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# 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