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# Es272 ch6

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### Es272 ch6

1. 1. Part 6: Numerical Integration and Differentiation – – – – – – – – Newton-Cotes Integration Formulas Trapezoidal Rule Simpson’s Rules Integration with Unequal Segments Integration of Equations Romberg Integration Gauss Quadrature Numerical Differentiation
2. 2. Newton-Cotes Integration Formula  Replacing the complicated integrand function or tabulated data by an approximating function such as a polynomial. f(x) f(x) Replacing by a straight line x Replacing by a parabola x  These formulas are applicable to both closed forms (i.e., data points at the beginning and end of the integration limits are known) and open forms (otherwise).  These formulas are applied for equally spaced intervals.
3. 3. Trapezoidal Rule  Complicated integrand function is replaced by a first order polynomial (straight line). b I f(x) b f ( x ) dx a f 1 ( x ) dx a where the first-order polynomial f1 ( x ) f (a ) f (b ) f (a ) b b I f (a ) f (b ) (b a I (b a) a (x a) a f (a ) (x a ) dx a) f (a ) f (b ) 2 Trapezoidal rule b x
4. 4. f(x) f(a)+f(b) 2 =average height f(a) I ( width ) ( average height ) f(b) b a x width  All the Newton-Cotes formulas can be formulated in general form above (only definition of average height changes). Error for trapezoidal rule: b '' f ( x ) dx Et 1 '' f ( )( b 12 Local truncation error a) 3 Ea 1 12 (b 3 '' a) f '' f a (b a) Average of the second derivative between the interval
5. 5. EX: Numerically integrate f ( x) 0 .2 25 x 200 x 2 675 x 3 900 x 4 400 x 5 from a=0 to b=0.8. Find the error. (Note that the exact solution is 1.6405). f (0) 0 .2 f ( 0 .8 ) Et I 0 .8 0 .2 0 . 1728 2 0 . 232 1 . 6405 0 . 232 0 . 1728 1 . 4677 t 89 . 5 % Normally, we don’t have the knowledge of the true error; we can calculate the approximate error. We need to calculate the second derivative of the function: '' f ( x) 400 4050 x 10800 x 2 8000 x 3 0 .8 400 '' f 4050 x 10800 x 0 Ea 12 8000 x 3 60 0 .8 1 2 ( 60 )( 0 . 8 ) 3 0 2 . 56 Et and Ea have the same order and sign
6. 6. Multiple-application trapezoidal rule: We can increase the accuracy of the integration by dividing the interval into many segments and apply the trapezoidal rule to each segment. f(x) For n+1 equally spaced base points, there are n segments of step size: h (b a) n I h 2 x0 h n 1 f ( x0 ) 2 f ( xi ) f ( xn ) i 1 Initial point x1 final point x2 x
7. 7. or n 1 f ( x0 ) I (b 2 f ( xi ) f ( xn ) i 1 a) 2n width average height Error for the multiple-application trapezoidal rule is found by summing individual errors. b Et a 12 n or 3 n '' f ( i) 3 i 1 n Ea b a 12 n 2 '' 3 '' f where f ( '' f i 1 n i ) Convergence: Error is inversely related to the square of n !
8. 8. EX: Use two-segment trapezoidal rule to integrate f ( x) 0 .2 2 675 x f ( 0 .4 ) 25 x 2 . 456 200 x 3 900 x 4 400 x 5 From a=0 to b=0.8. n=2 h=0.4 f (0) I 0 .8 0 .2 0 .2 2 ( 2 . 456 ) 0 . 232 f ( 0 .8 ) 0 . 232 1 . 0688 4 Ea 1 . 6405 1 12 ( 2 ) 2 ( 60 )( 0 . 8 ) 3 t 34 . 9 % 0 . 64 Error is inversely related to the square of n. 1.0688 34.9 3 1.3695 16.5 1.4848 9.5 5 0 . 57173 I 4 1 . 0688 n 2 Et 1.5399 6.1 6 1.5703 4.3 7 1.5887 3.2 8 1.6008 2.4 t
9. 9. Simpson’s Rules f(x) Instead of using a line segment, use higher-order polynomials to connect points increase the accuracy. Simpson’s 1/3 rule x1 x0 A parabola is substituted for the function. Lagrange polynomial forms are used for the replacement. Replacing Lagrange polynomials for the function at three points x0 , x1, and x2 : (x x2 I x0 x1 )( x ( x0 x2 ) x1 )( x 0 (x x 0 )( x ( x2 x 0 )( x 2 x2 ) f ( x0 ) x1 ) x1 ) f ( x2 ) (x x 0 )( x ( x1 x 0 )( x1 x2 ) x2 ) f ( x1 ) dx x2 x
10. 10. After integration and algebraic manipulations 1h I 3 f ( x0 ) 4 f ( x1 ) f x2 Simpson’s 1/3 rule. In another form: I (b f ( x0 ) a) 4 f ( x1 ) f x2 h=(b-a)/2 6 width average height Error for the single segment Simpon’s 1/3 rule: Et (b a) 2880 5 f (4) > It is more accurate than expected as the error is related to the forth order derivative (third order term is meaningful but is zero). > It yields exact results for cubic polynomials even though it is obtained from a parabola.
11. 11. EX: Use single application of Simpson’s 1/3 rule to integrate f ( x) 0 .2 2 675 x f ( 0 .4 ) 2 . 456 25 x 200 x 3 900 x 4 400 x 5 from a=0 to b=0.8. n=2 h=0.4 f (0) I 0 .8 0 .2 0 .2 4 ( 2 . 456 ) 0 . 232 f ( 0 .8 ) 0 . 232 1 . 3675 6 Et 1 . 6405 1 . 3675 0 . 273 t 16 . 6 % 0 .8 f (4) ( x) 21600 f 48000 x (4) f (4) 0 2400 0 .8 Ea (b a) 2880 5 (4) f ( 0 .8 ) 2880 5 ( 2400 ) ( x ) dx 0 . 273 0 Et and Ea are equal because the integrand is a fifth order polynomial
12. 12. Multiple-application Simpson’s 1/3 rule For n segments: h (b a) n Total integral can be represented x2 I xn x4 f ( x ) dx x0 f ( x ) dx ... x2 f ( x ) dx xn 2 Substituting Simpson’s 1/3 formula n 1 f ( x0 ) I (b a) 4 n 2 f ( xi ) 2 i 1, 3 , 5 f ( xi ) i 2,4,6 3n width average height f ( xn )
13. 13. Error for the multiple (n) segment Simpon’s 1/3 rule is calculated by adding error for individual segments. Hence, we get: Et (b a) 180 n 4 5 (4) f Multiple application Simpson’s rule returns very accurate results compare to trapezoidal rule. It is the method of choice for most applications.  As for all Newton-Cotes formulas, the intervals must be equally spaced. Because of the need for three points for applications, the method is limited to odd number of points (even number of segments).
14. 14. EX: Use multiple application Simpson’s 1/3 rule (n=4) to integrate f ( x) 2 675 x f ( 0 .2 ) 1 . 288 3 . 464 f ( 0 .8 ) 0 . 232 3 . 464 ) 2 ( 2 . 456 ) 0 .2 25 x 200 x 3 900 x 4 400 x 5 from a=0 to b=0.8. n=4 h=0.2 f (0) f ( 0 .6 ) I 0 .8 0 .2 4 (1 . 288 0 .2 f ( 0 .4 ) 0 . 232 2 . 456 1 . 6235 12 Et Ea 1 . 640533 1 . 623467 a) 5 180 ( n ) 4 (b (4) f 0 . 017067 ( 0 .8 ) t 1 . 04 % 5 180 ( 4 ) 4 ( 2400 ) 0 . 017067 Small error shows very accurate results are obtained.
15. 15. Simpson’s 3/8 rule Simpson’s 3/8 rule is used when the odd-number of segments are encountered. A third-order polynomial is used for replacing the function. b I b f ( x ) dx f 3 ( x ) dx a a Applying a third order Legendre polynomial to four points gives I 3h f ( x0 ) 8 I (b a) 3 f ( x1 ) f ( x0 ) 3 f ( x2 ) 3 f ( x1 ) 3 f ( x2 ) 8 width f x3 average height f x3 Simpson’s 3/8 formula. h=(b-a)/3
16. 16. Truncation error for Simpson’s 3/8 rule: Et (b a) 5 f (4) 6480 Simpson’s 3/8 rule is somewhat more accurate than 1/3 rule In general Simpson’s 1/3 rule is the method of choice because of obtaining third order accuracy and using three points instead of four. 3/8 rule has the utility when the number of segments is odd. In practice, use 1/3 rule for all the even number of segments and use 3/8 rule for the remaining last three segments. Higher-order Newton-Cotes formulas: Higher order (n=4,5) formulas have the same order error as n=1 (Trapezoid) or n=2,3 (Simpson’s) formulas. Higher-order formulas are rarely used in engineering practices. To increase the accuracy, just increase the number of segments!
17. 17. Integration with Unequal Segments There may be cases where the spacing between data points may not be even (e.g., experimentally derived data points) One way is to use trapezoidal rule for each segment I h1 f ( x0 ) f ( x1 ) 2 h2 f ( x1 ) f ( x2 ) 2 .. hn f ( xn 1 ) f ( xn ) 2 width of the segments (h) are not constant so the formula cannot be written in a compact form. A computer algorithm can easily be developed to do the integration for unequal-sized segments. The algorithm can be constructed such that, use Simpson’s 1/3 rule wherever two consecutive equal-sized segments are encountered, and use 3/8 rule wherever three consecutive equal-sized segments are encountered. When adjacent segments are unequal-sized , just use trapezoidal rule.
18. 18. Open Integration Formulas  There may be cases where integration limits are beyond the f(x) range of the data. General characteristics and order of error for open forms of the NewtonCotes formulas are similar to the closed forms. Even segment formulas are usually the method of choice as they require fewer points. a b x Open forms are not used for definite integration. They have utility for analyzing improper integrals (discussed later). They have connection to multi-step method for solving ordinary differential equations.
19. 19. Integration of Equations Use of multiple-application of trapezoidal and Simpson’s rules are not convenient for analyzing integration of functions. > Large number of operations required to evaluate the functions. > For large values of n, round of errors starts to dominate. Percent relative error Previously discussed methods are appropriate for tabulated data. If the function to be integrated is available, other modified methods are available. > Romberg integration (Richardson extrapolation) > Gauss quadrature Newton-Cotes method for functions: Trapezoidal rule Simpson’s rule n
20. 20. Romberg Integration It uses the trapezoidal rule, but much more efficient results are obtained through iterative refinement techniques . Richardson’s extrapolation: A sequence acceleration method to improve the rate of convergence. It offers a very practical tool for numerical integration and differentiation. Application of iterative refinement techniques to improve the error at each iteration. For each iteration I exact integral I (h) E (h) approximate integral truncation error In Richardson extrapolation, two approximate integrals are used to compute a third more accurate integral.
21. 21. Assume that two integrals with step sizes of h1 and h2 are available. Then, I ( h1 ) E ( h1 ) I ( h2 ) E ( h2 ) (b Error for multiple-equation trapezoidal rule ( n b E a 2 h '' h f 12 Assuming same f ’’ for two step sizes 2 1 2 2 E ( h1 ) h E ( h2 ) h 2 or E ( h1 ) E ( h2 ) h1 h2 a) )
22. 22. Inserting the error to the previous equation, and solving for E(h2) E ( h2 ) I ( h1 ) 1 I ( h2 ) h1 / h 2 2 Then, I 1 I ( h2 ) ( h1 / h 2 ) 2 1 I ( h2 ) I ( h1 ) It can be shown that the error for above approximation is O(h4) although the error from use of trapezoidal rule is O(h2). For the special case of interval being halved (h2=h1/2), we get I 4 3 O(h4) I ( h2 ) 1 3 I ( h1 ) O(h2)
23. 23. EX: For the numerical integration of f ( x) 0 .2 25 x 200 x 2 675 x 3 900 x 4 400 x 5 we previously obtained the following, using trapezoidal rule. n h I 1 0.8 0.1728 89.5 2 0.4 1.0688 34.9 4 0.2 1.4848 9.5 t Use Romberg integration approach to obtain improved approximations. use n=1 and n=2 for a improved approximation 4 1 I (1 . 0688 ) ( 0 . 1728 ) 1 . 367467 3 3 t 16 . 6 % If use n=2 and n=4 I 4 3 (1 . 4848 ) 1 3 (1 . 0688 ) 1 . 623467 t 1 .0 %
24. 24. The method can be generalized, e.g. combine two O(h4) approximation to obtain an O(h2) approximation, and so on… Using two improved O(h4) approximations in the previous examples, we can use the following formula I 16 15 Im 1 15 Il Im: more accurate approx. Il: less accurate approx. to obtain an O(h6) approximation. Similarly, two O(h6) approximation can be combined to obtain an O(h8) approximation I 64 63 Im 1 63 Il Im: more accurate approx. Il: less accurate approx. This sequential improvement convergence technique is named under the general approach of Richardson's extrapolation.
25. 25. EX: In the previous example use the two O(h4) approximations to obtain an O(h6) approximation for the integral. I 16 (1 . 623467 ) 15 1 (1 . 367467 ) 1 . 640533 15 True result is 1.640533 (correct answer to 7 S.F.!). Algorithm: O(h2) O(h4) … … … … … … … … … … Trapezoidal rule O(h6) … … … … … … … … O(h8) …
26. 26. Gauss Quadrature In Newton-Cotes formulation, predetermined and evenly spaced data are used. Points of integration are covered by the data. In Gauss quadrature, points of integration are undetermined. f(x) f(x) a b x a b x For the example above, integral approximation can be improved by using two wisely chosen interior points. Gauss quadrature offer a method to find these points.
27. 27. Method of undetermined coefficients: Assume two points (x0, x1) so that the integral can be written in the form I c0 f ( x0 ) f(x) f(x1) f(x0) c1 f ( x1 ) c‘s: constants -1 x0 x1 1 This requires calculation of four unknowns. We choose these four equations according to our preference. We can choose a function value that is exact to third order. x 1 c0 f ( x0 ) c1 f ( x1 ) 1dx 2 1 1 c0 f ( x0 ) c1 f ( x1 ) xdx 1 0 In general, integration limits can be made -1..1 for any integral by a simple change of variables.
28. 28. 1 c0 f ( x0 ) 2 c1 f ( x1 ) x dx 2 3 1 1 c0 f ( x0 ) 3 c1 f ( x1 ) x dx 0 1 Or, by evaluating the function values c0 c1 c0 x0 2 c1 x1 c x 3 c0 x 2 0 c1 x 1 c0 x0 2 1 1 3 0 2 3 0 By these deliberate choices, we force these two points give a an integral that is exact to cubic order.
29. 29. Solving for the unknowns c0 c1 1 1 x0 x1 0 . 5773503 ... 3 1 0 . 5773503 ... 3 Then, I f( 1 3 ) f( 1 3 ) Two-point GaussLegendre formula
30. 30. To change (normalize) the limits of integration, we do the following change of variable: x a0 a1 x d For the lower limit (x=a a xd=-1 ) a0 a1 ( 1) For the upper limit (x=b b (b a0 2 xd=1 ) a0 a1 a1 (1) a) (b a) 2 Substitute into the original formula: x (b a) (b a ) xd 2 Differentiate both sides: dx (b a) 2 dx d These equations are subsituted into the original integrand to transfrom into suitable form for application to Gauss-Legendre formula
31. 31. EX: Use two-point Gauss-Legendre quadrature formula to integrate f ( x) 0 .2 25 x 200 x 2 675 x 3 900 x 4 400 x 5 from x=0 to x=0.8 (remember that exact solution I=1.640533). First, perform a change of variable to convert the limits -1..1: x 0 .4 dx 0 .4 x d 0 . 4 dx d Substitute into the original equation to get: 0 .8 ( 0 .2 2 25 x 200 x 0 .2 25 ( 0 . 4 675 x 3 900 x 4 5 400 x ) dx 0 1 1 675 ( 0 . 4 0 .4 x d ) 0 .4 x d ) 3 200 ( 0 . 4 900 ( 0 . 4 0 .4 x d ) 0 .4 x d ) 2 4 400 ( 0 . 4 0 .4 x d ) 5 0 . 4 dx d RHS is suitable for evaluation using Gauss quadrature. Evaulating for -1/√3 (=0.516741) and for 1/√3 (=1.305837), and gives the following result: I 0 . 516741 1 . 305837 1 . 822578 t 11 . 1 % Equivalent to third order accuracy in Simpson’s method.
32. 32. Higher-point formulas: Higher-point versions of Gauss quadrature can be developed in the form I c0 f ( x0 ) c1 f ( x1 ) ... n: number of points cn 1 f ( xn 1 ) For example, for 3 point Gauss-Legendre formulas, c’s and x’s are: c0 c1 c2 x0 x1 0 . 774596669 0 .0 x2 0 . 5555556 0 . 8888889 0 . 5555556 0 . 774596669 Et ~ f(6)( ) Error for Gauss Quadrature: Et 2 (2n 2n 3 (n 3) ( 2 n 1)! 4 2 )! 3 f (2n 2) ( ) n= number of points minus one 1 1
33. 33. EX: Use three-point Gauss quadrature formula for the previous problem. Using the tabulated c and x values we write: I 0 . 5555556 f ( 0 . 7745967 ) 0 . 8888889 f ( 0 ) 0 . 5555556 f ( 0 . 7745967 ) which is I 0 . 2813013 0 . 8732444 0 . 4859876 1 . 640533 exact to 7 S.F.!
34. 34. Numerical Differentiation Types of Differentiation: Forward expansion of Taylor series: '' f ( xi 1 ) f ( xi ) ' f ( xi ) h f ( xi 1 ) ' f ( xi ) f ( xi ) h 2 ... h=step size 2! f ( xi ) O (h) h first forward difference Backward expansion of Taylor series '' f ( xi 1 ) f ( xi ) ' f ( xi ) ' f ( xi ) h f ( xi ) f ( xi ) f ( xi 1 ) h 2 ... 2! O (h) h first backward difference Subtract forward expansion from the backward expansion: ' f ( xi ) f ( xi 1 ) 2h f ( xi 1 ) 2 O (h ) centered difference
35. 35. Higher-accuracy Differentiation Formulas: High-accuracy diffentiation formulas can be obtained by adding more terms in Taylor expansion. Forward Taylor series expansion: '' f ( xi 1 ) f ( xi ) ' f ( xi ) f ( xi ) h h 2 ... 2 or ' f ( xi ) f ( xi 1 ) '' f ( xi ) f ( xi ) h h 2 O (h ) 2 Approximate the second derivative using finite difference formula '' f ( xi ) f ( xi 2 ) 2 f ( xi 1 ) h 2 f ( xi )
36. 36. Then, ' f ( xi ) f ( xi 1 ) f ( xi ) f ( xi 2 ) 2 f ( xi 1 ) h 2h f ( xi ) 2 h 2 O (h ) or, f ( xi ' f ( xi ) 2 ) 4 f ( xi 1 ) 3 f ( xi ) forward scheme 2 O (h ) 2h Backward and centered finite difference formulas can be derived in a similar way: 3 f ( xi ) ' f ( xi ) ' f ( xi ) f ( xi 4 f ( xi 1 ) f ( xi 2 ) backward scheme 2 O (h ) 2h 2 ) 8 f ( xi 1 ) 12 h 8 f ( xi 1 ) f ( xi 2 ) 4 O (h ) centered scheme
37. 37. EX: Calculate the approximate derivative of f ( x) 0 .1 x 4 0 . 15 x 3 0 .5 x 2 0 . 25 x 1 .2 at x=0.5 using a step size of h=0.25 (true value=-0.9125). First, evaluate the following data points: xi-2=0 ; f(xi-2)=1.2 xi+1=0.75 ; f(xi+1)=0.6363281 xi-1=0.25 ; f(xi-2)=1.103516 xi+2=1 ; f(xi+2)=0.2 xi=0.5 ; f(xi)=0.925 Forward difference scheme: 0 .2 ' f ( xi ) 4 ( 0 . 6363281 ) 3 ( 0 . 925 ) 0 . 859375 t 5 . 82 % t 3 . 77 % 2 ( 0 . 25 ) Backward difference scheme: ' f ( 0 .5 ) 3 ( 0 . 925 ) 4 (1 . 035156 ) 1 . 2 0 . 878125 2 ( 0 . 25 ) Centered difference scheme: ' f ( xi ) 0 .2 8 ( 0 . 6363281 ) 8 (1 . 035156 ) 1 . 2 12 ( 0 . 25 ) 0 . 9125 t 0%
38. 38. Richardson Extrapolation: As done for the integration, Richardson extrapolation uses two derivatives of different step sizes to obtain a more accurate derivative. In a similar fashion applied for the integration, use two step sizes such that h2=h1/2. Richardson extrapolation recursive formula: D 4 3 O(h4) D ( h2 ) 1 3 D ( h1 ) O(h2) [centered difference scheme] The approach can be iteratively used by Romberg algorithm to get higher accuracies.
39. 39. EX: Use Richardson extrapolation of step sizes h=0.5 and h=0.25 to calculate the derivative of the function f ( x) 0 .1 x 4 0 . 15 x 3 0 .5 x 2 0 . 25 x 1 .2 at x=0.5. Centered scheme finite difference approximation for h=0.5: 0 .2 D ( 0 .5 ) 1 .2 1 .0 t 1 9 .6 % Centered scheme finite difference approximation for h=0.25: D ( 0 .5 ) 0 . 6363281 1 . 103516 0 . 934375 t 0 .5 2 .4 % Apply Richardson extrapolation for improved accuracy: D ( 0 .5 ) 4 3 ( 0 . 934375 ) 1 3 ( 1) 0 . 9125 t 0%
40. 40. Derivatives of Unequally Spaced Data In the previous discussion, both finite difference approximations and Richardson extrapolation reqires evenly distributed data. So, these methods are more suitable to evaulate functions. Emprically derived data, experimental or from field surveys, are usually not even. One way is to fit a second-order Lagrange interpolating polynomial to each set of three data points. Derivative of the polynomial: 2x ' f ( x) ( xi 1 xi x i )( x i xi 1 1 xi 1 ) f ( xi 1 ) 2x ( xi xi 1 x i 1 )( x i xi 1 xi 1 ) f ( xi ) 2x ( xi 1 xi 1 x i 1 )( x i xi 1 xi ) f ( xi 1 ) Using above formula, any point in the range of three data points can be evaluated. This equation has the same accuracy of high-accuracy centered difference approximation even though data is not need to be equally spaced.
41. 41. Derivatives and Integrals for Data with Error: Differentiation process amplifies the error: y dy/dt x x As a remedy, fit a smoother function (low-order polynomial) to the uncertain data. On the other hand, integration process reduces the error. (Succusive negative and positive errors cancel out during integration). No further action is required.