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- 1. Numerical Methods - Interpolation Unequal Intervals Dr. N. B. Vyas Department of Mathematics, Atmiya Institute of Tech. and Science, Rajkot (Guj.) niravbvyas@gmail.com Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 2. Interpolation To 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Lagrange’s Interpolation This polynomial is given by the following formula: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 26. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 27. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 By Lagrange’s interpolation formula, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 28. Example Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156 y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490 By 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. 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. 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. 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. 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. 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. 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. 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. 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. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 38. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 39. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 40. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, y1 = 15 and y2 = 31 By Lagrange’s interpolation formula, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 41. Example Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3 These values are x0 = 1, x1 = 2 and x2 = 3 and y0 = 5, y1 = 15 and y2 = 31 By 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. 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. 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. 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. Example Thus 3x2 + x + 1 (x − 1)(x − 2)(x − 3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 46. Example Thus 3x2 + x + 1 (x − 1)(x − 2)(x − 3) = 2.5(x − 2)(x − 3) − 15(x − 1)(x − 3) + 15.5(x − 1)(x − 2) (x − 1)(x − 2)(x − 3) = 2.5 (x − 1) - 15 (x − 2) + 15.5 (x − 3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 47. Error in Interpolation Error in Interpolation: We assume that f(x) has continuous derivatives of order upto n + 1 for all x ∈ (a, b). Since, f(x) is approximated by Pn(x), the results contains errors. We deﬁne the error of interpolation or truncation error as Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 48. Error in Interpolation Error in Interpolation: We assume that f(x) has continuous derivatives of order upto n + 1 for all x ∈ (a, b). Since, f(x) is approximated by Pn(x), the results contains errors. We 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Newton’s Divided Difference Interpolation multiplying equation (2) by (x − x0) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 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. 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. 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. 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. 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. 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. 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. 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. 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. Example The Newtons divided diﬀerence formula gives: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 85. Example The Newtons divided diﬀerence formula gives: f(x) = f(x0) + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Example Ex. Find the missing term in the following table: x 0 1 2 3 4 y 1 3 9 - 81 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 100. Example Sol. Divided diﬀerence table: Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 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. 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. Spline Interpolation f(x) = (xi+1 − x)3Mi 6h + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 104. Spline Interpolation f(x) = (xi+1 − x)3Mi 6h + (x − xi)3Mi+1 6h Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 105. Spline Interpolation f(x) = (xi+1 − x)3Mi 6h + (x − xi)3Mi+1 6h + (xi+1 − x) h yi − h2 6 Mi + Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 106. Spline Interpolation f(x) = (xi+1 − x)3Mi 6h + (x − xi)3Mi+1 6h + (xi+1 − x) h yi − h2 6 Mi + (x − xi) h yi+1 − h2 6 Mi+1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 107. Spline Interpolation f(x) = (xi+1 − x)3Mi 6h + (x − xi)3Mi+1 6h + (xi+1 − x) h yi − h2 6 Mi + (x − xi) h yi+1 − h2 6 Mi+1 where Mi−1 + 4Mi + Mi+1 = 6 h2 (yi−1 − 2yi + yi+1), Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 108. Spline Interpolation f(x) = (xi+1 − x)3Mi 6h + (x − xi)3Mi+1 6h + (xi+1 − x) h yi − h2 6 Mi + (x − xi) h yi+1 − h2 6 Mi+1 where 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. Example Ex. Obtain cubic spline for the following data: x 0 1 2 3 y 2 -6 -8 2 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 110. Example Sol. Since points are equispaced with h = 1 and n = 3, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 111. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Example Ex. The following values of x and y are given: x 1 2 3 4 y 1 2 5 11 Find the cubic splines and evaluate y(1.5) and y (3) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 129. Example Sol. Since points are equispaced with h = 1 and n = 3, Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 130. Example Sol. Since points are equispaced with h = 1 and n = 3, the cubic spline can be determined from 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Example Ex. Find whether the following functions are cubic splines ? 1. f(x) = 5x3 − 3x2, −1 ≤ x ≤ 0 = −5x3 − 3x2, 0 ≤ x ≤ 1 2. f(x) = −2x3 − x2, −1 ≤ x ≤ 0 = 2x3 + 3x2, 0 ≤ x ≤ 1 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 148. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x→0+ f(x) Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 149. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim x→0+ f(x) = 0 Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
- 150. Example Sol. In both the examples, f(x) is a cubic polynomial in both intervals (1, 0) and (0, 1). 1. We have lim 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. 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. 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. 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. 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. 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. 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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