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- 1. NUMERICAL DIFFERENTIATION AND INTEGRATION ENGR 351 Numerical Methods for Engineers Southern Illinois University Carbondale College of Engineering Dr. L.R. Chevalier Dr. B.A. DeVantier
- 2. Copyright © 2003 by Lizette R. Chevalier Permission is granted to students at Southern Illinois University at Carbondale to make one copy of this material for use in the class ENGR 351, Numerical Methods for Engineers. No other permission is granted. All other rights are reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the copyright owner.
- 3. Specific Study Objectives • Understand the derivation of the Newton-Cotes formulas • Recognize that the trapezoidal and Simpson’s 1/3 and 3/8 rules represent the areas of 1st, 2nd, and 3rd order polynomials • Be able to choose the “best” among these formulas for any particular problem
- 4. Specific Study Objectives • Recognize the difference between open and closed integration formulas • Understand the theoretical basis of Richardson extrapolation and how it is applied in the Romberg integration algorithm and for numerical differentiation
- 5. Specific Study Objectives • Recognize why both Romberg integration and Gauss quadrature have utility when integrating equations (as opposed to tabular or discrete data). • Understand basic finite difference approximations • Understand the application of high-accuracy numerical-differentiation. • Recognize data error on the processes of integration and differentiation.
- 6. Numerical Differentiation and Integration • Calculus is the mathematics of change. • Engineers must continuously deal with systems and processes that change, making calculus an essential tool of our profession. • At the heart of calculus are the related mathematical concepts of differentiation and integration.
- 7. Differentiation • Dictionary definition of differentiate - “to mark off by differences, distinguish; ..to perceive the difference in or between” • Mathematical definition of derivative - rate of change of a dependent variable with respect to an independent variable ( ) ( ) x xfxxf x y ii ∆ −∆+ = ∆ ∆
- 8. ( ) ( )∆ ∆ ∆ ∆ y x f x x f x x i i = + − ( )f xi ( )f x xi + ∆ ∆y ∆x f(x) x
- 9. Integration • The inverse process of differentiation • Dictionary definition of integrate - “to bring together, as parts, into a whole; to unite; to indicate the total amount” • Mathematically, it is the total value or summation of f(x)dx over a range of x. In fact the integration symbol is actually a stylized capital S intended to signify the connection between integration and summation.
- 10. f(x) x ( )I f x dx a b = ∫
- 11. Mathematical Background d dx x d dx e d dx x d dx a d dx x d dx x d dx x if u and v are functions of x d dx u d dx uv x x n n sin ? ? cos ? ? tan ? ? ln ? ? ( ) ? = = = = = = = = =
- 12. udv u du a dx dx x e dx n bx ax = = = = = ∫ ∫ ∫ ∫ ∫ ? ? ? ? ? Mathematical Background
- 13. Overview • Newton-Cotes Integration Formulas • Trapezoidal rule • Simpson’s Rules • Unequal Segments • Open Integration • Integration of Equations • Romberg Integration • Gauss Quadrature • Improper Integrals
- 14. Overview • Numerical Differentiation • High accuracy formulas • Richardson’s extrapolation • Unequal spaced data • Uncertain data • Applied problems
- 15. Newton-Cotes Integration • Common numerical integration scheme • Based on the strategy of replacing a complicated function or tabulated data with some approximating function that is easy to integrate ( ) ( ) ( ) I f x dx f x dx f x a a x a x a b n a b n n n = ≅ = + + + ∫ ∫ 0 1 ....
- 16. Newton-Cotes Integration • Common numerical integration scheme • Based on the strategy of replacing a complicated function or tabulated data with some approximating function that is easy to integrate ( ) ( ) ( ) I f x dx f x dx f x a a x a x a b n a b n n n = ≅ = + + + ∫ ∫ 0 1 .... fn(x) is an nth order polynomial
- 17. 0 1 2 3 4 5 0 5 10 x f(x) 0 1 2 3 4 5 0 5 10 x f(x) The approximation of an integral by the area under - a first order polynomial - a second order polynomial
- 18. We can also approximated the integral by using a series of polynomials applied piece wise. 0 1 2 3 4 5 0 5 10 x f(x) 0 1 2 3 4 5 0 5 10 x f(x) The approximation of an integral by the area under - a first order polynomial - a second order polynomial
- 19. 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 x f(x) An approximation of an integral by the area under straight line segments.
- 20. 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 x f(x) An approximation of an integral by the area under straight line segments.
- 21. Newton-Cotes Formulas • Closed form - data is at the beginning and end of the limits of integration • Open form - integration limits extend beyond the range of data. 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 x f(x) 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 x f(x)
- 22. Trapezoidal Rule • First of the Newton-Cotes closed integration formulas • Corresponds to the case where the polynomial is a first order ( ) ( ) ( ) I f x dx f x dx f x a a x a b a b n = ≅ = + ∫ ∫ 1 0 1
- 23. ( ) ( ) ( ) I f x dx f x dx f x a a x a b a b n = ≅ = + ∫ ∫ 1 0 1 A straight line can be represented as: ( ) ( ) ( ) ( ) ( )f x f a f b f a b a x a1 = + − − − Trapezoidal Rule
- 24. ( ) ( ) ( ) ( ) ( ) ( ) I f x dx f x dx f a f b f a b a x a dx a b a b a b = ≅ = + − − − ∫ ∫ ∫ 1 Integrate this equation. Results in the trape zo idalrule . ( ) ( ) ( )I b a f a f b ≅ − + 2 Trapezoidal Rule
- 25. ( ) ( ) ( ) 2 bfaf abI + −≅ Recall the formula for computing the area of a trapezoid: height x (average of the bases) height base base Trapezoidal Rule
- 26. The concept is the same but the trapezoid is on its side. height base base widthheight height ( ) ( ) ( ) 2 bfaf abI + −≅ Trapezoidal Rule
- 27. Error of the Trapezoidal Rule ( )( )E f b a where a b t = − − < < 1 12 3 '' ξ ξ This indicates that is the function being integrated is linear, the trapezoidal rule will be exact. Otherwise, for section with second and higher order derivatives (that is with curvature) error can occur. A reasonable estimate of x is the average value of b and a
- 28. Multiple Application of the Trapezoidal Rule • Improve the accuracy by dividing the integration interval into a number of smaller segments • Apply the method to each segment • Resulting equations are called multiple-application or composite integration formulas
- 29. ( ) ( ) ( ) ( ) ( ) ( ) I f x dx f x dx f x dx I h f x f x h f x f x h f x f x x x x x x x n n n n = + + + ≅ + + + + + ∫ ∫ ∫ − − ( ) ( ) ( ) 0 1 1 2 1 0 1 1 2 1 2 2 2 where there are n+ 1 equally spaced base points. Multiple Application of the Trapezoidal Rule
- 30. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) I f x dx f x dx f x dx I h f x f x h f x f x h f x f x I b a f x f x f x n x x x x x x n n i n i n n n = + + + ≅ + + + + + ≅ − + + ∫ ∫ ∫ ∑ − − = − ( ) ( ) ( ) 0 1 1 2 1 0 1 1 2 1 0 1 1 2 2 2 2 2 We can group terms to express a general form } } width average height Multiple Application of the Trapezoidal Rule
- 31. ( ) ( ) ( ) ( ) I b a f x f x f x n i n i n ≅ − + + = − ∑0 1 1 2 2 } }width average height The average height represents a weighted average of the function values Note that the interior points are given twice the weight of the two end points ( ) E b a n fa = − − 3 2 12 '' Multiple Application of the Trapezoidal Rule
- 32. Example Evaluate the following integral using the trapezoidal rule and h = 0.1 ( ) ( ) ( ) ( ) I b a f x f x f x n i n i n ≅ − + + = − ∑0 1 1 2 2 I e dxx = ∫ 2 1 1 6. h b a n = − 0 5 10 15 20 25 30 0 0.3 0.6 0.9 1.2 1.5 1.8 x f(x)
- 33. Simpson’s 1/3 Rule • Corresponds to the case where the function is a second order polynomial ( ) ( ) ( ) I f x dx f x dx f x a a x a x a b a b n = ≅ = + + ∫ ∫ 2 0 1 2 2
- 34. Simpson’s 1/3 Rule • Designate a and b as x0 and x2, and estimate f2(x) as a second order Lagrange polynomial ( ) ( ) ( )( ) ( )( ) ( ) I f x dx f x dx x x x x x x x x f x dx a b a b x x = ≅ = − − − − + ∫ ∫ ∫ 2 1 2 0 1 0 2 0 0 2 .......
- 35. Simpson’s 1/3 Rule • After integration and algebraic manipulation, we get the following equations ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) I h f x f x f x b a f x f x f x ≅ + + ≅ − + + 3 4 4 6 0 1 2 0 1 2 } } width average height
- 36. Error ( )( )E f b a where a b t = − − < < 1 12 3 '' ξ ξ Single application of Trapezoidal Rule. Single application of Simpson’s 1/3 Rule ( )( )E f b at = − − 1 2880 4 5( ) ξ
- 37. Multiple Application of Simpson’s 1/3 Rule ( ) ( ) ( ) ( ) ( ) ( ) ( ) I f x dx f x dx f x dx I b a f x f x f x f x n E b a n f x x x x x x i j n j n i n a n n = + + + ≅ − + + + = − − ∫ ∫ ∫ ∑∑ − = − = − ( ) ( ) ( ) , , .., , .. 0 1 1 2 1 0 2 4 6 2 1 3 5 1 5 4 4 4 2 3 180
- 38. ( ) ( ) ( ) ( ) ( ) I b a f x f x f x f x n i j n j n i n ≅ − + + + = − = − ∑∑0 2 4 6 2 1 3 5 1 4 2 3 , , .., , .. The odd points represent the middle term for each application. Hence carry the weight 4. The even points are common to adjacent applications and are counted twice. f(x) x i=1 (odd) weight of 4 i=2 (even) weight of 2
- 39. Simpson’s 3/8 Rule • Corresponds to the case where the function is a third order polynomial ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]3210 3 3 2 210 3 33 8 3 xfxfxfxf h I xaxaxaaxf dxxfdxxfI n b a b a +++≅ +++= ≅= ∫∫
- 40. Integration of Unequal Segments • Experimental and field study data is often unevenly spaced • In previous equations we grouped the term (i.e. hi) which represented segment width. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) I b a f x f x f x n I h f x f x h f x f x h f x f x i n i n n n ≅ − + + ≅ + + + + + = − − ∑0 1 1 0 1 1 2 1 2 2 2 2 2
- 41. Integration of Unequal Segments • We should also consider alternately using higher order equations if we can find data in consecutively even segments trapezoidal rule
- 42. Integration of Unequal Segments trapezoidal rule 1/3 rule • We should also consider alternately using higher order equations if we can find data in consecutively even segments
- 43. Integration of Unequal Segments trapezoidal rule 1/3 rule 3/8 rule • We should also consider alternately using higher order equations if we can find data in consecutively even segments
- 44. Integration of Unequal Segments trapezoidal rule 1/3 rule 3/8 rule trapezoidal rule • We should also consider alternately using higher order equations if we can find data in consecutively even segments
- 45. Example Integrate the following using the trapezoidal rule, Simpson’s 1/3 Rule, a multiple application of the trapezoidal rule with n=2 and Simpson’s 3/8 Rule. Compare results with the analytical solution. xe dxx2 0 4 ∫ 0 2000 4000 6000 8000 10000 12000 14000 0 1 2 3 4 x f(x)
- 46. Integration of Equations • Integration of analytical as opposed to tabular functions • Romberg Integration • Richardson’s Extrapolation • Romberg Integration Algorithm • Gauss Quadrature • Improper Integrals
- 47. Richardson’s Extrapolation • Use two estimates of an integral to compute a third more accurate approximation • The estimate and error associated with a multiple application trapezoidal rule can be represented generally as: • I = I(h) + E(h) • where I is the exact value of the integral • I(h) is the approximation from an n-segment application • E(h) is the truncation error • h is the step size (b-a)/n
- 48. Make two separate estimates using step sizes of h1 and h2 . I(h1) + E(h1) = I(h2) + E(h2) Recall the error of the multiple-application of the trapezoidal rule E b a h f= − − 12 2 '' Assume that is constant regardless of the step sizef '' ( ) ( ) E h E h h h 1 2 1 2 2 2 ≅
- 49. ( ) ( ) ( ) ( ) E h E h h h E h E h h h 1 2 1 2 2 2 1 2 1 2 2 ≅ ≅ Substitute into previous equation: I(h1) + E(h1) = I(h2) + E(h2) ( ) ( ) ( ) E h I h I h h h 2 1 2 1 2 2 1 = − −
- 50. Thus we have developed an estimate of the truncation error in terms of the integral estimates and their step sizes. This estimate can then be substituted into: I = I(h2) + E(h2) to yield an improved estimate of the integral: ( ) ( ) ( )[ ]I I h h h I h I h≅ + − −2 1 2 2 2 1 1 1 ( ) ( ) ( ) E h I h I h h h 2 1 2 1 2 2 1 = − −
- 51. ( ) ( ) ( )[ ]I I h h h I h I h≅ + − −2 1 2 2 2 1 1 1 What is the equation for the special case where the interval is halved? i.e. h2 = h1 / 2
- 52. ( ) ( ) ( )[ ] ( ) ( ) h h h h I I h I h I h collecting terms I I h I h 1 2 2 2 2 2 2 1 2 1 2 2 1 2 1 4 3 1 3 = = ≅ + − − ≅ −
- 53. Example Use Richardson’s extrapolation to evaluate: xe dxx2 0 4 ∫ 0 2000 4000 6000 8000 10000 12000 14000 0 1 2 3 4 x f(x)
- 54. ( ) ( ) ( ) ( ) ( ) ( ) I I h I h I I h I h I I h I h I I I m l m l j k k j k j k k ≅ − ≅ − ≅ − ≅ − − − + − − − 4 3 1 3 16 15 1 15 64 63 1 63 4 4 1 2 1 1 1 1 1 1, , , We can continue to improve the estimate by successive halving of the step size to yield a general formula: k = 2; j = 1 Romberg Integration Note: the subscripts m and lrefer to m o re and le ss accurate estimates
- 55. Gauss Quadrature f(x) x f(x) x Extend the area under the straight line
- 56. Method of Undetermined Coefficients Recall the trapezoidal rule ( ) ( ) ( ) I b a f a f b ≅ − + 2 This can also be expressed as ( ) ( )I c f a c f b≅ +0 1 where the c’s are constant Before analyzing this method, answer this question. What are two functions that should be evaluated exactly by the trapezoidal rule?
- 57. The two cases that should be evaluated exactly by the trapezoidal rule: 1) y = constant 2) a straight line f(x) x y = 1 (b-a)/2-(b-a)/2 f(x) x y = x (b-a)/2 -(b-a)/2
- 58. Thus, the following equalities should hold. ( ) ( ) ( ) ( ) ( ) ( ) I c f a c f b c c dx c b a c b a xdx b a b a b a b a ≅ + + ≅ − − + − ≅ − − − − − − ∫ ∫ 0 1 0 1 2 2 0 1 2 2 1 2 2 FOR y=1 since f(a) = f(b) =1 FOR y =x since f(a) = x =-(b-a)/2 and f(b) = x =(b-a)/2
- 59. Evaluating both integrals c c b a c b a c b 0 1 0 1 2 1 2 0 + = − − − + − = For y = 1 For y = x Now we have two equations and two unknowns, c0 and c1. Solving simultaneously, we get : c0 = c1 = (b-a)/2 Substitute this back into: ( ) ( )I c f a c f b≅ +0 1
- 60. ( ) ( ) ( ) I b a f a f b ≅ − + 2 We get the equivalent of the trapezoidal rule. DERIVATION OF THE TWO-POINT GAUSS-LEGENDRE FORMULA ( ) ( )I c f x c f x≅ +0 0 1 1 Lets raise the level of sophistication by: - considering two points between -1 and 1 - i.e. “open integration”
- 61. f(x) x-1 x0 x1 1 Previously ,we assumed that the equation fit the integrals of a constant and linear function. Extend the reasoning by assuming that it also fits the integral of a parabolic and a cubic function.
- 62. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) c f x c f x dx c f x c f x xdx c f x c f x x dx c f x c f x x dx 0 0 1 1 1 1 0 0 1 1 1 1 0 0 1 1 2 1 1 0 0 1 1 3 1 1 1 2 0 2 3 0 + = = + = = + = = + = = − − − − ∫ ∫ ∫ ∫ / We now have four equations and four unknowns c0 c1 x0 and x1 What equations are you solving?
- 63. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) c f x c f x dx c c c f x c f x xdx c x c x c f x c f x x dx c x c x c f x c f x x dx c x c x 0 0 1 1 1 1 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 2 1 1 0 0 2 1 1 2 0 0 1 1 3 1 1 0 0 3 1 1 3 1 2 1 1 2 0 0 2 3 2 3 0 0 + = = + = + = = + = + = = + = + = = + = − − − − ∫ ∫ ∫ ∫ ` / / Solve these equations simultaneously f(xi) is either 1, xi, xi 2 or xi 3
- 64. c c x x I f f 0 1 0 1 1 1 3 1 3 1 3 1 3 = = = − = ≅ − + This results in the following The interesting result is that the integral can be estimated by the simple addition of the function values at −1 3 1 3 and
- 65. A simple change in variables can be use to translate other limits. Assume that the new variable xd is related to the original variable x in a linear fashion. x = a0 + a1xd Let the lower limit x = a correspond to xd = -1 and the upper limit x=b correspond to xd=1 a = a0 + a1(-1) b = a0 + a1(1) What if we aren’t integrating from –1 to 1?
- 66. a = a0 + a1(-1) b = a0 + a1(1) SOLVE THESE EQUATIONS SIMULTANEOUSLY a b a a b a 0 1 2 2 = + = − ( ) ( )x a a x b a b a x d d = + = + + − 0 1 2 substitute
- 67. ( ) ( )x a a x b a b a x dx b a dx d d d = + = + + − = − 0 1 2 2 These equations are substituted for x and dx respectively. Let’s do an example to appreciate the theory behind this numerical method.
- 68. Example Estimate the following using two-point Gauss Legendre: xe dxx2 0 4 ∫ 0 2000 4000 6000 8000 10000 12000 14000 0 1 2 3 4 x f(x)
- 69. Higher-Point Formulas ( ) ( ) ( )I c f x c f x c f xn n≅ + + + − −0 0 1 1 1 1 For two point, we determined that c0 = c1= 1 For three point: c0 = 0.556 x0=-0.775 c1= 0.889 x1=0.0 c2= 0.556 x2=0.775
- 70. Higher-Point Formulas Your text goes on to provide additional weighting factors (ci’s) and function arguments (xi’s) in Table 8.5 p. 593
- 71. Numerical Differentiation • Forward finite divided difference • Backward finite divided difference • Center finite divided difference • All based on the Taylor Series ( ) ( ) ( ) ( ) ......... !2 '' ' 2 1 +++=+ h xf hxfxfxf i iii
- 72. Forward Finite Difference ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )21 1 2 1 2 '' ' ' ......... !2 '' ' hOh xf h xfxf xf hO h xfxf xf h xf hxfxfxf iii i ii i i iii +− − = + − = +++= + + +
- 73. Forward Divided Difference ( ) ( ) ( ) ( ) ( )f x f x f x x x O x x f h O hi i i i i i i i ' = − − + − = ++ + + 1 1 1 ∆ f(x) x (xi, yi) What is derivative at this point?
- 74. Forward Divided Difference ( ) ( ) ( ) ( ) ( )f x f x f x x x O x x f h O hi i i i i i i i ' = − − + − = ++ + + 1 1 1 ∆ f(x) x (xi, yi) (x i+1,y i+1) Determine a second point base on Dx (h)
- 75. Forward Divided Difference ( ) ( ) ( ) ( ) ( )f x f x f x x x O x x f h O hi i i i i i i i ' = − − + − = ++ + + 1 1 1 ∆ f(x) x (xi, yi) (x i+1,y i+1) estimateHow does this compare to the actual first derivative at xi?
- 76. Forward Divided Difference ( ) ( ) ( ) ( ) ( )f x f x f x x x O x x f h O hi i i i i i i i ' = − − + − = ++ + + 1 1 1 ∆ f(x) x (xi, yi) (x i+1,y i+1) estimate actual
- 77. ( ) ( )f x f h O hi i ' = + ∆ first forward divided difference Error is proportional to the step size O(h2 ) error is proportional to the square of the step size O(h3 ) error is proportional to the cube of the step size Forward Divided Difference
- 78. f(x) x estimate actual (xi,yi) (xi-1,yi-1)
- 79. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) f x f x f x h f x h f x f x f x h f x h f x f x f x h f h i i i i i i i i i i i i + − − = + + + = − + + = − = ∇ 1 2 1 2 1 2 2 ' '' ! ...... ' '' ! ..... ' Backward Difference Approximation of the First Derivative Expand the Taylor series backwards The error is still O(h)
- 80. Centered Difference Approximation of the First Derivative Subtract backward difference approximation from forward Taylor series expansion ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )211 2 11 1 1 2 1 2 ' 6 ''' '2 ' .... !2 '' ' hO h xfxf xf h xf hxfxfxf xx xfxf xf h xf hxfxfxf ii i i iii ii ii i i iii − − = +++= − − = +++= −+ −− + + +
- 81. f(x) x estimate actual (xi,yi) (xi-1,yi-1) (xi+1,yi+1) ( ) ( ) ( ) ( )211 2 ' hO h xfxf xf ii i − − = −+
- 82. f(x) x f(x) x f(x) x f(x) x true derivative forward finite divided difference approx. backward finite divided difference approx. centered finite divided difference approx.
- 83. Numerical Differentiation • You should be familiar with the following Tables in your text • Table 7.1: Common Finite Difference Formulas • Table 7.2: Higher order finite difference formulas
- 84. Richardson Extrapolation • Two ways to improve derivative estimates ¤ decrease step size ¤ use a higher order formula that employs more points • Third approach, based on Richardson extrapolation, uses two derivatives estimates to compute a third, more accurate approximation
- 85. Richardson Extrapolation ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) I I h h h I h I h Special case where h h I I h I h In a similar fashion D D h D h ≅ + − − = ≅ − ≅ − 2 1 2 2 2 1 2 1 2 1 2 1 1 1 2 4 3 1 3 4 3 1 3 For a centered difference approximation with O(h2 ) the application of this formula will yield a new derivative estimate of O(h4 )
- 86. Example Given the following function, use Richardson’s extrapolation to determine the derivative at 0.5. f(x) = -0.1x4 - 0.15x3 - 0.5x2 - 0.25x +1.2 Note: f(0) = 1.2 f(0.25) =1.1035 f(0.75) = 0.636 f(1) = 0.2
- 87. Derivatives of Unequally Spaced Data • Common in data from experiments or field studies • Fit a second order Lagrange interpolating polynomial to each set of three adjacent points, since this polynomial does not require that the points be equi- spaced • Differentiate analytically ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) f x f x x x x x x x x f x x x x x x x x f x x x x x x x x i i i i i i i i i i i i i i i i i i i i i ' = − − − − + − − − − + − − − − − + − − + − + − + + − + + − 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2
- 88. Derivative and Integral Estimates for Data with Errors • In addition to unequal spacing, the other problem related to differentiating empirical data is measurement error • Differentiation amplifies error • Integration tends to be more forgiving • Primary approach for determining derivatives of imprecise data is to use least squares regression to fit a smooth, differentiable function to the data • In absence of other information, a lower order polynomial regression is a good first choice
- 89. 0 5 0 1 00 1 5 0 200 25 0 0 5 1 0 1 5 t y 0 1 0 20 30 0 1 0 t dy/dt 0 5 0 1 00 1 5 0 200 25 0 0 5 1 0 1 5 t y 0 1 0 20 30 40 0 1 0 t dy/dt
- 90. Specific Study Objectives • Understand the derivation of the Newton-Cotes formulas • Recognize that the trapezoidal and Simpson’s 1/3 and 3/8 rules represent the areas of 1st, 2nd, and 3rd order polynomials • Be able to choose the “best” among these formulas for any particular problem
- 91. Specific Study Objectives • Recognize the difference between open and closed integration formulas • Understand the theoretical basis of Richardson extrapolation and how it is applied in the Romberg integration algorithm and for numerical differentiation
- 92. Specific Study Objectives • Recognize why both Romberg integration and Gauss quadrature have utility when integrating equations (as opposed to tabular or discrete data). • Understand the application of high- accuracy numerical-differentiation. • Recognize data error on the processes of integration and differentiation.
- 93. …..end of lecture

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