1. Numerical Methods - Finite DifferencesN. B. VyasDepartment of Mathematics,Atmiya Institute of Tech. and Science,Rajkot (Guj.)niravbvyas@gmail.comN. B. Vyas Numerical Methods - Finite Differences

2. Finite DifferencesForward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.N. B. Vyas Numerical Methods - Finite Differences

3. Finite DifferencesForward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The constant dierence between two consecutive values of x iscalled the interval of dierences and is denoted by h.N. B. Vyas Numerical Methods - Finite Differences

4. Finite DifferencesForward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The constant dierence between two consecutive values of x iscalled the interval of dierences and is denoted by h.The operator dened byy0 = y1 y0N. B. Vyas Numerical Methods - Finite Differences

5. Finite DifferencesForward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The constant dierence between two consecutive values of x iscalled the interval of dierences and is denoted by h.The operator dened byy0 = y1 y0y1 = y2 y1N. B. Vyas Numerical Methods - Finite Differences

6. Finite DifferencesForward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The constant dierence between two consecutive values of x iscalled the interval of dierences and is denoted by h.The operator dened byy0 = y1 y0y1 = y2 y1..............yn1 = yn yn1N. B. Vyas Numerical Methods - Finite Differences

7. Finite DifferencesForward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The constant dierence between two consecutive values of x iscalled the interval of dierences and is denoted by h.The operator dened byy0 = y1 y0y1 = y2 y1..............yn1 = yn yn1is called Forward dierence operator.N. B. Vyas Numerical Methods - Finite Differences

8. Finite DifferencesThe rst forward dierence is yn = yn+1 ynN. B. Vyas Numerical Methods - Finite Differences

9. Finite DifferencesThe rst forward dierence is yn = yn+1 ynThe second forward dierence are dened as the dierence ofthe rst dierences.2y0 = (y0) = (y1 y0) = y1 y0= y2 y1 (y1 y0) = y2 2y1 + y02y1 = y2 y12yi = yi+1 yiIn general nth forward dierence of f is dened bynyi = n1yi+1 n1yiN. B. Vyas Numerical Methods - Finite Differences

10. Finite DifferencesForward Dierence Table:x y y 2y 3y 4y 5yx0 y0y0x1 y1 2y0y1 3y0x2 y2 2y1 4y0y2 3y1 5y0x3 y3 2y2 4y1y3 3y2x4 y4 2y3y4x5 y5N. B. Vyas Numerical Methods - Finite Differences

11. Finite DifferencesThe operator satises the following properties:1 [f(x) g(x)] = f(x) g(x)N. B. Vyas Numerical Methods - Finite Differences

12. Finite DifferencesThe operator satises the following properties:1 [f(x) g(x)] = f(x) g(x)2 [cf(x)] = cf(x)N. B. Vyas Numerical Methods - Finite Differences

13. Finite DifferencesThe operator satises the following properties:1 [f(x) g(x)] = f(x) g(x)2 [cf(x)] = cf(x)3 mnf(x) = m+nf(x), m, n are positive integers4 Since nyn is a constant, n+1yn = 0 , n+2yn = 0, . . .i.e. (n + 1)th and higher dierences are zero.N. B. Vyas Numerical Methods - Finite Differences

14. Finite DifferencesBackward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.N. B. Vyas Numerical Methods - Finite Differences

15. Finite DifferencesBackward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The operator dened byy1 = y1 y0N. B. Vyas Numerical Methods - Finite Differences

16. Finite DifferencesBackward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The operator dened byy1 = y1 y0y2 = y2 y1N. B. Vyas Numerical Methods - Finite Differences

17. Finite DifferencesBackward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The operator dened byy1 = y1 y0y2 = y2 y1..............yn = yn yn1N. B. Vyas Numerical Methods - Finite Differences

18. Finite DifferencesBackward dierenceSuppose that a function y = f(x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving thefunctional values y0, y1, y2, ..., yn.The operator dened byy1 = y1 y0y2 = y2 y1..............yn = yn yn1is called Backward dierence operator.N. B. Vyas Numerical Methods - Finite Differences

19. Finite DifferencesThe rst backward dierence is yn = yn yn1N. B. Vyas Numerical Methods - Finite Differences

20. Finite DifferencesThe rst backward dierence is yn = yn yn1The second backward dierence are obtain by the dierenceof the rst dierences.2y2 = ( y2) = (y2 y1) = y2 y1= y2 y1 (y1 y0) = y2 2y1 + y02y3 = y3 y22yn = yn yn1N. B. Vyas Numerical Methods - Finite Differences

21. Finite DifferencesThe rst backward dierence is yn = yn yn1The second backward dierence are obtain by the dierenceof the rst dierences.2y2 = ( y2) = (y2 y1) = y2 y1= y2 y1 (y1 y0) = y2 2y1 + y02y3 = y3 y22yn = yn yn1In general nth backward dierence of f is dened bynyi = n1yi n1yi1N. B. Vyas Numerical Methods - Finite Differences

22. Finite DifferencesBackward Dierence Table:x y y 2y 3y 4y 5yx0 y0y1x1 y12y2y23y3x2 y22y34y4y33y45y5x3 y32y44y5y43y5x4 y42y5y5x5 y5N. B. Vyas Numerical Methods - Finite Differences

23. Finite DifferencesCentral dierenceThe operator dened byy12= y1 y0N. B. Vyas Numerical Methods - Finite Differences

24. Finite DifferencesCentral dierenceThe operator dened byy12= y1 y0y32= y2 y1N. B. Vyas Numerical Methods - Finite Differences

25. Finite DifferencesCentral dierenceThe operator dened byy12= y1 y0y32= y2 y1..............yn12= yn yn1N. B. Vyas Numerical Methods - Finite Differences

26. Finite DifferencesCentral dierenceThe operator dened byy12= y1 y0y32= y2 y1..............yn12= yn yn1is called Central dierence operator.Similarly, higher order central dierences are dened as2y1 = y32 y12N. B. Vyas Numerical Methods - Finite Differences

27. Finite DifferencesCentral dierenceThe operator dened byy12= y1 y0y32= y2 y1..............yn12= yn yn1is called Central dierence operator.Similarly, higher order central dierences are dened as2y1 = y32 y122y2 = y52 y32N. B. Vyas Numerical Methods - Finite Differences

28. Finite DifferencesCentral dierenceThe operator dened byy12= y1 y0y32= y2 y1..............yn12= yn yn1is called Central dierence operator.Similarly, higher order central dierences are dened as2y1 = y32 y122y2 = y52 y32.......3y32= 2y2 2y1N. B. Vyas Numerical Methods - Finite Differences

29. Finite DifferencesCentral dierenceThe operator dened byy12= y1 y0y32= y2 y1..............yn12= yn yn1is called Central dierence operator.Similarly, higher order central dierences are dened as2y1 = y32 y122y2 = y52 y32.......3y32= 2y2 2y1N. B. Vyas Numerical Methods - Finite Differences

30. Finite DifferencesCentral Dierence Table:x y y 2y 3y 4y 5yx0 y0y12x1 y1 2y1y323y32x2 y2 2y2 4y2y523y525y52x3 y3 2y3 4y3y723y72x4 y4 2y4y92x5 y5N. B. Vyas Numerical Methods - Finite Differences

31. Finite DifferencesNOTE:From all three dierence tables, we can see that only thenotations changes not the dierences.y1 y0 = y0 = y1 = y12N. B. Vyas Numerical Methods - Finite Differences

32. Finite DifferencesNOTE:From all three dierence tables, we can see that only thenotations changes not the dierences.y1 y0 = y0 = y1 = y12Alternative notations for the function y = f(x).N. B. Vyas Numerical Methods - Finite Differences

33. Finite DifferencesNOTE:From all three dierence tables, we can see that only thenotations changes not the dierences.y1 y0 = y0 = y1 = y12Alternative notations for the function y = f(x).For two consecutive values of x diering by h.N. B. Vyas Numerical Methods - Finite Differences

34. Finite DifferencesNOTE:From all three dierence tables, we can see that only thenotations changes not the dierences.y1 y0 = y0 = y1 = y12Alternative notations for the function y = f(x).For two consecutive values of x diering by h.yx = yx+h yx = f(x + h) f(x)N. B. Vyas Numerical Methods - Finite Differences

35. Finite DifferencesNOTE:From all three dierence tables, we can see that only thenotations changes not the dierences.y1 y0 = y0 = y1 = y12Alternative notations for the function y = f(x).For two consecutive values of x diering by h.yx = yx+h yx = f(x + h) f(x)yx = yx yxh = f(x) f(x h)N. B. Vyas Numerical Methods - Finite Differences

36. Finite DifferencesNOTE:From all three dierence tables, we can see that only thenotations changes not the dierences.y1 y0 = y0 = y1 = y12Alternative notations for the function y = f(x).For two consecutive values of x diering by h.yx = yx+h yx = f(x + h) f(x)yx = yx yxh = f(x) f(x h)yx = yx+h2 yxh2= f(x + h2 ) f(xh2 )N. B. Vyas Numerical Methods - Finite Differences

37. Finite DifferencesEvaluate the following. The interval of dierence being h.1 nexN. B. Vyas Numerical Methods - Finite Differences

38. Finite DifferencesEvaluate the following. The interval of dierence being h.1 nex2 logf(x)N. B. Vyas Numerical Methods - Finite Differences

39. Finite DifferencesEvaluate the following. The interval of dierence being h.1 nex2 logf(x)3 (tan1x)N. B. Vyas Numerical Methods - Finite Differences

40. Finite DifferencesEvaluate the following. The interval of dierence being h.1 nex2 logf(x)3 (tan1x)4 2cos2xN. B. Vyas Numerical Methods - Finite Differences

41. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);N. B. Vyas Numerical Methods - Finite Differences

42. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);N. B. Vyas Numerical Methods - Finite Differences

43. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);N. B. Vyas Numerical Methods - Finite Differences

44. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);N. B. Vyas Numerical Methods - Finite Differences

45. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);The inverse operator E1 is dened asN. B. Vyas Numerical Methods - Finite Differences

46. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);The inverse operator E1 is dened asE1f(x) = f(x h);N. B. Vyas Numerical Methods - Finite Differences

47. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);The inverse operator E1 is dened asE1f(x) = f(x h);Enf(x) = f(x nh);N. B. Vyas Numerical Methods - Finite Differences

48. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);The inverse operator E1 is dened asE1f(x) = f(x h);Enf(x) = f(x nh);If yx is the function f(x), thenN. B. Vyas Numerical Methods - Finite Differences

49. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);The inverse operator E1 is dened asE1f(x) = f(x h);Enf(x) = f(x nh);If yx is the function f(x), thenEyx = yx+h;N. B. Vyas Numerical Methods - Finite Differences

50. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);The inverse operator E1 is dened asE1f(x) = f(x h);Enf(x) = f(x nh);If yx is the function f(x), thenEyx = yx+h;E1yx = yxh;N. B. Vyas Numerical Methods - Finite Differences

51. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);The inverse operator E1 is dened asE1f(x) = f(x h);Enf(x) = f(x nh);If yx is the function f(x), thenEyx = yx+h;E1yx = yxh;Enyx = yx+nh;N. B. Vyas Numerical Methods - Finite Differences

52. Finite DifferencesOther Dierence Operator1. Shift Operator E:E does the operation of increasing the argument x by h so thatEf(x) = f(x + h);E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);E3f(x) = f(x + 3h);Enf(x) = f(x + nh);The inverse operator E1 is dened asE1f(x) = f(x h);Enf(x) = f(x nh);If yx is the function f(x), thenEyx = yx+h;E1yx = yxh;Enyx = yx+nh;N. B. Vyas Numerical Methods - Finite Differences

53. Finite Differences2. Averaging Operator :It is dened asf(x) =12f x +h2+ f x h2;N. B. Vyas Numerical Methods - Finite Differences

54. Finite Differences2. Averaging Operator :It is dened asf(x) =12f x +h2+ f x h2;i.e. yx =12yx+h2+ yxh2;N. B. Vyas Numerical Methods - Finite Differences

55. Finite Differences2. Averaging Operator :It is dened asf(x) =12f x +h2+ f x h2;i.e. yx =12yx+h2+ yxh2;3. Dierential Operator D:N. B. Vyas Numerical Methods - Finite Differences

56. Finite Differences2. Averaging Operator :It is dened asf(x) =12f x +h2+ f x h2;i.e. yx =12yx+h2+ yxh2;3. Dierential Operator D:It is dened asN. B. Vyas Numerical Methods - Finite Differences

57. Finite Differences2. Averaging Operator :It is dened asf(x) =12f x +h2+ f x h2;i.e. yx =12yx+h2+ yxh2;3. Dierential Operator D:It is dened asDf(x) =ddxf(x) = f (x);N. B. Vyas Numerical Methods - Finite Differences

58. Finite DifferencesRelation between the operators1 = E 1N. B. Vyas Numerical Methods - Finite Differences

59. Finite DifferencesRelation between the operators1 = E 12 = 1 E1N. B. Vyas Numerical Methods - Finite Differences

60. Finite DifferencesRelation between the operators1 = E 12 = 1 E13 = E12 E12N. B. Vyas Numerical Methods - Finite Differences

61. Finite DifferencesRelation between the operators1 = E 12 = 1 E13 = E12 E124 =12{E12 E12 }N. B. Vyas Numerical Methods - Finite Differences

62. Finite DifferencesRelation between the operators1 = E 12 = 1 E13 = E12 E124 =12{E12 E12 }5 = E = E = E12N. B. Vyas Numerical Methods - Finite Differences

63. Finite DifferencesRelation between the operators1 = E 12 = 1 E13 = E12 E124 =12{E12 E12 }5 = E = E = E126 E = ehDN. B. Vyas Numerical Methods - Finite Differences

64. Finite DifferencesRelation between the operators1 = E 12 = 1 E13 = E12 E124 =12{E12 E12 }5 = E = E = E126 E = ehD7 (1 + )(1 ) = 1N. B. Vyas Numerical Methods - Finite Differences

65. Finite DifferencesRelation between the operators1 = E 12 = 1 E13 = E12 E124 =12{E12 E12 }5 = E = E = E126 E = ehD7 (1 + )(1 ) = 18 = = 2N. B. Vyas Numerical Methods - Finite Differences

66. Gregory - Newton Forward Interpolation FormulaTo estimate the value of a function near the beginning a table,the forward dierence interpolation formula in used.N. B. Vyas Numerical Methods - Finite Differences

67. Gregory - Newton Forward Interpolation FormulaTo estimate the value of a function near the beginning a table,the forward dierence interpolation formula in used.Let yx = f(x) be a function which takes the valuesyx0 , yx0+h, yx0+2h, . . . corresponding to the valuesx0, x0 + h, x0 + 2h, . . . of x.N. B. Vyas Numerical Methods - Finite Differences

68. Gregory - Newton Forward Interpolation FormulaTo estimate the value of a function near the beginning a table,the forward dierence interpolation formula in used.Let yx = f(x) be a function which takes the valuesyx0 , yx0+h, yx0+2h, . . . corresponding to the valuesx0, x0 + h, x0 + 2h, . . . of x.Suppose we want to evaluate yx when x = x0 + ph, where p isany real number.N. B. Vyas Numerical Methods - Finite Differences

69. Gregory - Newton Forward Interpolation FormulaTo estimate the value of a function near the beginning a table,the forward dierence interpolation formula in used.Let yx = f(x) be a function which takes the valuesyx0 , yx0+h, yx0+2h, . . . corresponding to the valuesx0, x0 + h, x0 + 2h, . . . of x.Suppose we want to evaluate yx when x = x0 + ph, where p isany real number.N. B. Vyas Numerical Methods - Finite Differences

70. Gregory - Newton Forward Interpolation FormulaLet it be yp. For any real number n, we have dened operator Esuch that Enf(x) = f(x + nh). yx = yx0+ph = f (x0 + ph) = Epyx0 = (1 + )py0N. B. Vyas Numerical Methods - Finite Differences

71. Gregory - Newton Forward Interpolation FormulaLet it be yp. For any real number n, we have dened operator Esuch that Enf(x) = f(x + nh). yx = yx0+ph = f (x0 + ph) = Epyx0 = (1 + )py0= 1 + p +p(p 1)2!2 +p(p 1)(p 2)3!3 + ... y0N. B. Vyas Numerical Methods - Finite Differences

72. Gregory - Newton Forward Interpolation FormulaLet it be yp. For any real number n, we have dened operator Esuch that Enf(x) = f(x + nh). yx = yx0+ph = f (x0 + ph) = Epyx0 = (1 + )py0= 1 + p +p(p 1)2!2 +p(p 1)(p 2)3!3 + ... y0yx = y0 + py0 +p(p 1)2!2y0 +p(p 1)(p 2)3!3y0 + ...is called Newtons forward interpolation formula.N. B. Vyas Numerical Methods - Finite Differences

73. Gregory - Newton Backward Interpolation FormulaTo estimate the value of a function near the end of a table, thebackward dierence interpolation formula in used.N. B. Vyas Numerical Methods - Finite Differences

74. Gregory - Newton Backward Interpolation FormulaTo estimate the value of a function near the end of a table, thebackward dierence interpolation formula in used.Let yx = f(x) be a function which takes the valuesyx0 , yx0+h, yx0+2h, . . . corresponding to the valuesx0, x0 + h, x0 + 2h, . . . of x.N. B. Vyas Numerical Methods - Finite Differences

75. Gregory - Newton Backward Interpolation FormulaTo estimate the value of a function near the end of a table, thebackward dierence interpolation formula in used.Let yx = f(x) be a function which takes the valuesyx0 , yx0+h, yx0+2h, . . . corresponding to the valuesx0, x0 + h, x0 + 2h, . . . of x.Suppose we want to evaluate yx when x = xn + ph, where p isany real number.N. B. Vyas Numerical Methods - Finite Differences

76. Gregory - Newton Backward Interpolation FormulaLet it be yp. For any real number n, we have dened operator Esuch that Enf(x) = f(x + nh). yx = yxn+ph = f (xn + ph) = Epyxn = (1 )pynN. B. Vyas Numerical Methods - Finite Differences

77. Gregory - Newton Backward Interpolation FormulaLet it be yp. For any real number n, we have dened operator Esuch that Enf(x) = f(x + nh). yx = yxn+ph = f (xn + ph) = Epyxn = (1 )pyn= 1 + p +p(p + 1)2!2 +p(p + 1)(p + 2)3!3 + ... ynN. B. Vyas Numerical Methods - Finite Differences

78. Gregory - Newton Backward Interpolation FormulaLet it be yp. For any real number n, we have dened operator Esuch that Enf(x) = f(x + nh). yx = yxn+ph = f (xn + ph) = Epyxn = (1 )pyn= 1 + p +p(p + 1)2!2 +p(p + 1)(p + 2)3!3 + ... ynyx = y0 + p yn +p(p + 1)2!2yn +p(p + 1)(p + 2)3!3yn + ...is called Newtons backward interpolation formulaN. B. Vyas Numerical Methods - Finite Differences

79. Stirlings Interpolation FormulaTo estimate the value of a function near the middle a table, thecentral dierence interpolation formula in used.Let yx = f(x) be a functional relation between x and y.If x takes the values x0 2h, x0 h, x0, x0 + h, x0 + 2h, . . . andthe corresponding values of y are y2, y1, y0, y1, y2 . . . then wecan form a central dierence table as follows:N. B. Vyas Numerical Methods - Finite Differences

80. x y1stdifference2nddifference3rddifferencex0 2h y2y2(= y3/2)x0 h y1 2y2(= 2y1)y1(= y1/2) 3y2(= 3y1/2)x0 y0 2y1(= 2y0) y0(= y1/2) 3y1(= 3y1/2)x0 + h y1 2y0(= 2y1)y1(= y3/2)x0 + 2h y2N. B. Vyas Numerical Methods - Finite Differences

81. Central DierenceGausss Forward interpolation formula:Pn(x) = y0 + py0 +p(p 1)2!2y1 +(p + 1)p(p 1)3!3y1+(p + 1)p(p 1)(p 2)4!4y2 + ...Gausss Backward interpolation formula:Pn(x) = y0 + py1 +p(p + 1)2!2y1 +(p + 1)p(p 1)3!3y2+(p + 2)(p + 1)p(p 1)4!4y2 + ...N. B. Vyas Numerical Methods - Finite Differences