2. 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

3. 5.1 Definition of interpolation. Definition

4. 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.

5. Table 1 t246810 m/second1014213050

6. Linear Interpolation

7. Quadratic Interpolation

8. Error

9. 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 2 +216a 3, v(8)= a 0 +8a 1 +64a 2 + 512a 3, v(10)= a 0 +10a 1 +100a 2 +1000a 3.

10. Continued………

11. Comparison table Table:-Comparison of different degree of the polynomials. Degree of polynomial 123 v(t=9) meter/sec4038.62538.0625 Absolute relative Approximate error ---------------3.55987051.4778325

12. Newton forward Interpolation formula

13. 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

14. Continued tv1 st difference2 nd difference3 rd difference4 th difference 210 4 7 9 20 3 2 11 4 14 621 830 1050

15. Result u=1.5. Putting these values in the Newton forward interpolation formula we get v(5)=17.2109 m/sec.

16. Newton Backward Interpolation formula

17. 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)=37.6719 m/sec.

18. 5.3 Newton’s Divided Difference method Table 2 t m/secon d 210 519 725 1152 1660

19. Linear Interpolation

20. Example

21. Quadratic Interpolation

22. Example

23. Error

24. Cubic Interpolation

25. Example

26. Error

27. Comparison Table Table:-Comparison of different degree of the polynomials. Degree of polynomial 123 v(t=12) meter/sec53.655.888958.06639 Absolute relative Approximate error ---------------4.09545 3.75

28. Lagrange’s Interpolation

29. Linear Interpolation

30. 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.

31. Quadratic Interpolation

32. 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)=55.8889 m/sec.

33. Error

34. Cubic Interpolation

35. 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)=58.0657 m/sec.

36. Error

37. Comparison table Table:-Comparison of different degree of the polynomials. Degree of polynomial123 v(t=12) meter/sec53.655.888958.0657 Absolute relative Approximate error ---------------4.09544653.75

38. 5.4 Applications of interpolation. Form of the function. To fill the gaps in a table. Computer graphics. Census.