficient of
x n−1 is nan h,
42 f (x ) is a polynomial of degree n − 2 where the coefficient of
x n−2 is n(n − 1)an h2
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Representing a Polynomial Using Factorial Notation
4n f (x ) is a polynomial of degree zero with coefficient of
(n(n − 1)(n − 2) . . . 1)hn an , and this can be written as n!an hn
∴ 4n f (x ) = n!an hn
The (n + 1)th and higher differences of polynomial of nth degree
will be zero
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Representing a Polynomial Using Factorial Notation
4n f (x ) is a polynomial of degree zero with coefficient of
(n(n − 1)(n − 2) . . . 1)hn an , and this can be written as n!an hn
∴ 4n f (x ) = n!an hn
The (n + 1)th and higher differences of polynomial of nth degree
will be zero
Representing a Polynomial Using Factorial Notation
4n f (x )
We can write
= an h n .
n!
Now for h=1, and x=0 we have
4f (0)
42 f (0)
43 f (0)
a0 = f (0), a1 =
, a2 =
, a3 =
, . . . an =
1!
2!
3!
n
4 f (0)
n!
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Representing a Polynomial Using Factorial Notation
Then
4f (0)
42 f (0) 2 43 f (0) 3
4n f (0) n
f (x ) = f (0) +
x+
x +
x +···+
x
1!
2!
3!
n!
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Representing a Polynomial Using Factorial Notation
Then
4f (0)
42 f (0) 2 43 f (0) 3
4n f (0) n
f (x ) = f (0) +
x+
x +
x +···+
x
1!
2!
3!
n!
Example -10
(a) Find 43 ((1 − 3x )(1 − 2x )(1 − x ))
(b) Express f (x ) = 3x 3 + x 2 + x + 1, in the factorial notation,
interval of differencing being unity
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
Interpolation is the approximate computation of the values of the
function f (x ) from several of its given values
f (x0 ), f (x1 ), f (x2 ), . . . , f (xn ), corresponding to the set of
x0 , x1 , x2 , . . . , xn equally spaced values of independent variable.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
Interpolation is the approximate computation of the values of the
function f (x ) from several of its given values
f (x0 ), f (x1 ), f (x2 ), . . . , f (xn ), corresponding to the set of
x0 , x1 , x2 , . . . , xn equally spaced values of independent variable.
Interpolation
Let y = f (x ) which takes the values y0 , y1 , y2 , . . . , yn
corresponding to set of equally spaced values of the independent
variable, x0 , x1 , x2 , . . . , xn that is xi = x0 + ih such that
i = 0, 1, 2, . . . , n where h is spacing.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
Let φ(x ) be a polynomial of the nth degree in x . Then φ(x )
represents the continuous function y = f (x ) such that
f (xi ) = φ(xi ) for i = 0, 1, 2, . . . , n and at all points
f (x ) = φ(x ) + R(x ) where R(x ) is called the error term of the
interpolation formula
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
Let φ(x ) be a polynomial of the nth degree in x . Then φ(x )
represents the continuous function y = f (x ) such that
f (xi ) = φ(xi ) for i = 0, 1, 2, . . . , n and at all points
f (x ) = φ(x ) + R(x ) where R(x ) is called the error term of the
interpolation formula
Interpolation
Let φ(x ) = a0 + a1 (x − x0 ) + a2 (x − x0 )(x − x1 ) + a3 (x − x0 )(x −
x1 )(x − x2 ) + · · · + an (x − x0 )(x − x1 )(x − x2 ) . . . (x − xn−1 ) and
φ(xi ) = yi , ∀ i = 0, 1, 2, 3, . . . , n
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
The constants a0 , a1 , a2 , . . . , an can be determined by
substituting x = x0 , x1 , x2 , . . . , xn successively in the equation
above, and get
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
The constants a0 , a1 , a2 , . . . , an can be determined by
substituting x = x0 , x1 , x2 , . . . , xn successively in the equation
above, and get
Interpolation
a0 = y0 when x = x0 , taking x = x1 we have
y1 = y0 + a1 (x1 − x0 )
y1 − y0
4y0
y1 − y0 = a1 (x1 − x0 ) =⇒ a1 =
=
x1 − x0
h
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
For x = x2 we have
y1 − y0
(x2 − x0 ) + a2 (x2 − x0 )(x2 − x1 )
y2 = y0 +
x1 − x0
y1 − y0
y2 − y0 −
(x2 − x0 ) = a2 (x2 − x0 )(x2 − x1 )
x1 − x0
y1 − y0
y2 − y0 −
2h = a2 (2h)(h)
h
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
For x = x2 we have
y1 − y0
(x2 − x0 ) + a2 (x2 − x0 )(x2 − x1 )
y2 = y0 +
x1 − x0
y1 − y0
y2 − y0 −
(x2 − x0 ) = a2 (x2 − x0 )(x2 − x1 )
x1 − x0
y1 − y0
y2 − y0 −
2h = a2 (2h)(h)
h
Interpolation
y2 − y0 − 2y1 + 2y0 = 2a2 h2
y2 − 2y1 + y0 = a2 (2h2 )
42 y0
42 y0 = a2 (2h2 =⇒ a2 =
2!h2
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
43 y0
Continuing in this way we have, a3 =
3!h3
4n y0
Generally an =
n!hn
42 y0
43 y0
4y0
(x − x0 ) +
(x
−
x
)(x
−
x
)
+
(x −
φ(x ) = y0 +
0
1
h
2!h2
3!h3
n
4 y0
x0 )(x −x1 (x −x2 )+· · ·+
(x −x0 )(x −x1 )(x −x2 ) . . . (x −xn−1 )
n!hn
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
43 y0
Continuing in this way we have, a3 =
3!h3
4n y0
Generally an =
n!hn
42 y0
43 y0
4y0
(x − x0 ) +
(x
−
x
)(x
−
x
)
+
(x −
φ(x ) = y0 +
0
1
h
2!h2
3!h3
n
4 y0
x0 )(x −x1 (x −x2 )+· · ·+
(x −x0 )(x −x1 )(x −x2 ) . . . (x −xn−1 )
n!hn
Interpolation
x − x0
, then
h
x − x1 = (x − x0 ) − (x1 − x0 ) = kh − h = (k − 1)h
x − x2 = (x − x1 ) − (x2 − x1 ) = (k − 1)h − h = (k − 2)h e.t.c
Let x = x0 + kh =⇒ k =
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
Now we have
k(k − 1) 2
k(k − 1)(k − 2) 3
4 y0 +
4 y0 +
2! n
3!
k(k − 1)(k − 2) . . . 1 4 y0
··· +
n!
φ(x ) = y0 + k 4 y0 +
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Interpolation
Now we have
k(k − 1) 2
k(k − 1)(k − 2) 3
4 y0 +
4 y0 +
2! n
3!
k(k − 1)(k − 2) . . . 1 4 y0
··· +
n!
φ(x ) = y0 + k 4 y0 +
Interpolation
This is called Newton’s Forward Interpolation Formula
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example- 11
(a) Find a polynomial of degree two which takes the following
value
x
f (x )
1
6
2
11
3
18
4
27
√
√
(b) Given that √ 15500 = 124.4990, √ 15510 =
124.5392, √
15520 = 124.5793, 15530 = 124.6194. Find
the value of 1516
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example- 11
(a) Find a polynomial of degree two which takes the following
value
x
f (x )
1
6
2
11
3
18
4
27
√
√
(b) Given that √ 15500 = 124.4990, √ 15510 =
124.5392, √
15520 = 124.5793, 15530 = 124.6194. Find
the value of 1516
Example- 11
(c) A second degree polynomial passes through the points
(1, −1), (2, −2), (3, −1), (4, 2). Find the polynomial
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example- 12
(a) Evaluate y = e 3x at x = 0.05. Using the following table
x
e 3x
0
1
0.1
1.3499
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0.2
1.8221
0.3
2.4596
0.4
3.3201
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example- 12
(a) Evaluate y = e 3x at x = 0.05. Using the following table
x
e 3x
0
1
0.1
1.3499
0.2
1.8221
0.3
2.4596
0.4
3.3201
Example- 12
(b) Find the value of sin 42◦ . Using the following table
x
sin x
40
0.6428
45
0.7071
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50
0.7660
55
0.8192
60
0.8660
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Exercise
Find the cubic polynomial which takes the following values
y (1) = 24, y (3) = 120, y (5) = 336, y (7) = 720. Hence obtain
the value of y (8), using Newton’s Forward Difference Interpolation.
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Newton’s Backward Interpolation Formula
Newton’s Backward Interpolation is given as
φ(x ) =
5yn
52 yn
53 yn
yn +
(x − xn ) +
(x
−
x
)(x
−
x
)
+
(x − xn )(x −
n
n−1
h
2!h2 n
3!h3
5 yn
xn−1 )(x − xn−2 ) + · · · +
(x − xn )(x − xn−1 . . . (x − x1 )(x − x0 )
n!hn
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Newton’s Backward Interpolation Formula
Newton’s Backward Interpolation is given as
φ(x ) =
5yn
52 yn
53 yn
yn +
(x − xn ) +
(x
−
x
)(x
−
x
)
+
(x − xn )(x −
n
n−1
h
2!h2 n
3!h3
5 yn
xn−1 )(x − xn−2 ) + · · · +
(x − xn )(x − xn−1 . . . (x − x1 )(x − x0 )
n!hn
Newton’s Backward Interpolation Formula
Now setting x = xn + vh, we obtain
x − xn = vh
x −xn−1 = (x −xn )−(xn−1 −xn ) = (x −xn )+(xn −xn−1 ) = (v +1)h
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Newton’s Backward Interpolation Formula
x − xn−2 = (x − xn ) + (xn − xn−2 ) = (v + 2)h
..
.
x − x0 = v + n − 1)h
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Newton’s Backward Interpolation Formula
x − xn−2 = (x − xn ) + (xn − xn−2 ) = (v + 2)h
..
.
x − x0 = v + n − 1)h
Newton’s Backward Interpolation Formula
Then
v (v + 1) 2
v (v + 1)(v + 2) 3
φ(x ) = yn + v 5 yn +
5 yn +
5 yn +
2!
3!
v (v + 1)(v + 2) . . . (v + n − 1) n
··· +
5 yn
n!
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Newton’s Backward Interpolation Formula
Newton’s forward interpolation formula is suitable for interpolating
values of y near the beginning of the table, while Newton’s
Backward Interpolation Formula is suitable for interpolation values
of y near the end of the table of values
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Newton’s Backward Interpolation Formula
Newton’s forward interpolation formula is suitable for interpolating
values of y near the beginning of the table, while Newton’s
Backward Interpolation Formula is suitable for interpolation values
of y near the end of the table of values
Example-13
Calculate the value of f (84) for the data given in the table below
x
f (x )
40
204
50
224
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60
246
70
270
80
296
90
324
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example-14
Values of x in degrees and sinx are given in the following table
x
sin x
15
0.2588
20
0.3420
25
0.4226
30
0.5
35
0.5736
40
0.6428
Determine the value of sin 38◦
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GSB 3204: Statistics & Numerical Analysis
Finite Differences
Example-14
Values of x in degrees and sinx are given in the following table
x
sin x
15
0.2588
20
0.3420
25
0.4226
30
0.5
35
0.5736
40
0.6428
Determine the value of sin 38◦
Example-15
From the following table estimate the number of students who
obtained marks in GSB 3204 between 75 and 80
Marks
No. of Students
35-45
20
Yasin K. Kowa Email:kowayasin@gmail.com
45-55
40
55-65
60
65-75
60
75-85
20
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Difference Interpolation Formulas
The central difference interpolation formulas are most suitable for
interpolation near the middle of a tabulated set
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Difference Interpolation Formulas
The central difference interpolation formulas are most suitable for
interpolation near the middle of a tabulated set
Central Difference Interpolation Formulas
The most important central difference formulas are Gauss’s
Formulas, Bessel, Stirling, and Everett
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Difference Interpolation Formulas
Let’s start with Gauss’s formulas, let the function y = f (x ) be
given for 2n + 1 equispaced values of argument
x0 , x0 ± h, x0 ± 2h, . . . , x0 ± nh
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Difference Interpolation Formulas
Let’s start with Gauss’s formulas, let the function y = f (x ) be
given for 2n + 1 equispaced values of argument
x0 , x0 ± h, x0 ± 2h, . . . , x0 ± nh
Central Difference Interpolation Formulas
The corresponding values of y be yi for i = 0, ‘ ± 1, ±2, . . . , ±n
Also let y = y0 denote the central ordinate corresponding to
x = x0 . We can then form the difference table as shown below
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Central Difference Interpolation Formulas
x
x0 − 3h
x0 − 2h
y
4y
y−3
4y−3
y−2
42 y
42 y−2
y−1
42
y0
x0 + 3h
y2
4y2
y−3
44 y−2
42 y0
y1
45 y−3
43 y−1
46 y−3
45 y−2
44 y−1
43 y0
4y1
x0 + 2h
46 y
44 y−3
43 y−2
4y0
x0 + h
45 y
43 y−3
4y−1
x0
44 y
42 y−3
4y−2
x0 − h
43 y
42 y1
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
Consider the Newton’s Forward Interpolation formula
u(u − 1) 2
u(u − 1)(u − 2) 3
y = f (x ) = y0 +u4y0 +
4 y0 +
4 y0 +. . .
2!
3!
x − x0
where u =
and x = x0 is the origin. Now for Gauss’s
h
Forward Interpolation Formula, we assume the differences lies on
the bottom solid lines and they are of the form
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
Consider the Newton’s Forward Interpolation formula
u(u − 1) 2
u(u − 1)(u − 2) 3
y = f (x ) = y0 +u4y0 +
4 y0 +
4 y0 +. . .
2!
3!
x − x0
where u =
and x = x0 is the origin. Now for Gauss’s
h
Forward Interpolation Formula, we assume the differences lies on
the bottom solid lines and they are of the form
Gauss’s Forward Interpolation Formulas
y = y0 + G1 4 y0 + G2 42 y−1 + G3 43 y−1 + G4 44 y−2 + . . .
Where G1 , G2 , G3 , G4 , . . . , Gn are coefficients to be determined.
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
From the Newton’s Forward Interpolation formula we have
yp = E p y0 = (1 + 4)p y0 =
u(u − 1) 2
u(u − 1)(u − 2) 3
y0 + u 4 y0 +
4 y0 +
4 y0 + . . .
2!
3!
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
From the Newton’s Forward Interpolation formula we have
yp = E p y0 = (1 + 4)p y0 =
u(u − 1) 2
u(u − 1)(u − 2) 3
y0 + u 4 y0 +
4 y0 +
4 y0 + . . .
2!
3!
Gauss’s Forward Interpolation Formulas
Now 42 y−1 = 42 E −1 y0 = 42 (1 + 4)−1 y0
42 (1 + 4)−1 y0 = 42 (1 − 4 + 42 − 43 + . . . )y0
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
42 y−1 = 42 y0 − 43 y0 + 44 y0 − 45 y0 + . . .
43 y−1 = 43 y0 − 44 y0 + 45 y0 − 46 y0 + . . .
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
42 y−1 = 42 y0 − 43 y0 + 44 y0 − 45 y0 + . . .
43 y−1 = 43 y0 − 44 y0 + 45 y0 − 46 y0 + . . .
Gauss’s Forward Interpolation Formulas
44 y−2 = 44 E −2 y0 = 44 (y0 − 2 4 y0 + 3 42 y0 − 4 43 y0 + . . . )
44 y2 = 44 y0 − 2 45 y0 + 3 46 y0 − 4 47 + . . .
and so on.
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
Now
yp =
y0 + G1 4 y0 + G2 (42 y0 − 43 y0 + 44 y0 − 45 y0 + . . . ) + G3 (43 y0 −
44 y0 +45 y0 −46 y0 +. . . )+G4 (44 y0 −245 y0 +346 y0 −447 + . . . )
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
Now
yp =
y0 + G1 4 y0 + G2 (42 y0 − 43 y0 + 44 y0 − 45 y0 + . . . ) + G3 (43 y0 −
44 y0 +45 y0 −46 y0 +. . . )+G4 (44 y0 −245 y0 +346 y0 −447 + . . . )
Gauss’s Forward Interpolation Formulas
Comparing the equation above to Newton’s Forward Interpolation
Formula we have
u(u − 1)
(u + 1)u(u − 1)
G1 = u, G2 =
, G3 =
,
2!
3!
(u + 1)u(u − 1)(u − 2)
G4 =
4!
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
Hence, the Gauss’s Forward Interpolation Formula can be written
as
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Forward Interpolation Formulas
Hence, the Gauss’s Forward Interpolation Formula can be written
as
Gauss’s Forward Interpolation Formulas
u(u − 1) 2
(u + 1)u(u − 1) 3
4 y0 +
4 y0 +
2!
3!
(u + 1)u(u − 1)(u − 2) 4
4 y0 + . . .
4!
yp = y0 + u 4 y0 +
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Backward Interpolation Formulas
The Gauss’s backward Interpolation Formula uses the differences
which lies on the upper dash line and can be assumed of the form
yp = y0 + J1 4 y−1 + J2 42 y−1 + J3 43 y−2 + J4 44 y−2 + . . .
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Backward Interpolation Formulas
The Gauss’s backward Interpolation Formula uses the differences
which lies on the upper dash line and can be assumed of the form
yp = y0 + J1 4 y−1 + J2 42 y−1 + J3 43 y−2 + J4 44 y−2 + . . .
Gauss’s Backward Interpolation Formulas
Where J1 , J2 , J3 , J4 , . . . are to be determined.
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Backward Interpolation Formulas
Using the same procedure as in the Gauss’s Forward Interpolation
Formula we determine
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Backward Interpolation Formulas
Using the same procedure as in the Gauss’s Forward Interpolation
Formula we determine
Gauss’s Backward Interpolation Formulas
J1 = u
u(u + 1)
J2 =
2!
(u + 2)(u + 1)u(u − 1)
J3 =
3!
(u + 1)u(u − 1)(u − 2)
J4 =
4!
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Backward Interpolation Formulas
u(u + 1) 2
(u + 1)u(u − 1) 3
4 y−1 +
4 y−2 +
2!
3!
(u + 2)(u + 1)u(u − 1) 4
4 y−2 + . . .
4!
y = y0 + u 4 y−1 +
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Finite Differences
Gauss’s Backward Interpolation Formulas
u(u + 1) 2
(u + 1)u(u − 1) 3
4 y−1 +
4 y−2 +
2!
3!
(u + 2)(u + 1)u(u − 1) 4
4 y−2 + . . .
4!
y = y0 + u 4 y−1 +
Example- 16
Use Gauss’s Forward Interpolation Formula to find y for x = 20
given that
x
y
11
19.5673
15
18.8243
19
18.2173
Yasin K. Kowa Email:kowayasin@gmail.com
23
17.1236
27
16.6162
GSB 3204: Statistics & Numerical Analysis
Central Differences Interpolation Formula
Laplace-Everett’s Formula
Eliminating odd differences in Gauss’s Forward formula, by using
the relation
4y0 = y1 −y0 , 43 y−1 = 42 y0 −42 y−1 , 45 y−2 = 44 y−1 −44 y−2
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Central Differences Interpolation Formula
Laplace-Everett’s Formula
Eliminating odd differences in Gauss’s Forward formula, by using
the relation
4y0 = y1 −y0 , 43 y−1 = 42 y0 −42 y−1 , 45 y−2 = 44 y−1 −44 y−2
Laplace-Everett’s Formula
u(u − 1) 2
(u + 1)u(u − 1) 2
4 y−1 +
(4 y0 −
2!
3!
(u + 1)u(u − 1)(u − 2) 4
42 y−1 ) +
4 y−2 +
4!
(u + 2)(u + 1)u(u − 1)(u − 2) 4
(4 y−1 − 44 y−2 ) + . . .
5!
y = y0 + u(y1 − y0 ) +
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Central Differences Interpolation Formula
Laplace-Everett’s Formula
y = (1 − u)y0 + uy1 + u(u − 1)
1
u+1
−
2!
3!
42 y−1 +
(u + 1)u(u − 1) 2
1
u+2
4 y0 + (u + 1)u(u − 1)(u − 2)
−
44
3!
4!
6!
(u + 2)(u + 1)u(u − 1)(u − 2) 4
y−2 +
4 y−1 + . . .
5!
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Central Differences Interpolation Formula
Laplace-Everett’s Formula
y = (1 − u)y0 + uy1 + u(u − 1)
1
u+1
−
2!
3!
42 y−1 +
(u + 1)u(u − 1) 2
1
u+2
4 y0 + (u + 1)u(u − 1)(u − 2)
−
44
3!
4!
6!
(u + 2)(u + 1)u(u − 1)(u − 2) 4
y−2 +
4 y−1 + . . .
5!
Laplace-Everett’s Formula
u(u − 1)(u − 2) 2
4 y−1 +
3!
(u + 1)u(u − 1) 2
(u + 1)u(u − 1)(u − 2)(u − 3) 4
4 y0 −
4 y−2 +
3!
5!
(u + 2)(u + 1)u(u − 1)(u − 2) 4
4 y−1 + . . .
5!
y = (1 − u)y0 + uy1 −
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Central Differences Interpolation Formula
Laplace-Everett’s Formula
Let v = 1 − u =⇒ u = 1 − v , u − 2 = −v − 1
(1 − v )v (v + 1) 2
(u + 1)u(u − 1) 2
y = vy0 + uy1 +
4 y−1 +
4
3!
3!
(v + 2)(v + 1)v (v − 1)(v − 2) 4
y0 +
4 y−2 +
5!
(u + 2)(u + 1)u(u − 1)(u − 2) 4
4 y−1 + . . .
5!
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Central Differences Interpolation Formula
Laplace-Everett’s Formula
Let v = 1 − u =⇒ u = 1 − v , u − 2 = −v − 1
(1 − v )v (v + 1) 2
(u + 1)u(u − 1) 2
y = vy0 + uy1 +
4 y−1 +
4
3!
3!
(v + 2)(v + 1)v (v − 1)(v − 2) 4
y0 +
4 y−2 +
5!
(u + 2)(u + 1)u(u − 1)(u − 2) 4
4 y−1 + . . .
5!
Laplace-Everett’s Formula
v (v 2 − 1) 2
v (v 2 − 1)(v 2 − 4) 4
4 y−1 +
4 y−2 +. . . +uy1 +
3!
5!
(u + 1)u(u − 1) 2
(u + 2)(u + 1)u(u − 1)(u − 2) 4
4 y0 +
4 y−1 +. . .
3!
5!
y = vy0 +
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Laplace-Everett’s Formula
Example
Use Everett’s Interpolation formula to find the value of y given
that x = 1.60 from the following table
x
y
1.0
1.0543
1.25
1.1281
1.5
1.2247
Yasin K. Kowa Email:kowayasin@gmail.com
1.75
1.3219
2
1.4243
2.25
1.4987
GSB 3204: Statistics & Numerical Analysis
Interpolation
Choice of Interpolation Formula
The selection of an interpolation formula depends to a great extent
on the position of the interpolated value in the given data
(a) Use Newton’s Forward Interpolation formula to find a
tabulated value near the beginning of the table
(b) Use Newton’s backward interpolation formula to find a value
near the end of the table
(c) Use either Stirling’s or Bessel’s or Laplace-Everrett’s formula
to find an interpoated value near the center of the table. if u
1
1
is lying between −1
4 and 4 ., Stirling’s formula is preffered, 4
3
and 4 use Bessel’s or Everett’s Formula
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Interpolation with Unequal Intervals
Lagrange’s Interpolation Formula
Let y = f (x ) be areal valued continuous function defined in the
interval [a, b], let x0 , x1 , ..., xn be (n+1) distinct points which are
not necessary equally spaced and the corresponding values of the
function are y0 , y1 , y2 , ...., yn
Lagrange’s Interpolation Formula
We can represent the function y = f (x ) as polynomial in x of
degree n.
Let the polynomial be represented as
f (x ) = a0 (x − x1 )(x − x2 )...(x − xn ) + a1 (x − x0 )(x − x2 )...(x −
xn ) + a2 (x − x0 )(x − x1 )(x − x3 )...(x − xn ) + .......... + an (x −
x0 )(x − x1 )(x − x2 )...(x − xn−1 )
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Interpolation with Unequal Intervals
Lagrange’s Interpolation Formula
If we put x = x0 in the above we get,
f (x0 )
a0 =
(x0 − x1 ))(x0 − x2 )....(x0 − xn )
Putting x = x1
f (x1 )
a1 =
(x1 − x0 ))(x1 − x2 )....(x1 − xn )
Putting x = x2
f (x2 )
a2 =
(x2 − x0 ))(x2 − x1 )....(x2 − xn )
Putting x = xn
f (xn )
an =
(xn − x0 ))(xn − x1 )....(xn − xn−1 )
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Interpolation with Unequal Intervals
Lagrange’s Interpolation Formula
y=
(x − x1 )(x − x2 )(x − x3 ) . . . (x − xn )
f (x0 )+
(x0 − x1 )(x0 − x2 )(x0 − x3 ) . . . (x0 − xn )
(x − x0 )(x − x2 )(x − x3 ) . . . (x − xn )
f (x1 ) . . . +
(x1 − x0 )(x1 − x2 )(x1 − x3 ) . . . (x1 − xn )
(x − x1 )(x − x2 )(x − x3 )....(x − xn−1 )
f (xn )
(xn − x1 )(xn − x2 )(xn − x3 ) . . . (xn − xn−1 )
This is called as Lagrange’s Interpolation Formula
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Lagrange’s Interpolation Formula
Example
Apply Lagrange’s Interpolation Formula to find a polynomial of
which passes through the point (0, -20), (1,-12), (3, -20) and (4,
-24)
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Lagrange’s Interpolation Formula
Example
Apply Lagrange’s Interpolation Formula to find a polynomial of
which passes through the point (0, -20), (1,-12), (3, -20) and (4,
-24)
Example
(a) Use Lagrange’s Interpolation Formula to find a polynomial
which passes through the points (0, -12), (1, 0), (3, 6), (4,
12)
(b) Use Lagrange’s Interpolation Formula to find the value of y
corresponding to x = 10 from the following table
x
y
5
380
Yasin K. Kowa Email:kowayasin@gmail.com
6
-2
9
196
11
508
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Numerical Differentiation
Numerical differentiation is the process of calculating the derivative
of a function at some particular value of independent variable by
means of a set of given values of that function.
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Numerical Differentiation
Numerical differentiation is the process of calculating the derivative
of a function at some particular value of independent variable by
means of a set of given values of that function.
Numerical Differentiation
Consider the function y=f(x) which is tabulated for the values
x = a + nh, n = 0, 1, 2, ....
Numerical Differentiation are based on numerical interpolation as
follows:
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Numerical Differentiation
(a) Derivatives based on Newton’s Forward interpolation Formula.
This formula is used to find the derivative for some given x
lying near the beginning of the data table/points.
(b) Derivatives based on Newton’s Forward interpolation Formula.
This formula is used to find the derivative for some given x
lying near the end of the data table/points.
(c) Derivatives based on Stirling’s Interpolation Formula. This
formula is used to find the derivative for some point lying near
the middle of tabulated value
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Using Newton’s Forward Difference Formula
Recall the Newton’s Forward Interpolation Formula
u(u − 1)(u − 2) 3
u(u − 1) 2
4 y0 +
4 y0 +
2!
3!
u(u − 1)(u − 2)(u − 3) 4
4 y0 + ....
4!
y = y0 + u 4 y0 +
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Using Newton’s Forward Difference Formula
Recall the Newton’s Forward Interpolation Formula
u(u − 1)(u − 2) 3
u(u − 1) 2
4 y0 +
4 y0 +
2!
3!
u(u − 1)(u − 2)(u − 3) 4
4 y0 + ....
4!
y = y0 + u 4 y0 +
Derivatives Using Newton’s Forward Difference Formula
x − x0
where u =
⇒ x = x0 + uh
h
u2 − u 2
u 3 − u 2 + 2u 3
y = y0 + u 4 y0 +
4 y0 +
4 y0 +
2
6
4
3
2
u − 6u + 11u − 6u 4
4 y0 + ....
24
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Using Newton’s Forward Difference Formula
2u − 1 2
3u 2 − 6u + 2 3
dy
= 4y0 +
4 y0 +
4 y0 +
du
2
6
4u 3 − 18u 2 + 22u − 6 4
4 y0 + ....
24
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Using Newton’s Forward Difference Formula
2u − 1 2
3u 2 − 6u + 2 3
dy
= 4y0 +
4 y0 +
4 y0 +
du
2
6
4u 3 − 18u 2 + 22u − 6 4
4 y0 + ....
24
Derivatives Using Newton’s Forward Difference Formula
dx
=h
and
du
dy
dy
du
=
×
dx
du dx
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Using Newton’s Forward Difference Formula
dy
1
2u − 1 2
3u 2 − 6u + 2 3
= [4y0 +
4 y0 +
4 y0 +
dx
h
2
6
4u 3 − 18u 2 + 22u − 6 4
4 y0 + ....]
24
If x = x0 , u = 0
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Using Newton’s Forward Difference Formula
dy
1
2u − 1 2
3u 2 − 6u + 2 3
= [4y0 +
4 y0 +
4 y0 +
dx
h
2
6
4u 3 − 18u 2 + 22u − 6 4
4 y0 + ....]
24
If x = x0 , u = 0
Derivatives Using Newton’s Forward Difference Formula
dy
1
42 y0 43 y0 44 y0
= [4y0 −
+
−
+ ....]
dx
h
2
3
4
and
2
d 2y
1
2 y + 6u − 6 43 y + 12u − 36u + 22 44 y + ...]
=
[4
0
0
0
dx 2
h2
6
24
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Using Newton’s Forward Difference Formula
Foru = 0
d 2y
5
1
11 4
4 y0 − 45 y0 + ...]
= 2 [42 y0 − 43 y0 +
2
dx
h
24
6
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Using Newton’s Forward Difference Formula
Foru = 0
d 2y
5
1
11 4
4 y0 − 45 y0 + ...]
= 2 [42 y0 − 43 y0 +
2
dx
h
24
6
Example
dy
d 2y
From the following table find the value of
and
at the
dx
dx 2
point x = 1.0
x
y
1
5.4680
1.1
5.6665
1.2
5.9264
Yasin K. Kowa Email:kowayasin@gmail.com
1.3
6.2551
1.4
6.6601
1.5
7.1488
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Example
d 2y
dy
and
From the following table of values of x and y, obtain
dx
dx 2
for x = 1.1 and x = 1.2
x
y
1.0
2.7183
1.2
3.3201
1.4
4.0562
Yasin K. Kowa Email:kowayasin@gmail.com
1.6
4.9530
1.8
6.0496
2.0
7.3891
GSB 3204: Statistics & Numerical Analysis
2.2
9.0250
Numerical Differentiation
Example
d 2y
dy
and
From the following table of values of x and y, obtain
dx
dx 2
for x = 1.1 and x = 1.2
x
y
1.0
2.7183
1.2
3.3201
1.4
4.0562
1.6
4.9530
1.8
6.0496
2.0
7.3891
2.2
9.0250
Derivatives Based on Newton’s Backward Interpolation Formula
Here, we assume that the function y=f(x) is known at (n+1)
points x0 , x1 , x2 , . .
., xn
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Based on Newton’s Backward Interpolation Formula
x − xn
Let xi = x0 + ih i=0, 1, 2, . . .,n and u =
h
Recall the Newton’s Backward Interpolation formula
u(u + 1) 2
u(u + 1)(u + 2) 3
5 yn +
5 yn +
2!
3!
u(u + 1)(u + 2)(u + 3) 4
5 yn +.....
4!
f (x ) = yn + u 5 yn +
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Based on Newton’s Backward Interpolation Formula
x − xn
Let xi = x0 + ih i=0, 1, 2, . . .,n and u =
h
Recall the Newton’s Backward Interpolation formula
u(u + 1) 2
u(u + 1)(u + 2) 3
5 yn +
5 yn +
2!
3!
u(u + 1)(u + 2)(u + 3) 4
5 yn +.....
4!
f (x ) = yn + u 5 yn +
Derivatives Based on Newton’s Backward Interpolation Formula
1
2u + 1 2
3u 2 + 6u + 2 3
[5yn +
5 yn +
5 yn +
h
2!
3!
3
2
4u + 18u + 22u + 6 4
5 yn + ......]
4!
f 0 (x ) =
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Based on Newton’s Backward Interpolation Formula
f 00 (x ) =
2
1
2 y + 6u + 6 53 y + 12u + 36u + 22 54 y + ......]
[5
n
n
n
h2
3!
4!
For x = xn , u = 0 we have
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Based on Newton’s Backward Interpolation Formula
f 00 (x ) =
2
1
2 y + 6u + 6 53 y + 12u + 36u + 22 54 y + ......]
[5
n
n
n
h2
3!
4!
For x = xn , u = 0 we have
Derivatives Based on Newton’s Backward Interpolation Formula
f 0 (x ) =
1
52 yn 53 yn 54 yn
[5yn +
+
+
+ ......]
h
2
3
4
f 00 (x ) =
1
11 4
5 yn + ......]
[52 yn + 53 yn +
2
h
12
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Example
A slider in a machine moves a long a fixed straight rod. Its
distance xm along the rod are given in the following table for
various values of the time t(seconds).
t(sec)
x(m)
1
0.0201
2
0.0844
3
0.3444
4
1.0100
5
2.3660
6
4.7719
Find the velocity and acceleration of the slider at time t=6 sec
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Example
A slider in a machine moves a long a fixed straight rod. Its
distance xm along the rod are given in the following table for
various values of the time t(seconds).
t(sec)
x(m)
1
0.0201
2
0.0844
3
0.3444
4
1.0100
5
2.3660
6
4.7719
Find the velocity and acceleration of the slider at time t=6 sec
Derivatives Based on Stirling’s Interpolation Formula
Suppose y±i = f (x±i ), i=0, 1, 2, ......,n are given for (2n+1)
points x0 , x±1 , x±2 , ......, x±n where x±i = x0 ± ih, i=0, 1, .....n
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Based on Stirling’s Interpolation Formula
The Stirling’s Interpolation formula is given by
f (x ) =
u2 2
u 3 − u 43 y−2 + 43 y−1
4y−1 + 4y0
]+
4 y−1 +
[
]+
2
2!
3!
2
u4 − u2 4
u 5 − 5u 3 + 4u 45 y−3 + 45 y−2
4 y−2 +
[
+ ....]
4!
5!
2
y0 + u[
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Based on Stirling’s Interpolation Formula
The Stirling’s Interpolation formula is given by
f (x ) =
u2 2
u 3 − u 43 y−2 + 43 y−1
4y−1 + 4y0
]+
4 y−1 +
[
]+
2
2!
3!
2
u4 − u2 4
u 5 − 5u 3 + 4u 45 y−3 + 45 y−2
4 y−2 +
[
+ ....]
4!
5!
2
y0 + u[
Derivatives Based on Stirling’s Interpolation Formula
x − x0
where u =
h
Then
1 4y−1 + 4y0
3u 2 − 1 43 y−2 + 43 y−1
f 0 (x ) = [
+ u 42 y−1 +
[
]+
h
2
6
2
3
4
2
5
5
4u − 2u 4
5u − 15u + 4 4 y−3 + 4 y−2
4 y−2 +
[
] + .......]
4!
5!
2
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Based on Stirling’s Interpolation Formula
and
12u 2 − 2 4
1
43 y−2 + 43 y−2
]+
4 y−2 +
f 00 (x ) = 2 [42 y−1 + u[
h
2
4!
20u 3 − 30u 45 y−3 + 45 y−2
[
]+..........]
5!
2
at x = x0
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Derivatives Based on Stirling’s Interpolation Formula
and
12u 2 − 2 4
1
43 y−2 + 43 y−2
]+
4 y−2 +
f 00 (x ) = 2 [42 y−1 + u[
h
2
4!
20u 3 − 30u 45 y−3 + 45 y−2
[
]+..........]
5!
2
at x = x0
Derivatives Based on Stirling’s Interpolation Formula
f 0 (x ) =
1 4y−1 +4y0 1 43 y−2 + 43 y−1
1 45 y−3 + 45 y−2
[
−
[
]
+
[
] + .....]
2
h
6
2
30
2
f 00 (x ) =
1
1 4
4 y−2 + ..........]
[42 y−1 −
h2
12
Yasin K. Kowa Email:kowayasin@gmail.com
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Example
dy
d 2y
Find
and
for x=0.2 for the data given in the following
dx
dx 2
table.
x
y
0
0
0.1
0.10017
0.2
0.20134
Yasin K. Kowa Email:kowayasin@gmail.com
0.3
0.30452
0.4
0.41076
0.5
0.52115
GSB 3204: Statistics & Numerical Analysis
Numerical Differentiation
Example
dy
d 2y
Find
and
for x=0.2 for the data given in the following
dx
dx 2
table.
x
y
0
0
0.1
0.10017
0.2
0.20134
0.3
0.30452
0.4
0.41076
0.5
0.52115
Exercise
Compute the values of f 0 (3.1) and f 0 (3.2) using the following table
x
f(x)
1
0
2
1.4
Yasin K. Kowa Email:kowayasin@gmail.com
3
3.3
4
5.6
5
8.1
GSB 3204: Statistics & Numerical Analysis