Numerical Analysis: Interpolation: Forward, Backward, Lagrange's and Linear Interpolation
AnupGiri
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About This Presentation
Numerical Analysis: Interpolation
GENERAL INTERPOLATION PROBLEM
WEIERSTRASS THEOREM
NEWTON’S FORWARD INTERPOLATION FORMULA
NEWTON’S BACKWARD INTERPOLATION FORMULA
LAGRANGE'S INTERPOLATION POLYNOMIAL, Related RESULTS
LINEAR INTERPOLATION
Slide Content
Multivariate Analysis || Lecture Notes || Anup Kumar Giri
NUMERICAL ANALYSIS:
INTERPOLATION
NUMERICAL ANALYSIS:
INTERPOLATION
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 1
INTERPOLATION
It is well known that the population of India are known for the years 1951, 1961, 1971, 1981, 1991, 2001, 2011. What is
the population of India in the year 2008? Exact result is not known for this question. What is the approximate value? By
guessing we can say an approximate figure. But, guessing gives different results for different persons, in different times,
etc. To avoid this ambiguity, a method is developed known as interpolation which gives an approximate value of this
problem. Obviously, this value is not exact, contains some error. Also, there is a method to estimate such error.
Interpolation has been defined as the art of reading between the lines of a table, and in elementary mathematics, the term
usually denotes the process of computing intermediate values of a function from a set of given or tabulator values of that
function. In higher mathematics, we frequently deal with functions whose analytical form is either completely unknown
or is such of a nature (complicated) that the function cannot be easily subjected to such operations as may we required.
In either cases, it is desirable to replace the given function by another which can readily handle. This operation of replacing
or representing a given function by a simpler one constitutes interpolation in the broad sense of term.
GENERAL INTERPOLATION PROBLEM
Let �=�(�) be a function, whose explicit expression is not known, but a table of functional values of � (=�(�)) or
entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? is known for a given set of (??????+1) values or arguments �
0,�
1,�
2,⋯,�
?????? of �. There
is no other information available about the function �(�).
The problem of interpolation is to find the value of � (=�(�)) for a given value of � say, �̃ within the minimum and
maximum values of �
0,�
1,�
2,⋯,�
??????. Obviously, the value of � at �̃ is unknown.
Many different methods are available to find the value of �=�(�) at the given point �=�̃. The main step of
interpolation is to find an approximate function, say, ??????(�), for the given function �(�) based on the given tabulated
values. The approximate function should be simple and easy to handle. The constructed function ??????(�) may be a
polynomial, exponential, geometric function, Taylor's series, Fourier series, etc. If the function ??????(�) is a polynomial,
then the corresponding interpolation is called polynomial interpolation. Polynomial interpolation is used in most of the
situations as polynomial is easy to evaluate, continuous, differentiable and integrable in any range.
The term extrapolation is used to find data points outside the range of known data points.
A polynomial ??????(�) is called interpolating polynomial if �
??????=�(�
??????)=??????(�
??????), ∀ ??????=0,1,2,⋯,?????? and (
??????
??????
??????
????????????
??????
)
??????̂
=(
??????
??????
??????
????????????
??????
)
??????̂
for some finite ??????, and �̂ is one of the values of �
0,�
1,�
2,⋯,�
??????.
Every interpolating polynomial must satisfy the following condition.
WEIERSTRASS THEOREM
If the function �(�) is continuous on [�,�], then for any pre-assigned positive number ??????>0, there exists a polynomial
??????(�) such that |�(�)−??????(�)|<?????? for all �∈(�,�).
Numerical Analysis || Lecture Notes || Anup Kumar Giri 1
INTERPOLATION
It is well known that the population of India are known for the years 1951, 1961, 1971, 1981, 1991, 2001, 2011. What is
the population of India in the year 2008? Exact result is not known for this question. What is the approximate value? By
guessing we can say an approximate figure. But, guessing gives different results for different persons, in different times,
etc. To avoid this ambiguity, a method is developed known as interpolation which gives an approximate value of this
problem. Obviously, this value is not exact, contains some error. Also, there is a method to estimate such error.
Interpolation has been defined as the art of reading between the lines of a table, and in elementary mathematics, the term
usually denotes the process of computing intermediate values of a function from a set of given or tabulator values of that
function. In higher mathematics, we frequently deal with functions whose analytical form is either completely unknown
or is such of a nature (complicated) that the function cannot be easily subjected to such operations as may we required.
In either cases, it is desirable to replace the given function by another which can readily handle. This operation of replacing
or representing a given function by a simpler one constitutes interpolation in the broad sense of term.
GENERAL INTERPOLATION PROBLEM
Let �=�(�) be a function, whose explicit expression is not known, but a table of functional values of � (=�(�)) or
entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? is known for a given set of (??????+1) values or arguments �
0,�
1,�
2,⋯,�
?????? of �. There
is no other information available about the function �(�).
The problem of interpolation is to find the value of � (=�(�)) for a given value of � say, �̃ within the minimum and
maximum values of �
0,�
1,�
2,⋯,�
??????. Obviously, the value of � at �̃ is unknown.
Many different methods are available to find the value of �=�(�) at the given point �=�̃. The main step of
interpolation is to find an approximate function, say, ??????(�), for the given function �(�) based on the given tabulated
values. The approximate function should be simple and easy to handle. The constructed function ??????(�) may be a
polynomial, exponential, geometric function, Taylor's series, Fourier series, etc. If the function ??????(�) is a polynomial,
then the corresponding interpolation is called polynomial interpolation. Polynomial interpolation is used in most of the
situations as polynomial is easy to evaluate, continuous, differentiable and integrable in any range.
The term extrapolation is used to find data points outside the range of known data points.
A polynomial ??????(�) is called interpolating polynomial if �
??????=�(�
??????)=??????(�
??????), ∀ ??????=0,1,2,⋯,?????? and (
??????
??????
??????
????????????
??????
)
??????̂
=(
??????
??????
??????
????????????
??????
)
??????̂
for some finite ??????, and �̂ is one of the values of �
0,�
1,�
2,⋯,�
??????.
Every interpolating polynomial must satisfy the following condition.
WEIERSTRASS THEOREM
If the function �(�) is continuous on [�,�], then for any pre-assigned positive number ??????>0, there exists a polynomial
??????(�) such that |�(�)−??????(�)|<?????? for all �∈(�,�).
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 2
Geometrically, the theorem implies that the interpolating polynomial ??????(�) is bounded by the functions �(�)−?????? and
�(�)+??????, for some given ??????>0.
The relation between the function �=�(�), interpolating polynomial ??????(�) and two boundaries functions are shown in
the following figure.
Several interpolation methods are available in literature. Among them Newton’s forward interpolation, Newton’s
backward interpolation and Lagrange's interpolation are mostly useful methods.
NEWTON’S FORWARD INTERPOLATION FORMULA
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing
or spacing.
With these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation function such that
�
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
Suppose the n-th degree polynomial is taken as
??????(�)=�
0+�
1(�−�
0)+�
2(�−�
0)(�−�
1)+�
3(�−�
0)(�−�
1)(�−�
2)+⋯
+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)+⋯+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1) ⋯⋯⋯ (∗)
Now, the constants �
0,�
1,⋯,�
?????? are estimate by ??????(�
??????)=�
??????, ??????=0,1,2,⋯,??????.
Putting �=�
0 in (∗), we get
??????(�
0)=�
0 or,�
0=�
0.
Again, for �=�
1,
??????(�
1)=�
0+�
1(�
1−�
0)=�
0+�
1(�
1−�
0)
or, �
1=�
0+�
1(�
1−�
0)=�
0+�
1ℎ
or, �
1=
�
1−�
0
ℎ
=
∆�
0
ℎ
.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 2
Geometrically, the theorem implies that the interpolating polynomial ??????(�) is bounded by the functions �(�)−?????? and
�(�)+??????, for some given ??????>0.
The relation between the function �=�(�), interpolating polynomial ??????(�) and two boundaries functions are shown in
the following figure.
Several interpolation methods are available in literature. Among them Newton’s forward interpolation, Newton’s
backward interpolation and Lagrange's interpolation are mostly useful methods.
NEWTON’S FORWARD INTERPOLATION FORMULA
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing
or spacing.
With these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation function such that
�
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
Suppose the n-th degree polynomial is taken as
??????(�)=�
0+�
1(�−�
0)+�
2(�−�
0)(�−�
1)+�
3(�−�
0)(�−�
1)(�−�
2)+⋯
+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)+⋯+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1) ⋯⋯⋯ (∗)
Now, the constants �
0,�
1,⋯,�
?????? are estimate by ??????(�
??????)=�
??????, ??????=0,1,2,⋯,??????.
Putting �=�
0 in (∗), we get
??????(�
0)=�
0 or,�
0=�
0.
Again, for �=�
1,
??????(�
1)=�
0+�
1(�
1−�
0)=�
0+�
1(�
1−�
0)
or, �
1=�
0+�
1(�
1−�
0)=�
0+�
1ℎ
or, �
1=
�
1−�
0
ℎ
=
∆�
0
ℎ
.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 3
Similarly, when �=�
2,
??????(�
2)=�
0+�
1(�
2−�
0)+�
2(�
2−�
0)(�
2−�
1)
or, �
2=�
0+
∆�
0
ℎ
∙2ℎ+�
2∙2ℎ∙ℎ=�
0+2(�
1−�
0)+2�
2ℎ
2
or, �
2=
�
2−2�
1+�
0
2ℎ
2
=
∆
2
�
0
2!ℎ
2
.
Proceeding in the same way, we can find
�
??????=
∆
??????
�
0
??????!ℎ
??????
; ??????=1,2,3,⋯,??????.
Substituting the values of �
0,�
1,⋯,�
?????? as
�
0=�
0,�
1=
∆�
0
ℎ
,�
2=
∆
2
�
0
2!ℎ
2
,�
3=
∆
3
�
0
3!ℎ
3
,�
4=
∆
4
�
0
4!ℎ
4
,⋯,�
??????=
∆
??????
�
0
??????!ℎ
??????
,⋯,�
??????=
∆
??????
�
0
??????!ℎ
??????
in (∗), the polynomial ??????(�) becomes
??????(�)=�
0+
∆�
0
ℎ
(�−�
0)+
∆
2
�
0
2!ℎ
2
(�−�
0)(�−�
0+�
0−�
1)+
∆
3
�
0
3!ℎ
3
(�−�
0)(�−�
1)(�−�
2)+⋯
+
∆
??????
�
0
??????!ℎ
??????
(�−�
0)(�−�
1)⋯(�−�
??????−1)+⋯+
∆
??????
�
0
??????!ℎ
??????
(�−�
0)(�−�
1)⋯(�−�
??????−1)
Let us consider
�=
(�−�
0)
ℎ
i.e., �=�
0+�ℎ.
Since �
??????=�
0+??????ℎ, we get �−�
??????=(�−??????)ℎ ; ??????=0,1,2,⋯,??????.
Thus, the polynomial ??????(�) reduces to the following form:
??????(�)=??????(�
0+�ℎ)
=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+⋯+
�(�−1)(�−2)⋯(�−??????+1)
??????!
∆
??????
�
0
=(
�
0
)�
0+(
�
1
)∆�
0+(
�
2
)∆
2
�
0+(
�
3
)∆
3
�
0+⋯+(
�
??????
)∆
??????
�
0
=∑(
�
??????
)∆
??????
�
0
??????
??????=0
This is known as Newton or Newton-Gregory forward difference interpolating polynomial.
NOTE:
This formula contains values of the function beginning from �
0 forward (to the right) and none backward and thus named
forward interpolation formula. It is suitable mainly for interpolation near the beginning of a set of tabulated values. In
general, Newton's forward difference formula is used to compute the approximate value of �(�) when the value of � is
near to �
0 of the given table. But, if the value of � is at the end of the table, then this formula gives more error.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 3
Similarly, when �=�
2,
??????(�
2)=�
0+�
1(�
2−�
0)+�
2(�
2−�
0)(�
2−�
1)
or, �
2=�
0+
∆�
0
ℎ
∙2ℎ+�
2∙2ℎ∙ℎ=�
0+2(�
1−�
0)+2�
2ℎ
2
or, �
2=
�
2−2�
1+�
0
2ℎ
2
=
∆
2
�
0
2!ℎ
2
.
Proceeding in the same way, we can find
�
??????=
∆
??????
�
0
??????!ℎ
??????
; ??????=1,2,3,⋯,??????.
Substituting the values of �
0,�
1,⋯,�
?????? as
�
0=�
0,�
1=
∆�
0
ℎ
,�
2=
∆
2
�
0
2!ℎ
2
,�
3=
∆
3
�
0
3!ℎ
3
,�
4=
∆
4
�
0
4!ℎ
4
,⋯,�
??????=
∆
??????
�
0
??????!ℎ
??????
,⋯,�
??????=
∆
??????
�
0
??????!ℎ
??????
in (∗), the polynomial ??????(�) becomes
??????(�)=�
0+
∆�
0
ℎ
(�−�
0)+
∆
2
�
0
2!ℎ
2
(�−�
0)(�−�
0+�
0−�
1)+
∆
3
�
0
3!ℎ
3
(�−�
0)(�−�
1)(�−�
2)+⋯
+
∆
??????
�
0
??????!ℎ
??????
(�−�
0)(�−�
1)⋯(�−�
??????−1)+⋯+
∆
??????
�
0
??????!ℎ
??????
(�−�
0)(�−�
1)⋯(�−�
??????−1)
Let us consider
�=
(�−�
0)
ℎ
i.e., �=�
0+�ℎ.
Since �
??????=�
0+??????ℎ, we get �−�
??????=(�−??????)ℎ ; ??????=0,1,2,⋯,??????.
Thus, the polynomial ??????(�) reduces to the following form:
??????(�)=??????(�
0+�ℎ)
=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+⋯+
�(�−1)(�−2)⋯(�−??????+1)
??????!
∆
??????
�
0
=(
�
0
)�
0+(
�
1
)∆�
0+(
�
2
)∆
2
�
0+(
�
3
)∆
3
�
0+⋯+(
�
??????
)∆
??????
�
0
=∑(
�
??????
)∆
??????
�
0
??????
??????=0
This is known as Newton or Newton-Gregory forward difference interpolating polynomial.
NOTE:
This formula contains values of the function beginning from �
0 forward (to the right) and none backward and thus named
forward interpolation formula. It is suitable mainly for interpolation near the beginning of a set of tabulated values. In
general, Newton's forward difference formula is used to compute the approximate value of �(�) when the value of � is
near to �
0 of the given table. But, if the value of � is at the end of the table, then this formula gives more error.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 4
EXAMPLE
Given below is a table of values of the probability integral �=�(�)=
2
??????
∫�
−??????
2
��
??????
0
.
Determine the value of �=�(�) when the value of � is 0.456.
�: 0.45 0.46 0.47 0.48 0.49
�: 0.475482 0.484656 0.497452 0.502750 0.511668
Solution:
Since the given value of �=0.456 is at the beginning of the table, we apply the Newton's forward interpolation formula
to determine the value of �=�(�) when the value of � is 0.456.
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing.
The Newton’s forward difference interpolating polynomial ??????(�) with is given by
??????(�)=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+⋯+
�(�−1)(�−2)⋯(�−??????+1)
??????!
∆
??????
�
0
where �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and �=
(??????−??????0
)
ℎ
.
The forward difference table is as follows:
� �=??????(�) ∆� ∆
�
� ∆
�
� ∆
�
�
0.45 0.475482
∆�
0=0.009174
0.46 0.484656 ∆
2
�
0=0.003622
∆�
1=0.012796 ∆
3
�
0=−0.011120
0.47 0.497452 ∆
2
�
1=−0.007498 ∆
4
�
0=0.022238
∆�
2=0.005298 ∆
3
�
1=0.011118
0.48 0.502750 ∆
2
�
2=0.003620
∆�
3=0.008918
0.49 0.511668
Here, �
0=0.45 ; �=0.456 ; ℎ=0.01 ; �=
(??????−??????0
)
ℎ
=
(0.456−0.45)
0.01
=0.6 and ∆
??????
�
0=0 ∀ ??????>4.
Now,
�
0.456=�(0.456)=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+
�(�−1)(�−2)(�−3)
4!
∆
4
�
0
=0.475482+0.6×0.009174+
0.6(0.6−1)
2!
×0.003622+
0.6(0.6−1)(0.6−2)
3!
×(−0.011120)
+
0.6(0.6−1)(0.6−2)(0.6−3)
4!
×0.022238
=0.475482+0.0055044−0.00043464−0.00062272−0.0007471968
=0.4791818432
Hence, the value of �=�(�) when �=0.456 is 0.4791818432.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 4
EXAMPLE
Given below is a table of values of the probability integral �=�(�)=
2
??????
∫�
−??????
2
��
??????
0
.
Determine the value of �=�(�) when the value of � is 0.456.
�: 0.45 0.46 0.47 0.48 0.49
�: 0.475482 0.484656 0.497452 0.502750 0.511668
Solution:
Since the given value of �=0.456 is at the beginning of the table, we apply the Newton's forward interpolation formula
to determine the value of �=�(�) when the value of � is 0.456.
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing.
The Newton’s forward difference interpolating polynomial ??????(�) with is given by
??????(�)=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+⋯+
�(�−1)(�−2)⋯(�−??????+1)
??????!
∆
??????
�
0
where �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and �=
(??????−??????0
)
ℎ
.
The forward difference table is as follows:
� �=??????(�) ∆� ∆
�
� ∆
�
� ∆
�
�
0.45 0.475482
∆�
0=0.009174
0.46 0.484656 ∆
2
�
0=0.003622
∆�
1=0.012796 ∆
3
�
0=−0.011120
0.47 0.497452 ∆
2
�
1=−0.007498 ∆
4
�
0=0.022238
∆�
2=0.005298 ∆
3
�
1=0.011118
0.48 0.502750 ∆
2
�
2=0.003620
∆�
3=0.008918
0.49 0.511668
Here, �
0=0.45 ; �=0.456 ; ℎ=0.01 ; �=
(??????−??????0
)
ℎ
=
(0.456−0.45)
0.01
=0.6 and ∆
??????
�
0=0 ∀ ??????>4.
Now,
�
0.456=�(0.456)=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+
�(�−1)(�−2)(�−3)
4!
∆
4
�
0
=0.475482+0.6×0.009174+
0.6(0.6−1)
2!
×0.003622+
0.6(0.6−1)(0.6−2)
3!
×(−0.011120)
+
0.6(0.6−1)(0.6−2)(0.6−3)
4!
×0.022238
=0.475482+0.0055044−0.00043464−0.00062272−0.0007471968
=0.4791818432
Hence, the value of �=�(�) when �=0.456 is 0.4791818432.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 5
NEWTON’S BACKWARD INTERPOLATION FORMULA
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ,??????=0,1,2,⋯,??????, where h being the interval of differencing
or spacing.
With these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation function such that
�
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
Suppose the n-th degree polynomial is taken as
??????(�)=�
0+�
1(�−�
??????)+�
2(�−�
??????)(�−�
??????−1)+�
3(�−�
??????)(�−�
??????−1)(�−�
??????−2)+⋯
+�
??????(�−�
??????)(�−�
??????−1)⋯(�−�
??????−??????+1)+⋯+�
??????(�−�
??????)(�−�
??????−1)⋯(�−�
1) ⋯⋯⋯ (∗∗)
Now the constants �
0,�
1,⋯,�
?????? are estimate by ??????(�
??????)=�
??????, ??????=0,1,2,⋯,??????.
Putting �=�
?????? in (∗∗), we get
??????(�
??????)=�
0 or,�
0=�
??????.
Again, for �=�
??????−1, ??????(�
??????−1)=�
0+�
1(�
??????−1−�
??????)=�
??????+�
1(�
??????−1−�
??????)
or, �
??????−1=�
??????+�
1(�
??????−1−�
??????)
or, �
1=
�
??????−�
??????−1
(�
??????−�
??????−1)
=
∆�
??????−1
ℎ
[=
∇�
??????
ℎ
] .
Similarly, when �=�
??????−2,
??????(�
??????−2)=�
0+�
1(�
??????−2−�
??????)+�
2(�
??????−2−�
??????)(�
??????−2−�
??????−1)
or, �
??????−2=�
??????+
∆�
??????−1
ℎ
∙(−2ℎ)+�
2∙2ℎ∙ℎ=�
??????+2(�
??????−1−�
??????)+2�
2ℎ
2
or, �
2=
�
??????−2�
??????−1+�
??????−2
2ℎ
2
=
∆
2
�
??????−2
2!ℎ
2
[=
∇
2
�
??????
2!ℎ
2
] .
Proceeding in the same way, we can find
�
??????=
∆
??????
�
??????−??????
??????!ℎ
??????
[=
∇
??????
�
??????
??????!ℎ
??????
] ; ??????=1,2,3,⋯,??????.
Putting the values of �
0,�
1,⋯,�
?????? as
�
0=�
??????,�
1=
∆�
??????−1
ℎ
,�
2=
∆
2
�
??????−2
2!ℎ
2
,�
3=
∆
3
�
??????−3
3!ℎ
3
,�
4=
∆
4
�
??????−4
4!ℎ
4
,⋯,�
??????=
∆
??????
�
??????−??????
??????!ℎ
??????
,⋯,�
??????=
∆
??????
�
0
??????!ℎ
??????
in (∗∗), he polynomial ??????(�) becomes
??????(�)=�
??????+
∆�
??????−1
ℎ
(�−�
??????)+
∆
2
�
??????−2
2!ℎ
2
(�−�
??????)(�−�
??????−1)+
∆
3
�
??????−3
3!ℎ
3
(�−�
??????)(�−�
??????−1)(�−�
??????−2)+⋯
+
∆
??????
�
??????−??????
??????!ℎ
??????
(�−�
??????)(�−�
??????−1)⋯(�−�
??????−??????+1)+⋯+
∆
??????
�
0
??????!ℎ
??????
(�−�
??????)(�−�
??????−1)⋯(�−�
1)
Numerical Analysis || Lecture Notes || Anup Kumar Giri 5
NEWTON’S BACKWARD INTERPOLATION FORMULA
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ,??????=0,1,2,⋯,??????, where h being the interval of differencing
or spacing.
With these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation function such that
�
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
Suppose the n-th degree polynomial is taken as
??????(�)=�
0+�
1(�−�
??????)+�
2(�−�
??????)(�−�
??????−1)+�
3(�−�
??????)(�−�
??????−1)(�−�
??????−2)+⋯
+�
??????(�−�
??????)(�−�
??????−1)⋯(�−�
??????−??????+1)+⋯+�
??????(�−�
??????)(�−�
??????−1)⋯(�−�
1) ⋯⋯⋯ (∗∗)
Now the constants �
0,�
1,⋯,�
?????? are estimate by ??????(�
??????)=�
??????, ??????=0,1,2,⋯,??????.
Putting �=�
?????? in (∗∗), we get
??????(�
??????)=�
0 or,�
0=�
??????.
Again, for �=�
??????−1, ??????(�
??????−1)=�
0+�
1(�
??????−1−�
??????)=�
??????+�
1(�
??????−1−�
??????)
or, �
??????−1=�
??????+�
1(�
??????−1−�
??????)
or, �
1=
�
??????−�
??????−1
(�
??????−�
??????−1)
=
∆�
??????−1
ℎ
[=
∇�
??????
ℎ
] .
Similarly, when �=�
??????−2,
??????(�
??????−2)=�
0+�
1(�
??????−2−�
??????)+�
2(�
??????−2−�
??????)(�
??????−2−�
??????−1)
or, �
??????−2=�
??????+
∆�
??????−1
ℎ
∙(−2ℎ)+�
2∙2ℎ∙ℎ=�
??????+2(�
??????−1−�
??????)+2�
2ℎ
2
or, �
2=
�
??????−2�
??????−1+�
??????−2
2ℎ
2
=
∆
2
�
??????−2
2!ℎ
2
[=
∇
2
�
??????
2!ℎ
2
] .
Proceeding in the same way, we can find
�
??????=
∆
??????
�
??????−??????
??????!ℎ
??????
[=
∇
??????
�
??????
??????!ℎ
??????
] ; ??????=1,2,3,⋯,??????.
Putting the values of �
0,�
1,⋯,�
?????? as
�
0=�
??????,�
1=
∆�
??????−1
ℎ
,�
2=
∆
2
�
??????−2
2!ℎ
2
,�
3=
∆
3
�
??????−3
3!ℎ
3
,�
4=
∆
4
�
??????−4
4!ℎ
4
,⋯,�
??????=
∆
??????
�
??????−??????
??????!ℎ
??????
,⋯,�
??????=
∆
??????
�
0
??????!ℎ
??????
in (∗∗), he polynomial ??????(�) becomes
??????(�)=�
??????+
∆�
??????−1
ℎ
(�−�
??????)+
∆
2
�
??????−2
2!ℎ
2
(�−�
??????)(�−�
??????−1)+
∆
3
�
??????−3
3!ℎ
3
(�−�
??????)(�−�
??????−1)(�−�
??????−2)+⋯
+
∆
??????
�
??????−??????
??????!ℎ
??????
(�−�
??????)(�−�
??????−1)⋯(�−�
??????−??????+1)+⋯+
∆
??????
�
0
??????!ℎ
??????
(�−�
??????)(�−�
??????−1)⋯(�−�
1)
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 6
Let us consider
�=
(�−�
??????)
ℎ
, i.e., �=�
??????+�ℎ.
Since �
??????=�
0+??????ℎ, we get �−�
??????=(�+??????−??????)ℎ ; ??????=0,1,2,⋯,??????.
Thus, the polynomial ??????(�) reduces to the following form:
??????(�)=??????(�
??????+�ℎ)
=�
??????+�∆�
??????−1+
�(�+1)
2!
∆
2
�
??????−2+
�(�+1)(�+2)
3!
∆
3
�
??????−3+⋯+
�(�+1)(�+2)⋯(�+??????−1)
??????!
∆
??????
�
0
=�
??????+(
�
1
)∆�
??????−1+(
�+1
2
)∆
2
�
??????−2+(
�+2
3
)∆
3
�
??????−3+⋯+(
�+??????−1
??????
)∆
??????
�
0
=∑(
�+??????−1
??????
)∆
??????
�
??????−??????
??????
??????=0
This is known as Newton's backward or Newton-Gregory backward interpolation formula.
In terms of backward differencing operator (∇), the Newton's backward interpolating polynomial ??????(�) becomes
??????(�)=??????(�
??????+�ℎ)
=�
??????+
∇�
??????
ℎ
(�−�
??????)+
∇
2
�
??????
2!ℎ
2
(�−�
??????)(�−�
??????−1)+
∇
3
�
??????
3!ℎ
3
(�−�
??????)(�−�
??????−1)(�−�
??????−2)+⋯
+
∇
??????
�
??????
??????!ℎ
??????
(�−�
??????)(�−�
??????−1)⋯(�−�
??????−??????+1)+⋯+
∇
??????
�
??????
??????!ℎ
??????
(�−�
??????)(�−�
??????−1)⋯(�−�
1)
=�
??????+�∇�
??????+
�(�+1)
2!
∇
2
�
??????+
�(�+1)(�+2)
3!
∇
3
�
??????+⋯+
�(�+1)(�+2)⋯(�+??????−1)
??????!
∇
??????
�
??????
=�
??????+(
�
1
)∇�
??????+(
�+1
2
)∇
2
�
??????+(
�+2
3
)∇
3
�
??????+⋯+(
�+??????−1
??????
)∇
??????
�
??????=∑(
�+??????−1
??????
)∇
??????
�
??????
??????
??????=0
NOTE
The Newton's backward difference interpolation formula is used to compute the value of �(�) when � is near to �
??????, i.e.
when � is at the end of the table.
This formula contains values of the function beginning from �
?????? backward (to the left) and none forward and thus named
backward interpolation formula. But, if the value of � is at the begining of the table, then this formula gives more error.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 6
Let us consider
�=
(�−�
??????)
ℎ
, i.e., �=�
??????+�ℎ.
Since �
??????=�
0+??????ℎ, we get �−�
??????=(�+??????−??????)ℎ ; ??????=0,1,2,⋯,??????.
Thus, the polynomial ??????(�) reduces to the following form:
??????(�)=??????(�
??????+�ℎ)
=�
??????+�∆�
??????−1+
�(�+1)
2!
∆
2
�
??????−2+
�(�+1)(�+2)
3!
∆
3
�
??????−3+⋯+
�(�+1)(�+2)⋯(�+??????−1)
??????!
∆
??????
�
0
=�
??????+(
�
1
)∆�
??????−1+(
�+1
2
)∆
2
�
??????−2+(
�+2
3
)∆
3
�
??????−3+⋯+(
�+??????−1
??????
)∆
??????
�
0
=∑(
�+??????−1
??????
)∆
??????
�
??????−??????
??????
??????=0
This is known as Newton's backward or Newton-Gregory backward interpolation formula.
In terms of backward differencing operator (∇), the Newton's backward interpolating polynomial ??????(�) becomes
??????(�)=??????(�
??????+�ℎ)
=�
??????+
∇�
??????
ℎ
(�−�
??????)+
∇
2
�
??????
2!ℎ
2
(�−�
??????)(�−�
??????−1)+
∇
3
�
??????
3!ℎ
3
(�−�
??????)(�−�
??????−1)(�−�
??????−2)+⋯
+
∇
??????
�
??????
??????!ℎ
??????
(�−�
??????)(�−�
??????−1)⋯(�−�
??????−??????+1)+⋯+
∇
??????
�
??????
??????!ℎ
??????
(�−�
??????)(�−�
??????−1)⋯(�−�
1)
=�
??????+�∇�
??????+
�(�+1)
2!
∇
2
�
??????+
�(�+1)(�+2)
3!
∇
3
�
??????+⋯+
�(�+1)(�+2)⋯(�+??????−1)
??????!
∇
??????
�
??????
=�
??????+(
�
1
)∇�
??????+(
�+1
2
)∇
2
�
??????+(
�+2
3
)∇
3
�
??????+⋯+(
�+??????−1
??????
)∇
??????
�
??????=∑(
�+??????−1
??????
)∇
??????
�
??????
??????
??????=0
NOTE
The Newton's backward difference interpolation formula is used to compute the value of �(�) when � is near to �
??????, i.e.
when � is at the end of the table.
This formula contains values of the function beginning from �
?????? backward (to the left) and none forward and thus named
backward interpolation formula. But, if the value of � is at the begining of the table, then this formula gives more error.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 7
EXAMPLE
Given below is a table of values of the probability integral �=�(�)=
2
??????
∫�
−??????
2
��
??????
0
.
Determine the value of �=�(�) when the value of � is 0.485.
�: 0.45 0.46 0.47 0.48 0.49
�: 0.475482 0.484656 0.497452 0.502750 0.511668
Solution:
Since the given value of �=0.456 is at the end of the table, we apply the Newton's backward interpolation formula to
determine the value of �=�(�) when the value of � is 0.485.
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing.
The Newton’s backward difference interpolating polynomial ??????(�) with is given by
??????(�)=�
??????+�∇�
??????+
�(�+1)
2!
∇
2
�
??????+
�(�+1)(�+2)
3!
∇
3
�
??????+⋯+
�(�+1)(�+2)⋯(�+??????−1)
??????!
∇
??????
�
??????
where �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and �=
(??????−????????????
)
ℎ
.
The backward difference table is as follows:
� �=??????(�) ??????� ??????
�
� ??????
�
� ??????
�
�
0.45 0.475482
0.46 0.484656 ∇�
1=0.009174
0.47 0.497452 ∇�
2=0.012796 ∇
2
�
2=0.003622
0.48 0.502750 ∇�
3=0.005298 ∇
2
�
3=−0.007498 ∇
3
�
3=−0.011120
0.49 0.511668 ∇�
4=0.008918 ∇
2
�
4=0.003620 ∇
3
�
4=0.011118 ∇
4
�
4=0.022238
Here, ??????=4 ; �
??????=0.49 ; �=0.485 ; ℎ=0.01 ; �=
(??????−????????????
)
ℎ
=
(0.485−0.49)
0.01
=−0.5 and ∇
??????
�
4=0 ∀ ??????>4.
Now,
�
0.485=�(0.485)=�
4+�∇�
4+
�(�+1)
2!
∇
2
�
4+
�(�+1)(�+2)
3!
∇
3
�
4+
�(�+1)(�+2)(�+3)
4!
∇
4
�
4
=0.511668+(−0.5)×0.008918+
(−0.5)(−0.5+1)
2!
×0.003620+
(−0.5)(−0.5+1)(−0.5+2)
3!
×0.011118+
(−0.5)(−0.5+1)(−0.5+2)(−0.5+3)
4!
×0.022238
=0.511668−0.004459−0.0004525−0.000694875−0.0008686719
=0.5051929531
Hence, the value of �=�(�) when �=0.485 is 0.5051929531.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 7
EXAMPLE
Given below is a table of values of the probability integral �=�(�)=
2
??????
∫�
−??????
2
��
??????
0
.
Determine the value of �=�(�) when the value of � is 0.485.
�: 0.45 0.46 0.47 0.48 0.49
�: 0.475482 0.484656 0.497452 0.502750 0.511668
Solution:
Since the given value of �=0.456 is at the end of the table, we apply the Newton's backward interpolation formula to
determine the value of �=�(�) when the value of � is 0.485.
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing.
The Newton’s backward difference interpolating polynomial ??????(�) with is given by
??????(�)=�
??????+�∇�
??????+
�(�+1)
2!
∇
2
�
??????+
�(�+1)(�+2)
3!
∇
3
�
??????+⋯+
�(�+1)(�+2)⋯(�+??????−1)
??????!
∇
??????
�
??????
where �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and �=
(??????−????????????
)
ℎ
.
The backward difference table is as follows:
� �=??????(�) ??????� ??????
�
� ??????
�
� ??????
�
�
0.45 0.475482
0.46 0.484656 ∇�
1=0.009174
0.47 0.497452 ∇�
2=0.012796 ∇
2
�
2=0.003622
0.48 0.502750 ∇�
3=0.005298 ∇
2
�
3=−0.007498 ∇
3
�
3=−0.011120
0.49 0.511668 ∇�
4=0.008918 ∇
2
�
4=0.003620 ∇
3
�
4=0.011118 ∇
4
�
4=0.022238
Here, ??????=4 ; �
??????=0.49 ; �=0.485 ; ℎ=0.01 ; �=
(??????−????????????
)
ℎ
=
(0.485−0.49)
0.01
=−0.5 and ∇
??????
�
4=0 ∀ ??????>4.
Now,
�
0.485=�(0.485)=�
4+�∇�
4+
�(�+1)
2!
∇
2
�
4+
�(�+1)(�+2)
3!
∇
3
�
4+
�(�+1)(�+2)(�+3)
4!
∇
4
�
4
=0.511668+(−0.5)×0.008918+
(−0.5)(−0.5+1)
2!
×0.003620+
(−0.5)(−0.5+1)(−0.5+2)
3!
×0.011118+
(−0.5)(−0.5+1)(−0.5+2)(−0.5+3)
4!
×0.022238
=0.511668−0.004459−0.0004525−0.000694875−0.0008686719
=0.5051929531
Hence, the value of �=�(�) when �=0.485 is 0.5051929531.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 8
EXAMPLE
The upward velocity of a rocket is given below:
� (sec) : 10 15 20 25 30
�(�) (m/sec) : 126.75 350.5 510.8 650.4 920.25
Determine the value of the velocity at �=26 sec using Newton's backward formula. Also determine the required velocity
using Newton's forward formula by considering data in reverse order and compare the results.
Solution:
Since the given value of �=26 sec is at the end of the table, we apply the Newton's backward interpolation formula to
determine the value of �=�(�) when the value of � is 26 sec.
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing.
The Newton’s backward difference interpolating polynomial ??????(�) with is given by
??????(�)=�
??????+�∇�
??????+
�(�+1)
2!
∇
2
�
??????+
�(�+1)(�+2)
3!
∇
3
�
??????+⋯+
�(�+1)(�+2)⋯(�+??????−1)
??????!
∇
??????
�
??????
where �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and �=
(??????−????????????
)
ℎ
.
The backward difference table is as follows:
?????? �=??????(??????) ??????� ??????
�
� ??????
�
� ??????
�
�
10 126.75
15 350.5 223.75
20 510.8 160.3 -63.45
25 650.4 139.6 -20.7 42.75
30 920.25 269.85 130.25 150.95 108.2
Here, ??????=4 ; �
??????=30 ; �=26 ; ℎ=5 ; �=
(??????−????????????
)
ℎ
=
(26−30)
5
=−0.8 and ∇
??????
�
4=0 ∀ ??????>4.
Now,
�
26=�(26)=�
4+�∇�
4+
�(�+1)
2!
∇
2
�
4+
�(�+1)(�+2)
3!
∇
3
�
4+
�(�+1)(�+2)(�+3)
4!
∇
4
�
4
=920.25+(−0.8)×269.85+
(−0.8)(−0.8+1)
2!
×130.25+
(−0.8)(−0.8+1)(−0.8+2)
3!
×150.95
+
(−0.8)(−0.8+1)(−0.8+2)(−0.8+3)
4!
×108.2
=920.25−215.88−10.42−4.8304−1.90432
=687.21528
Hence, the velocity of the rocket �=�(�) when �=26 sec is 687.21528 m/sec.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 8
EXAMPLE
The upward velocity of a rocket is given below:
� (sec) : 10 15 20 25 30
�(�) (m/sec) : 126.75 350.5 510.8 650.4 920.25
Determine the value of the velocity at �=26 sec using Newton's backward formula. Also determine the required velocity
using Newton's forward formula by considering data in reverse order and compare the results.
Solution:
Since the given value of �=26 sec is at the end of the table, we apply the Newton's backward interpolation formula to
determine the value of �=�(�) when the value of � is 26 sec.
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing.
The Newton’s backward difference interpolating polynomial ??????(�) with is given by
??????(�)=�
??????+�∇�
??????+
�(�+1)
2!
∇
2
�
??????+
�(�+1)(�+2)
3!
∇
3
�
??????+⋯+
�(�+1)(�+2)⋯(�+??????−1)
??????!
∇
??????
�
??????
where �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and �=
(??????−????????????
)
ℎ
.
The backward difference table is as follows:
?????? �=??????(??????) ??????� ??????
�
� ??????
�
� ??????
�
�
10 126.75
15 350.5 223.75
20 510.8 160.3 -63.45
25 650.4 139.6 -20.7 42.75
30 920.25 269.85 130.25 150.95 108.2
Here, ??????=4 ; �
??????=30 ; �=26 ; ℎ=5 ; �=
(??????−????????????
)
ℎ
=
(26−30)
5
=−0.8 and ∇
??????
�
4=0 ∀ ??????>4.
Now,
�
26=�(26)=�
4+�∇�
4+
�(�+1)
2!
∇
2
�
4+
�(�+1)(�+2)
3!
∇
3
�
4+
�(�+1)(�+2)(�+3)
4!
∇
4
�
4
=920.25+(−0.8)×269.85+
(−0.8)(−0.8+1)
2!
×130.25+
(−0.8)(−0.8+1)(−0.8+2)
3!
×150.95
+
(−0.8)(−0.8+1)(−0.8+2)(−0.8+3)
4!
×108.2
=920.25−215.88−10.42−4.8304−1.90432
=687.21528
Hence, the velocity of the rocket �=�(�) when �=26 sec is 687.21528 m/sec.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 9
Since the Newton's forward interpolation formula is used when the given value is at the beginning of the table. For this
purpose, the table is written in reverse order as
� (sec) : 10 15 20 25 30
�(�) (m/sec) : 126.75 350.5 510.8 650.4 920.25
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing.
The Newton’s forward difference interpolating polynomial ??????(�) with is given by
??????(�)=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+⋯+
�(�−1)(�−2)⋯(�−??????+1)
??????!
∆
??????
�
0
where �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and �=
(??????−??????0
)
ℎ
.
The forward difference table is as follows:
� �=??????(??????) ∆� ∆
�
� ∆
�
� ∆
�
�
30 920.25
-269.85
25 650.4 130.25
-139.6 -150.95
20 510.8 -20.7 108.2
-160.3 -42.75
15 350.5 -63.45
-223.75
10 126.75
Here, �
0=30 ; �=26 ; ℎ=−5 ; �=
(??????−??????0
)
ℎ
=
(26−30)
−5
=0.8 and ∆
??????
�
0=0 ∀ ??????>4.
Now,
�
26=�(26)=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+
�(�−1)(�−2)(�−3)
4!
∆
4
�
0
=920.25+0.8×(−269.85)+
0.8(0.8−1)
2!
×130.25+
0.8(0.8−1)(0.8−2)
3!
×(−150.95)
+
0.8(0.8−1)(0.8−2)(0.8−3)
4!
×108.2
=920.25−215.88−10.42−4.8304−1.90432
=687.21528
Hence, the velocity of the rocket �=�(�) when �=26 sec is 687.21528 m/sec.
Thus, the value of �(26) obtained by Newton's forward formula is 687.21528 m/sec which is same as the value obtained
by Newton's backward formula.
From the above example one can conclude that if a problem is solved by Newton's forward formula, then it can be solved
by Newton's backward formula also.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 9
Since the Newton's forward interpolation formula is used when the given value is at the beginning of the table. For this
purpose, the table is written in reverse order as
� (sec) : 10 15 20 25 30
�(�) (m/sec) : 126.75 350.5 510.8 650.4 920.25
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ, ??????=0,1,2,⋯,??????, where h being the interval of differencing.
The Newton’s forward difference interpolating polynomial ??????(�) with is given by
??????(�)=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+⋯+
�(�−1)(�−2)⋯(�−??????+1)
??????!
∆
??????
�
0
where �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and �=
(??????−??????0
)
ℎ
.
The forward difference table is as follows:
� �=??????(??????) ∆� ∆
�
� ∆
�
� ∆
�
�
30 920.25
-269.85
25 650.4 130.25
-139.6 -150.95
20 510.8 -20.7 108.2
-160.3 -42.75
15 350.5 -63.45
-223.75
10 126.75
Here, �
0=30 ; �=26 ; ℎ=−5 ; �=
(??????−??????0
)
ℎ
=
(26−30)
−5
=0.8 and ∆
??????
�
0=0 ∀ ??????>4.
Now,
�
26=�(26)=�
0+�∆�
0+
�(�−1)
2!
∆
2
�
0+
�(�−1)(�−2)
3!
∆
3
�
0+
�(�−1)(�−2)(�−3)
4!
∆
4
�
0
=920.25+0.8×(−269.85)+
0.8(0.8−1)
2!
×130.25+
0.8(0.8−1)(0.8−2)
3!
×(−150.95)
+
0.8(0.8−1)(0.8−2)(0.8−3)
4!
×108.2
=920.25−215.88−10.42−4.8304−1.90432
=687.21528
Hence, the velocity of the rocket �=�(�) when �=26 sec is 687.21528 m/sec.
Thus, the value of �(26) obtained by Newton's forward formula is 687.21528 m/sec which is same as the value obtained
by Newton's backward formula.
From the above example one can conclude that if a problem is solved by Newton's forward formula, then it can be solved
by Newton's backward formula also.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 10
LAGRANGE'S INTERPOLATION POLYNOMIAL
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) values or arguments �
??????,??????=0,1,2,⋯,?????? (need not necessarily equidistance).
With these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation function such that
�
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
Suppose the n-th degree polynomial is taken as
??????(�)=�
0(�−�
1)(�−�
2)⋯(�−�
??????)+�
1(�−�
0)(�−�
2)⋯(�−�
??????)+�
2(�−�
0)(�−�
1)(�−�
3)⋯(�−�
??????)
+⋯+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)+⋯
+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1) ⋯⋯⋯ (∗∗∗)
Now, the constants �
0,�
1,⋯,�
?????? are estimate by ??????(�
??????)=�
??????, ??????=0,1,2,⋯,??????.
When �=�
0, from (∗∗∗) we get
??????(�
0)=�
0 or,�
0=
�
0
(�
0−�
1)(�
0−�
2)⋯(�
0−�
??????)
.
In general, when �=�
??????, we get
??????(�
??????)=�
?????? or,�
??????=
�
??????
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
; ??????=0,1,2,⋯,??????
Substituting the values of �
0,�
1,⋯,�
?????? in (∗∗∗), the polynomial ??????(�) becomes
??????(�)=
(�−�
1)(�−�
2)⋯(�−�
??????)
(�
0−�
1)(�
0−�
2)⋯(�
0−�
??????)
�
0+
(�−�
0)(�−�
2)⋯(�−�
??????)
(�
1−�
0)(�
1−�
2)⋯(�
1−�
??????)
�
1+⋯
+
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????+⋯
+
(�−�
0)(�−�
1)⋯(�−�
??????−1)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)
�
??????
=∑
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????
??????
??????=0
=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
The polynomial ??????(�) is called the Lagrange's interpolating polynomial. This is also known as Lagrange's interpolation
formula.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 10
LAGRANGE'S INTERPOLATION POLYNOMIAL
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) values or arguments �
??????,??????=0,1,2,⋯,?????? (need not necessarily equidistance).
With these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation function such that
�
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
Suppose the n-th degree polynomial is taken as
??????(�)=�
0(�−�
1)(�−�
2)⋯(�−�
??????)+�
1(�−�
0)(�−�
2)⋯(�−�
??????)+�
2(�−�
0)(�−�
1)(�−�
3)⋯(�−�
??????)
+⋯+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)+⋯
+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1) ⋯⋯⋯ (∗∗∗)
Now, the constants �
0,�
1,⋯,�
?????? are estimate by ??????(�
??????)=�
??????, ??????=0,1,2,⋯,??????.
When �=�
0, from (∗∗∗) we get
??????(�
0)=�
0 or,�
0=
�
0
(�
0−�
1)(�
0−�
2)⋯(�
0−�
??????)
.
In general, when �=�
??????, we get
??????(�
??????)=�
?????? or,�
??????=
�
??????
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
; ??????=0,1,2,⋯,??????
Substituting the values of �
0,�
1,⋯,�
?????? in (∗∗∗), the polynomial ??????(�) becomes
??????(�)=
(�−�
1)(�−�
2)⋯(�−�
??????)
(�
0−�
1)(�
0−�
2)⋯(�
0−�
??????)
�
0+
(�−�
0)(�−�
2)⋯(�−�
??????)
(�
1−�
0)(�
1−�
2)⋯(�
1−�
??????)
�
1+⋯
+
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????+⋯
+
(�−�
0)(�−�
1)⋯(�−�
??????−1)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)
�
??????
=∑
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????
??????
??????=0
=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
The polynomial ??????(�) is called the Lagrange's interpolating polynomial. This is also known as Lagrange's interpolation
formula.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 11
ALTERNATIVE
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) values or arguments �
??????,??????=0,1,2,⋯,?????? (need not necessarily equidistance).
With these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation function such that
�
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
Suppose the polynomial ??????(�) is the following form
??????(�)=∑�
??????(�)�
??????
??????
??????=0
where each �
??????(�) is polynomial in x, of degree less than or equal to ??????. The function �
??????(�) called the Lagrangian function.
The polynomial ??????(�) satisfies the condition �
??????=??????(�
??????),∀ ??????=0,1,2,⋯,?????? if
�
??????(�
??????)={
0 for ??????≠??????
1 for ??????=??????
or, �
??????(�)={
0 for �=�
0,�
1,⋯,�
??????−1,�
??????+1,⋯,�
??????
1 for �=�
??????
Let us consider the function �
??????(�) be
�
??????(�)=�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????) ; ??????=0,1,2,⋯,??????,
where �
?????? be a constant.
The constants �
0,�
1,⋯,�
?????? are estimate by �
??????(�
??????)=1, ??????=0,1,2,⋯,??????.
∴ �
??????=
1
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
and
�
??????(�)=
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
; ??????=0,1,2,⋯,??????,
The Lagrange’s interpolating polynomial ??????(�) becomes
??????(�)=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
.
MERIT AND DEMERITS OF LAGRANGE’S INTERPOLATION FORMULA
Lagrange's interpolation formula is used for any type of data as in this formula there is no restriction on equidistance
argument or the equal spacing ℎ. This formula is also applicable to find the value of �(�) at any point � within the
minimum and maximum values of �
0,�
1,�
2,⋯,�
??????.
There are also some disadvantages of this formula. If the number of interpolating points decreases or increases, then a
fresh calculation is needed to find all Lagrange's functions or weights (�
??????(�)) of the functional values. Again, each �
??????(�)
is a polynomial of degree ?????? if the table contains (??????+1) points. Thus we have to calculate ?????? such polynomials each of
degree ?????? and it is very laborious.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 11
ALTERNATIVE
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) values or arguments �
??????,??????=0,1,2,⋯,?????? (need not necessarily equidistance).
With these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation function such that
�
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
Suppose the polynomial ??????(�) is the following form
??????(�)=∑�
??????(�)�
??????
??????
??????=0
where each �
??????(�) is polynomial in x, of degree less than or equal to ??????. The function �
??????(�) called the Lagrangian function.
The polynomial ??????(�) satisfies the condition �
??????=??????(�
??????),∀ ??????=0,1,2,⋯,?????? if
�
??????(�
??????)={
0 for ??????≠??????
1 for ??????=??????
or, �
??????(�)={
0 for �=�
0,�
1,⋯,�
??????−1,�
??????+1,⋯,�
??????
1 for �=�
??????
Let us consider the function �
??????(�) be
�
??????(�)=�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????) ; ??????=0,1,2,⋯,??????,
where �
?????? be a constant.
The constants �
0,�
1,⋯,�
?????? are estimate by �
??????(�
??????)=1, ??????=0,1,2,⋯,??????.
∴ �
??????=
1
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
and
�
??????(�)=
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
; ??????=0,1,2,⋯,??????,
The Lagrange’s interpolating polynomial ??????(�) becomes
??????(�)=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
.
MERIT AND DEMERITS OF LAGRANGE’S INTERPOLATION FORMULA
Lagrange's interpolation formula is used for any type of data as in this formula there is no restriction on equidistance
argument or the equal spacing ℎ. This formula is also applicable to find the value of �(�) at any point � within the
minimum and maximum values of �
0,�
1,�
2,⋯,�
??????.
There are also some disadvantages of this formula. If the number of interpolating points decreases or increases, then a
fresh calculation is needed to find all Lagrange's functions or weights (�
??????(�)) of the functional values. Again, each �
??????(�)
is a polynomial of degree ?????? if the table contains (??????+1) points. Thus we have to calculate ?????? such polynomials each of
degree ?????? and it is very laborious.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 12
EXAMPLE
Given that �
0=0.4,�
2=0.132,�
3=0.254 and �
4=0.376. Find the value of �
1 by using Lagrange’s interpolating
polynomial.
Solution:
We have 4 functional values of �
?????? corresponding to �=0,2,3,4. We want to find the value of �
?????? when �=1.
Given that (�,�
??????),�=0,2,3,4 are 4 pairs of data points of argument � and the entry �
??????.
Let us consider a 3-rd degree polynomial ??????(�) such that �
??????=??????(�), for each �=0,2,3,4.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=
(�−�
2)(�−�
3)(�−�
4)
(�
0−�
2)(�
0−�
3)(�
0−�
4)
�
0+
(�−�
0)(�−�
3)(�−�
4)
(�
2−�
0)(�
2−�
3)(�
2−�
4)
�
2+
(�−�
0)(�−�
2)(�−�
4)
(�
3−�
0)(�
3−�
3)(�
2−�
4)
�
3
+
(�−�
0)(�−�
2)(�−�
3)
(�
4−�
0)(�
4−�
2)(�
4−�
3)
�
4
�
1=??????(1)=
(1−2)(1−3)(1−4)
(0−2)(0−3)(0−4)
×0.4+
(1−0)(1−3)(1−4)
(2−0)(2−3)(2−4)
×0.132+
(1−0)(1−2)(1−4)
(3−0)(3−2)(3−4)
×0.254+
(1−0)(1−2)(1−3)
(4−0)(4−2)(4−3)
×0.376
=0.1+0.198−0.254+0.094=0.138
Hence, the value of �
1 obtained by Lagrange’s interpolation formula is �
1=0.138.
EXAMPLE
By using Lagrange’s interpolation formula, show that, approximately, �
0=�
2−0.3 (�
4−�
−4)+0.2 (�
−4−�
−6).
Solution:
We have 4 functional values of �
?????? corresponding to �=−6,−4,2,4 in the right hand side of the given expression.
Let us assume �
?????? be a polynomial of degree 3.
By using Lagrange’s interpolation formula, we can write
�
0=??????(0)=
(0+4)(0−2)(0−4)
(−6+4)(−6−2)(−6−4)
�
−6+
(0+6)(0−2)(0−4)
(−4+6)(−4−2)(−4−4)
�
−4
+
(0+6)(0+4)(0−4)
(2+6)(2+4)(2−4)
�
2+
(0+6)(0+4)(0−2)
(4+6)(4+4)(4−2)
�
4
=−0.2 �
−6+0.5 �
−4+�
2−0.3 �
4
=�
2−0.3 (�
4−�
−4)+0.2 (�
−4−�
−6)
Hence, �
0=�
2−0.3 (�
4−�
−4)+0.2 (�
−4−�
−6).
Numerical Analysis || Lecture Notes || Anup Kumar Giri 12
EXAMPLE
Given that �
0=0.4,�
2=0.132,�
3=0.254 and �
4=0.376. Find the value of �
1 by using Lagrange’s interpolating
polynomial.
Solution:
We have 4 functional values of �
?????? corresponding to �=0,2,3,4. We want to find the value of �
?????? when �=1.
Given that (�,�
??????),�=0,2,3,4 are 4 pairs of data points of argument � and the entry �
??????.
Let us consider a 3-rd degree polynomial ??????(�) such that �
??????=??????(�), for each �=0,2,3,4.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=
(�−�
2)(�−�
3)(�−�
4)
(�
0−�
2)(�
0−�
3)(�
0−�
4)
�
0+
(�−�
0)(�−�
3)(�−�
4)
(�
2−�
0)(�
2−�
3)(�
2−�
4)
�
2+
(�−�
0)(�−�
2)(�−�
4)
(�
3−�
0)(�
3−�
3)(�
2−�
4)
�
3
+
(�−�
0)(�−�
2)(�−�
3)
(�
4−�
0)(�
4−�
2)(�
4−�
3)
�
4
�
1=??????(1)=
(1−2)(1−3)(1−4)
(0−2)(0−3)(0−4)
×0.4+
(1−0)(1−3)(1−4)
(2−0)(2−3)(2−4)
×0.132+
(1−0)(1−2)(1−4)
(3−0)(3−2)(3−4)
×0.254+
(1−0)(1−2)(1−3)
(4−0)(4−2)(4−3)
×0.376
=0.1+0.198−0.254+0.094=0.138
Hence, the value of �
1 obtained by Lagrange’s interpolation formula is �
1=0.138.
EXAMPLE
By using Lagrange’s interpolation formula, show that, approximately, �
0=�
2−0.3 (�
4−�
−4)+0.2 (�
−4−�
−6).
Solution:
We have 4 functional values of �
?????? corresponding to �=−6,−4,2,4 in the right hand side of the given expression.
Let us assume �
?????? be a polynomial of degree 3.
By using Lagrange’s interpolation formula, we can write
�
0=??????(0)=
(0+4)(0−2)(0−4)
(−6+4)(−6−2)(−6−4)
�
−6+
(0+6)(0−2)(0−4)
(−4+6)(−4−2)(−4−4)
�
−4
+
(0+6)(0+4)(0−4)
(2+6)(2+4)(2−4)
�
2+
(0+6)(0+4)(0−2)
(4+6)(4+4)(4−2)
�
4
=−0.2 �
−6+0.5 �
−4+�
2−0.3 �
4
=�
2−0.3 (�
4−�
−4)+0.2 (�
−4−�
−6)
Hence, �
0=�
2−0.3 (�
4−�
−4)+0.2 (�
−4−�
−6).
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 13
RESULT
Suppose (�
??????,�
??????),??????=0,1,2,⋯,?????? are (??????+1) pairs of data points of argument � and the entry �=�(�). Show that the
Lagrange’s interpolating polynomial can be represented in the form,
??????(�)=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????) is the value of �
′
(�) at �=�
??????,??????=0,1,2,⋯,??????.
PROOF
Let (�
??????,�
??????),??????=0,1,2,⋯,?????? be (??????+1) pairs of data points of argument � and the entry �=�(�).
Let us consider a n-th degree polynomial ??????(�) such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
=∑
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????
??????
??????=0
. ⋯(∗)
Let �(�)=∏(�−�
??????)
??????
??????=0
i.e.,ln�(�)=∑ln(�−�
??????)
??????
??????=0
. ⋯ (∗∗)
By differentiating both side of (∗∗) with respect to �, we get
�
′
(�)
�(�)
=∑
1
(�−�
??????)
??????
??????=0
or, �
′
(�)=∑
�(�)
(�−�
??????)
??????
??????=0
=∑(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
??????
??????=0
or, �
′
(�)=(�−�
1)(�−�
2)⋯(�−�
??????)+(�−�
0)(�−�
2)⋯(�−�
??????)+⋯
+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)+⋯+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)
At �=�
??????, �
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????) , ??????=0,1,2,⋯,??????
The coefficients of �
?????? in (∗) is given by
�
??????(�)=
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=
�(�)
(�−�
??????)�
′
(�
??????)
; ??????=0,1,2,⋯,??????.
Hence, the Lagrange’s interpolating polynomial can be represented in the form,
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????) is the value of �
′
(�) at �=�
??????,??????=0,1,2,⋯,??????.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 13
RESULT
Suppose (�
??????,�
??????),??????=0,1,2,⋯,?????? are (??????+1) pairs of data points of argument � and the entry �=�(�). Show that the
Lagrange’s interpolating polynomial can be represented in the form,
??????(�)=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????) is the value of �
′
(�) at �=�
??????,??????=0,1,2,⋯,??????.
PROOF
Let (�
??????,�
??????),??????=0,1,2,⋯,?????? be (??????+1) pairs of data points of argument � and the entry �=�(�).
Let us consider a n-th degree polynomial ??????(�) such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
=∑
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????
??????
??????=0
. ⋯(∗)
Let �(�)=∏(�−�
??????)
??????
??????=0
i.e.,ln�(�)=∑ln(�−�
??????)
??????
??????=0
. ⋯ (∗∗)
By differentiating both side of (∗∗) with respect to �, we get
�
′
(�)
�(�)
=∑
1
(�−�
??????)
??????
??????=0
or, �
′
(�)=∑
�(�)
(�−�
??????)
??????
??????=0
=∑(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
??????
??????=0
or, �
′
(�)=(�−�
1)(�−�
2)⋯(�−�
??????)+(�−�
0)(�−�
2)⋯(�−�
??????)+⋯
+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)+⋯+�
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)
At �=�
??????, �
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????) , ??????=0,1,2,⋯,??????
The coefficients of �
?????? in (∗) is given by
�
??????(�)=
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=
�(�)
(�−�
??????)�
′
(�
??????)
; ??????=0,1,2,⋯,??????.
Hence, the Lagrange’s interpolating polynomial can be represented in the form,
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????) is the value of �
′
(�) at �=�
??????,??????=0,1,2,⋯,??????.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 14
NOTE
The coefficients of �
?????? in the Lagrange’s interpolating polynomial is
�
??????(�)=
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
=
�(�)
(�−�
??????)�
′
(�
??????)
; ??????=0,1,⋯,??????
The Lagrange’s interpolating polynomial can be represented in the form,
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????) is the value of �
′
(�) at �=�
??????,??????=0,1,2,⋯,??????.
RESULT
The weights of the functional values in Lagrange’s interpolating polynomial are invariant under linear transformation or,
the weights of the functional values in Lagrange’s interpolating polynomial are independent of change of origin and scale.
PROOF
Let (�
??????,�
??????),??????=0,1,2,⋯,?????? be (??????+1) pairs of data points of argument � and the entry �=�(�).
Let us consider a n-th degree polynomial ??????(�) such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
=∑
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????
??????
??????=0
. ⋯(∗)
Let �=��+�, where �,� are constants.
Therefore, �
??????=��
??????+�, (�−�
??????)=�(�−�
??????) and (�
??????−�
??????)=�(�
??????−�
??????) when ??????≠??????.
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)=�
??????
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)=�
??????
(�
??????−�
0)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
Now, �
??????(�)=∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
=∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
=�
??????(�) ; ??????=0,1,2,⋯,??????.
Alternatively, �
??????(�)=
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=
�
??????
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
�
??????
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=�
??????(�) ; ??????=0,1,2,⋯,??????.
Hence, the weights of the functional values in Lagrange’s interpolating polynomial are invariant under linear
transformation or, the weights of the functional values in Lagrange’s interpolating polynomial are independent of change
of origin and scale.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 14
NOTE
The coefficients of �
?????? in the Lagrange’s interpolating polynomial is
�
??????(�)=
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
=
�(�)
(�−�
??????)�
′
(�
??????)
; ??????=0,1,⋯,??????
The Lagrange’s interpolating polynomial can be represented in the form,
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????) is the value of �
′
(�) at �=�
??????,??????=0,1,2,⋯,??????.
RESULT
The weights of the functional values in Lagrange’s interpolating polynomial are invariant under linear transformation or,
the weights of the functional values in Lagrange’s interpolating polynomial are independent of change of origin and scale.
PROOF
Let (�
??????,�
??????),??????=0,1,2,⋯,?????? be (??????+1) pairs of data points of argument � and the entry �=�(�).
Let us consider a n-th degree polynomial ??????(�) such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
�
??????
??????
??????=0
=∑
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????
??????
??????=0
. ⋯(∗)
Let �=��+�, where �,� are constants.
Therefore, �
??????=��
??????+�, (�−�
??????)=�(�−�
??????) and (�
??????−�
??????)=�(�
??????−�
??????) when ??????≠??????.
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)=�
??????
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)=�
??????
(�
??????−�
0)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
Now, �
??????(�)=∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
=∏
(�−�
??????)
(�
??????−�
??????)
??????
??????=0
??????≠??????
=�
??????(�) ; ??????=0,1,2,⋯,??????.
Alternatively, �
??????(�)=
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=
�
??????
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
�
??????
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=�
??????(�) ; ??????=0,1,2,⋯,??????.
Hence, the weights of the functional values in Lagrange’s interpolating polynomial are invariant under linear
transformation or, the weights of the functional values in Lagrange’s interpolating polynomial are independent of change
of origin and scale.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 15
ALTERNATIVE
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????) is the value of �
′
(�) at �=�
??????,??????=0,1,2,⋯,??????.
Let �=��+�, where �,� are constants.
Therefore, �
??????=��
??????+�, (�−�
??????)=�(�−�
??????) and (�
??????−�
??????)=�(�
??????−�
??????) when ??????≠??????.
�(�)=∏(�−�
??????)
??????
??????=0
=(�−�
0)(�−�
1)⋯(�−�
??????)=�
??????+1
(�−�
0)(�−�
1)⋯(�−�
??????)=�
??????+1
�(�)
�
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=�
??????
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)=�
??????
�
′
(�
??????), ??????=0,1,2,⋯,??????
Now, �
??????(�)=
�(�)
(�−�
??????)�
′
(�
??????)
=
�
??????+1
�(�)
�(�−�
??????)�
??????
�
′
(�
??????)
=
�(�)
(�−�
??????)�
′
(�
??????)
=�
??????(�) ; ??????=0,1,2,⋯,??????.
NOTE
The Lagrange’s interpolating polynomial are invariant under linear transformation of the argument or, Lagrange’s
interpolating polynomial are independent of change of origin and scale of argument.
RESULT
The sum of weights of the functional values in Lagrange’s interpolating polynomial is unity, or, sum of Lagrangian
functions in Lagrange’s interpolating polynomial is 1, i.e., ∑�
??????(�)
??????
??????=0 =1.
PROOF
Let (�
??????,�
??????),??????=0,1,2,⋯,?????? be (??????+1) pairs of data points of argument � and the entry �=�(�).
Let us consider a n-th degree polynomial ??????(�) such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????) , ??????=0,1,2,⋯,??????.
Let us consider the expression
1
�(�)
=
1
∏(�−�
??????)
??????
??????=0
=
??????
0
(�−�
0)
+
??????
1
(�−�
1)
⋯+
??????
??????
(�−�
??????)
+⋯+
??????
??????
(�−�
??????)
⋯⋯(∗)
where ??????
0,??????
1,⋯,??????
?????? are constants.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 15
ALTERNATIVE
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????) is the value of �
′
(�) at �=�
??????,??????=0,1,2,⋯,??????.
Let �=��+�, where �,� are constants.
Therefore, �
??????=��
??????+�, (�−�
??????)=�(�−�
??????) and (�
??????−�
??????)=�(�
??????−�
??????) when ??????≠??????.
�(�)=∏(�−�
??????)
??????
??????=0
=(�−�
0)(�−�
1)⋯(�−�
??????)=�
??????+1
(�−�
0)(�−�
1)⋯(�−�
??????)=�
??????+1
�(�)
�
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=�
??????
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)=�
??????
�
′
(�
??????), ??????=0,1,2,⋯,??????
Now, �
??????(�)=
�(�)
(�−�
??????)�
′
(�
??????)
=
�
??????+1
�(�)
�(�−�
??????)�
??????
�
′
(�
??????)
=
�(�)
(�−�
??????)�
′
(�
??????)
=�
??????(�) ; ??????=0,1,2,⋯,??????.
NOTE
The Lagrange’s interpolating polynomial are invariant under linear transformation of the argument or, Lagrange’s
interpolating polynomial are independent of change of origin and scale of argument.
RESULT
The sum of weights of the functional values in Lagrange’s interpolating polynomial is unity, or, sum of Lagrangian
functions in Lagrange’s interpolating polynomial is 1, i.e., ∑�
??????(�)
??????
??????=0 =1.
PROOF
Let (�
??????,�
??????),??????=0,1,2,⋯,?????? be (??????+1) pairs of data points of argument � and the entry �=�(�).
Let us consider a n-th degree polynomial ??????(�) such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑�
??????(�)�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????) , ??????=0,1,2,⋯,??????.
Let us consider the expression
1
�(�)
=
1
∏(�−�
??????)
??????
??????=0
=
??????
0
(�−�
0)
+
??????
1
(�−�
1)
⋯+
??????
??????
(�−�
??????)
+⋯+
??????
??????
(�−�
??????)
⋯⋯(∗)
where ??????
0,??????
1,⋯,??????
?????? are constants.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 16
From (∗), we get
1=??????
0(�−�
1)(�−�
2)⋯(�−�
??????)+??????
1(�−�
0)(�−�
2)⋯(�−�
??????)+⋯
+??????
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)+⋯
+??????
??????(�−�
0)(�−�
1)⋯(�−�
??????−1) ⋯⋯ (∗∗)
When �=�
0, we get
??????
0=
1
(�
0−�
1)(�
0−�
2)⋯(�
0−�
??????)
=
1
�
′
(�
0)
In general, when �=�
??????, we get
??????
??????=
1
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=
1
�
′
(�
??????)
, ??????=0,1,2,⋯,??????
Substituting the values of ??????
0,??????
1,⋯,??????
?????? in (∗), we get
1
�(�)
=∑
??????
??????
(�−�
??????)
??????
??????=0
=∑
1
(�−�
??????)�
′
(�
??????)
??????
??????=0
or, ∑
�(�)
(�−�
??????)�
′
(�
??????)
??????
??????=0
=1
or, ∑�
??????(�)
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
??????
??????=0
=1, where �
??????(�)=
�(�)
(�−�
??????)�
′
(�
??????)
=Lagrangian functions,??????=0,1,⋯,??????
Hence, the sum of weights of the functional values in Lagrange’s interpolating polynomial is unity, or, sum of Lagrangian
functions in Lagrange’s interpolating polynomial is 1, i.e., ∑�
??????(�)
??????
??????=0 =1.
LAGRANGE’S INTERPOLATION FORMULA FOR EQUALLY SPACED POINTS
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ,??????=0,1,2,⋯,??????, where h being the interval of differencing
or spacing.
Let us consider a n-th degree polynomial ??????(�) such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????) , ??????=0,1,2,⋯,??????.
Let us consider a new variable � is introduced which is related with � as �=�
0+�ℎ.
Then (�−�
??????)=(�−??????)ℎ and (�
??????−�
??????)=(??????−??????)ℎ.
Now, �(�)=(�−�
0)(�−�
1)⋯(�−�
??????)=�ℎ(�−1)ℎ(�−2)ℎ⋯(�−??????)ℎ=ℎ
??????+1
�(�−1)(�−2)⋯(�−??????)
Numerical Analysis || Lecture Notes || Anup Kumar Giri 16
From (∗), we get
1=??????
0(�−�
1)(�−�
2)⋯(�−�
??????)+??????
1(�−�
0)(�−�
2)⋯(�−�
??????)+⋯
+??????
??????(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)+⋯
+??????
??????(�−�
0)(�−�
1)⋯(�−�
??????−1) ⋯⋯ (∗∗)
When �=�
0, we get
??????
0=
1
(�
0−�
1)(�
0−�
2)⋯(�
0−�
??????)
=
1
�
′
(�
0)
In general, when �=�
??????, we get
??????
??????=
1
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
=
1
�
′
(�
??????)
, ??????=0,1,2,⋯,??????
Substituting the values of ??????
0,??????
1,⋯,??????
?????? in (∗), we get
1
�(�)
=∑
??????
??????
(�−�
??????)
??????
??????=0
=∑
1
(�−�
??????)�
′
(�
??????)
??????
??????=0
or, ∑
�(�)
(�−�
??????)�
′
(�
??????)
??????
??????=0
=1
or, ∑�
??????(�)
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
??????
??????=0
=1, where �
??????(�)=
�(�)
(�−�
??????)�
′
(�
??????)
=Lagrangian functions,??????=0,1,⋯,??????
Hence, the sum of weights of the functional values in Lagrange’s interpolating polynomial is unity, or, sum of Lagrangian
functions in Lagrange’s interpolating polynomial is 1, i.e., ∑�
??????(�)
??????
??????=0 =1.
LAGRANGE’S INTERPOLATION FORMULA FOR EQUALLY SPACED POINTS
Let �=�(�) be a function which assumes the functional values or entries �
??????=�(�
??????),??????=0,1,2,⋯,?????? corresponding
to the (??????+1) equidistant values or arguments �
??????=�
0+??????ℎ,??????=0,1,2,⋯,??????, where h being the interval of differencing
or spacing.
Let us consider a n-th degree polynomial ??????(�) such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,??????.
The Lagrange’s interpolating polynomial ??????(�) is given by
??????(�)=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
, where �(�)=∏(�−�
??????)
??????
??????=0
and
�
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????) , ??????=0,1,2,⋯,??????.
Let us consider a new variable � is introduced which is related with � as �=�
0+�ℎ.
Then (�−�
??????)=(�−??????)ℎ and (�
??????−�
??????)=(??????−??????)ℎ.
Now, �(�)=(�−�
0)(�−�
1)⋯(�−�
??????)=�ℎ(�−1)ℎ(�−2)ℎ⋯(�−??????)ℎ=ℎ
??????+1
�(�−1)(�−2)⋯(�−??????)
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 17
�
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)(�
??????−�
??????+2)⋯(�
??????−�
??????)
=??????ℎ(??????−1)ℎ(??????−2)ℎ⋯{??????−(??????−1)}ℎ{??????−(??????+1)}ℎ{??????−(??????+2)}ℎ⋯(??????−??????)ℎ
=ℎ
??????
??????(??????−1)(??????−2)⋯2∙1∙(−1)(−2)⋯{−(??????−??????)}
=(−1)
??????−??????
∙ℎ
??????
∙??????!∙(??????−??????)! , ??????=0,1,2,⋯,??????
With these values of �(�) and �
′
(�
??????); ??????=0,1,2,⋯,?????? the Lagrange's interpolation formula for equal spaced points is
given by
??????(�)=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
=∑
ℎ
??????+1
�(�−1)(�−2)⋯(�−??????)
{(�−??????)ℎ}∙{(−1)
??????−??????
∙ℎ
??????
∙??????!∙(??????−??????)!}
�
??????
??????
??????=0
=∑(−1)
??????−??????
�(�−1)(�−2)⋯(�−??????)
??????!∙(??????−??????)!∙(�−??????)
�
??????
??????
??????=0
, where �=�
0+�ℎ.
Note that, in general, � may be a fraction and each time an integer is subtracted from it.
INVERSE INTERPOLATION AND INVERSE INTERPOLATION BASED ON LAGRANGE'S FORMULA
Let (�
??????,�
??????),??????=0,1,2,⋯,?????? be (??????+1) pairs of data points of argument � and the entry �=�(�).
In interpolation, with these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation
function of independent variable � such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and we find the required
approximate value of dependent variable �=�(�) corresponding to a specified value of independent variable � within
the minimum and maximum values of �
0,�
1,�
2,⋯,�
??????.
The process of finding the value of � for a certain value of � within the minimum and maximum values of �
0,�
1,⋯,�
??????
is called the inverse interpolation.
Since Lagrange’s interpolation formula is merely a relation between two variables, either of which may take as
independent variable, we can consider the Lagrange’s interpolating polynomial as a function of � for the given values of
(�
??????,�
??????) ; ??????=0,1,2,⋯,??????.
Thus, the inverse interpolation formula by using the Lagrange’s interpolating polynomial for given values of (�
??????,�
??????) ; ??????=
0,1,2,⋯,?????? is given by
??????(�)=∑
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
,
where �(�)=∏(�−�
??????)
??????
??????=0
and �
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????), ??????=0,1,2,⋯,??????.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 17
�
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)(�
??????−�
??????+2)⋯(�
??????−�
??????)
=??????ℎ(??????−1)ℎ(??????−2)ℎ⋯{??????−(??????−1)}ℎ{??????−(??????+1)}ℎ{??????−(??????+2)}ℎ⋯(??????−??????)ℎ
=ℎ
??????
??????(??????−1)(??????−2)⋯2∙1∙(−1)(−2)⋯{−(??????−??????)}
=(−1)
??????−??????
∙ℎ
??????
∙??????!∙(??????−??????)! , ??????=0,1,2,⋯,??????
With these values of �(�) and �
′
(�
??????); ??????=0,1,2,⋯,?????? the Lagrange's interpolation formula for equal spaced points is
given by
??????(�)=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
=∑
ℎ
??????+1
�(�−1)(�−2)⋯(�−??????)
{(�−??????)ℎ}∙{(−1)
??????−??????
∙ℎ
??????
∙??????!∙(??????−??????)!}
�
??????
??????
??????=0
=∑(−1)
??????−??????
�(�−1)(�−2)⋯(�−??????)
??????!∙(??????−??????)!∙(�−??????)
�
??????
??????
??????=0
, where �=�
0+�ℎ.
Note that, in general, � may be a fraction and each time an integer is subtracted from it.
INVERSE INTERPOLATION AND INVERSE INTERPOLATION BASED ON LAGRANGE'S FORMULA
Let (�
??????,�
??????),??????=0,1,2,⋯,?????? be (??????+1) pairs of data points of argument � and the entry �=�(�).
In interpolation, with these (??????+1) values, we consider a n-th degree polynomial ??????(�) as a polynomial interpolation
function of independent variable � such that �
??????=�(�
??????)=??????(�
??????), for each ??????=0,1,2,⋯,?????? and we find the required
approximate value of dependent variable �=�(�) corresponding to a specified value of independent variable � within
the minimum and maximum values of �
0,�
1,�
2,⋯,�
??????.
The process of finding the value of � for a certain value of � within the minimum and maximum values of �
0,�
1,⋯,�
??????
is called the inverse interpolation.
Since Lagrange’s interpolation formula is merely a relation between two variables, either of which may take as
independent variable, we can consider the Lagrange’s interpolating polynomial as a function of � for the given values of
(�
??????,�
??????) ; ??????=0,1,2,⋯,??????.
Thus, the inverse interpolation formula by using the Lagrange’s interpolating polynomial for given values of (�
??????,�
??????) ; ??????=
0,1,2,⋯,?????? is given by
??????(�)=∑
(�−�
0)(�−�
1)⋯(�−�
??????−1)(�−�
??????+1)⋯(�−�
??????)
(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????)
�
??????
??????
??????=0
=∑
�(�)
(�−�
??????)�
′
(�
??????)
�
??????
??????
??????=0
,
where �(�)=∏(�−�
??????)
??????
??????=0
and �
′
(�
??????)=(�
??????−�
0)(�
??????−�
1)⋯(�
??????−�
??????−1)(�
??????−�
??????+1)⋯(�
??????−�
??????), ??????=0,1,2,⋯,??????.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 18
LINEAR INTERPOLATION
There is a major disadvantage of Lagrange's interpolation. The degree of the interpolating polynomial is high for a large
data set. To remove this drawback different interpolating formulae are developed. One of them is linear interpolation. In
this method, for each interval [�
??????,�
??????+1
]; ??????=0,1,2,⋯,??????−1, one interpolating polynomial is constructed. But, for two
given points there is a polynomial whose degree must be one, i.e. linear.
Let (�
0,�
0) and (�
1,�
1) be two points.
Now,??????(�)=�
0(�)�
0+�
1(�)�
1=
(�−�
1)
(�
0−�
1)
�
0+
(�−�
0)
(�
1−�
0)
�
1=�
0+
(�−�
0)
(�
1−�
0)
(�
1−�
0)
This polynomial is known as linear interpolation polynomial for the interval [�
0,�
1
].
For the given (??????+1) points (�
??????,�
??????) ; ??????=0,1,2,⋯,??????, the linear interpolating polynomials are
??????(�)=�
??????+
(�−�
??????)
(�
??????+1−�
??????)
(�
??????+1−�
??????), when �
??????≤�≤�
??????+1; ??????=0,1,2,⋯,??????−1.
NOTE
Generally, we think that if the number of points increases then there is a significant improvement in the approximate
value, but it is not true for all functions. For example, if we consider �(�)=cos�, then for 11 points the Lagrange's
interpolating polynomial exactly matches with �(�) within the interval [−5,5]. But, this is not happened for all functions.
In case of Lorentz function �(�)=1/(1+�
2
), if the number of points increases, the approximate function oscillates
more rapidly.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 18
LINEAR INTERPOLATION
There is a major disadvantage of Lagrange's interpolation. The degree of the interpolating polynomial is high for a large
data set. To remove this drawback different interpolating formulae are developed. One of them is linear interpolation. In
this method, for each interval [�
??????,�
??????+1
]; ??????=0,1,2,⋯,??????−1, one interpolating polynomial is constructed. But, for two
given points there is a polynomial whose degree must be one, i.e. linear.
Let (�
0,�
0) and (�
1,�
1) be two points.
Now,??????(�)=�
0(�)�
0+�
1(�)�
1=
(�−�
1)
(�
0−�
1)
�
0+
(�−�
0)
(�
1−�
0)
�
1=�
0+
(�−�
0)
(�
1−�
0)
(�
1−�
0)
This polynomial is known as linear interpolation polynomial for the interval [�
0,�
1
].
For the given (??????+1) points (�
??????,�
??????) ; ??????=0,1,2,⋯,??????, the linear interpolating polynomials are
??????(�)=�
??????+
(�−�
??????)
(�
??????+1−�
??????)
(�
??????+1−�
??????), when �
??????≤�≤�
??????+1; ??????=0,1,2,⋯,??????−1.
NOTE
Generally, we think that if the number of points increases then there is a significant improvement in the approximate
value, but it is not true for all functions. For example, if we consider �(�)=cos�, then for 11 points the Lagrange's
interpolating polynomial exactly matches with �(�) within the interval [−5,5]. But, this is not happened for all functions.
In case of Lorentz function �(�)=1/(1+�
2
), if the number of points increases, the approximate function oscillates
more rapidly.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 19
ERROR IN INTERPOLATING POLYNOMIAL
Suppose (�
??????,�(�
??????)),??????=0,1,2,⋯,?????? are (??????+1) pairs of data points of argument � and the entry or function �(�). Let
us assume �
0=min{�
?????? ; ??????=0(1)??????} and �
??????=max{�
?????? ; ??????=0(1)??????}.
In the problem of polynomial interpolation with (??????+1) data points (�
??????,�(�
??????)) ; ??????=0,1,2,⋯,??????, we approximate the
function �(�) by a polynomial ??????(�) of degree at most ?????? within [�
0,�
??????
] such that �(�
??????)=??????(�
??????) for all ??????=0,1,2,⋯,??????.
At all other points except �
0,�
1,�
2,⋯,�
?????? in [�
0,�
??????
], �(�) may not be equal to ??????(�).
Let us consider �(�)=??????(�)+??????
??????(�), where ??????
??????(�) be the remainder or error term at �∈[�
0,�
??????
] and assume that the
unknown function �(�) be continuous and possesses continuous derivatives up to order (??????+1) within the closed interval
[�
0,�
??????
].
At �=�
??????, the error term is ??????
??????(�)=�(�)−??????(�)=0 for all ??????=0,1,2,⋯,??????.
Let us define an arbitrary function �(�), by considering �∈[�
0,�
??????
] fixed, as
�(�)={�(�)−??????(�)}−{�(�)−??????(�)}
(�−�
0)(�−�
1)⋯(�−�
??????)
(�−�
0)(�−�
1)⋯(�−�
??????)
. ⋯⋯(∗)
The function �(�) vanishes at �=�,�
0,�
1,�
2,⋯,�
??????. Thus, �(�)=0 has (??????+2) roots in [�
0,�
??????
].
As �(�) satisfies all the conditions of Rolle’s theorem in each of the (??????+1) sub-intervals in [�
0,�
??????
], �
′
(�) vanishes at
least one time in each of the (??????+1) sub-intervals. Therefore, �
′
(�) vanishes at least (??????+1) times in (�
0,�
??????). By
repeated application of Rolle’s theorem, �
′′
(�) vanishes at least ?????? times within (�
0,�
??????) and so on. Thus, �
(??????+1)
(�)
vanishes at least ones within (�
0,�
??????).
Let �
(??????+1)
(??????)=0 ; ??????∈(�
0,�
??????).
The expression (�−�
0)(�−�
1)⋯(�−�
??????) is a polynomial of degree (??????+1) in � and the coefficient of �
??????+1
is 1.
Therefore, (??????+1)-th derivative of the expression (�−�
0)(�−�
1)⋯(�−�
??????) will be (??????+1)!.
∴ �
(??????+1)
(??????)=0 ⟹ {�
(??????+1)
(??????)−0}−{�(�)−??????(�)}
(??????+1)!
(�−�
0)(�−�
1)⋯(�−�
??????)
; ??????∈(�
0,�
??????)
or, Remainder term=??????
??????(�)=�(�)−??????(�)=(�−�
0)(�−�
1)⋯(�−�
??????)
�
(??????+1)
(??????)
(??????+1)!
; ??????∈(�
0,�
??????)
This is an analytical form of the required remainder or error term related to the Newton’s forward, Newton’s backward
and Lagrange’s interpolating polynomial.
NOTE:
Practically, this formula is not much useful because in many situations �
(??????+1)
(??????); ??????∈(�
0,�
??????) cannot be determined.
But, this expression gives an upper bound of the error.
Let �(??????+1) be the upper bound of �
(??????+1)
(??????); ??????∈(�
0,�
??????). Then
|??????
??????(�)|≤|(�−�
0)(�−�
1)⋯(�−�
??????)|
�(??????+1)
(??????+1)!
.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 19
ERROR IN INTERPOLATING POLYNOMIAL
Suppose (�
??????,�(�
??????)),??????=0,1,2,⋯,?????? are (??????+1) pairs of data points of argument � and the entry or function �(�). Let
us assume �
0=min{�
?????? ; ??????=0(1)??????} and �
??????=max{�
?????? ; ??????=0(1)??????}.
In the problem of polynomial interpolation with (??????+1) data points (�
??????,�(�
??????)) ; ??????=0,1,2,⋯,??????, we approximate the
function �(�) by a polynomial ??????(�) of degree at most ?????? within [�
0,�
??????
] such that �(�
??????)=??????(�
??????) for all ??????=0,1,2,⋯,??????.
At all other points except �
0,�
1,�
2,⋯,�
?????? in [�
0,�
??????
], �(�) may not be equal to ??????(�).
Let us consider �(�)=??????(�)+??????
??????(�), where ??????
??????(�) be the remainder or error term at �∈[�
0,�
??????
] and assume that the
unknown function �(�) be continuous and possesses continuous derivatives up to order (??????+1) within the closed interval
[�
0,�
??????
].
At �=�
??????, the error term is ??????
??????(�)=�(�)−??????(�)=0 for all ??????=0,1,2,⋯,??????.
Let us define an arbitrary function �(�), by considering �∈[�
0,�
??????
] fixed, as
�(�)={�(�)−??????(�)}−{�(�)−??????(�)}
(�−�
0)(�−�
1)⋯(�−�
??????)
(�−�
0)(�−�
1)⋯(�−�
??????)
. ⋯⋯(∗)
The function �(�) vanishes at �=�,�
0,�
1,�
2,⋯,�
??????. Thus, �(�)=0 has (??????+2) roots in [�
0,�
??????
].
As �(�) satisfies all the conditions of Rolle’s theorem in each of the (??????+1) sub-intervals in [�
0,�
??????
], �
′
(�) vanishes at
least one time in each of the (??????+1) sub-intervals. Therefore, �
′
(�) vanishes at least (??????+1) times in (�
0,�
??????). By
repeated application of Rolle’s theorem, �
′′
(�) vanishes at least ?????? times within (�
0,�
??????) and so on. Thus, �
(??????+1)
(�)
vanishes at least ones within (�
0,�
??????).
Let �
(??????+1)
(??????)=0 ; ??????∈(�
0,�
??????).
The expression (�−�
0)(�−�
1)⋯(�−�
??????) is a polynomial of degree (??????+1) in � and the coefficient of �
??????+1
is 1.
Therefore, (??????+1)-th derivative of the expression (�−�
0)(�−�
1)⋯(�−�
??????) will be (??????+1)!.
∴ �
(??????+1)
(??????)=0 ⟹ {�
(??????+1)
(??????)−0}−{�(�)−??????(�)}
(??????+1)!
(�−�
0)(�−�
1)⋯(�−�
??????)
; ??????∈(�
0,�
??????)
or, Remainder term=??????
??????(�)=�(�)−??????(�)=(�−�
0)(�−�
1)⋯(�−�
??????)
�
(??????+1)
(??????)
(??????+1)!
; ??????∈(�
0,�
??????)
This is an analytical form of the required remainder or error term related to the Newton’s forward, Newton’s backward
and Lagrange’s interpolating polynomial.
NOTE:
Practically, this formula is not much useful because in many situations �
(??????+1)
(??????); ??????∈(�
0,�
??????) cannot be determined.
But, this expression gives an upper bound of the error.
Let �(??????+1) be the upper bound of �
(??????+1)
(??????); ??????∈(�
0,�
??????). Then
|??????
??????(�)|≤|(�−�
0)(�−�
1)⋯(�−�
??????)|
�(??????+1)
(??????+1)!
.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 20
REMAINDER TERM OF NEWTON’S FORWARD INTERPOLAT ION FORMULA
Let us consider �=
(??????−??????0
)
ℎ
i.e., �=�
0+�ℎ.
Since �
??????=�
0+??????ℎ, we get �−�
??????=(�−??????)ℎ ; ??????=0,1,2,⋯,??????.
The error term is given by
??????
??????(�)=(�−�
0)(�−�
1)⋯(�−�
??????)
�
(??????+1)
(??????)
(??????+1)!
=�(�−1)(�−2)⋯(�−??????)
ℎ
??????+1
∙�
(??????+1)
(??????)
(??????+1)!
; ??????∈(�
0,�
??????).
In general, the form of �(�) is unknown. When �
(??????+1)
(??????); ??????∈(�
0,�
??????) does not vary too rapidly in the interval, an
estimate of the derivative can be obtained as follows:
�
′
(�)≅
�(�+ℎ)−�(�)
ℎ
or, ??????≡
∆
ℎ
(approximately) and
??????
??????+1
≡
∆
??????+1
ℎ
??????+1
(approximately)
Thus, �
(??????+1)
(??????)≅
1
ℎ
??????+1
∆
??????+1
�(�
0) ; ??????∈(�
0,�
??????).
Thus, the error term in Newton’s forward interpolation formula becomes
??????
??????(�)≅�(�−1)(�−2)⋯(�−??????)
∆
??????+1
�(�
0)
(??????+1)!
,where �=
(�−�
0)
ℎ
.
REMAINDER TERM OF NEWTON’S BACKWARD INTERPOLATION F ORMULA
Let us consider �=
(??????−????????????
)
ℎ
i.e., �=�
??????+�ℎ.
Since �
??????=�
0+??????ℎ, we get �−�
??????=(�+??????−??????)ℎ ; ??????=0,1,2,⋯,??????.
The error term is given by
??????
??????(�)=(�−�
0)(�−�
1)⋯(�−�
??????)
�
(??????+1)
(??????)
(??????+1)!
=�(�+1)(�+2)⋯(�+??????)
ℎ
??????+1
∙�
(??????+1)
(??????)
(??????+1)!
; ??????∈(�
0,�
??????).
In general, the form of �(�) is unknown. When �
(??????+1)
(??????); ??????∈(�
0,�
??????) does not vary too rapidly in the interval, an
estimate of the derivative can be obtained as follows:
�
′
(�)≅
�(�+ℎ)−�(�)
ℎ
or, ??????≡
∆
ℎ
(approximately) and ??????
??????+1
≡
∆
??????+1
ℎ
??????+1
(approximately)
Thus, �
(??????+1)
(??????)≅
1
ℎ
??????+1
∆
??????+1
�(�
0)=
1
ℎ
??????+1
∇
??????+1
�(�
??????) ; ??????∈(�
0,�
??????).
Thus, the error term in Newton’s backward interpolation formula becomes
??????
??????(�)≅�(�+1)(�+2)⋯(�+??????)
∇
??????+1
�(�
??????)
(??????+1)!
,where �=
(�−�
??????)
ℎ
.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 20
REMAINDER TERM OF NEWTON’S FORWARD INTERPOLAT ION FORMULA
Let us consider �=
(??????−??????0
)
ℎ
i.e., �=�
0+�ℎ.
Since �
??????=�
0+??????ℎ, we get �−�
??????=(�−??????)ℎ ; ??????=0,1,2,⋯,??????.
The error term is given by
??????
??????(�)=(�−�
0)(�−�
1)⋯(�−�
??????)
�
(??????+1)
(??????)
(??????+1)!
=�(�−1)(�−2)⋯(�−??????)
ℎ
??????+1
∙�
(??????+1)
(??????)
(??????+1)!
; ??????∈(�
0,�
??????).
In general, the form of �(�) is unknown. When �
(??????+1)
(??????); ??????∈(�
0,�
??????) does not vary too rapidly in the interval, an
estimate of the derivative can be obtained as follows:
�
′
(�)≅
�(�+ℎ)−�(�)
ℎ
or, ??????≡
∆
ℎ
(approximately) and
??????
??????+1
≡
∆
??????+1
ℎ
??????+1
(approximately)
Thus, �
(??????+1)
(??????)≅
1
ℎ
??????+1
∆
??????+1
�(�
0) ; ??????∈(�
0,�
??????).
Thus, the error term in Newton’s forward interpolation formula becomes
??????
??????(�)≅�(�−1)(�−2)⋯(�−??????)
∆
??????+1
�(�
0)
(??????+1)!
,where �=
(�−�
0)
ℎ
.
REMAINDER TERM OF NEWTON’S BACKWARD INTERPOLATION F ORMULA
Let us consider �=
(??????−????????????
)
ℎ
i.e., �=�
??????+�ℎ.
Since �
??????=�
0+??????ℎ, we get �−�
??????=(�+??????−??????)ℎ ; ??????=0,1,2,⋯,??????.
The error term is given by
??????
??????(�)=(�−�
0)(�−�
1)⋯(�−�
??????)
�
(??????+1)
(??????)
(??????+1)!
=�(�+1)(�+2)⋯(�+??????)
ℎ
??????+1
∙�
(??????+1)
(??????)
(??????+1)!
; ??????∈(�
0,�
??????).
In general, the form of �(�) is unknown. When �
(??????+1)
(??????); ??????∈(�
0,�
??????) does not vary too rapidly in the interval, an
estimate of the derivative can be obtained as follows:
�
′
(�)≅
�(�+ℎ)−�(�)
ℎ
or, ??????≡
∆
ℎ
(approximately) and ??????
??????+1
≡
∆
??????+1
ℎ
??????+1
(approximately)
Thus, �
(??????+1)
(??????)≅
1
ℎ
??????+1
∆
??????+1
�(�
0)=
1
ℎ
??????+1
∇
??????+1
�(�
??????) ; ??????∈(�
0,�
??????).
Thus, the error term in Newton’s backward interpolation formula becomes
??????
??????(�)≅�(�+1)(�+2)⋯(�+??????)
∇
??????+1
�(�
??????)
(??????+1)!
,where �=
(�−�
??????)
ℎ
.
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 21
EXAMPLE
Show that the linear interpolation formula can be expressed in the form
??????(�)=
(�
2−�)�(�
1)+(�−�
1)�(�
2)
(�
2−�
1)
and the corresponding remainder term ??????(�)=�(�)−??????(�) has the following bound:
|??????(�)|≤
�(�
2−�
1)
2
8
, where |�
′′
(�)|≤� and �
1<�<�
2.
A table of natural sines is given with entries for every degree. What is the maximum error of linear interpolation in such
a table?
Solution:
The Lagrange’s interpolation formula based on the pairs of data points (�
1,�(�
1)) and (�
2,�(�
2)) is given by
??????(�)=
(�−�
2)
(�
1−�
2)
�(�
1)+
(�−�
1)
(�
2−�
1)
�(�
2)
=
(�
2−�)�(�
1)+(�−�
1)�(�
2)
(�
2−�
1)
For two given points, the degree of the Lagrange’s interpolating polynomial must be one, i.e. linear. This polynomial is
known as linear interpolation polynomial in the interval [�
1,�
2
].
Suppose (�
??????,�(�
??????)),??????=0,1,2,⋯,?????? are (??????+1) pairs of data points of argument � and the entry or function �(�). Let
us assume �
0=min{�
?????? ; ??????=0(1)??????} and �
??????=max{�
?????? ; ??????=0(1)??????}.
The remainder term in a polynomial interpolation formula of degree at most ?????? with (??????+1) data points is given by
??????
??????(�)=�(�)−??????(�)=(�−�
0)(�−�
1)⋯(�−�
??????)
�
(??????+1)
(??????)
(??????+1)!
; ??????∈(�
0,�
??????)
For ??????=1, the remainder term in the linear interpolation formula based on the pairs of data points (�
1,�(�
1)) and
(�
2,�(�
2)) is
??????(�)=(�−�
1)(�−�
2)
�
′′
(??????)
2!
; ??????∈(�
1,�
2).
Let us consider (�
2−�
1)=ℎ and �=
(??????−??????1
)
ℎ
i.e., �
2=�
1+�ℎ.
∴ ??????(�)=�(�−1)ℎ
2
�
′′
(??????)
2!
; ??????∈(�
1,�
2).
Since �
1<�<�
2, 0<�<1 and |�(�−1)|=�(1−�)=�−�
2
=
1
4
−(
1
2
−�)
2
≤
1
4
.
∴ |??????(�)|=|�(�)−??????(�)|≤
ℎ
2
4
�
′′
(??????)
2!
=
ℎ
2
�
′′
(??????)
8
; ??????∈(�
1,�
2).
Given that, ??????∈(�
1,�
2) and |�
′′
(�)|≤�,�
1<�<�
2 i.e., the second derivative of �(�) is bonded in (�
1,�
2).
Numerical Analysis || Lecture Notes || Anup Kumar Giri 21
EXAMPLE
Show that the linear interpolation formula can be expressed in the form
??????(�)=
(�
2−�)�(�
1)+(�−�
1)�(�
2)
(�
2−�
1)
and the corresponding remainder term ??????(�)=�(�)−??????(�) has the following bound:
|??????(�)|≤
�(�
2−�
1)
2
8
, where |�
′′
(�)|≤� and �
1<�<�
2.
A table of natural sines is given with entries for every degree. What is the maximum error of linear interpolation in such
a table?
Solution:
The Lagrange’s interpolation formula based on the pairs of data points (�
1,�(�
1)) and (�
2,�(�
2)) is given by
??????(�)=
(�−�
2)
(�
1−�
2)
�(�
1)+
(�−�
1)
(�
2−�
1)
�(�
2)
=
(�
2−�)�(�
1)+(�−�
1)�(�
2)
(�
2−�
1)
For two given points, the degree of the Lagrange’s interpolating polynomial must be one, i.e. linear. This polynomial is
known as linear interpolation polynomial in the interval [�
1,�
2
].
Suppose (�
??????,�(�
??????)),??????=0,1,2,⋯,?????? are (??????+1) pairs of data points of argument � and the entry or function �(�). Let
us assume �
0=min{�
?????? ; ??????=0(1)??????} and �
??????=max{�
?????? ; ??????=0(1)??????}.
The remainder term in a polynomial interpolation formula of degree at most ?????? with (??????+1) data points is given by
??????
??????(�)=�(�)−??????(�)=(�−�
0)(�−�
1)⋯(�−�
??????)
�
(??????+1)
(??????)
(??????+1)!
; ??????∈(�
0,�
??????)
For ??????=1, the remainder term in the linear interpolation formula based on the pairs of data points (�
1,�(�
1)) and
(�
2,�(�
2)) is
??????(�)=(�−�
1)(�−�
2)
�
′′
(??????)
2!
; ??????∈(�
1,�
2).
Let us consider (�
2−�
1)=ℎ and �=
(??????−??????1
)
ℎ
i.e., �
2=�
1+�ℎ.
∴ ??????(�)=�(�−1)ℎ
2
�
′′
(??????)
2!
; ??????∈(�
1,�
2).
Since �
1<�<�
2, 0<�<1 and |�(�−1)|=�(1−�)=�−�
2
=
1
4
−(
1
2
−�)
2
≤
1
4
.
∴ |??????(�)|=|�(�)−??????(�)|≤
ℎ
2
4
�
′′
(??????)
2!
=
ℎ
2
�
′′
(??????)
8
; ??????∈(�
1,�
2).
Given that, ??????∈(�
1,�
2) and |�
′′
(�)|≤�,�
1<�<�
2 i.e., the second derivative of �(�) is bonded in (�
1,�
2).
NUMERICAL ANALYSIS: INTERPOLATION
Numerical Analysis || Lecture Notes || Anup Kumar Giri 22
Thus, the remainder term becomes
|??????(�)|≤
ℎ
2
�
′′
(??????)
8
≤
�(�
2−�
1)
2
8
, where |�
′′
(�)|≤� and �
1<�<�
2.
The maximum error of a linear interpolating polynomial is
�(�
2−�
1)
2
8
, where |�
′′
(�)|≤� and �
1<�<�
2.
In a table of natural sines with entries for every degree, we have
(�
2−�
1)=1
0
=(
??????
180
)
??????
, �(�)=sin�, �
′
(�)=cos�, �
′′
(�)=−sin� and
�=max{ |�
′′
(�)|}=max{ |−sin�|}=1.
Thus, the maximum error of a linear interpolating polynomial of natural sines with entries for every degree is
�(�
2−�
1)
2
8
=
1∙(
??????
180
)
2
8
=
1
8
(
??????
180
)
2
.
Numerical Analysis || Lecture Notes || Anup Kumar Giri 22
Thus, the remainder term becomes
|??????(�)|≤
ℎ
2
�
′′
(??????)
8
≤
�(�
2−�
1)
2
8
, where |�
′′
(�)|≤� and �
1<�<�
2.
The maximum error of a linear interpolating polynomial is
�(�
2−�
1)
2
8
, where |�
′′
(�)|≤� and �
1<�<�
2.
In a table of natural sines with entries for every degree, we have
(�
2−�
1)=1
0
=(
??????
180
)
??????
, �(�)=sin�, �
′
(�)=cos�, �
′′
(�)=−sin� and
�=max{ |�
′′
(�)|}=max{ |−sin�|}=1.
Thus, the maximum error of a linear interpolating polynomial of natural sines with entries for every degree is
�(�
2−�
1)
2
8
=
1∙(
??????
180
)
2
8
=
1
8
(
??????
180
)
2
.
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