Numerical Analysis_Approximation Error & Operator.pdf

Numerical Analysis: Approximation Error & Operator:
ROUNDING-OFF NUMBERS AND ERRORS
OPERATORS: FORWARD, BACKWARD, SHIFTING OPERATORS, DIFFERENTIAL OPERATOR
RELATIONS AMONG OPERATORS

1. Multivariate Analysis || Lecture Notes || Anup Kumar Giri


NUMERICAL ANALYSIS:
APPROXIMATION , ERRORS & OPERATORS

2. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 1
ROUNDING-OFF NUMBERS AND ERRORS
Two major techniques are used to solve any mathematical problem analytical and numerical. The analytical solution is
obtained in a compact form and generally it is free from error. On the other hand, numerical method is a technique which
is used to solve a problem with the help of computer or calculator. In general, the solution obtained by this method
contains some error. But, for some class of problems it is very difficult to obtain an analytical solution. For these problems
we generally use numerical methods.
Numerical methods are to provide practical procedures for obtaining the numerical solutions of problems to a specified
degree of accuracy. In numerical analysis, besides the study of the methods, one studies the errors involving in the methods
and in the final results.
APPROXIMATE NUMBER AND SIGNIFICANT FIGURES
In the decimal representation of a number, an approximate number may be defined as a number which is used as an
approximation to an exact number. Some numbers like ??????,� and √2 are exact numbers, but they cannot be represented
exactly by a finite number of digits. If we express ??????, for example, in decimal form as 3.1416, then the number 3.1416 is
only an approximation to the true value of ?????? and it is termed an approximate number.
In the decimal representation of a number, a digit is said to be significant figure if it is either a non-zero digit or any
zero(s) lying between two non-zero digits are used as a placeholder, to indicate a retrained place. All other zeros used to
fix-up the position of the decimal point are not to be counted as significant digit or figure.
The number of significant digits in a number will be counted from the left-most non-zero digit towards right. Thus, the
numbers 0.7452 and 0.007452 both have four significant digits. Similarly, 0.00400307 has 6 significant digits, 0.4003001
has 7 significant digits and 0.30040000 has 8 significant digits.
EXAMPLE:
Find the number of significant figures in the numbers: (�) 6.01203 (�) 0.38002 (�) 0.000032401 (�) 82010000
Solution:
(a) 6.01203=0.601203×10
1

The number of significant figures is 6.
(b) 0.38002=0.38002×10
0

The number of significant figures is 5.
(c) 0.000032401=0.32401×10
−4

The number of significant digits is 5.
(d) 82010000=0.8201×10
8
=0.82010×10
8
=0.820100×10
8

The number of significant figures is 4 or 5 or 6. In such cases, the number of significant figures is not definite.

3. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 2
RULES OF ROUNDING-OFF
In rounding-off process the number is approximated to a very close number consisting of a smaller number of digits. In
that case, one or more digits are kept with the number, taken from left to right, and all other digits are discarded.
Let ??????=�
��
�−1⋯�
2�
1�
0∙�
−1�
−2⋯�
−??????�
−(??????+1)⋯ be a number in the decimal system, where each �
?????? assumes one of
the values 0,1,2,⋯,9 and �
�≠0.
Suppose we want to round-off the number ?????? to ?????? places of decimals.
RULE I:
If �
−(??????+1) is one of the digits 0,1,2,3,4, then ?????? is rounded to �
��
�−1⋯�
2�
1�
0∙�
−1�
−2⋯�
−??????,
i.e. if the discarded digits constitute a number which is smaller than half the unit in the last decimal place that remains,
then the digits that remain do not change.
RULE II:
If �
−(??????+1) is one of the digits 6,7,8,9, then ?????? is rounded to �
��
�−1⋯�
2�
1�
0∙�
−1�
−2⋯(�
−??????+1),
i.e. if the discarded digits constitute a number which is larger than half the unit in the last decimal place that remains, then
the last digit that is left is increased by one.
RULE III:
If �
−(??????+1)=5 which is followed by at least one non-zero digit, ?????? is rounded to �
��
�−1⋯�
2�
1�
0∙�
−1�
−2⋯(�
−??????+1),
and if �
−(??????+1)=5, being the last non-zero digit, P is rounded to
(??????) �
��
�−1⋯�
2�
1�
0∙�
−1�
−2⋯�
−?????? when �
−?????? is even,and
(????????????) �
��
�−1⋯�
2�
1�
0∙�
−1�
−2⋯(�
−??????+1) when �
−?????? is odd.
i.e. If the discarded digits constitute a number which is equal to half the unit in the last decimal place that remains, then
the last digit that is half is increased by one, if it is odd, and is unchanged if it is even. This rule is often called a rule of
an even digit.
ILLUSTRATION:
Exact Number
Round-off number to six
significant figures
Remark
26.0123728 26.0124 (added 1 in the last digit)
23.12432605 23.1243 (last digit remains unchanged)
30.455354 30.4554 (added 1 in the last digit)
19.652453 19.6525 (added 1 in the last digit)
126.3545 126.354 (last digit remains unchanged)
34.44275 34.4428 (added 1 in the last digit to make even digit)

4. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 3
SOURCES OF ERROR
It is well known that the solution of a problem obtained by numerical method contains some errors. But our intension is
to minimize the error. To minimize it, the most essential thing is to identify the causes or sources of the error. Three
sources of errors, viz. inherent errors, round-off errors and truncation errors occur to find a solution of a problem by using
numerical method.
INHERENT ERROR:
These types of errors occur due to the simplified assumptions made during mathematical modelling of the problem. These
errors also occur when the data is obtained from certain physical measurements of the parameters of the proposed problem.
When performing computations with algebraic operations among approximate numbers, we naturally carry to some extent
the errors of the original data into final result. Such errors are called inherent error/error of the operation.
For example, let �=0.3333 and �=3.1416 be two approximate numbers for the exact number 1/3 and ??????. Obviously,
if we perform an algebraic operation between these two approximate numbers, the error will introduce in the final result
accordingly.
ROUND-OFF ERROR:
When the rational numbers like 1/3; 22/7; 5/9; 8/9 etc, whose decimal representation involve infinite number of digits,
are involved in our calculations, we are forced to take only a few number of digits from their decimal expression and thus
an error named round-off error gets involved. So, in arithmetic computation, some errors will occur due to the finite
representation of the numbers; these errors are called round-off errors. There are universal rules for rounding a number
as rounding rules.
RESIDUAL ERROR OR TRUNCATION ERROR:
This error occurs when mathematical functions like
cos�=1−
�
2
2!
+
�
4
4!
−
�
6
6!
+⋯ and �
??????
=1+�+
�
2
2!
+
�
3
3!
+⋯
whose infinite series expansion exist, are used in the calculations.
These errors occur due to the finite representation of an inherently infinite process. If we consider only a finite number
of terms to calculate the value of cos� for a given �, then we obtained an approximate value. The error occurs due to the
truncation of the remaining terms of the series and it is called the truncation of error.

5. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 4
ABSOLUTE, RELATIVE AND PERCENTAGE ERROR
The absolute error of a number is defined as the absolute difference between the exact value (�
??????) and its approximate
value (�
??????). Let �
?????? be the approximate value of the exact number �
?????? (not necessarily known). Then the absolute error is
denoted by �
?????? and defined as
�
??????=|�
??????−�
??????|.
The relative error of a number is the absolute error divided by its absolute true value. The relative error is the error per
unit measurement. The relative error is denoted by �
?????? and is defined by
�
??????=
|�
??????−�
??????|
|�
??????|
.
The percentage error is the relative error multiplied by 100. The relative error is measured in 1-unit scale while the
percentage error is measured in 100-unit scale. The percentage error denoted by �
?????? and is measured by
�
??????=�
??????×100%=
|�
??????−�
??????|
|�
??????|
×100%.
NOTE:
(??????) The absolute error satisfies the relation �
�≥|�
??????−�
??????|. The exact value �
?????? lies between �
??????−�
� and �
??????+�
�. It
can be written as �
??????=�
??????±�
�.
(????????????) The upper bound of the absolute error is
1
2
×10
−�
??????.�. absolute error≤
1
2
×10
−�
, when the number is rounded
to m decimal places.
(??????????????????) Sometimes, the relative error of a number approximately measures by the absolute error divided by its absolute
approximate value i.e.
�
??????=
|�
??????−�
??????|
|�
??????|
≃
|�
??????−�
??????|
|�
??????|
.
(????????????) The relative and percentage errors are free from the unit of measurement, while absolute error depends on the
measuring unit.
EXAMPLE:
Find the absolute, relative and percentage error in �
?????? when �
??????=
1
7
and �
??????= 0.1429.
Solution:
The absolute error is �
??????=|�
??????−�
??????|=|
1
7
−0.1429|=|
1−1.0003
7
|=
0.0003
7
=0.0000428571⋍0.000043, (rounded up
to two significant figures).
The relative error is �
??????=
|????????????−????????????|
|????????????|
=
|
1−1.0003
7
|
|
1
7
|
⋍
0.000043
1
7
=0.000301⋍0.0003.
The percentage error is �
??????=�
??????×100%⋍0.0301%.

6. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 5
OPERATORS
Lot of operators are used in numerical analysis/computation. Some of the frequently used operators are shifting operator
(�), forward difference (∆), backward difference (∇), central difference operator (�), etc.
Suppose the values of � and � are arguments and entries respectively. Let us consider the functional values or entries
�
??????=�(�
??????),??????=0,1,2,⋯,� be given for (�+1) equidistant arguments �
??????=�
0+??????ℎ,??????=0,1,2,⋯,�, where ℎ being
the interval of differencing or spacing.
Arguments (�) Entries �=??????(�)
�
0 �
0=�(�
0)
�
1=�
0+ℎ �
1=�(�
1)=�(�
0+ℎ)
�
2=�
1+ℎ=�
0+2ℎ �
2=�(�
2)=�(�
1+ℎ)=�(�
0+2ℎ)
⋮ ⋮
�
??????=�
??????−1+ℎ=�
0+??????ℎ �
??????=�(�
??????)=�(�
??????−1+ℎ)=�(�
0+??????ℎ)
⋮ ⋮
�
�=�
�−1+ℎ=�
0+�ℎ �
�=�(�
�)=�(�
�−1+ℎ)=�(�
0+�ℎ)
FINITE DIFFERENCE OPERATOR
Different types of finite difference operators are defined, among them forward difference, backward difference and central
difference operators are widely used.
FORWARD DIFFERENCE OPERATOR (∆)
Let us consider the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,� be given for (�+1) equidistant arguments
�
??????=�
0+??????ℎ,??????=0,1,2,⋯,�, where ℎ being the interval of differencing or spacing.
The forward difference operator is denoted by ∆ and is defined as
∆�(�)=�(�+ℎ)−�(�) i.e.,∆�
??????=�
??????+ℎ−�
??????
When �=�
??????, then ∆�
??????=�
??????+1−�
?????? i.e., ∆�(�
??????)=�(�
??????+ℎ)−�(�
??????) ; ??????=0,1,2,⋯,(�−1).
In particular, ∆�
0=�
1−�
0,∆�
1=�
2−�
1,⋯,∆�
�−1=�
�−�
�−1. These are called first order forward differences.
The differences of the first order forward differences are called second order differences. The second order forward
difference is denoted by ∆
2
and is defined as
∆
�
�(�)=∆(∆�(�))=∆�(�+ℎ)−∆�(�)
={�(�+2ℎ)−�(�+ℎ)}−{�(�+ℎ)−�(�)}
=�(�+2ℎ)−2�(�+ℎ)+�(�)

7. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 6
The second order forward differences are
∆
2
�
??????=∆(∆�
??????)=∆�
??????+1−∆�
??????=�
??????+2−2�
??????+1+�
??????=∑(−1)
??????
(
2
??????
)�
??????+2−??????
2
??????=0
; ??????=0,1,2,⋯,(�−2)
The third order forward differences are also defined in similar manner, i.e.,
∆
3
�
??????=∆(∆
2
�
??????)=∆
2
�
??????+1−∆
2
�
??????=(�
??????+3−2�
??????+2+�
??????+1)−(�
??????+2−2�
??????+1+�
??????)
=�
??????+3−3�
??????+2+3�
??????+1−�
??????=∑(−1)
??????
(
3
??????
)�
??????+3−??????
3
??????=0
; ??????=0,1,2,⋯,(�−3)
Similarly, higher order differences can be defined.
In general,
∆
??????
�
??????=∆(∆
??????−1
�
??????)=∆
??????−1
�
??????+1−∆
??????−1
�
?????? or, [∆
??????
�
??????=∆
??????−1
(∆�
??????)]
=(
??????
0
)�
??????+??????−(
??????
1
)�
??????+??????−1+⋯+(−1)
??????
(
??????
??????
)�
??????+??????−??????+⋯+(−1)
??????−1
(
??????
??????−1
)�
??????+1+(−1)
??????
(
??????
??????
)�
??????
=∑(−1)
??????
(
??????
??????
)�
??????+??????−??????
??????
??????=0
; ??????=0,1,2,⋯,(�−??????) ; ??????=1,2,⋯,�.
We may write
∆
�
�
0=(
�
0
)�
�−(
�
1
)�
�−1+⋯+(−1)
??????
(
�
??????
)�
�−??????+⋯+(−1)
�
(
�
�
)�
0=∑(−1)
??????
(
�
??????
)�
�−??????
�
??????=0

NOTE:
∆
0
�(�)=�(�) and ∆
??????
�(�)=0 for all ??????>�.
FINITE / FORWARD DIFFERENCE TABLE
All the forward differences can be represented in a tabular form, called the forward difference or diagonal difference
table.
� �=??????(�) ∆� ∆
�
� ∆
�
�
�
0 �
0=�(�
0)
∆�
0=�
1−�
0
�
1=�
0+ℎ �
1=�(�
1) ∆
2
�
0=∆�
1−∆�
0
∆�
1=�
2−�
1 ∆
3
�
0=∆
2
�
1−∆
2
�
0
�
2=�
0+2ℎ �
2=�(�
2) ∆
2
�
1=∆�
2−∆�
1
∆�
2=�
3−�
2
�
3=�
0+3ℎ �
3=�(�
3)

8. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 7
PROPERTIES OF ∆
Some common properties of forward difference operator are as follows:
(??????) ∆�=�−�=0, where � is a constant.
(????????????) ∆{��(�)}=��(�+ℎ)−��(�)=�{�(�+ℎ)−�(�)}=�∆�(�).
(??????????????????) ∆{�
1(�)+�
2(�)+⋯+�
�(�)}={�
1(�+ℎ)+�
2(�+ℎ)+⋯+�
�(�+ℎ)}−{�
1(�)+�
2(�)+⋯+�
�(�)}
=∆�
1(�)+∆�
2(�)+⋯+∆�
�(�).
(????????????) ∆{�
1�
1(�)+�
2�
2(�)+⋯+�
��
�(�)}=�
1∆�
1(�)+�
2∆�
2(�)+⋯+�
�∆�
�(�).
(??????) For any two positive integers � and �,
∆
�
∆
�
�(�)=∆
�+�
�(�)=∆
�
∆
�
�(�)=∆
??????
∆
�+�−??????
�(�) ; ??????=0,1,2,⋯,� or �.
(????????????) ∆
??????
�(�
??????)=(
??????
0
)�(�
??????+??????ℎ)−(
??????
1
)�{�
??????+(??????−1)ℎ}+⋯+(−1)
??????
(
??????
??????
)�{�
??????+(??????−??????)ℎ}+⋯+(−1)
??????
(
??????
??????
)�(�
??????)
=∑(−1)
??????
(
??????
??????
)�{�
??????+(??????−??????)ℎ}
??????
??????=0
; ??????=0,1,2,⋯,(�−??????), ??????=1,2,⋯,�
(??????????????????) ∆{�(�)�(�)}=�(�+ℎ)�(�+ℎ)−�(�)�(�)
=�(�+ℎ)�(�+ℎ)−�(�+ℎ)�(�)+�(�+ℎ)�(�)−�(�)�(�)
=�(�+ℎ){�(�+ℎ)−�(�)}+�(�){�(�+ℎ)−�(�)}
=�(�+ℎ)∆�(�)+�(�)∆�(�).
Also, ∆{�(�)�(�)}=�(�)∆�(�)+�(�+ℎ)∆�(�)=�(�)∆�(�)+�(�)∆�(�)+∆�(�)∆�(�).
(????????????????????????) ∆{
�(�)
�(�)
}=
�(�)∆�(�)−�(�)∆�(�)
�(�+ℎ)�(�)
, �(�)≠0.
∆{
�(�)
�(�)
}=
�(�+ℎ)
�(�+ℎ)
−
�(�)
�(�)
=
�(�+ℎ)�(�)−�(�)�(�+ℎ)
�(�+ℎ)�(�)

=
�(�){�(�+ℎ)−�(�)}−�(�){�(�+ℎ)−�(�)}
�(�+ℎ)�(�)

=
�(�)∆�(�)−�(�)∆�(�)
�(�+ℎ)�(�)
, �(�)≠0.
In particular, when the numerator is 1, then
∆{
1
�(�)
}=−
∆�(�)
�(�+ℎ)�(�)
, �(�)≠0 [∵ ∆�(�)=∆(1)=0]

9. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 8
PROPAGATION OF ERROR IN A DIFFERENCE TABLE
If any entry of the difference table is erroneous, then this error spread over the table in convex manner.
Let us assumed that �
3 be erroneous and the amount of the error be �.
The propagation of error in a difference table is illustrated in the following table.
� � ∆� ∆
�
� ∆
�
� ∆
�
� ∆
�
� ∆
�
�
�
0 �
0
∆�
0
�
1 �
1 ∆
2
�
0
∆�
1 ∆
3
�
0+�
�
2 �
2 ∆
2
�
1+� ∆
4
�
0−4�
∆�
2+� ∆
3
�
1−3� ∆
5
�
0+10�
�
3 �
3+� ∆
2
�
2−2� ∆
4
�
1+6� ∆
6
�
0−20�
∆�
3−� ∆
3
�
2+3� ∆
5
�
1−10�
�
4 �
4 ∆
2
�
3+� ∆
4
�
2−4�
∆�
4 ∆
3
�
3−�
�
5 �
5 ∆
2
�
4
∆�
5
�
6 �
6

(i) The error increases with the order of the differences.
(ii) The error is maximum (in magnitude) along the horizontal line through the erroneous tabulated value.
(iii) In the ??????-th difference column, the coefficients of errors are the binomial coefficients in the expansion of (1−�)
??????
.
In particular, the errors in the second difference column are �,−2�,�, in the third difference column these are
�,−3�,3�,�, and so on.
(iv) The algebraic sum of errors in any complete column is zero.
If there is any error in a single entry of the table, then we can detect and correct it from the difference table.

10. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 9
BACKWARD DIFFERENCE OPERATOR (??????)
Let us consider the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,� be given for (�+1) equidistant arguments
�
??????=�
0+??????ℎ,??????=0,1,2,⋯,�, where ℎ being the interval of differencing or spacing.
The backward difference operator is denoted by ∇ and is defined as
∇�(�)=�(�)−�(�−ℎ) i.e.,∇�
??????=�
??????−�
??????−ℎ
When �=�
??????, then ∇�
??????=�
??????−�
??????−1 i.e., ∇�(�
??????)=�(�
??????)−�(�
??????−ℎ) ; ??????=�,( �−1),⋯,3,2,1.
In particular, ∇�
1=�
1−�
0,∇�
2=�
2−�
1,⋯,∇�
�=�
�−�
�−1. These are called first order backward differences.
The differences of the first order backward differences are called second order backward differences. The second order
backward difference is denoted by ∇
2
and is defined as
∇
2
�(�)=∇(∇�(�))=∇�(�)−∇�(�−ℎ)={�(�)−�(�−ℎ)}−{�(�−ℎ)−�(�−2ℎ)}
=�(�)−2�(�−ℎ)+�(�−2ℎ)
The second order backward differences are
∇
2
�
??????=∇(∇�
??????)=∇�
??????−∇�
??????−1=�
??????−2�
??????−1+�
??????−2=∑(−1)
??????
(
2
??????
)�
??????−??????
2
??????=0
; ??????=�,( �−1),⋯,3,2.
The third order backward differences are also defined in similar manner, i.e.,
∇
3
�
??????=∇(∇
2
�
??????)=∇
2
�
??????−∇
2
�
??????−1=(�
??????−2�
??????−1+�
??????−2)−(�
??????−1−2�
??????−2+�
??????−3)
=�
??????−3�
??????−1+3�
??????−2−�
??????−3=∑(−1)
??????
(
3
??????
)�
??????−??????
3
??????=0
; ??????=�,( �−1),⋯,4,3.
Similarly, higher order differences can be defined.
In general,
∇
??????
�
??????=∇(∇
??????−1
�
??????)=∇
??????−1
�
??????−∇
??????−1
�
??????−1 or, [∇
??????
�
??????=∇
??????−1
(∆�
??????)]
=(
??????
0
)�
??????−(
??????
1
)�
??????−1+⋯+(−1)
??????
(
??????
??????
)�
??????−??????+⋯+(−1)
??????−1
(
??????
??????−1
)�
??????−??????+1+(−1)
??????
(
??????
??????
)�
??????−??????
=∑(−1)
??????
(
??????
??????
)�
??????−??????
??????
??????=0
; ??????=�,( �−1),⋯,(??????+1),?????? ; ??????=1,2,⋯,�.
We may write
∇
�
�
�=(
�
0
)�
�−(
�
1
)�
�−1+⋯+(−1)
??????
(
�
??????
)�
�−??????+⋯+(−1)
�
(
�
�
)�
0=∑(−1)
??????
(
�
??????
)�
�−??????
�
??????=0

NOTE:
∇
0
�(�)=�(�) and ∇
??????
�(�)=0 for all ??????>�.

11. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 10
BACKWARD DIFFERENCE TABLE
Backward differences can be represented in a tabular form, called the backward difference or horizontal difference table.
� �=??????(�) ??????� ??????
�
� ??????
�
�
�
0 �
0=�(�
0)
�
1=�
0+ℎ �
1=�(�
1) ∇�
1=�
1−�
0
�
2=�
0+2ℎ �
2=�(�
2) ∇�
2=�
2−�
1 ∇
2
�
2=∇�
2−∇�
1
�
3=�
0+3ℎ �
3=�(�
3) ∇�
3=�
3−�
2 ∇
2
�
3=∇�
3−∇�
2 ∇
3
�
3=∇
2
�
3−∇
2
�
2

It is observed from the forward and backward difference tables that for a given table of values both the tables are same.
Practically, there are no differences among the values of the tables, but theoretically they have separate significant.
SHIFTING OPERATOR (�)
Let us consider the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,� be given for (�+1) equidistant arguments
�
??????=�
0+??????ℎ,??????=0,1,2,⋯,�, where ℎ being the interval of differencing or spacing.
The shifting operator is denoted by � and is defined as
��(�)=�(�+ℎ)
When �=�
??????, ��(�
??????)=�(�
??????+ℎ) or, �(�
??????)=�
??????+1 ; ??????=0,1,2,⋯,(�−1).
Now, �
2
�(�)=�{��(�)}=�{�(�+ℎ)}=�(�+2ℎ) or, �
2
(�
??????)=�
??????+2 ; ??????=0,1,2,⋯,(�−2).
In general,
�
??????
�(�)=�
??????−1
[��(�)]=�
??????−1
[�(�+ℎ)]=⋯=�
??????−??????
[�(�+??????ℎ)]=⋯=��[�+(??????−1)ℎ]=�(�+??????ℎ)
i.e., �
??????
(�
??????)=�
??????+?????? ; ??????=0,1,2,⋯,(�−??????); ??????=1,2,⋯,�.
We may write �
�
�(�)=�(�+�ℎ) or, �
�
(�
0)=�
�.
The inverse shifting operator denoted by �
−1
and is defined as
�
−1
�(�)=�(�−ℎ).
When �=�
??????, �
−1
�(�
??????)=�(�
??????−ℎ) or, �
−1
(�
??????)=�
??????−1 ; ??????=�,( �−1),⋯,3,2,1.
Now, �
−2
�(�)=�
−1
{�
−1
�(�)}=�
−1
{�(�−ℎ)}=�(�−2ℎ) or, �
−2
(�
??????)=�
??????−2 ; ??????=�,( �−1),⋯,3,2.
In general,
�
−??????
�(�)=�
−(??????−1)
[�(�−ℎ)]=⋯=�
−(??????−??????)
[�(�−??????ℎ)]=⋯=�
−1
�{�−(??????−1)ℎ}=�(�−??????ℎ)
i.e., �
−??????
(�
??????)=�
??????−?????? ; ??????=�,( �−1),⋯,(??????+1),?????? ; ??????=1,2,⋯,�.
We may write �
−�
�(�)=�(�−�ℎ) or, �
−�
(�
�)=�
0.

12. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 11
PROPERTIES OF �
Some common properties of shifting operator are as follows:
(??????) ��=�−�=0, where � is a constant.
(????????????) �{��(�)}=��(�+ℎ)=���(�).
(??????????????????) �{�
1(�)+�
2(�)+⋯+�
�(�)}={�
1(�+ℎ)+�
2(�+ℎ)+⋯+�
�(�+ℎ)}
=��
1(�)+��
2(�)+⋯+��
�(�).
(????????????) �{�
1�
1(�)+�
2�
2(�)+⋯+�
��
�(�)}=�
1��
1(�)+�
2��
2(�)+⋯+�
���
�(�).
(??????) For any two positive integers � and �,
�
�
�
�
�(�)=�
�+�
�(�)=�
�
�
�
�(�)=�
??????
�
�+�−??????
�(�) ; ??????=0,1,2,⋯,� or �.
(????????????) (�
�
)
�
�(�)=(�
�
)
�
�(�)=�
��
�(�)
(??????????????????) �
�
�
−�
�(�)=�
−�
�
�
�(�)=�(�)
(????????????????????????) �
??????
�(�
??????)=�(�
??????+??????ℎ) ; ??????=0,1,⋯,(�−??????)
and �
−??????
�(�
??????)=�(�
??????−??????ℎ); ??????=�,⋯,(??????+1),?????? ; ??????=1,2,⋯,�
(??????�) �{�(�)�(�)}=�(�+ℎ)�(�+ℎ)=��(�)��(�)
(�) �{
�(�)
�(�)
}=
�(�+ℎ)
�(�+ℎ)
=
��(�)
��(�)
, �(�)≠0.

DIFFERENTIAL OPERATOR (�)
The differential operator gives the derivative of function. The differential operator is denoted by � and is defined as
��(�)=
�
��
�(�)=�
(1)
(�)
�
2
�(�)=
�
2
��
2
�(�)=�
(2)
(�)
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
�
�
�(�)=
�
�
��
�
�(�)=�
(�)
(�)

13. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 12
RELATIONS AMONG OPERATORS
RELATIONSHIP BETWEEN ∆ AND ??????
Let us consider the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,� be given for (�+1) equidistant arguments
�
??????=�
0+??????ℎ,??????=0,1,2,⋯,�, where ℎ being the interval of differencing or spacing.
The forward difference operator of �(�) and backward difference operator of �(�+ℎ) are
∆�(�)=�(�+ℎ)−�(�) and ∇�(�+ℎ)=�(�+ℎ)−�(�)
∴ ∆�(�)=�(�+ℎ)−�(�)=∇�(�+ℎ)
Now, ∆
�
�(�)=∆(∆�(�))=∆�(�+ℎ)−∆�(�)={�(�+2ℎ)−�(�+ℎ)}−{�(�+ℎ)−�(�)}
=�(�+2ℎ)−2�(�+ℎ)+�(�)
∇
2
�(�+2ℎ)=∇(∇�(�+2ℎ))=∇�(�+2ℎ)−∇�(�+ℎ)={�(�+2ℎ)−�(�+ℎ)}−{�(�+ℎ)−�(�)}
=�(�+2ℎ)−2�(�+ℎ)+�(�)
∴ ∆
�
�(�)=�(�+2ℎ)−2�(�+ℎ)+�(�)=∇
2
�(�+2ℎ)
In general,
∆
??????
�(�)=(
??????
0
)�(�+??????ℎ)−(
??????
1
)�{�+(??????−1)ℎ}+⋯+(−1)
??????
(
??????
??????
)�{�+(??????−??????)ℎ}+⋯+(−1)
??????
(
??????
??????
)�(�)
=∑(−1)
??????
(
??????
??????
)�{�+(??????−??????)ℎ}
??????
??????=0
=∇
??????
�(�+??????ℎ); ??????=1,2,⋯,�.
∴ ∆
??????
�(�)=∑(−1)
??????
(
??????
??????
)�{�+(??????−??????)ℎ}
??????
??????=0
=∇
??????
�(�+??????ℎ); ??????=1,2,⋯,� and
∆
??????
�(�)=∑(−1)
??????
(
�
??????
)�{�+(�−??????)ℎ}
�
??????=0
=∇
�
�(�+�ℎ)
When �=�
??????, ∆�
??????=�
??????+1−�
??????=∇�
??????+1 ; ??????=0,1,2,⋯,(�−1).
∆
2
�
??????=�
??????+2−2�
??????+1+�
??????=∑(−1)
??????
(
2
??????
)�
??????+2−??????
2
??????=0
=∇
2
�
??????+2 ; ??????=0,1,2,⋯,(�−2)
In general, ∆
??????
�
??????=(
??????
0
)�
??????+??????−(
??????
1
)�
??????+??????−1+⋯+(−1)
??????
(
??????
??????
)�
??????+??????−??????+⋯+(−1)
??????
(
??????
??????
)�
??????
=∑(−1)
??????
(
??????
??????
)�
??????+??????−??????
??????
??????=0
=∇
p
�
??????+?????? ; ??????=0,1,2,⋯,(�−??????) ; ??????=1,2,⋯,�.
∴ ∆
??????
�
??????=∇
p
�
??????+?????? ; ??????=0,1,2,⋯,(�−??????) ; ??????=1,2,⋯,� and
∆
�
�
0=∇
n
�
� .

14. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 13
RELATIONSHIP BETWEEN ∆ AND �
Let us consider the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,� be given for (�+1) equidistant arguments
�
??????=�
0+??????ℎ,??????=0,1,2,⋯,�, where ℎ being the interval of differencing or spacing.
The shifting and forward difference operator of �(�) are ��(�)=�(�+ℎ) and ∆�(�)=�(�+ℎ)−�(�).
Now, ∆�(�)=�(�+ℎ)−�(�)=��(�)−�(�)=(�−1)�(�) and
(∆+1)�(�)=∆�(�)+�(�)=�(�+ℎ)−�(�)+�(�)=�(�+ℎ)=��(�)
∴ ∆≡�−1 or, �≡∆+1
Again, ∆
�
�(�)=∆(∆�(�))=∆�(�+ℎ)−∆�(�)={�(�+2ℎ)−�(�+ℎ)}−{�(�+ℎ)−�(�)}
=�(�+2ℎ)−2�(�+ℎ)+�(�)
=�
2
�(�)−2��(�)+�(�)=(�−1)
2
�(�)
(∆+1)
2
�(�)=∆
�
�(�)+2∆�(�)+�(�)
={�(�+2ℎ)−2�(�+ℎ)+�(�)}+2{�(�+ℎ)−�(�)}+�(�)
=�(�+2ℎ)=�
2
�(�)
∴ ∆
2
≡(�−1)
2
or, �
2
≡(∆+1)
2

In general,
∆
??????
�(�)=(
??????
0
)�(�+??????ℎ)−(
??????
1
)�{�+(??????−1)ℎ}+⋯+(−1)
??????
(
??????
??????
)�{�+(??????−??????)ℎ}+⋯+(−1)
??????
(
??????
??????
)�(�)
=(
??????
0
)�
??????
�(�)−(
??????
1
)�
(??????−1)
�(�)+⋯+(−1)
??????
(
??????
??????
)�
(??????−??????)
�(�)+⋯+(−1)
??????
(
??????
??????
)�(�)
=∑(−1)
??????
(
??????
??????
)�{�+(??????−??????)ℎ}
??????
??????=0
=∑(−1)
??????
(
??????
??????
)�
(??????−??????)
�(�)
??????
??????=0
=(�−1)
??????
�(�); ??????=1,2,⋯,�.
∴ ∆
??????
≡(�−1)
??????
or, �
??????
≡(∆+1)
??????
; ??????=1,2,⋯,� and
∆
??????
≡(�−1)
�
or, �
�
≡(∆+1)
�

When �=�
??????, ∆�
??????=�
??????+1−�
??????=��
??????−�
??????=(�−1)�
?????? ; ??????=0,1,2,⋯,(�−1).
∆
2
�
??????=�
??????+2−2�
??????+1+�
??????=∑(−1)
??????
(
2
??????
)�
(2−??????)
�
??????
2
??????=0
=(�−1)
2
�
?????? ; ??????=0,1,2,⋯,(�−2)
In general, ∆
??????
�
??????=(
??????
0
)�
??????+??????−(
??????
1
)�
??????+??????−1+⋯+(−1)
??????
(
??????
??????
)�
??????+??????−??????+⋯+(−1)
??????
(
??????
??????
)�
??????
=∑(−1)
??????
(
??????
??????
)�
(??????−??????)
�
??????
??????
??????=0
=(�−1)
??????
�
?????? ; ??????=0,1,2,⋯,(�−??????) ; ??????=1,2,⋯,�.
∴ ∆
??????
≡(�−1)
??????
or, �
??????
≡(∆+1)
??????
; ??????=1,2,⋯,� and ∆
??????
≡(�−1)
�
or, �
�
≡(∆+1)
�

15. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 14
RELATIONSHIP BETWEEN ?????? AND �
Let us consider the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,� be given for (�+1) equidistant arguments
�
??????=�
0+??????ℎ,??????=0,1,2,⋯,�, where ℎ being the interval of differencing or spacing.
The shifting and backward difference operator of �(�) are �
−1
�(�)=�(�−ℎ) and ∇�(�)=�(�)−�(�−ℎ).
Now, ∇�(�)=�(�)−�(�−ℎ)=�(�)−�
−1
�(�)=(1−�
−1
)�(�)
∴ ∇≡1−�
−1
or, �∇≡�−1 or, �∇≡∆
Again, ∇
2
�(�)=∇(∇�(�))=∇�(�)−∇�(�−ℎ)={�(�)−�(�−ℎ)}−{�(�−ℎ)−�(�−2ℎ)}
=�(�)−2�(�−ℎ)+�(�−2ℎ)
=�(�)−2�
−1
�(�)+�
−2
�(�)=(1−�
−1
)
2
�(�)
∴ ∇
2
≡(1−�
−1
)
2
or, �
2
∇
2
≡(�−1)
2
or, �
2
∇
2
≡∆
2

In general,
∇
??????
�(�)=(
??????
0
)�(�)−(
??????
1
)�(�−ℎ)+⋯+(−1)
??????
(
??????
??????
)�(�−??????ℎ)+⋯+(−1)
??????
(
??????
??????
)�(�−??????ℎ)
=(
??????
0
)�(�)−(
??????
1
)�
−1
�(�)+⋯+(−1)
??????
(
??????
??????
)�
−??????
�(�)+⋯+(−1)
??????
(
??????
??????
)�
−??????
�(�)
=∑(−1)
??????
(
??????
??????
)�(�−??????ℎ)
??????
??????=0
=∑(−1)
??????
(
??????
??????
)�
−??????
�(�)
??????
??????=0
=(1−�
−1
)
??????
�(�); ??????=1,2,⋯,�.
∴ ∇
??????
≡(1−�
−1
)
??????
or, �
??????
∇
??????
≡(�−1)
??????
or, �
??????
∇
??????
≡∆
??????
; ??????=1,2,⋯,� and
∇
�
≡(1−�
−1
)
�
or, �
�
∇
�
≡(�−1)
�
or, �
�
∇
�
≡∆
�

When �=�
??????, ∇�
??????=�
??????−�
??????−1=�
??????−�
−1
�
??????=(1−�
−1
)�
?????? ; ??????=�,( �−1),⋯,3,2,1.
∇
2
�
??????=∇�
??????−∇�
??????−1=�
??????−2�
??????−1+�
??????−2=∑(−1)
??????
(
2
??????
)�
−??????
�
??????
2
??????=0
=(1−�
−1
)
2
�
?????? ; ??????=�,( �−1),⋯,3,2
In general, ∆
??????
�
??????=(
??????
0
)�
??????−(
??????
1
)�
??????−1+⋯+(−1)
??????
(
??????
??????
)�
??????−??????+⋯+(−1)
??????
(
??????
??????
)�
??????−??????
=∑(−1)
??????
(
??????
??????
)�
−??????
�
??????
??????
??????=0
=(1−�
−1
)
??????
�
?????? ; ??????=�,( �−1),⋯,(??????+1),?????? ; ??????=1,2,⋯,�.
∴ ∇
??????
≡(1−�
−1
)
??????
or, �
??????
∇
??????
≡(�−1)
??????
or, �
??????
∇
??????
≡∆
??????
; ??????=1,2,⋯,� and
∇
�
≡(1−�
−1
)
�
or, �
�
∇
�
≡(�−1)
�
or, �
�
∇
�
≡∆
�

16. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 15
RELATIONSHIP BETWEEN �, � AND ∆
Let us consider the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,� be given for (�+1) equidistant arguments
�
??????=�
0+??????ℎ,??????=0,1,2,⋯,�, where ℎ being the interval of differencing or spacing.
The shifting operator of �(�) is given by ��(�)=�(�+ℎ). By Taylor's series, we get
��(�)=�(�+ℎ)=�(�)+ℎ�
(1)
(�)+
ℎ
2
2!
�
(2)
(�)+
ℎ
3
3!
�
(3)
(�)+⋯
=�(�)+ℎ��(�)+
ℎ
2
2!
�
2
�(�)+
ℎ
3
3!
�
3
�(�)+⋯
=[1+ℎ�+
ℎ
2
2!
�
2
+
ℎ
3
3!
�
3
+⋯]�(�)
=�
ℎ�
�(�)
∴ �≡�
ℎ�
or, ℎ�≡ln� or, ℎ�≡ln(∆+1) or, ℎ�≡−ln(1−∇)
�≡
1
ℎ
ln(1+∆)≡
1
ℎ
[∆−
∆
2
2
+
∆
3
3
−
∆
4
4
+⋯]
∴ ��(�)=
�
��
�(�)=
1
ℎ
[∆�(�)−
∆
2
�(�)
2
+
∆
3
�(�)
3
−
∆
4
�(�)
4
+⋯]
FORWARD DIFFERENCE OF A POLYNOMIAL
Let �(�)=�
��
�
+�
�−1�
�−1
+⋯+�
2�
2
+�
1�+�
0 (�
�≠0) be a polynomial in � of degree �, where �
??????'s be the
given coefficients. Let us consider ℎ be the interval of differencing.
Now, ∆�(�)=�(�+ℎ)−�(�)
=�
�
[(�+ℎ)
�
−�
�
]+�
�−1
[(�+ℎ)
�−1
−�
�−1
]+⋯+�
2
[(�+ℎ)
2
−�
2
]+�
1
[(�+ℎ)−�]
=�
�[�
�
+(
�
1
)�
�−1
ℎ+(
�
2
)�
�−2
ℎ
2
+⋯+(
�
�−1
)�ℎ
�−1
+ℎ
�
−�
�
]
+�
�−1[�
�−1
+(
�−1
1
)�
�−2
ℎ+(
�−1
2
)�
�−3
ℎ
2
+⋯+(
�−1
�−2
)�ℎ
�−2
+ℎ
�−1
−�
�−1
]
+⋯+�
2[�
2
+(
2
1
)�ℎ+ℎ
2
−�
2
]+�
1ℎ
=�
��ℎ�
�−1
+[�
�(
�
2
)ℎ
2
+�
�−1(
�−1
1
)ℎ]�
�−2
+⋯
+[�
�(
�
�−1
)ℎ
�−1
+�
�−1(
�−1
�−2
)ℎ
�−2
+⋯+�
2(
2
1
)ℎ]�
+[�
�ℎ
�
+�
�−1ℎ
�−1
+⋯+�
2ℎ
2
+�
1ℎ]
=�
�−1
(1)
�
�−1
+�
�−2
(1)
�
�−2
+⋯+�
2
(1)
�
2
+�
1
(1)
�+�
0
(1)

The coefficients �
??????
(1)
's, ??????=1,2,⋯,(�−1) are suitable constants and �
�−1
(1)
=�
��ℎ.
The first difference of a polynomial of the �-th degree is thus another polynomial of degree (�−1).

17. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 16
Similarly,
∆
2
�(�)=�
�−1
(1)
[(�+ℎ)
�−1
−�
�−1
]+�
�−2
(1)
[(�+ℎ)
�−2
−�
�−2
]+⋯+�
2
(1)
[(�+ℎ)
2
−�
2
]+�
1
(1)
[(�+ℎ)−�]
=�
�−2
(2)
�
�−2
+�
�−3
(2)
�
�−3
+⋯+�
2
(2)
�
2
+�
1
(2)
�+�
0
(2)

The coefficients �
??????
(2)
's, ??????=1,2,⋯,(�−2) are suitable constants and �
�−2
(2)
=�
�−1
(1)
(�−1)ℎ=�
�
{�(�−1)}ℎ
2
.
The second difference of a polynomial of the �-th degree is thus another polynomial of degree (�−2).
In general, ∆
??????
�(�)=�
�−??????+1
(??????−1)
[(�+ℎ)
�−??????+1
−�
�−??????+1
]+⋯+�
2
(??????−1)
[(�+ℎ)
2
−�
2
]+�
1
(??????−1)
[(�+ℎ)−�]
=�
�−??????
(??????)
�
�−??????
+�
�−??????−1
(??????)
�
�−3
+⋯+�
2
(??????)
�
2
+�
1
(??????)
�+�
0
(??????)

The coefficients �
??????
(??????)
's, ??????=1,2,⋯,(�−??????) are suitable constants and
�
�−??????
(??????)
=�
�−??????+1
(??????−1)
(�−??????+1)ℎ=�
�
{�(�−1)⋯(�−??????+1)}ℎ
??????
.
The ??????-th difference of a polynomial of the �-th degree is thus another polynomial of degree (�−??????).
In this manner we arrive at a polynomial of zero degree for the �-th difference; that is，
∆
�
�(�)=�
0
(�)
=�
1
(�−1)
∙1∙ℎ=�
�
{�(�−1)⋯3∙2∙1}ℎ
�
=�
��!ℎ
�

The �-th difference is therefore constant, and all higher differences are zero.
Hence, ∆
??????
&#3627408467;(&#3627408485;),??????<&#3627408475; is a polynomial of degree (&#3627408475;−??????),
∆
&#3627408475;
&#3627408467;(&#3627408485;)=&#3627408462;
&#3627408475;
{&#3627408475;(&#3627408475;−1)⋯3∙2∙1}ℎ
&#3627408475;
=&#3627408462;
&#3627408475;&#3627408475;!ℎ
&#3627408475;
=constant, and
∆
??????
&#3627408467;(&#3627408485;)=0 if ??????>&#3627408475;.
NOTE:
The &#3627408475;-th differences of a polynomial of the &#3627408475;-th degree are constant when the values of the independent variable are taken
in arithmetic progression or with equal interval of differencing.
The converse of this proposition is also true, that is, if the &#3627408475;-th differences of a tabulated function are constant when the
values of the independent variable are taken in arithmetic progression, the function is a polynomial of degree &#3627408475;.

18. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 17
BACKWARD DIFFERENCE OF A POLYNOMIAL
Let &#3627408467;(&#3627408485;)=&#3627408462;
&#3627408475;&#3627408485;
&#3627408475;
+&#3627408462;
&#3627408475;−1&#3627408485;
&#3627408475;−1
+⋯+&#3627408462;
2&#3627408485;
2
+&#3627408462;
1&#3627408485;+&#3627408462;
0 (&#3627408462;
&#3627408475;≠0) be a polynomial in &#3627408485; of degree &#3627408475;, where &#3627408462;
??????'s be the
given coefficients. Let us consider ℎ be the interval of differencing.
Now, ∇&#3627408467;(&#3627408485;)=&#3627408467;(&#3627408485;)−&#3627408467;(&#3627408485;−ℎ)
=&#3627408462;
&#3627408475;
[&#3627408485;
&#3627408475;
−(&#3627408485;−ℎ)
&#3627408475;
]+&#3627408462;
&#3627408475;−1
[&#3627408485;
&#3627408475;−1
−(&#3627408485;−ℎ)
&#3627408475;−1
]+⋯+&#3627408462;
2
[&#3627408485;
2
−(&#3627408485;−ℎ)
2
]+&#3627408462;
1
[&#3627408485;−(&#3627408485;−ℎ)]
=&#3627408462;
&#3627408475;[&#3627408485;
&#3627408475;
−&#3627408485;
&#3627408475;
+(
&#3627408475;
1
)&#3627408485;
&#3627408475;−1
ℎ−(
&#3627408475;
2
)&#3627408485;
&#3627408475;−2
ℎ
2
+⋯+(−1)
&#3627408475;
(
&#3627408475;
&#3627408475;−1
)&#3627408485;ℎ
&#3627408475;−1
+(−1)
&#3627408475;+1
(
&#3627408475;
&#3627408475;
)ℎ
&#3627408475;
]
+&#3627408462;
&#3627408475;−1[&#3627408485;
&#3627408475;−1
−&#3627408485;
&#3627408475;−1
+(
&#3627408475;−1
1
)&#3627408485;
&#3627408475;−2
ℎ+⋯+(−1)
&#3627408475;−1
(
&#3627408475;−1
&#3627408475;−2
)&#3627408485;ℎ
&#3627408475;−2
+(−1)
&#3627408475;
ℎ
&#3627408475;−1
]+⋯
+&#3627408462;
2[&#3627408485;
2
−&#3627408485;
2
+(
2
1
)&#3627408485;ℎ−ℎ
2
]+&#3627408462;
1ℎ
=&#3627408462;
&#3627408475;&#3627408475;ℎ&#3627408485;
&#3627408475;−1
+[−&#3627408462;
&#3627408475;(
&#3627408475;
2
)ℎ
2
+&#3627408462;
&#3627408475;−1(
&#3627408475;−1
1
)ℎ]&#3627408485;
&#3627408475;−2
+⋯
+[&#3627408462;
&#3627408475;(−1)
&#3627408475;
(
&#3627408475;
&#3627408475;−1
)ℎ
&#3627408475;−1
+&#3627408462;
&#3627408475;−1(−1)
&#3627408475;−1
(
&#3627408475;−1
&#3627408475;−2
)ℎ
&#3627408475;−2
+⋯+&#3627408462;
2(−1)
2−1
(
2
1
)ℎ]&#3627408485;
+[&#3627408462;
&#3627408475;(−1)
&#3627408475;+1
ℎ
&#3627408475;
+&#3627408462;
&#3627408475;−1(−1)
&#3627408475;
ℎ
&#3627408475;−1
+⋯−&#3627408462;
2ℎ
2
+&#3627408462;
1ℎ]
=&#3627408462;
&#3627408475;−1
(1)
&#3627408485;
&#3627408475;−1
+&#3627408462;
&#3627408475;−2
(1)
&#3627408485;
&#3627408475;−2
+⋯+&#3627408462;
2
(1)
&#3627408485;
2
+&#3627408462;
1
(1)
&#3627408485;+&#3627408462;
0
(1)

The coefficients &#3627408462;
??????
(1)
's are suitable constants and &#3627408462;
&#3627408475;−1
(1)
=&#3627408462;
&#3627408475;&#3627408475;ℎ.
The first backward difference of a polynomial of the &#3627408475;-th degree is thus another polynomial of degree (&#3627408475;−1).
Similarly,
∇
2
&#3627408467;(&#3627408485;)=&#3627408462;
&#3627408475;−1
(1)
[&#3627408485;
&#3627408475;−1
−(&#3627408485;−ℎ)
&#3627408475;−1
]+&#3627408462;
&#3627408475;−2
(1)
[&#3627408485;
&#3627408475;−2
−(&#3627408485;−ℎ)
&#3627408475;−2
]+⋯+&#3627408462;
2
(1)
[&#3627408485;
2
−(&#3627408485;−ℎ)
2
]+&#3627408462;
1
(1)
[&#3627408485;−(&#3627408485;−ℎ)]
=&#3627408462;
&#3627408475;−2
(2)
&#3627408485;
&#3627408475;−2
+&#3627408462;
&#3627408475;−3
(2)
&#3627408485;
&#3627408475;−3
+⋯+&#3627408462;
2
(2)
&#3627408485;
2
+&#3627408462;
1
(2)
&#3627408485;+&#3627408462;
0
(2)

The coefficients &#3627408462;
??????
(2)
's are suitable constants and &#3627408462;
&#3627408475;−2
(2)
=&#3627408462;
&#3627408475;−1
(1)
(&#3627408475;−1)ℎ=&#3627408462;
&#3627408475;
{&#3627408475;(&#3627408475;−1)}ℎ
2
.
The second backward difference of a polynomial of the &#3627408475;-th degree is thus another polynomial of degree (&#3627408475;−2).
In general, ∇
??????
&#3627408467;(&#3627408485;)=&#3627408462;
&#3627408475;−??????+1
(??????−1)
[&#3627408485;
&#3627408475;−??????+1
−(&#3627408485;−ℎ)
&#3627408475;−??????+1
]+⋯+&#3627408462;
2
(??????−1)
[&#3627408485;
2
−(&#3627408485;−ℎ)
2
]+&#3627408462;
1
(??????−1)
[&#3627408485;−(&#3627408485;−ℎ)]
=&#3627408462;
&#3627408475;−??????
(??????)
&#3627408485;
&#3627408475;−??????
+&#3627408462;
&#3627408475;−??????−1
(??????)
&#3627408485;
&#3627408475;−3
+⋯+&#3627408462;
2
(??????)
&#3627408485;
2
+&#3627408462;
1
(??????)
&#3627408485;+&#3627408462;
0
(??????)

The coefficients &#3627408462;
??????
(??????)
's are suitable constants and
&#3627408462;
&#3627408475;−??????
(??????)
=&#3627408462;
&#3627408475;−??????+1
(??????−1)
(&#3627408475;−??????+1)ℎ=&#3627408462;
&#3627408475;
{&#3627408475;(&#3627408475;−1)⋯(&#3627408475;−??????+1)}ℎ
??????
.
The ??????-th backward difference of a polynomial of the &#3627408475;-th degree is thus another polynomial of degree (&#3627408475;−??????).
In this manner we arrive at a polynomial of zero degree for the &#3627408475;-th backward difference; that is，
∇
&#3627408475;
&#3627408467;(&#3627408485;)=&#3627408462;
0
(&#3627408475;)
=&#3627408462;
1
(&#3627408475;−1)
∙1∙ℎ=&#3627408462;
&#3627408475;
{&#3627408475;(&#3627408475;−1)⋯3∙2∙1}ℎ
&#3627408475;
=&#3627408462;
&#3627408475;&#3627408475;!ℎ
&#3627408475;

The &#3627408475;-th backward difference is therefore constant, and all higher backward differences are zero.

19. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 18
Hence, ∇
??????
&#3627408467;(&#3627408485;),??????<&#3627408475; is a polynomial of degree (&#3627408475;−??????),
∇
&#3627408475;
&#3627408467;(&#3627408485;)=&#3627408462;
&#3627408475;
{&#3627408475;(&#3627408475;−1)⋯3∙2∙1}ℎ
&#3627408475;
=&#3627408462;
&#3627408475;&#3627408475;!ℎ
&#3627408475;
=constant, and
∇
??????
&#3627408467;(&#3627408485;)=0 if ??????>&#3627408475;.
NOTE:
The &#3627408475;-th backward differences of a polynomial of the &#3627408475;-th degree are constant when the values of the independent variable
are taken in arithmetic progression or with equal interval of differencing.
The converse of this proposition is also true, that is, if the &#3627408475;-th backward differences of a tabulated function are constant
when the values of the independent variable are taken in arithmetic progression, the function is a polynomial of degree &#3627408475;.
PROBLEM
Show that &#3627408440;∆≡∆&#3627408440; and ∆∇≡∇∆.
Solution:
&#3627408440;∆&#3627408467;(&#3627408485;)=&#3627408440;[∆&#3627408467;(&#3627408485;)]=&#3627408440;[&#3627408467;(&#3627408485;+ℎ)−&#3627408467;(&#3627408485;)]=&#3627408440;&#3627408467;(&#3627408485;+ℎ)−&#3627408440;&#3627408467;(&#3627408485;)=&#3627408467;(&#3627408485;+2ℎ)−&#3627408467;(&#3627408485;+ℎ)=∆&#3627408467;(&#3627408485;+ℎ)=∆&#3627408440;&#3627408467;(&#3627408485;)
∆∇&#3627408467;(&#3627408485;)=∆[&#3627408467;(&#3627408485;)−&#3627408467;(&#3627408485;−ℎ)]=∆&#3627408467;(&#3627408485;)−∆&#3627408467;(&#3627408485;−ℎ)
=[&#3627408467;(&#3627408485;+ℎ)−&#3627408467;(&#3627408485;)]−[&#3627408467;(&#3627408485;)−&#3627408467;(&#3627408485;−ℎ)]
=∇&#3627408467;(&#3627408485;+ℎ)−∇&#3627408467;(&#3627408485;)=∇[&#3627408467;(&#3627408485;+ℎ)−&#3627408467;(&#3627408485;)]=∇∆&#3627408467;(&#3627408485;)
∴ &#3627408440;∆≡∆&#3627408440; and ∆∇≡∇∆.
PROBLEM
Suppose ??????
0=−3,??????
1=6,??????
2=8 and ??????
3=12. Find ??????
6.
Solution:
From the given data, we get the following difference table.
Argument (&#3627408485;) Entry (??????
??????) ∆??????
?????? ∆
2
??????
?????? ∆
3
??????
??????
0 −3
∆??????
0=9
1 6 ∆
2
??????
0=−7
∆??????
1=2 ∆
3
??????
0=9
2 8 ∆
2
??????
1=2
∆??????
2=4
3 12
Since four entries ??????
?????? for the arguments &#3627408485;=0,1,2,3 are given, we may assume that the fourth and higher order forward
differences are zero i.e., ∆
??????
??????
0=0 ; ??????≥4.
Now,??????
6=&#3627408440;
6
??????
0=(∆+1)
6
??????
0=(
6
0
)∆
6
??????
0+(
6
1
)∆
5
??????
0+(
6
2
)∆
4
??????
0+(
6
3
)∆
3
??????
0+(
6
4
)∆
2
??????
0+(
6
5
)∆??????
0+(
6
6
)??????
0
=20∆
3
??????
0+15∆
2
??????
0+6∆??????
0+??????
0=20×9+15×(−7)+6×9+(−3)=126

20. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 19
PROBLEM
Find the missing term in the following table.
&#3627408485; 0 1 2 3 4
&#3627408486; 1 3 9 --- 81
Solution:
Since four entries are available, we may assume that the third order forward difference is constant and, fourth and higher
order forward differences are zero i.e., ∆
3
&#3627408486;
0=&#3627408464;,∆
??????
&#3627408486;
0=0 ; ??????≥4, where &#3627408464; is a constant.
Now, ∆
4
&#3627408486;
0=0 or, (&#3627408440;−1)
4
&#3627408486;
0=0 or, &#3627408440;
4
&#3627408486;
0−4&#3627408440;
3
&#3627408486;
0+6&#3627408440;
2
&#3627408486;
0−4&#3627408440;&#3627408486;
0+&#3627408486;
0=0
or, &#3627408486;
4−4&#3627408486;
3+6&#3627408486;
2−4&#3627408486;
1+&#3627408486;
0=0 or, &#3627408486;
3=
1
4
[&#3627408486;
4+6&#3627408486;
2−4&#3627408486;
1+&#3627408486;
0
]=31
PROBLEM
Find ∆
&#3627408475;
??????
0, where ??????
??????=&#3627408466;
&#3627408462;??????+&#3627408463;
.
Solution:
Let interval of differencing being ℎ.
∆??????
??????=&#3627408466;
&#3627408462;(??????+ℎ)+&#3627408463;
−&#3627408466;
&#3627408462;??????+&#3627408463;
=&#3627408466;
&#3627408462;??????+&#3627408463;
(&#3627408466;
&#3627408462;ℎ
−1)=(&#3627408466;
&#3627408462;ℎ
−1)??????
??????
Again, ∆
2
??????
??????=∆(∆??????
??????)=∆[(&#3627408466;
&#3627408462;ℎ
−1)??????
??????
]=(&#3627408466;
&#3627408462;ℎ
−1)∆??????
??????=(&#3627408466;
&#3627408462;ℎ
−1)
2
??????
??????
Similarly, ∆
&#3627408475;
??????
??????=(&#3627408466;
&#3627408462;ℎ
−1)
&#3627408475;
??????
??????=(&#3627408466;
&#3627408462;ℎ
−1)
&#3627408475;
&#3627408466;
&#3627408462;??????+&#3627408463;
.
PROBLEM
Show that &#3627408467;(&#3627408462;)+&#3627408485;&#3627408467;(&#3627408462;+ℎ)+
&#3627408485;
2
2!
&#3627408467;(&#3627408462;+2ℎ)+
&#3627408485;
3
3!
&#3627408467;(&#3627408462;+3ℎ)+⋯=&#3627408466;
??????
[&#3627408467;(&#3627408462;)+&#3627408485;∆&#3627408467;(&#3627408462;)+
&#3627408485;
2
2!
∆
2
&#3627408467;(&#3627408462;)+⋯]
Solution:
&#3627408467;(&#3627408462;)+&#3627408485;&#3627408467;(&#3627408462;+ℎ)+
&#3627408485;
2
2!
&#3627408467;(&#3627408462;+2ℎ)+
&#3627408485;
3
3!
&#3627408467;(&#3627408462;+3ℎ)+⋯=&#3627408467;(&#3627408462;)+&#3627408485;&#3627408440;&#3627408467;(&#3627408462;)+
&#3627408485;
2
2!
&#3627408440;
2
&#3627408467;(&#3627408462;)+
&#3627408485;
3
3!
&#3627408440;
3
&#3627408467;(&#3627408462;)+⋯
=&#3627408466;
??????&#3627408440;
&#3627408467;(&#3627408462;)=&#3627408466;
??????(1+∆)
&#3627408467;(&#3627408462;) [∵ &#3627408440;≡1+∆]
=&#3627408466;
??????
∙&#3627408466;
??????∆
&#3627408467;(&#3627408462;)
=&#3627408466;
??????
[1+&#3627408485;∆+
&#3627408485;
2
2!
∆
2
+
&#3627408485;
3
3!
∆
3
+⋯]&#3627408467;(&#3627408462;)
=&#3627408466;
??????
[&#3627408467;(&#3627408462;)+&#3627408485;∆&#3627408467;(&#3627408462;)+
&#3627408485;
2
2!
∆
2
&#3627408467;(&#3627408462;)+
&#3627408485;
3
3!
∆
3
&#3627408467;(&#3627408462;)+⋯] (Proved)

21. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 20
PROBLEM
Show that ∆
&#3627408475;
??????
??????−&#3627408475;=(
&#3627408475;
0
)??????
??????−(
&#3627408475;
1
)??????
??????−1+(
&#3627408475;
2
)??????
??????−2−(
&#3627408475;
3
)??????
??????−3+⋯+(−1)
&#3627408475;
(
&#3627408475;
&#3627408475;
)??????
??????−&#3627408475;.
Solution:
∆
&#3627408475;
??????
??????−&#3627408475;=(&#3627408440;−1)
&#3627408475;
&#3627408440;
−&#3627408475;
??????
?????? [∵ &#3627408440;≡1+∆ and ??????
??????−&#3627408475;=&#3627408440;
−&#3627408475;
??????
??????,interval of differencing ℎ=1]
=(1−&#3627408440;
−1
)
&#3627408475;
??????
??????
=[(
&#3627408475;
0
)−(
&#3627408475;
1
)&#3627408440;
−1
+(
&#3627408475;
2
)&#3627408440;
−2
−(
&#3627408475;
3
)&#3627408440;
−3
+⋯+(−1)
&#3627408475;
(
&#3627408475;
&#3627408475;
)&#3627408440;
−&#3627408475;
]??????
??????
=(
&#3627408475;
0
)??????
??????−(
&#3627408475;
1
)??????
??????−1+(
&#3627408475;
2
)??????
??????−2−(
&#3627408475;
3
)??????
??????−3+⋯+(−1)
&#3627408475;
(
&#3627408475;
&#3627408475;
)??????
??????−&#3627408475;
PROBLEM
Show that &#3627408485;??????
1+&#3627408485;
2
??????
2+&#3627408485;
3
??????
3+⋯=(
&#3627408485;
1−&#3627408485;
)??????
1+(
&#3627408485;
1−&#3627408485;
)
2
∆??????
1+(
&#3627408485;
1−&#3627408485;
)
3
∆
2
??????
1+⋯.
Solution:
&#3627408453;??????&#3627408454;=(
&#3627408485;
1−&#3627408485;
)??????
1+(
&#3627408485;
1−&#3627408485;
)
2
∆??????
1+(
&#3627408485;
1−&#3627408485;
)
3
∆
2
??????
1+⋯=(
&#3627408485;
1−&#3627408485;
)[1+(
&#3627408485;
1−&#3627408485;
)∆+(
&#3627408485;
1−&#3627408485;
)
2
∆
2
+⋯]??????
1
=(
&#3627408485;
1−&#3627408485;
)[
1
1−(
&#3627408485;
1−&#3627408485;
)∆
]??????
1=(
&#3627408485;
1−&#3627408485;
)[
1−&#3627408485;
1−&#3627408485;(1+∆)
]??????
1=(
&#3627408485;
1−&#3627408485;&#3627408440;
)??????
1 [∵ &#3627408440;≡1+∆]
=&#3627408485;(1+&#3627408485;&#3627408440;+&#3627408485;
2
&#3627408440;
2
+&#3627408485;
3
&#3627408440;
3
+⋯)??????
1
=&#3627408485;??????
1+&#3627408485;
2
??????
2+&#3627408485;
3
??????
3+⋯=????????????&#3627408454; (Proved)
PROBLEM
Show that ??????
2&#3627408475;−(
&#3627408475;
1
)2??????
2&#3627408475;−1+(
&#3627408475;
2
)2
2
??????
2&#3627408475;−2+⋯+(−2)
&#3627408475;
??????
&#3627408475;=(−1)
&#3627408475;
(&#3627408464;−2&#3627408462;&#3627408475;),where ??????
??????=&#3627408462;&#3627408485;
2
+&#3627408463;&#3627408485;+&#3627408464;.
Solution:
??????
2&#3627408475;−(
&#3627408475;
1
)2??????
2&#3627408475;−1+(
&#3627408475;
2
)2
2
??????
2&#3627408475;−2+⋯+(−2)
&#3627408475;
??????
&#3627408475;=&#3627408440;
&#3627408475;
??????
&#3627408475;−(
&#3627408475;
1
)2&#3627408440;
&#3627408475;−1
??????
&#3627408475;+(
&#3627408475;
2
)2
2
&#3627408440;
&#3627408475;−2
??????
&#3627408475;+⋯+(−2)
&#3627408475;
??????
&#3627408475;
=[&#3627408440;
&#3627408475;
−(
&#3627408475;
1
)2&#3627408440;
&#3627408475;−1
+(
&#3627408475;
2
)2
2
&#3627408440;
&#3627408475;−2
+⋯+(−2)
&#3627408475;
]??????
&#3627408475;
=(&#3627408440;−2)
&#3627408475;
??????
&#3627408475;=(−1)
&#3627408475;
(1−∆)
&#3627408475;
??????
&#3627408475;=(−1)
&#3627408475;
(1−∆)
&#3627408475;
&#3627408440;
&#3627408475;
??????
0 [∵ &#3627408440;≡1+∆]
=(−1)
&#3627408475;
(1−∆
2
)
&#3627408475;
??????
0=(−1)
&#3627408475;
(??????
0−&#3627408475;∆
2
??????
0) [∵ ??????
??????=&#3627408462;&#3627408485;
2
+&#3627408463;&#3627408485;+&#3627408464;,∆
??????
??????
0=0 ; ??????>2 ]
=(−1)
&#3627408475;
{??????
0−&#3627408475;(??????
2−2??????
1+??????
0)}
=(−1)
&#3627408475;
{&#3627408464;−&#3627408475;(4&#3627408462;+2&#3627408463;+&#3627408464;−2&#3627408462;−2&#3627408463;−2&#3627408464;+&#3627408464;)}
=(−1)
&#3627408475;
(&#3627408464;−2&#3627408462;&#3627408475;)

22. NUMERICAL ANALYSIS: APPROXIMATION, ERRORS & OPERATORS
Numerical Analysis || Lecture Notes || Anup Kumar Giri 21
PROBLEM
If &#3627408467;(0)=1,&#3627408467;(1)+&#3627408467;(2)=10, &#3627408467;(3)+&#3627408467;(4)+ &#3627408467;(5)=65,find &#3627408467;(4).
Solution:
Let us assume &#3627408467;(&#3627408485;) be a polynomial of degree 2 in &#3627408485; and &#3627408467;(&#3627408485;)=&#3627408462;&#3627408485;
2
+&#3627408463;&#3627408485;+&#3627408464;.
Now, &#3627408467;(0)=&#3627408464;=1, &#3627408467;(1)+&#3627408467;(2)=&#3627408462;+&#3627408463;+&#3627408464;+4&#3627408462;+2&#3627408463;+&#3627408464;=10 or, 5&#3627408462;+3&#3627408463;=8
&#3627408467;(3)+&#3627408467;(4)+ &#3627408467;(5)=50&#3627408462;+12&#3627408463;+3&#3627408464;=65 or, 50&#3627408462;+12&#3627408463;=62 or, 25&#3627408462;+6&#3627408463;=31
By solving the above equations, we get &#3627408462;=1,&#3627408463;=1.
Hence &#3627408467;(4)=16+4+1=21.