SRIT / UICM004 / Numerical Methods / Interpolation and ...
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SRIT / UICM004 / Numerical Methods / Interpolation and Approximation
SRIT / M & H / M. Vijaya Kumar 1
SRI RAMAKRISHNA INSTITUTE OF TECHNOLOGY
(AN AUTONOMOUS INSTITUTION)
COIMBATORE- 641010
UICM004 & NUMERICAL METHODS
Interpolation and Approximation
One of the oldest problems in mathematics and at the same time, one of the most
applied is the problem of constructing an approximation to a given function from among
simpler functions, typically (but not always) polynomials. A slight variation of this problem
is that of constructing a smooth function from a discrete set of data points.
Lagrange polynomial
In numerical analysis, Lagrange polynomials are used for polynomial interpolation.
For a given set of points ( ) with no two values equal, the Lagrange polynomial is the
polynomial of lowest degree that assumes at each value the corresponding value
(i.e. the functions coincide at each point). The interpolating polynomial of the least degree
is unique, however, and since it can be arrived at through multiple methods, referring to
"the Lagrange polynomial" is perhaps not as correct as referring to "the Lagrange form" of
that unique polynomial.
1. State Lagrange’s interpolation formula.
Answer:
Let ( ) be a function which takes the values corresponding to
The Lagrange’s interpolation formula is
( ) ( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
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2. What is the Lagrange’s interpolation formula to find , if three sets of values
( ) ( ) ( ) are given.
Answer:
( ) ( )( )
( )( )
( )( )
( )( )
( )( )
( )( )
3. What is the assumption we make when Lagrange’s formula is used?
Answer:
Lagrange’s interpolation formula can be used whether the vales of , the
independent variable are equally spaced or not whether the difference of become
smaller or not.
4. What advantages has Lagrange’s interpolation formula over Newton?
Answer:
The forward and backward interpolation formulae of Newton can be used only when
the values of the independent variable are equally spaced can also be used when the
differences of the independent variable become smaller ultimately. But Lagrange’s
interpolation formula can be used whether the values of , the independent variable are
equally spaced or not and whether the difference of become smaller or not.
5. What is the disadvantage in practice in applying Lagrange’s interpolation
formula?
Answer:
Through Lagrange’s interpolation formula is simple and easy to remember, its
application is not speedy. It requires close attention to sign and there is always a chance of
committing some error due to a number of positive and negative signs in the numerator
and the denominator.
6. What is inverse interpolation?
Answer:
Suppose we are given a table of vales of and . Direct interpolation is the process
of finding the values of corresponding to a value of , not present in the table. Inverse
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interpolation is the process of finding the values of corresponding to a value not present
in the table.
7. State Lagrange’s inverse interpolation formula.
Answer:
( ) ( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
Problem: 1
Find the third degree polynomial satisfying the following data
: 1 3 5 7
( ) : 24 120 336 720
Answer:
: 1 ( ) 3 ( ) 5 ( ) 7 ( )
( ) : 24 ( ) 120 ( ) 336 ( ) 720 ( )
Lagrange’s interpolation formula, we have
( ) ( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
[( )( )( )]
[( )( )( )]
[( )( )( )] [( )( )( )]
[( )( )]
[( )( )]
[( )( )] [( )( )]
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[ ]
[ ]
[ ] [ ]
(
) (
)
(
) (
)
( )
Verification:
( ) ( ) ( ) ( )
( )
Problem: 2
Find the polynomial f (x) by using Lagrange’s formula and hence find f (3) for the
following values of and
: 0 1 2 5
( ) : 2 3 12 147
(or)
Find the interpolation polynomial ( ) by Lagrange’s formula and hence find ( )
for ( ) ( ) ( ) and ( ).
Answer:
: 0 ( ) 1 ( ) 2 ( ) 5 ( )
( ) : 2 ( ) 3 ( ) 12 ( ) 147 ( )
Lagrange’s interpolation formula, we have
( ) ( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
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[( )( )]
[( )( )]
[( )( )]
[( )( )]
[( )]
[ ]
[ ]
[ ]
(
) (
)
(
) (
)
( )
( )
Verification:
( )
( ( ) ( ) ( ) )
( )
Value of ( ):
( )
( ( ) ( ) ( ) )
( )
Problem: 3
Find the value ( ) by using Lagrange’s formula from
: 0 1 2 5
( ) : 2 3 12 147
Answer:
Lagrange’s interpolation formula, we have
( ) ( )( )( )
( )( )( )
( )( )( )
( )( )( )
( )( )( )
( )( )( )
( )( )( )
( )( )( )
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Here
also
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
Problem: 4
Using Lagrange’s formula find ( ) from the following data.
Answer: : 0 ( ) 1 ( ) 3 ( ) 4 ( ) 5 ( )
( ) : 0 ( ) 1 ( ) 81 ( ) 256 ( ) 625 ( )
Lagrange’s interpolation formula, we have
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( ) ( )
( )( )( )( )
( )( )( )( ) ( )
( )( )( )( )
( )( )( )( ) ( )
( )( )( )( )
( )( )( )( ) ( )
( )( )( )( )
( )( )( )( ) ( )
: 0 1 3 4 5
( ) : 0 1 81 256 625
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Problem: 5
Using Lagrange’s interpolation formula find ( ) given that ( ) ( )
( ) ( ) .
Answer:
: 0 ( ) 1 ( ) 2 ( ) 15 ( ) ( ) : 2 ( ) 3 ( ) 12 ( ) 3587 ( )
Lagrange’s interpolation formula, we have
( ) ( )( )( )
( )( )( )
( )( )( )
( )( )( )
( )( )( )
( )( )( )
( )( )( )
( )( )( )
( ) ( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
( )( )( )
( )( )( ) ( )
Divided Differnces
Newton's Divided Difference is a way of finding an interpolation polynomial (a
polynomial that fits a particular set of points or data). Similar to Lagrange's method for
finding an interpolation polynomial, it finds the same interpolation polynomial due to the
uniqueness of interpolation polynomials.
1. Define ‘Divided Differences’.
Answer:
Let the function ( ) take the values ( ) ( ) ( ) corresponding to the
values of the argument where need
not necessarily be equal.
The first divided difference of ( ) for the arguments ( ) is
( ) ( ) ( )
( )
( ) ( )
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2. Give the Newton’s divided difference formula.
Solution:
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )( ) ( ) ( )
3. Form the divided difference table for the following data
:
:
Solution:
The divided difference table is as follows
4. Form the divided difference table for the following data
:
:
Solution:
The divided difference table is as follows
: ( ) ( ) ( )
5
15
22
7
36
160
: ( ) ( ) ( )
2
5
10
5
29
109
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5. State any properties of divided differences.
Solution:
The divide differences are symmetrical in all their arguments. That is the value of any
difference is independent of the order of the arguments.
The divided difference of the sum or difference of two functions is equal to the sum or
difference of the corresponding separate divided differences.
Problem: 6
Using Newton’s divided difference formula, find ( ) given ( ) ( )
( ) ( ) .
Answer:
( ) ( ) ( ) ( )
By Newton’s divided difference interpolation formula
( ) ( ) ( ) ( ) ( )( ) ( )
( )( )( ) ( )
Here and .
and ( ) ( ) ( ) ( )
( ) ( ) ( )( ) ( )( )( )
( ) ( ) ( )( ) ( )( )( )
( ) ( )( ) ( )( )( )
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Problem: 7
Find ( ) as a polynomial in from the following data by Newton’s divided
difference formula
( )
Answer:
First we form the divided difference table
( ) ( ) ( ) ( ) ( )
By Newton’s divided difference interpolation formula
( ) ( ) ( ) ( ) ( )( ) ( )
( )( )( ) ( )
( )( )( )( ) ( )
Here
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )
( )( )( )( )( )
( ) ( )[ ] [ ]
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[( )( )]
[ ]
[ ]
( )
Verification:
( ) ( ) ( ) ( ) ( )
Problem: 8
Using Newton’s divided difference formula, find the equation ( ) of least
degree and passing through the points ( ) ( ) ( ) ( ). Find also at
.
Answer:
( )
First we form the divided difference table
( ) ( ) ( ) ( )
By Newton’s divided difference interpolation formula
( ) ( ) ( ) ( ) ( )( ) ( )
( )( )( ) ( )
Here
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and ( ) ( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )( )( ) ( )
[ ] [ ]( )
Verification:
( ) ( ) ( )
( )
Value of ( ):
( )
Problem: 9
Find ( ) by Newton’s divided difference formula for the data,
4 5 7 10 11 13
( ) 48 100 294 900 1210 2028
Answer:
First we form the divided difference table
( ) ( ) ( ) ( )
By Newton’s divided difference interpolation formula
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( ) ( ) ( ) ( ) ( )( ) ( )
( )( )( ) ( )
Here and
and ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )
( ) ( ) ( )( ) ( ) ( )( )( ) ( )
Problem: 10
Using Newton’s divided difference formula, find the missing term from the following data
1 2 4 5 6
( ) 14 15 5 -- 9
Answer:
First we form the divided difference table
( ) ( ) ( ) ( )
By Newton’s divided difference interpolation formula
( ) ( ) ( ) ( ) ( )( ) ( )
( )( )( ) ( )
Here and
and ( ) ( ) ( ) ( )
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( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )
( ) ( ) ( )( ) ( ) ( )( )( ) ( )
Newtons Forward & Backward Interpolation Formula
Newton forward interpolation formula:
( )
( )
( )( )
( )( )( )
Newton backward interpolation formula:
( )
( )
( )( )
( )( )( )
1. Derive Newton’s Backward difference formula by using operator method.
Answer:
( ) ( ) ( )
( ) [ ( ) ]
* ( )
( )( )
+
( )
( )( )
2. Derive Newton’s Forward difference formula by using operator method.
(or) State Gregory – Newton Forward difference interpolation formula.
Answer:
( ) ( ) ( )
( )
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*
( )
( )( )
+
( )
( )( )
3. When Newton’s Backward interpolation formula used.
Answer:
The formula is used mainly to interpolate the values of near the end of a set of
tabular values and also for extrapolating the values of a short distance ached (to the
right) of .
4. Say True or False.
Newton’s interpolation formulae are not suited to estimate the value of a function
near the middle of a table.
Answer:
True.
5. Say True or False.
Newton’s forward and Newton’s backward interpolation formulae are applicable for
interpolation near the beginning and end respectively of tabulated values.
Answer:
The statement is true.
6. When will we use Newton’s forward interpolation formula?
Answer:
The formula is used to interpolate the values of near the beginning of the table
value and also for extrapolating the values of short distance (to the left) ahead of
7. Newton’s Backward interpolation formula used only for ………………………?
Answer:
Equidistant intervals (or) Equal intervals.
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Problem : 11
Find the cubic polynomial which takes the following values:
0 1 2 3
( ) 1 2 1 10
Answer:
We form the difference table
There are only four data are given. Hence the polynomial is of order 3
( )
( )
( )( )
( )
( )
( )
( )
( )( )
( )
( ) ( )( )
( )
( )
( )
Verification:
( ) ( ) ( ) ( )
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Problem: 12
Using Newton’s forward interpolation formula, find a polynomial ( ) satisfying the
following data. Hence evaluate .
4 6 8 10
1 3 8 10
Answer:
We form the difference table
There are only four data are given. Hence the polynomial is of order 3
( )
( )
( )( )
( ) ( )
( )
( ) *(
) +
( )
( ) *(
) + *(
) +
( )
( ) (
)
( )
( ) (
) (
)
( )
(
) (
) (
) (
) (
) (
) (
)
(
) ( ) (
) ( )( )
(
) ( ) (
) ( )( )
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(
) ( ) (
) ( )
(
) ( ) (
) ( )
(
) [ ( ) ( )]
(
) [ ]
(
) [ ]
( ) (
) ( )
To find :
( ) (
) [ ( ) ( ) ( ) ]
Verification:
( ) (
) [ ( ) ( ) ( ) ]
Therefore the polynomial is correct.
Problem : 13
Using Newton’s forward formula, find a polynomial ( ) satisfying the following
data. Hence find ( ).
0 5 10 15
14 379 1444 3584
Answer:
First we form the forward difference table
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There are only four data are given. Hence the polynomial is of order 3
( )
( )
( )( )
( ) ( )
( )
( ) *(
) +
( )
( ) *(
) + *(
) +
( )
(
) ( )
( ) ( )
( )
( ) ( ) (
)
( )
(
) [ ] (
) [ ]( )
[ ]
[ ]
( )
( )
( )
To find ( ) :
( ) (
) [( ) ( ) ( ) ]
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Verification:
( ) (
) [( ) ( ) ( ) ]
Problem : 14
A Third degree polynomial passes through the points ( ) ( ) ( ) & ( )
using Newton’s forward interpolation formulae find the polynomial. Hence evaluate the
value at 1.5
Answer:
Given the values are
First we form the forward difference table
There are only four data are given. Hence the polynomial is of order 3
( ) ( )
( )
( )( )
( )
( )
( )
( )
( )( )
( )
( )
[ ]( )
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(
) [ ]
(
) [ ]
(
) [ ]
( ) (
) [ ]
To find y at 1.5
( ) (
) [( ) ( ) ( ) ]
Problem : 15
From the following data, find the number of students whose weight is between 60 to 70.
0 – 40 40 – 60 60 – 80 80 - 100 100 – 120
No. of Students 250 120 100 70 50
Answer:
We form the difference table
Now let us calculate the no. of students whose weight is between 70 and then
subtract weight from 60 we will get the required amount.
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The Newton’s forward formula is
( )
( )
( )( )
( )
( )
( )[ ]
( )
( )[ ][ ]
( )
( )[ ][ ][ ]
( )
( ) ( )
Students weight is between 60 to 70 is
Problem : 16
The population of a town is as follows:
Year 1941 1951 1961 1971 1981 1991
Population in
thoushands 20 24 29 36 46 51
Estimate the population increase during the period 1946 to 1976.
Answer:
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We form the difference table
To find the population increase during the period 1946:
For 1946 we need to use forward difference formula.
The Newton’s forward formula is
( )
( )
( )( )
( )
( )
( )
( )
( )( )
( )
( ) ( )
To find the population increase during the period 1976:
For 1976 we need to use backward difference formula.
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The Newton’s forward formula is
( )
( )
( )( )
( )
( )
( )
( )
( )( )
( )
( )( )( )
( )
( ) ( )
Problem : 17
Find the value of at and from the data given .
20 23 26 29
0.3420 0.3907 0.4384 0.4848
Answer:
We form the forward difference table
To find thevalue of at :
For 21 we need to use forward difference formula.
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The Newton’s forward formula is
( )
( )
( )( )
( ) ( )
( )
( ) (
)
( )
( ) (
) (
)
( )
To find thevalue of at :
For 28 we need to use backward difference formula.
The Newton’s forward formula is
( )
( )
( )( )
( ) ( )
( )
( ) (
)
( )
( ) (
) (
)
( )
( ) ( )
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CENTRAL DIFFERENCE INTREPOLATION FORMULAE
Bessel’s formula
The Bessels central difference intrtapolation formula is
( ) (
) (
)
( )
(
) (
) ( )
Problem : 18
From the following table, find ( ) using Bessel’s Formula.
20 25 30 35 40
11.4699 12.7834 13.7648 14.4982 15.0463
Answer:
We form the forward difference table
Let us choose the origin as
The Bessels central difference intrtapolation formula is
( ) (
) (
)
( )
(
) (
) ( )
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( ) (
) (
) ( )
( )
(
)
(
) ( )
( )
Problem: 19
Given the following table, find ( ), by using Bessel’s Formula.
20 30 40 50
512 439 346 243
Answer:
Let us form the forward difference table as follows:
Let us choose the origin as
The Bessels central difference intrtapolation formula is
( ) (
) (
)
( )
(
) (
) ( )
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( ) (
) (
) ( )
( )
(
)
Problem: 20
From the following table estimate correct to five decimals using Bessel’s formula.
0.61 0.62 0.63 0.64 0.65 0.66 0.67
1.840431 1.858928 1.877610 1.896481 1.915541 1.934792 1.954237
Answer:
0.61 1.840431
0.018497
0.62 1.858928 0.000185
0.018682 0.000004
0.63 1.87761 0.000189
0.018871 0.000000
0.64 1.896481 0.000189
0.01906 0.000002
0.65 1.915541 0.000191
0.019251 0.000003
0.66 1.934792 0.000194
0.019445
0.67 1.954237
Let us choose the origin as
The Bessels central difference intrtapolation formula is
( ) (
) (
)
( )
(
) ( )(
)
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( ) (
) (
) ( )
( )
(
)
( ) ( )
( )
Problem: 21
The following table gives the values of the probability integral ( )
√ ∫
for
certain values of . Find the value of this integral when using Bessel’s Formula
0.51 0.52 0.53 0.54 0.55 0.56 0.57
0.5292437 0.5378987 0.5464641 0.5549392 0.5633232 0.5716157 0.5798158
Answer:
Let us form the forward difference as follows:
Let us choose the origin as
The Bessels central difference intrtapolation formula is
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( ) (
) (
)
( )
(
) ( )(
)
( ) (
) (
) ( )
( )
(
)
( ) (
)
( )
Stirling's Interpolation
The Stirling's central difference intrtapolation formula is
( )
(
)
( )
(
)
Problem: 22 Using Stirling's formula compute ( ) from the data
10 11 12 13 14
0.23967 0.28060 0.31788 0.35209 0.38368
Answer:
Let us form the forward difference table as follows:
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Let us choose the origin as
The Stirling's central difference intrtapolation formula is
( )
(
)
( )
(
)
( )
(
)
( )
( )
(
)
Problem: 23
( )
342 343 344 345 346 347
6.993191 7.000000 7.006796 7.013579 7.020349 7.027106
Answer:
Let us form the forward difference table as follows:
Let us choose the origin as
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The Stirling's central difference intrtapolation formula is
( )
(
)
( )
(
)
( )
(
)
( )
Problem: 24
Use strilings formula to find the value of for from the table given below.
20 30 40 50
512 439 346 243
Answer:
Let us form the forward difference table as follows:
Let us choose the origin as
The Stirling's central difference intrtapolation formula is
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( )
(
)
( )
(
)
( )
(
)
( )
( )
(
)
Problem: 25
Use strilings formula to find the value of for from the table given below.
0.1 0.2 0.3 0.4 0.5 0.6 0.7
0.9900 0.9608 0.9139 0.8521 0.7788 0.6177 0.6126
Answer:
Let us form the forward difference table as follows:
Let us choose the origin as
The Stirling's central difference intrtapolation formula is
SRIT / UICM004 / Numerical Methods / Interpolation and Approximation
SRIT / M & H / M. Vijaya Kumar 34
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“Newton was the greatest genius who ever lived, and once added that
Newton was also "the most fortunate, for we cannot find more than
once a system of the world to establish"
Joseph-Louis Lagrange