Interpolation with unequal interval

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Interpolation methods, Lagrange's interpolation formula, Lagrange's inverse interpolation formula, Newton's divided difference forumla, Cubic Spline

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Interpolation with unequal interval

  1. 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. 2. Interpolation To find the value of y for an x between different x - values x0, x1, . . . , xn is called problem of interpolation. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  3. 3. Interpolation To find the value of y for an x between different x - values x0, x1, . . . , xn is called problem of interpolation. To find 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. 4. Interpolation To find the value of y for an x between different x - values x0, x1, . . . , xn is called problem of interpolation. To find 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. 5. Interpolation To find the value of y for an x between different x - values x0, x1, . . . , xn is called problem of interpolation. To find 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. 6. Interpolation To find the value of y for an x between different x - values x0, x1, . . . , xn is called problem of interpolation. To find 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. 7. Interpolation To find the value of y for an x between different x - values x0, x1, . . . , xn is called problem of interpolation. To find 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. 8. Interpolation To find the value of y for an x between different x - values x0, x1, . . . , xn is called problem of interpolation. To find 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. 9. Interpolation with unequal intervals Lagrange’s interpolation formula with unequal intervals: Let y = f(x) be continuous and differentiable in the interval (a, b). Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  10. 10. Interpolation with unequal intervals Lagrange’s interpolation formula with unequal intervals: Let y = f(x) be continuous and differentiable 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. 11. Interpolation with unequal intervals Lagrange’s interpolation formula with unequal intervals: Let y = f(x) be continuous and differentiable 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 find 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. 12. Interpolation with unequal intervals Lagrange’s interpolation formula with unequal intervals: Let y = f(x) be continuous and differentiable 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 find 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. 13. Lagrange’s Interpolation This polynomial is given by the following formula: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  14. 14. Lagrange’s 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. 15. Lagrange’s 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. 16. Lagrange’s 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 − xn−1) (xn − x0)(xn − x1) . . . (xn − xn−1) yn Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  17. 17. Lagrange’s 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 − xn−1) (xn − x0)(xn − x1) . . . (xn − xn−1) 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. 18. Lagrange’s Inverse Interpolation In the Lagrange’s 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. 19. Lagrange’s Inverse Interpolation In the Lagrange’s 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 Lagrange’s interpolation formula becomes Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  20. 20. Lagrange’s Inverse Interpolation In the Lagrange’s 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 Lagrange’s 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. 21. Lagrange’s Inverse Interpolation In the Lagrange’s 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 Lagrange’s 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. 22. Lagrange’s Inverse Interpolation In the Lagrange’s 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 Lagrange’s 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 − yn−1) (yn − y0)(yn − y1) . . . (yn − yn−1) xn Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  23. 23. Lagrange’s Inverse Interpolation In the Lagrange’s 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 Lagrange’s 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 − yn−1) (yn − y0)(yn − y1) . . . (yn − yn−1) xn This relation is referred as Lagrange’s inverse interpolation formula. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  24. 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 Lagrange’s interpolation formula. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  25. 25. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  26. 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. 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 Lagrange’s interpolation formula, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  28. 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 Lagrange’s 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. 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 Lagrange’s 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. 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 Lagrange’s 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. 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 Lagrange’s 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. 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 Lagrange’s 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. 33. Example Ex. Compute f(0.4) for the table below by the Lagrange’s 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. 34. Example Ex. Using Lagrange’s formula, find 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. 35. Example Ex. Using Lagrange’s formula, find 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. 36. Example Ex. Using Lagrange’s formula, express the function 3x2 + x + 1 (x − 1)(x − 2)(x − 3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  37. 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. 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. 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. 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 Lagrange’s interpolation formula, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  41. 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 Lagrange’s 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. 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 Lagrange’s 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. 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 Lagrange’s 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. 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 Lagrange’s 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. 45. Example Thus 3x2 + x + 1 (x − 1)(x − 2)(x − 3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  46. 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. 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 define the error of interpolation or truncation error as Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  48. 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 define 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)(ξ) where min(x0, x1, . . . , xn, x) < ξ < min(x0, x1, . . . , xn, x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  49. 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 define 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)(ξ) where min(x0, x1, . . . , xn, x) < ξ < min(x0, x1, . . . , xn, x) since, ξ is an unknown, it is difficult to find the value of error. However, we can find a bound of the error. The bound of the error is obtained as Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  50. 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 define 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)(ξ) where min(x0, x1, . . . , xn, x) < ξ < min(x0, x1, . . . , xn, x) since, ξ is an unknown, it is difficult to find the value of error. However, we can find 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 a≤ξ≤b |f(n+1)(ξ)| Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  51. 51. Example Ex. Using the data sin(0.1) = 0.09983 and sin(0.2) = 0.19867, find 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. 52. Lagrange’s 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. 53. Lagrange’s 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. 54. Lagrange’s 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 Pk−1(x). Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  55. 55. Lagrange’s 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 Pk−1(x). That means we need to calculate entirely new polynomial. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  56. 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 − xn−1 are not necessarily equally spaced. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  57. 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 − xn−1 are not necessarily equally spaced. Then the first divided difference of f for the arguments x0, x1, . . . , xn are defined by , Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  58. 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 − xn−1 are not necessarily equally spaced. Then the first divided difference of f for the arguments x0, x1, . . . , xn are defined by , f(x0, x1) = f(x1) − f(x0) x1 − x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  59. 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 − xn−1 are not necessarily equally spaced. Then the first divided difference of f for the arguments x0, x1, . . . , xn are defined 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. 60. Divided Difference The second divided difference of f for three arguments x0, x1, x2 is defined by f(x0, x1, x2) = f(x1, x2) − f(x0, x1) x2 − x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  61. 61. Divided Difference The second divided difference of f for three arguments x0, x1, x2 is defined by f(x0, x1, x2) = f(x1, x2) − f(x0, x1) x2 − x0 and similarly the divided difference of order n is defined by f(x0, x1, . . . , xn) = f(x1, x2, . . . , xn) − f(x0, x1, . . . , xn−1) xn − x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  62. 62. Divided Difference Properties: The divided differences are symmetrical in all their arguments; that is, the value of any divided difference is independent of the order of the arguments. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  63. 63. Divided Difference Properties: The divided differences are symmetrical in all their arguments; that is, the value of any divided difference is independent of the order of the arguments. The divided difference operator is linear. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  64. 64. Divided Difference Properties: The divided differences are symmetrical in all their arguments; that is, the value of any divided difference is independent of the order of the arguments. The divided difference operator is linear. The nth order divided differences of a polynomial of degree n are constant, equal to the coefficient of xn. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  65. 65. Newton’s 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. 66. Newton’s 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. Newton’s general interpolation formula is one such formula and terms in it are called divided differences. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  67. 67. Newton’s 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. Newton’s general interpolation formula is one such formula and terms in it are called divided differences. 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 − xn−1 are not necessarily equally spaced. By the definition of divided difference, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  68. 68. Newton’s 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. Newton’s general interpolation formula is one such formula and terms in it are called divided differences. 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 − xn−1 are not necessarily equally spaced. By the definition of divided difference, f(x, x0) = f(x) − f(x0) x − x0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  69. 69. Newton’s 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. Newton’s general interpolation formula is one such formula and terms in it are called divided differences. 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 − xn−1 are not necessarily equally spaced. By the definition of divided difference, 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. 70. Newton’s 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. 71. Newton’s 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. 72. Newton’s 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. 73. Newton’s 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, ..., xn−1) = f(x0, x1, ..., xn) + (x − xn)f(x, x0, x1, ..., xn) − −(4) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  74. 74. Newton’s Divided Difference Interpolation multiplying equation (2) by (x − x0) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  75. 75. Newton’s 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. 76. Newton’s Divided Difference Interpolation multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1) and so on, and finally the last term (4) by (x − x0) (x − x1) ... (x − xn−1) and Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  77. 77. Newton’s Divided Difference Interpolation multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1) and so on, and finally the last term (4) by (x − x0) (x − x1) ... (x − xn−1) and adding (1), (2) , (3) up to (4) we obtain Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  78. 78. Newton’s Divided Difference Interpolation multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1) and so on, and finally the last term (4) by (x − x0) (x − x1) ... (x − xn−1) 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 − xn−1) f (x0, x1, ..., xn) This formula is called Newton’s divided difference formula. Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  79. 79. Newton’s Divided Difference Interpolation The divided difference upto third order x y 1stdiv.diff. 2nddiv.diff. 3rddiv.diff. 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. 80. Example Ex. Obtain the divided difference table for the data: x -1 0 2 3 y -8 3 1 12 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  81. 81. Example Sol. We have the following divided difference 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. 82. Example Ex. Find f(x) as a polynomial in x for the following data by Newtons divided difference formula: x -4 -1 0 2 5 f(x) 1245 33 5 9 1335 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  83. 83. Example Sol. We have the following divided difference 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. 84. Example The Newtons divided difference formula gives: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  85. 85. Example The Newtons divided difference formula gives: f(x) = f(x0) + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  86. 86. Example The Newtons divided difference 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. 87. Example The Newtons divided difference 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. 88. Example The Newtons divided difference 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. 89. Example The Newtons divided difference 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. 90. Example The Newtons divided difference 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. 91. Example The Newtons divided difference 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. 92. Example Ex. Find f(x) as a polynomial in x for the following data by Newtons divided difference 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. 93. Example Sol. We form the divided difference 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. 94. Example Since, the fourth order differences are zeros, the data represents a third degree polynomial. Newtons divided difference formula gives the polynomial as f(x) = f(x0) + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  95. 95. Example Since, the fourth order differences are zeros, the data represents a third degree polynomial. Newtons divided difference 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. 96. Example Since, the fourth order differences are zeros, the data represents a third degree polynomial. Newtons divided difference 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. 97. Example Since, the fourth order differences are zeros, the data represents a third degree polynomial. Newtons divided difference 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. 98. Example Since, the fourth order differences are zeros, the data represents a third degree polynomial. Newtons divided difference 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. 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. 100. Example Sol. Divided difference table: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  101. 101. Example Sol. Divided difference table: By Newton’s divided difference 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. 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 fitting. 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. 103. Spline Interpolation f(x) = (xi+1 − x)3Mi 6h + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  104. 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. 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. 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. 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 Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 2yi + yi+1), Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  108. 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 Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 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. 110. Example Sol. Since points are equispaced with h = 1 and n = 3, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  111. 111. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 2yi + yi+1), i = 1, 2 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  112. 112. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  113. 113. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 ∴ for i = 1, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  114. 114. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 115. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 116. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 117. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 118. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 119. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 120. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 121. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 122. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 123. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 124. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 125. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 126. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 127. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 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. 129. Example Sol. Since points are equispaced with h = 1 and n = 3, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  130. 130. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 2yi + yi+1), i = 1, 2 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  131. 131. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  132. 132. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 2yi + yi+1), i = 1, 2 also M0 = 0 , M3 = 0 ∴ for i = 1, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  133. 133. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 134. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 135. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 136. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 137. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 138. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 139. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 140. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 141. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 142. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 143. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 144. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 145. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 146. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 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. 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. 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 x→0+ f(x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  149. 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 x→0+ f(x) = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  150. 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 x→0+ f(x) = 0 = lim x→0− 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. 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 x→0+ f(x) = 0 = lim x→0− 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. 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 x→0+ f(x) = 0 = lim x→0− 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 x→0+ f (x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  153. 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 x→0+ f(x) = 0 = lim x→0− 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 x→0+ f (x) = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  154. 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 x→0+ f(x) = 0 = lim x→0− 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 x→0+ f (x) = 0 = lim x→0− f (x) therefore, the function f (x) is continuous on (−1, 1). f (x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  155. 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 x→0+ f(x) = 0 = lim x→0− 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 x→0+ f (x) = 0 = lim x→0− f (x) therefore, the function f (x) is continuous on (−1, 1). f (x) = 30x−6, −1 ≤ x ≤ 0 = −30x − 6, 0 ≤ x ≤ 1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
  156. 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 x→0+ f(x) = 0 = lim x→0− 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 x→0+ f (x) = 0 = lim x→0− f (x) therefore, the function f (x) is continuous on (−1, 1). f (x) = 30x−6, −1 ≤ x ≤ 0 = −30x − 6, 0 ≤ x ≤ 1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals

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