srit / uicm004 / numerical methods / interpolation and

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

( ) ( )( ) ( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

https://en.wikipedia.org/wiki/Numerical_analysis

https://en.wikipedia.org/wiki/Polynomial_interpolation

https://en.wikipedia.org/wiki/Degree_of_a_polynomial



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



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

( ) ( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

[( )( )( )]

[( )( )( )]

[( )( )( )] [( )( )( )]

[( )( )]

[( )( )]

[( )( )] [( )( )]



[ ]

[ ]

[ ] [ ]

(

) (

)

(

) (

)

( )

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 ( )


( ) ( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )



[( )( )]

[( )( )]

[( )( )]

[( )( )]

[( )]

[ ]

[ ]

[ ]

(

) (

)

(

) (

)

( )

( )

Verification:

( )

( ( ) ( ) ( ) )

( )

Value of ( ):

( )

( ( ) ( ) ( ) )

( )

Problem: 3

Find the value ( ) by using Lagrange’s formula from

: 0 1 2 5

( ) : 2 3 12 147

Answer:


( ) ( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )



Here

also

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

Problem: 4

Using Lagrange’s formula find ( ) from the following data.

Answer: : 0 ( ) 1 ( ) 3 ( ) 4 ( ) 5 ( )

( ) : 0 ( ) 1 ( ) 81 ( ) 256 ( ) 625 ( )


( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( ) ( )

( )( )( )( )

( )( )( )( ) ( )

( )( )( )( )

( )( )( )( ) ( )

( )( )( )( )

( )( )( )( ) ( )

( )( )( )( )

( )( )( )( ) ( )

: 0 1 3 4 5

( ) : 0 1 81 256 625



Problem: 5

Using Lagrange’s interpolation formula find ( ) given that ( ) ( )

( ) ( ) .

Answer:

: 0 ( ) 1 ( ) 2 ( ) 15 ( ) ( ) : 2 ( ) 3 ( ) 12 ( ) 3587 ( )


( ) ( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( ) ( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

( )( )( )

( )( )( ) ( )

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

( ) ( ) ( )

( )

( ) ( )



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



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 ( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )( )( )

( ) ( ) ( )( ) ( )( )( )

( ) ( )( ) ( )( )( )



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

( ) ( ) ( ) ( ) ( )


( ) ( ) ( ) ( ) ( )( ) ( )

( )( )( ) ( )

( )( )( )( ) ( )

Here

( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )

( )( )( )( )( )

( ) ( )[ ] [ ]



[( )( )]

[ ]

[ ]

( )

Verification:

( ) ( ) ( ) ( ) ( )

Problem: 8

Using Newton’s divided difference formula, find the equation ( ) of least

degree and passing through the points ( ) ( ) ( ) ( ). Find also at

.

Answer:

( )


( ) ( ) ( ) ( )


( ) ( ) ( ) ( ) ( )( ) ( )

( )( )( ) ( )

Here



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:


( ) ( ) ( ) ( )




( ) ( ) ( ) ( ) ( )( ) ( )

( )( )( ) ( )

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:


( ) ( ) ( ) ( )


( ) ( ) ( ) ( ) ( )( ) ( )

( )( )( ) ( )

Here and

and ( ) ( ) ( ) ( )



( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )

( ) ( ) ( )( ) ( ) ( )( )( ) ( )

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:

( ) ( ) ( )

( )



*

( )

( )( )

+

( )

( )( )

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.



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:

( ) ( ) ( ) ( )



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:



( )

( )

( )( )

( ) ( )

( )

( ) *(

) +

( )

( ) *(

) + *(

) +

( )

( ) (

)

( )

( ) (

) (

)

( )

(

) (

) (

) (

) (

) (

) (

)

(

) ( ) (

) ( )( )

(

) ( ) (

) ( )( )



(

) ( ) (

) ( )

(

) ( ) (

) ( )

(

) [ ( ) ( )]

(

) [ ]

(

) [ ]

( ) (

) ( )

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




( )

( )

( )( )

( ) ( )

( )

( ) *(

) +

( )

( ) *(

) + *(

) +

( )

(

) ( )

( ) ( )

( )

( ) ( ) (

)

( )

(

) [ ] (

) [ ]( )

[ ]

[ ]

( )

( )

( )

To find ( ) :

( ) (

) [( ) ( ) ( ) ]



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


( ) ( )

( )

( )( )

( )

( )

( )

( )

( )( )

( )

( )

[ ]( )



(

) [ ]

(

) [ ]

(

) [ ]

( ) (

) [ ]

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:


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.



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:




To find the population increase during the period 1946:

For 1946 we need to use forward difference formula.


( )

( )

( )( )

( )

( )

( )

( )

( )( )

( )

( ) ( )

To find the population increase during the period 1976:

For 1976 we need to use backward difference formula.




( )

( )

( )( )

( )

( )

( )

( )

( )( )

( )

( )( )( )

( )

( ) ( )

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.




( )

( )

( )( )

( ) ( )

( )

( ) (

)

( )

( ) (

) (

)

( )

To find thevalue of at :

For 28 we need to use backward difference formula.


( )

( )

( )( )

( ) ( )

( )

( ) (

)

( )

( ) (

) (

)

( )

( ) ( )



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


( ) (

) (

)

( )

(

) (

) ( )



( ) (

) (

) ( )

( )

(

)

(

) ( )

( )

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:



( ) (

) (

)

( )

(

) (

) ( )



( ) (

) (

) ( )

( )

(

)

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



( ) (

) (

)

( )

(

) ( )(

)



( ) (

) (

) ( )

( )

(

)

( ) ( )

( )

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:





( ) (

) (

)

( )

(

) ( )(

)

( ) (

) (

) ( )

( )

(

)

( ) (

)

( )

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:






( )

(

)

( )

(

)

( )

(

)

( )

( )

(

)

Problem: 23

( )

342 343 344 345 346 347

6.993191 7.000000 7.006796 7.013579 7.020349 7.027106

Answer:






( )

(

)

( )

(

)

( )

(

)

( )

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:






( )

(

)

( )

(

)

( )

(

)

( )

( )

(

)

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:






( )

(

)

( )

(

)

( )

(

)

( )

( )

(

)

“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

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