1. Numerical Methods - Interpolation Unequal Intervals Dr. N. B. Vyas Department of Mathematics, Atmiya Institute of Tech. and Science, Rajkot (Guj.) niravbvyas@gmail.com Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals

2. Interpolation To nd the value of y for an x between dierent x - values x0, x1, . . . , xn is called problem of interpolation. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 3. Interpolation To nd the value of y for an x between dierent x - values x0, x1, . . . , xn is called problem of interpolation. To nd the value of y for an x which falls outside the range of x (x < x0 or x > xn) is called the problem of extrapolation. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 4. Interpolation To nd the value of y for an x between dierent x - values x0, x1, . . . , xn is called problem of interpolation. To nd the value of y for an x which falls outside the range of x (x < x0 or x > xn) is called the problem of extrapolation. Theorem by Weierstrass in 1885, Every continuous function in an interval (a,b) can be represented in that interval to any desired accuracy by a polynomial. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 5. Interpolation To nd the value of y for an x between dierent x - values x0, x1, . . . , xn is called problem of interpolation. To nd the value of y for an x which falls outside the range of x (x < x0 or x > xn) is called the problem of extrapolation. Theorem by Weierstrass in 1885, Every continuous function in an interval (a,b) can be represented in that interval to any desired accuracy by a polynomial. Let us assign polynomial Pn of degree n (or less) that assumes the given data values Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 6. Interpolation To nd the value of y for an x between dierent x - values x0, x1, . . . , xn is called problem of interpolation. To nd the value of y for an x which falls outside the range of x (x < x0 or x > xn) is called the problem of extrapolation. Theorem by Weierstrass in 1885, Every continuous function in an interval (a,b) can be represented in that interval to any desired accuracy by a polynomial. Let us assign polynomial Pn of degree n (or less) that assumes the given data values Pn(x0) = y0, Pn(x1) = y1, . . ., Pn(xn) = yn Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 7. Interpolation To nd the value of y for an x between dierent x - values x0, x1, . . . , xn is called problem of interpolation. To nd the value of y for an x which falls outside the range of x (x < x0 or x > xn) is called the problem of extrapolation. Theorem by Weierstrass in 1885, Every continuous function in an interval (a,b) can be represented in that interval to any desired accuracy by a polynomial. Let us assign polynomial Pn of degree n (or less) that assumes the given data values Pn(x0) = y0, Pn(x1) = y1, . . ., Pn(xn) = yn This polynomial Pn is called interpolation polynomial. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 8. Interpolation To nd the value of y for an x between dierent x - values x0, x1, . . . , xn is called problem of interpolation. To nd the value of y for an x which falls outside the range of x (x < x0 or x > xn) is called the problem of extrapolation. Theorem by Weierstrass in 1885, Every continuous function in an interval (a,b) can be represented in that interval to any desired accuracy by a polynomial. Let us assign polynomial Pn of degree n (or less) that assumes the given data values Pn(x0) = y0, Pn(x1) = y1, . . ., Pn(xn) = yn This polynomial Pn is called interpolation polynomial. x0, x1, . . . , xn is called the nodes ( tabular points, pivotal points or arguments). Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 9. Interpolation with unequal intervals Lagranges interpolation formula with unequal intervals: Let y = f(x) be continuous and dierentiable in the interval (a, b). Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 10. Interpolation with unequal intervals Lagranges interpolation formula with unequal intervals: Let y = f(x) be continuous and dierentiable in the interval (a, b). Given the set of n + 1 values (x0, y0), (x1, y1), . . . , (xn, yn) of x and y, where the values of x need not necessarily be equally spaced. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 11. Interpolation with unequal intervals Lagranges interpolation formula with unequal intervals: Let y = f(x) be continuous and dierentiable in the interval (a, b). Given the set of n + 1 values (x0, y0), (x1, y1), . . . , (xn, yn) of x and y, where the values of x need not necessarily be equally spaced. It is required to nd Pn(x), a polynomial of degree n such that y and Pn(x) agree at the tabulated points. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 12. Interpolation with unequal intervals Lagranges interpolation formula with unequal intervals: Let y = f(x) be continuous and dierentiable in the interval (a, b). Given the set of n + 1 values (x0, y0), (x1, y1), . . . , (xn, yn) of x and y, where the values of x need not necessarily be equally spaced. It is required to nd Pn(x), a polynomial of degree n such that y and Pn(x) agree at the tabulated points. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 13. Lagranges Interpolation This polynomial is given by the following formula: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 14. Lagranges Interpolation This polynomial is given by the following formula: y = f(x) Pn(x) = (x x1)(x x2) . . . (x xn) (x0 x1)(x0 x2) . . . (x0 xn) y0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 15. Lagranges Interpolation This polynomial is given by the following formula: y = f(x) Pn(x) = (x x1)(x x2) . . . (x xn) (x0 x1)(x0 x2) . . . (x0 xn) y0 + (x x0)(x x2) . . . (x xn) (x1 x0)(x1 x2) . . . (x1 xn) y1 + . . . Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 16. Lagranges Interpolation This polynomial is given by the following formula: y = f(x) Pn(x) = (x x1)(x x2) . . . (x xn) (x0 x1)(x0 x2) . . . (x0 xn) y0 + (x x0)(x x2) . . . (x xn) (x1 x0)(x1 x2) . . . (x1 xn) y1 + . . . + (x x0)(x x1) . . . (x xn1) (xn x0)(xn x1) . . . (xn xn1) yn Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 17. Lagranges Interpolation This polynomial is given by the following formula: y = f(x) Pn(x) = (x x1)(x x2) . . . (x xn) (x0 x1)(x0 x2) . . . (x0 xn) y0 + (x x0)(x x2) . . . (x xn) (x1 x0)(x1 x2) . . . (x1 xn) y1 + . . . + (x x0)(x x1) . . . (x xn1) (xn x0)(xn x1) . . . (xn xn1) yn NOTE: The above formula can be used irrespective of whether the values x0, x1, . . . , xn are equally spaced or not. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 18. Lagranges Inverse Interpolation In the Lagranges interpolation formula y is treated as dependent variable and expressed as function of independent variable x. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 19. Lagranges Inverse Interpolation In the Lagranges interpolation formula y is treated as dependent variable and expressed as function of independent variable x. Instead if x is treated as dependent variable and expressed as the function of independent variable y, then Lagranges interpolation formula becomes Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 20. Lagranges Inverse Interpolation In the Lagranges interpolation formula y is treated as dependent variable and expressed as function of independent variable x. Instead if x is treated as dependent variable and expressed as the function of independent variable y, then Lagranges interpolation formula becomes x = g(y) Pn(y) = (y y1)(y y2) . . . (y yn) (y0 y1)(y0 y2) . . . (y0 yn) x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 21. Lagranges Inverse Interpolation In the Lagranges interpolation formula y is treated as dependent variable and expressed as function of independent variable x. Instead if x is treated as dependent variable and expressed as the function of independent variable y, then Lagranges interpolation formula becomes x = g(y) Pn(y) = (y y1)(y y2) . . . (y yn) (y0 y1)(y0 y2) . . . (y0 yn) x0 + (y y0)(y y2) . . . (y yn) (y1 y0)(y1 y2) . . . (y1 yn) x1 + . . . Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 22. Lagranges Inverse Interpolation In the Lagranges interpolation formula y is treated as dependent variable and expressed as function of independent variable x. Instead if x is treated as dependent variable and expressed as the function of independent variable y, then Lagranges interpolation formula becomes x = g(y) Pn(y) = (y y1)(y y2) . . . (y yn) (y0 y1)(y0 y2) . . . (y0 yn) x0 + (y y0)(y y2) . . . (y yn) (y1 y0)(y1 y2) . . . (y1 yn) x1 + . . . + (y y0)(y y1) . . . (y yn1) (yn y0)(yn y1) . . . (yn yn1) xn Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 23. Lagranges Inverse Interpolation In the Lagranges interpolation formula y is treated as dependent variable and expressed as function of independent variable x. Instead if x is treated as dependent variable and expressed as the function of independent variable y, then Lagranges interpolation formula becomes x = g(y) Pn(y) = (y y1)(y y2) . . . (y yn) (y0 y1)(y0 y2) . . . (y0 yn) x0 + (y y0)(y y2) . . . (y yn) (y1 y0)(y1 y2) . . . (y1 yn) x1 + . . . + (y y0)(y y1) . . . (y yn1) (yn y0)(yn y1) . . . (yn yn1) xn This relation is referred as Lagranges inverse interpolation formula. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 24. Example Ex. Given the table of values: x 150 152 154 156 y = x 12.247 12.329 12.410 12.490 Evaluate 155 using Lagranges interpolation formula. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 25. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 26. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 27. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 By Lagranges interpolation formula, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 28. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 By Lagranges interpolation formula, f(x) Pn(x) = (x x1)(x x2)(x x3) (x0 x1)(x0 x2)(x0 x3) y0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 29. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 By Lagranges interpolation formula, f(x) Pn(x) = (x x1)(x x2)(x x3) (x0 x1)(x0 x2)(x0 x3) y0 + (x x0)(x x2)(x x3) (x1 x0)(x1 x2)(x1 x3) y1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 30. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 By Lagranges interpolation formula, f(x) Pn(x) = (x x1)(x x2)(x x3) (x0 x1)(x0 x2)(x0 x3) y0 + (x x0)(x x2)(x x3) (x1 x0)(x1 x2)(x1 x3) y1 + (x x0)(x x1)(x x3) (x2 x0)(x2 x1)(x2 x3) y2 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 31. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 By Lagranges interpolation formula, f(x) Pn(x) = (x x1)(x x2)(x x3) (x0 x1)(x0 x2)(x0 x3) y0 + (x x0)(x x2)(x x3) (x1 x0)(x1 x2)(x1 x3) y1 + (x x0)(x x1)(x x3) (x2 x0)(x2 x1)(x2 x3) y2 + (x x0)(x x1)(x x2) (x3 x0)(x3 x1)(x3 x2) y3 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 32. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 By Lagranges interpolation formula, f(x) Pn(x) = (x x1)(x x2)(x x3) (x0 x1)(x0 x2)(x0 x3) y0 + (x x0)(x x2)(x x3) (x1 x0)(x1 x2)(x1 x3) y1 + (x x0)(x x1)(x x3) (x2 x0)(x2 x1)(x2 x3) y2 + (x x0)(x x1)(x x2) (x3 x0)(x3 x1)(x3 x2) y3 for x = 155 f(155) = Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 33. Example Ex. Compute f(0.4) for the table below by the Lagranges interpolation: x 0.3 0.5 0.6 f(x) 0.61 0.69 0.72 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 34. Example Ex. Using Lagranges formula, nd the form of f(x) for the following data: x 0 1 2 5 f(x) 2 3 12 147 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 35. Example Ex. Using Lagranges formula, nd x for y = 7 for the following data: x 1 3 4 y 4 12 19 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 36. Example Ex. Using Lagranges formula, express the function 3x2 + x + 1 (x 1)(x 2)(x 3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 37. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 38. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 39. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 40. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, y1 = 15 and y2 = 31 By Lagranges interpolation formula, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 41. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, y1 = 15 and y2 = 31 By Lagranges interpolation formula, y = (x x1)(x x2) (x0 x1)(x0 x2) y0 + (x x0)(x x2) (x1 x0)(x1 x2) y1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 42. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, y1 = 15 and y2 = 31 By Lagranges interpolation formula, y = (x x1)(x x2) (x0 x1)(x0 x2) y0 + (x x0)(x x2) (x1 x0)(x1 x2) y1 + (x x0)(x x1) (x2 x0)(x2 x1) y2 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 43. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, y1 = 15 and y2 = 31 By Lagranges interpolation formula, y = (x x1)(x x2) (x0 x1)(x0 x2) y0 + (x x0)(x x2) (x1 x0)(x1 x2) y1 + (x x0)(x x1) (x2 x0)(x2 x1) y2 substituting above values, we get y = 2.5(x 2)(x 3) 15(x 1)(x 3) + 15.5(x 1)(x 2) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 44. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, y1 = 15 and y2 = 31 By Lagranges interpolation formula, y = (x x1)(x x2) (x0 x1)(x0 x2) y0 + (x x0)(x x2) (x1 x0)(x1 x2) y1 + (x x0)(x x1) (x2 x0)(x2 x1) y2 substituting above values, we get y = 2.5(x 2)(x 3) 15(x 1)(x 3) + 15.5(x 1)(x 2) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 45. Example Thus 3x2 + x + 1 (x 1)(x 2)(x 3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 46. Example Thus 3x2 + x + 1 (x 1)(x 2)(x 3) = 2.5(x 2)(x 3) 15(x 1)(x 3) + 15.5(x 1)(x 2) (x 1)(x 2)(x 3) = 2.5 (x 1) - 15 (x 2) + 15.5 (x 3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 47. Error in Interpolation Error in Interpolation: We assume that f(x) has continuous derivatives of order upto n + 1 for all x (a, b). Since, f(x) is approximated by Pn(x), the results contains errors. We dene the error of interpolation or truncation error as Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 48. Error in Interpolation Error in Interpolation: We assume that f(x) has continuous derivatives of order upto n + 1 for all x (a, b). Since, f(x) is approximated by Pn(x), the results contains errors. We dene the error of interpolation or truncation error as E(f, x) = f(x) Pn(x) = (x x0)(x x1) . . . (x xn) (n + 1)! f(n+1)() where min(x0, x1, . . . , xn, x) < < min(x0, x1, . . . , xn, x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 49. Error in Interpolation Error in Interpolation: We assume that f(x) has continuous derivatives of order upto n + 1 for all x (a, b). Since, f(x) is approximated by Pn(x), the results contains errors. We dene the error of interpolation or truncation error as E(f, x) = f(x) Pn(x) = (x x0)(x x1) . . . (x xn) (n + 1)! f(n+1)() where min(x0, x1, . . . , xn, x) < < min(x0, x1, . . . , xn, x) since, is an unknown, it is dicult to nd the value of error. However, we can nd a bound of the error. The bound of the error is obtained as Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 50. Error in Interpolation Error in Interpolation: We assume that f(x) has continuous derivatives of order upto n + 1 for all x (a, b). Since, f(x) is approximated by Pn(x), the results contains errors. We dene the error of interpolation or truncation error as E(f, x) = f(x) Pn(x) = (x x0)(x x1) . . . (x xn) (n + 1)! f(n+1)() where min(x0, x1, . . . , xn, x) < < min(x0, x1, . . . , xn, x) since, is an unknown, it is dicult to nd the value of error. However, we can nd a bound of the error. The bound of the error is obtained as |E(f, x)| |(x x0)(x x1) . . . (x xn)| (n + 1)! max ab |f(n+1)()| Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 51. Example Ex. Using the data sin(0.1) = 0.09983 and sin(0.2) = 0.19867, nd an approximate value of sin(0.15) by Lagrange interpolation. Obtain a bound on the error at x = 0.15. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 52. Lagranges Interpolation Disadvantages: In practice, we often do not know the degree of the interpolation polynomial that will give the required accuracy, so we should be prepared to increase the degree if necessary. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 53. Lagranges Interpolation Disadvantages: In practice, we often do not know the degree of the interpolation polynomial that will give the required accuracy, so we should be prepared to increase the degree if necessary. To increase the degree the addition of another interpolation point leads to re-computation. i.e. no previous work is useful. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 54. Lagranges Interpolation Disadvantages: In practice, we often do not know the degree of the interpolation polynomial that will give the required accuracy, so we should be prepared to increase the degree if necessary. To increase the degree the addition of another interpolation point leads to re-computation. i.e. no previous work is useful. E.g: In calculating Pk(x), no obvious advantage can be taken of the fact that one already has calculated Pk1(x). Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 55. Lagranges Interpolation Disadvantages: In practice, we often do not know the degree of the interpolation polynomial that will give the required accuracy, so we should be prepared to increase the degree if necessary. To increase the degree the addition of another interpolation point leads to re-computation. i.e. no previous work is useful. E.g: In calculating Pk(x), no obvious advantage can be taken of the fact that one already has calculated Pk1(x). That means we need to calculate entirely new polynomial. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 56. Divided Difference Let f(x0), f(x1), . . . , f(xn) be the values of a function f corresponding to the arguments x0, x1, . . . , xn where the intervals x1 x0, x2 x1, . . . , xn xn1 are not necessarily equally spaced. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 57. Divided Difference Let f(x0), f(x1), . . . , f(xn) be the values of a function f corresponding to the arguments x0, x1, . . . , xn where the intervals x1 x0, x2 x1, . . . , xn xn1 are not necessarily equally spaced. Then the rst divided dierence of f for the arguments x0, x1, . . . , xn are dened by , Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 58. Divided Difference Let f(x0), f(x1), . . . , f(xn) be the values of a function f corresponding to the arguments x0, x1, . . . , xn where the intervals x1 x0, x2 x1, . . . , xn xn1 are not necessarily equally spaced. Then the rst divided dierence of f for the arguments x0, x1, . . . , xn are dened by , f(x0, x1) = f(x1) f(x0) x1 x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 59. Divided Difference Let f(x0), f(x1), . . . , f(xn) be the values of a function f corresponding to the arguments x0, x1, . . . , xn where the intervals x1 x0, x2 x1, . . . , xn xn1 are not necessarily equally spaced. Then the rst divided dierence of f for the arguments x0, x1, . . . , xn are dened by , f(x0, x1) = f(x1) f(x0) x1 x0 f(x1, x2) = f(x2) f(x1) x2 x1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 60. Divided Difference The second divided dierence of f for three arguments x0, x1, x2 is dened by f(x0, x1, x2) = f(x1, x2) f(x0, x1) x2 x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 61. Divided Difference The second divided dierence of f for three arguments x0, x1, x2 is dened by f(x0, x1, x2) = f(x1, x2) f(x0, x1) x2 x0 and similarly the divided dierence of order n is dened by f(x0, x1, . . . , xn) = f(x1, x2, . . . , xn) f(x0, x1, . . . , xn1) xn x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 62. Divided Difference Properties: The divided dierences are symmetrical in all their arguments; that is, the value of any divided dierence is independent of the order of the arguments. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 63. Divided Difference Properties: The divided dierences are symmetrical in all their arguments; that is, the value of any divided dierence is independent of the order of the arguments. The divided dierence operator is linear. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 64. Divided Difference Properties: The divided dierences are symmetrical in all their arguments; that is, the value of any divided dierence is independent of the order of the arguments. The divided dierence operator is linear. The nth order divided dierences of a polynomial of degree n are constant, equal to the coecient of xn. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 65. Newtons Divided Difference Interpolation An interpolation formula which has the property that a polynomial of higher degree may be derived from it by simply adding new terms. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 66. Newtons Divided Difference Interpolation An interpolation formula which has the property that a polynomial of higher degree may be derived from it by simply adding new terms. Newtons general interpolation formula is one such formula and terms in it are called divided dierences. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 67. Newtons Divided Difference Interpolation An interpolation formula which has the property that a polynomial of higher degree may be derived from it by simply adding new terms. Newtons general interpolation formula is one such formula and terms in it are called divided dierences. Let f(x0), f(x1), . . . , f(xn) be the values of a function f corresponding to the arguments x0, x1, . . . , xn where the intervals x1 x0, x2 x1, . . . , xn xn1 are not necessarily equally spaced. By the denition of divided dierence, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 68. Newtons Divided Difference Interpolation An interpolation formula which has the property that a polynomial of higher degree may be derived from it by simply adding new terms. Newtons general interpolation formula is one such formula and terms in it are called divided dierences. Let f(x0), f(x1), . . . , f(xn) be the values of a function f corresponding to the arguments x0, x1, . . . , xn where the intervals x1 x0, x2 x1, . . . , xn xn1 are not necessarily equally spaced. By the denition of divided dierence, f(x, x0) = f(x) f(x0) x x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 69. Newtons Divided Difference Interpolation An interpolation formula which has the property that a polynomial of higher degree may be derived from it by simply adding new terms. Newtons general interpolation formula is one such formula and terms in it are called divided dierences. Let f(x0), f(x1), . . . , f(xn) be the values of a function f corresponding to the arguments x0, x1, . . . , xn where the intervals x1 x0, x2 x1, . . . , xn xn1 are not necessarily equally spaced. By the denition of divided dierence, f(x, x0) = f(x) f(x0) x x0 f(x) = f(x0) + (x x0)f(x, x0) (1) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 70. Newtons Divided Difference Interpolation Further f(x, x0, x1) = f(x, x0) f(x0, x1) x x1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 71. Newtons Divided Difference Interpolation Further f(x, x0, x1) = f(x, x0) f(x0, x1) x x1 which yields f(x, x0) = f(x0, x1) + (x x1)f(x, x0, x1) (2) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 72. Newtons Divided Difference Interpolation Further f(x, x0, x1) = f(x, x0) f(x0, x1) x x1 which yields f(x, x0) = f(x0, x1) + (x x1)f(x, x0, x1) (2) Similarly f(x, x0, x1) = f(x0, x1, x2) + (x x2)f(x, x0, x1, x2) (3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 73. Newtons Divided Difference Interpolation Further f(x, x0, x1) = f(x, x0) f(x0, x1) x x1 which yields f(x, x0) = f(x0, x1) + (x x1)f(x, x0, x1) (2) Similarly f(x, x0, x1) = f(x0, x1, x2) + (x x2)f(x, x0, x1, x2) (3) and in general f(x, x0, ..., xn1) = f(x0, x1, ..., xn) + (x xn)f(x, x0, x1, ..., xn) (4) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 74. Newtons Divided Difference Interpolation multiplying equation (2) by (x x0) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 75. Newtons Divided Difference Interpolation multiplying equation (2) by (x x0) , (3) by (x x0) (x x1) and so on, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 76. Newtons Divided Difference Interpolation multiplying equation (2) by (x x0) , (3) by (x x0) (x x1) and so on, and nally the last term (4) by (x x0) (x x1) ... (x xn1) and Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 77. Newtons Divided Difference Interpolation multiplying equation (2) by (x x0) , (3) by (x x0) (x x1) and so on, and nally the last term (4) by (x x0) (x x1) ... (x xn1) and adding (1), (2) , (3) up to (4) we obtain Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 78. Newtons Divided Difference Interpolation multiplying equation (2) by (x x0) , (3) by (x x0) (x x1) and so on, and nally the last term (4) by (x x0) (x x1) ... (x xn1) and adding (1), (2) , (3) up to (4) we obtain f(x) = f (x0) + (x x0) f (x0, x1) + (x x0) (x x1) f (x0, x1, x2) + ... + (x x0) (x x1) ... (x xn1) f (x0, x1, ..., xn) This formula is called Newtons divided dierence formula. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 79. Newtons Divided Difference Interpolation The divided dierence upto third order x y 1stdiv.di. 2nddiv.di. 3rddiv.di. x0 y0 [x0, x1] x1 y1 [x0, x1, x2] [x1, x2] [x0, x1, x2, x3] x2 y2 [x1, x2, x3] [x2, x3] x3 y3 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 80. Example Ex. Obtain the divided dierence table for the data: x -1 0 2 3 y -8 3 1 12 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 81. Example Sol. We have the following divided dierence table for the data: x y First d.d Second d.d Third d.d -1 -8 3 + 8 0 + 1 = 11 0 3 1 11 2 + 1 = 4 1 3 2 0 = 1 4 + 4 3 + 1 = 2 2 1 11 + 1 3 0 = 4 12 1 3 2 = 11 3 12 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 82. Example Ex. Find f(x) as a polynomial in x for the following data by Newtons divided dierence formula: x -4 -1 0 2 5 f(x) 1245 33 5 9 1335 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 83. Example Sol. We have the following divided dierence table for the data: x y 1st d.d 2nd d.d 3rd d.d 4th d.d -4 1245 404 -1 33 94 28 14 0 5 10 3 2 13 2 9 88 442 5 1335 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 84. Example The Newtons divided dierence formula gives: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 85. Example The Newtons divided dierence formula gives: f(x) = f(x0) + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 86. Example The Newtons divided dierence formula gives: f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 87. Example The Newtons divided dierence formula gives: f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 88. Example The Newtons divided dierence formula gives: f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] + (x x0)(x x1)(x x2)f[x0, x1, x2, x3] Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 89. Example The Newtons divided dierence formula gives: f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] + (x x0)(x x1)(x x2)f[x0, x1, x2, x3] + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 90. Example The Newtons divided dierence formula gives: f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] + (x x0)(x x1)(x x2)f[x0, x1, x2, x3] + (x x0)(x x1)(x x2)(x x3)f[x0, x1, x2, x3, x4] Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 91. Example The Newtons divided dierence formula gives: f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] + (x x0)(x x1)(x x2)f[x0, x1, x2, x3] + (x x0)(x x1)(x x2)(x x3)f[x0, x1, x2, x3, x4] = ... = 3x4 5x3 + 6x2 14x + 5 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 92. Example Ex. Find f(x) as a polynomial in x for the following data by Newtons divided dierence formula: x -2 -1 0 1 3 4 f(x) 9 16 17 18 44 81 Hence, interpolate at x = 0.5 and x = 3.1. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 93. Example Sol. We form the divided dierence table for the given data. x f(x) 1st d.d 2nd d.d 3rd d.d 4th d.d 2 9 7 1 16 3 1 1 0 17 0 0 1 1 1 18 4 0 13 1 3 44 8 37 4 81 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 94. Example Since, the fourth order dierences are zeros, the data represents a third degree polynomial. Newtons divided dierence formula gives the polynomial as f(x) = f(x0) + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 95. Example Since, the fourth order dierences are zeros, the data represents a third degree polynomial. Newtons divided dierence formula gives the polynomial as f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 96. Example Since, the fourth order dierences are zeros, the data represents a third degree polynomial. Newtons divided dierence formula gives the polynomial as f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 97. Example Since, the fourth order dierences are zeros, the data represents a third degree polynomial. Newtons divided dierence formula gives the polynomial as f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] + (x x0)(x x1)(x x2)f[x0, x1, x2, x3] Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 98. Example Since, the fourth order dierences are zeros, the data represents a third degree polynomial. Newtons divided dierence formula gives the polynomial as f(x) = f(x0) + (x x0)f[x0, x1] + (x x0)(x x1)f[x0, x1, x2] + (x x0)(x x1)(x x2)f[x0, x1, x2, x3] = ... = x3 + 17 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 99. Example Ex. Find the missing term in the following table: x 0 1 2 3 4 y 1 3 9 - 81 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 100. Example Sol. Divided dierence table: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 101. Example Sol. Divided dierence table: By Newtons divided dierence formula f(x) = f (x0) + (x x0) f (x0, x1) + (x x0) (x x1) f (x0, x1, x2) + ... Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 102. Spline Interpolation Spline interpolation is a form of interpolation where the interpolant is a special type of piecewise polynomial called a spline Consider the problem of interpolating between the data points (x0, y0), (x1, y1), . . . , (xn, yn) by means of spline tting. Then the cubic spline f(x) is such that (i) f(x) is a linear polynomial outside the interval (x0, xn) (ii) f(x) is a cubic polynomial in each of the subintervals, (iii) f (x) and f (x) are continuous at each point. Since f(x) is cubic in each of the subintervals f (x) shall be linear. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 103. Spline Interpolation f(x) = (xi+1 x)3Mi 6h + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 104. Spline Interpolation f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 105. Spline Interpolation f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 106. Spline Interpolation f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + (x xi) h yi+1 h2 6 Mi+1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 107. Spline Interpolation f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + (x xi) h yi+1 h2 6 Mi+1 where Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 108. Spline Interpolation f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + (x xi) h yi+1 h2 6 Mi+1 where Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2, 3, ..., (n 1) and M0 = 0, Mn = 0, xi+1 xi = h. which gives n + 1 equations in n + 1 unknowns Mi(i = 0, 1, ..., n) which can be solved. Substituting the value of Mi gives the concerned cubic spline. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 109. Example Ex. Obtain cubic spline for the following data: x 0 1 2 3 y 2 -6 -8 2 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 110. Example Sol. Since points are equispaced with h = 1 and n = 3, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 111. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 112. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 113. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 114. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 115. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 116. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 117. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 118. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 119. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 = Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 120. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 =4.8 and M2 = Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 121. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 =4.8 and M2 =16.8 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 122. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 =4.8 and M2 =16.8 Now the cubic spline in (xi x xi+1) is Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 123. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 =4.8 and M2 =16.8 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 124. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 =4.8 and M2 =16.8 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 125. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 =4.8 and M2 =16.8 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 126. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 =4.8 and M2 =16.8 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + (x xi) h yi+1 h2 6 Mi+1 (3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 127. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 36; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 72 (2) solving these, we get M1 =4.8 and M2 =16.8 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + (x xi) h yi+1 h2 6 Mi+1 (3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 128. Example Ex. The following values of x and y are given: x 1 2 3 4 y 1 2 5 11 Find the cubic splines and evaluate y(1.5) and y (3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 129. Example Sol. Since points are equispaced with h = 1 and n = 3, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 130. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 131. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 132. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 133. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 134. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 135. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 136. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 137. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 138. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 = Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 139. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 =2 and M2 = Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 140. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 =2 and M2 =4 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 141. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 =2 and M2 =4 Now the cubic spline in (xi x xi+1) is Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 142. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 =2 and M2 =4 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 143. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 =2 and M2 =4 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 144. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 =2 and M2 =4 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 145. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 =2 and M2 =4 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + (x xi) h yi+1 h2 6 Mi+1 (3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 146. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi1 + 4Mi + Mi+1 = 6 h2 (yi1 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 for i = 1, M0 + 4M1 + M2 = 6(y0 2y1 + y2) therefore, 4M1 + M2 = 12; (1) for i = 2, M1 + 4M2 + M3 = 6(y1 2y2 + y3) M1 + 4M2 = 18 (2) solving these, we get M1 =2 and M2 =4 Now the cubic spline in (xi x xi+1) is f(x) = (xi+1 x)3Mi 6h + (x xi)3Mi+1 6h + (xi+1 x) h yi h2 6 Mi + (x xi) h yi+1 h2 6 Mi+1 (3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 147. Example Ex. Find whether the following functions are cubic splines ? 1. f(x) = 5x3 3x2, 1 x 0 = 5x3 3x2, 0 x 1 2. f(x) = 2x3 x2, 1 x 0 = 2x3 + 3x2, 0 x 1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 148. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 149. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 150. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) = 0 = lim x0 f(x) therefore, given function f(x) is continuous on (1, 1). Now f (x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 151. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) = 0 = lim x0 f(x) therefore, given function f(x) is continuous on (1, 1). Now f (x) = 15x2 6x, 1 x 0 = 15x2 6x, 0 x 1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 152. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) = 0 = lim x0 f(x) therefore, given function f(x) is continuous on (1, 1). Now f (x) = 15x2 6x, 1 x 0 = 15x2 6x, 0 x 1 we have, lim x0+ f (x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 153. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) = 0 = lim x0 f(x) therefore, given function f(x) is continuous on (1, 1). Now f (x) = 15x2 6x, 1 x 0 = 15x2 6x, 0 x 1 we have, lim x0+ f (x) = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 154. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) = 0 = lim x0 f(x) therefore, given function f(x) is continuous on (1, 1). Now f (x) = 15x2 6x, 1 x 0 = 15x2 6x, 0 x 1 we have, lim x0+ f (x) = 0 = lim x0 f (x) therefore, the function f (x) is continuous on (1, 1). f (x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 155. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) = 0 = lim x0 f(x) therefore, given function f(x) is continuous on (1, 1). Now f (x) = 15x2 6x, 1 x 0 = 15x2 6x, 0 x 1 we have, lim x0+ f (x) = 0 = lim x0 f (x) therefore, the function f (x) is continuous on (1, 1). f (x) = 30x6, 1 x 0 = 30x 6, 0 x 1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals 156. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x0+ f(x) = 0 = lim x0 f(x) therefore, given function f(x) is continuous on (1, 1). Now f (x) = 15x2 6x, 1 x 0 = 15x2 6x, 0 x 1 we have, lim x0+ f (x) = 0 = lim x0 f (x) therefore, the function f (x) is continuous on (1, 1). f (x) = 30x6, 1 x 0 = 30x 6, 0 x 1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals