5. Interpolation 5.1 Definition of interpolation. 5.2 Formulas for Interpolation. 5.3 Formulas for Interpolation for unequal interval. 5.4 Applications of interpolation

5.1 Definition of interpolation. Definition

5.2 DIRECT METHOD Now we consider a polynomial of degree n, i.e. y = a 0 +a 1 x+a 2 x 2 +……..+a n x n, where a 0,a 1,a 2,….a n are constants, on which we suppose (n+1) points like (x 0,y 0 ),(x 1,y 1 ),….,(x n,y n ). (i) Here we have to find (n+1) constants i.e. a 0,a 1,a 2,…..a n, for that we construct (n+1) equations. (ii) Substitute the value of x for the corresponding value of y in the above polynomial.

Table 1 t m/second

Linear Interpolation

Quadratic Interpolation

Error

Cubic Interpolation Example: - The velocity of a train is given in table 1. Find the velocity at t = 9 second using cubic Interpolation method. Solution:- Let v(t) =a 0 +a 1 t+a 2 t 2 +a 3 t 3. v(4)= a 0 +4a 1 +16a 2 +64a 3, v(6)= a 0 +6a 1 +36a a 3, v(8)= a 0 +8a 1 +64a a 3, v(10)= a 0 +10a a a 3.

Continued………

Comparison table Table:-Comparison of different degree of the polynomials. Degree of polynomial 123 v(t=9) meter/sec Absolute relative Approximate error

Newton forward Interpolation formula

Example The velocity of a train is given in table 1. Find the velocity at t= 5 second using Newton forward interpolation formula. Solution:- Forward Difference Table

Continued tv1 st difference2 nd difference3 rd difference4 th difference

Result u=1.5. Putting these values in the Newton forward interpolation formula we get v(5)= m/sec.

Newton Backward Interpolation formula

Example The velocity of a train is given in table 1. Find the velocity at t= 9 second using Newton backward interpolation formula. Solution: - u=-0.5. Putting these values in the Newton backward interpolation formula we get v(9)= m/sec.

5.3 Newton’s Divided Difference method Table 2 t m/secon d

Linear Interpolation

Example

Quadratic Interpolation

Example

Error

Cubic Interpolation

Example

Error

Comparison Table Table:-Comparison of different degree of the polynomials. Degree of polynomial 123 v(t=12) meter/sec Absolute relative Approximate error

Lagrange’s Interpolation

Linear Interpolation

Example The velocity of a train is given in table 2. Find the velocity at t= 12 second using linear method for interpolation. Solution:- Here t 0 =11sec, t 1 =16sec and t=12 sec. Putting these values in the v(t)= L 0 (t)v(t 0 )+ L 1 (t)v(t 1 ) we get v(12)=53.6 m/sec.

Quadratic Interpolation

Example The velocity of a train is given in table 2. Find the velocity at t= 12 second using quadratic method for interpolation. Solution:- Here t 0 =7sec, t 1 =11sec, t 2 =16 sec and t=12 sec. Putting these values in the v(t)= L 0 (t)v(t 0 )+ L 1 (t)v(t 1 ) + L 2 (t)v(t 2 ), we get v(12)= m/sec.

Error

Cubic Interpolation

Example The velocity of a train is given in table 2. Find the velocity at t= 12 second using cubic method for interpolation. Solution:- Here t 0 =5sec, t 1 =7sec, t 2 =11sec, t 3 =16 sec and t=12 sec. Putting the values in the v(t)= L 0 (t)v(t 0 )+ L 1 (t)v(t 1 ) + L 2 (t)v(t 2 ) + L 3 (t)v(t 3 ), we get v(12)= m/sec.

Error

Comparison table Table:-Comparison of different degree of the polynomials. Degree of polynomial123 v(t=12) meter/sec Absolute relative Approximate error

5.4 Applications of interpolation. Form of the function. To fill the gaps in a table. Computer graphics. Census.