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# Numerical differentiation integration

Numerical differentiation integration

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### Numerical differentiation integration

1. 1. NUMERICAL DIFFERENTIATION The derivative of f (x) at x0 is: f ′( x 0 ) = Limit h→ 0 f ( x0 + h) − f ( x0 ) h An approximation to this is: f ( x0 + h) − f ( x0 ) f ′( x 0 ) ≈ h for small values of h. Forward Difference Formula
2. 2. Let f ( x ) = ln x and x0 = 1.8 Find an approximate value for f ′(1.8) h f (1.8) f (1.8 + h) 0.1 0.5877867 0.6418539 f (1.8 + h) − f (1.8) h 0.5406720 0.01 0.5877867 0.5933268 0.5540100 0.001 0.5877867 0.5883421 0.5554000 The exact value of f ′(1.8) = 0.55 5
3. 3. Assume that a function goes through three points: ( x0 , f ( x0 ) ) , ( x1 , f ( x1 ) ) and ( x2 , f ( x2 ) ) . f ( x) ≈ P( x) P ( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 ) Lagrange Interpolating Polynomial
4. 4. P ( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x 2 ) ( x − x1 )( x − x2 ) P( x) = f ( x0 ) ( x0 − x1 )( x0 − x2 ) ( x − x0 )( x − x2 ) + f ( x1 ) ( x1 − x0 )( x1 − x2 ) ( x − x0 )( x − x1 ) + f ( x2 ) ( x2 − x0 )( x2 − x1 )
5. 5. f ′( x ) ≈ P ′( x ) 2 x − x1 − x2 P ′( x ) = f ( x0 ) ( x0 − x1 )( x0 − x2 ) 2 x − x0 − x2 + f ( x1 ) ( x1 − x0 )( x1 − x2 ) 2 x − x0 − x1 + f ( x2 ) ( x2 − x0 )( x2 − x1 )
6. 6. If the points are equally spaced, i.e., x1 = x0 + h and x 2 = x0 + 2h 2 x0 − ( x0 + h) − ( x0 + 2h) P ′( x0 ) = f ( x0 ) { x0 − ( x0 + h)}{ x0 − ( x0 + 2h)} 2 x0 − x0 − ( x0 + 2h) + f ( x1 ) { ( x0 + h) − x0 }{ ( x0 + h) − ( x0 + 2h)} 2 x0 − x0 − ( x0 + h) + f ( x2 ) { ( x0 + 2h) − x0 }{ ( x0 + 2h) − ( x0 + h)}
7. 7. − 3h − 2h −h P ′( x 0 ) = f ( x0 ) + f ( x1 ) + 2 f ( x 2 ) 2 2 2h −h 2h 1 { − 3 f ( x0 ) + 4 f ( x1 ) − f ( x2 ) } P ′( x0 ) = 2h Three-point formula: 1 { − 3 f ( x0 ) + 4 f ( x0 + h) − f ( x0 + 2h) } f ′( x0 ) ≈ 2h
8. 8. If the points are equally spaced with x0 in the middle: x1 = x0 − h and x 2 = x0 + h 2 x0 − ( x0 − h) − ( x0 + h) P ′( x0 ) = f ( x0 ) { x0 − ( x0 − h)}{ ( x0 − ( x0 + h)} 2 x0 − x0 − ( x0 + h) + f ( x1 ) { ( x0 − h) − x0 }{ ( x0 − h) − ( x0 + h)} 2 x0 − x0 − ( x0 − h) + f ( x2 ) { ( x0 + h) − x0 }{ ( x0 + h) − ( x0 − h)}
9. 9. 0 −h h P ′( x 0 ) = f ( x0 ) + 2 f ( x1 ) + 2 f ( x 2 ) 2 −h 2h 2h Another Three-point formula: 1 { f ( x0 + h) − f ( x0 − h) } f ′( x0 ) ≈ 2h
10. 10. Alternate approach (Error estimate) Take Taylor series expansion of f(x+h) about x: 2 3 h ( 2) h ( 3) f ( x + h) = f ( x ) + hf ′( x ) + f ( x) + f ( x) +  2 3! h2 ( 2 ) h3 ( 3 ) f ( x + h) − f ( x ) = hf ′( x ) + f ( x) + f ( x) +  2 3! f ( x + h) − f ( x ) h ( 2) h2 ( 3 ) = f ′( x ) + f ( x ) + f ( x) +  h 2 3! .............. (1)
11. 11. f ( x + h) − f ( x ) = f ′( x ) + O ( h) h f ( x + h) − f ( x ) f ′( x ) = − O ( h) h f ( x + h) − f ( x ) f ′( x ) ≈ h Forward Difference Formula h ( 2) h2 ( 3 ) O ( h) = f ( x ) + f ( x) + 2 3!
12. 12. 4h 2 ( 2 ) 8h 3 ( 3 ) f ( x + 2h) = f ( x ) + 2hf ′( x ) + f ( x) + f ( x) +  2 3! 4h 2 ( 2 ) 8h 3 ( 3 ) f ( x + 2h) − f ( x ) = 2hf ′( x ) + f ( x) + f ( x) +  2 3! f ( x + 2h ) − f ( x ) 2h ( 2 ) 4h 2 ( 3 ) = f ′( x ) + f ( x) + f ( x) +  2h 2 3! ................. (2)
13. 13. f ( x + h) − f ( x ) h ( 2) h ( 3) = f ′( x ) + f ( x ) + f ( x) +  h 2 3! .............. (1) 2 f ( x + 2h ) − f ( x ) 2h ( 2 ) 4h 2 ( 3 ) = f ′( x ) + f ( x) + f ( x) +  2h 2 3! ................. (2) 2 X Eqn. (1) – Eqn. (2)
14. 14. f ( x + h) − f ( x ) f ( x + 2h) − f ( x ) 2 − h 2h 2 h2 ( 3 ) 6 h3 ( 4 ) = f ′( x ) − f ( x) − f ( x) −  3! 4! − f ( x + 2 h) + 4 f ( x + h) − 3 f ( x ) 2h 2 3 2h ( 3 ) 6h ( 4 ) ′( x ) − = f f ( x) − f ( x) −  3! 4! = f ′ ( x ) + O h2 ( )
15. 15. − f ( x + 2h ) + 4 f ( x + h ) − 3 f ( x ) = f ′( x ) + O h2 2h ( ) − f ( x + 2h ) + 4 f ( x + h ) − 3 f ( x ) f ′( x ) = − O h2 2h ( ) − f ( x + 2h ) + 4 f ( x + h ) − 3 f ( x ) f ′( x ) ≈ 2h Three-point Formula ( ) 2 h2 ( 3 ) 6 h3 ( 4 ) O h2 = − f ( x) − f ( x) −  3! 4!
16. 16. The Second Three-point Formula Take Taylor series expansion of f(x+h) about x: h2 ( 2 ) h3 ( 3 ) f ( x + h) = f ( x ) + hf ′( x ) + f ( x) + f ( x) + 2 3! Take Taylor series expansion of f(x-h) about x: 2 3 h ( 2) h ( 3) f ( x − h) = f ( x ) − hf ′( x ) + f ( x) − f ( x) +  2 3! Subtract one expression from another 2h 3 ( 3 ) 2h 6 ( 6 ) f ( x + h) − f ( x − h) = 2hf ′( x ) + f ( x) + f ( x) +  3! 6!
17. 17. 2h 3 ( 3 ) 2h 6 ( 6 ) f ( x + h) − f ( x − h) = 2hf ′( x ) + f ( x) + f ( x) +  3! 6! f ( x + h) − f ( x − h) h2 ( 3) h5 ( 6 ) = f ′( x ) + f ( x) + f ( x) +  2h 3! 6! f ( x + h) − f ( x − h) h 2 ( 3 ) h5 ( 6 ) f ′( x ) = − f ( x) − f ( x) −  2h 3! 6!
18. 18. f ′( x ) = f ( x + h ) − f ( x − h) + O h2 2h ( ) ( ) h2 ( 3) h5 ( 6 ) O h2 = − f ( x) − f ( x) −  3! 6! f ( x + h) − f ( x − h ) f ′( x ) ≈ 2h Second Three-point Formula
19. 19. Summary of Errors f ( x + h) − f ( x ) f ′( x ) ≈ h Error term Forward Difference Formula h ( 2) h2 ( 3 ) O ( h) = f ( x ) + f ( x) + 2 3!
20. 20. Summary of Errors continued First Three-point Formula − f ( x + 2h ) + 4 f ( x + h ) − 3 f ( x ) f ′( x ) ≈ 2h ( ) 2 h2 ( 3 ) 6 h3 ( 4 ) Error term O h2 = − f ( x) − f ( x) −  3! 4!
21. 21. Summary of Errors continued Second Three-point Formula f ( x + h) − f ( x − h ) f ′( x ) ≈ 2h ( ) h2 ( 3) h5 ( 6 ) O h2 = − f ( x) − f ( x) −  Error term 3! 6!
22. 22. Example: f ( x ) = xe x Find the approximate value of x f ( x) 1.9 12.703199 2.0 14.778112 2.1 2.2 17.148957 19.855030 f ′( 2 ) with h = 0.1
23. 23. Using the Forward Difference formula: 1 f ′ ( x0 ) ≈ { f ( x0 + h ) − f ( x0 ) } h 1 f ′ ( 2) ≈ { f ( 2.1) − f ( 2 ) } 0.1 1 = { 17.148957 − 14.778112} 0.1 = 23.708450
24. 24. Using the 1st Three-point formula: 1 { − 3 f ( x0 ) + 4 f ( x0 + h) − f ( x0 + 2h) } f ′( x0 ) ≈ 2h 1 [ − 3 f ( 2) + 4 f ( 2.1) − f ( 2.2)] f ′( 2 ) ≈ 2 × 0.1 1 [ − 3 × 14.778112 + 4 × 17.148957 = 0.2 − 19.855030 ] = 22.032310
25. 25. Using the 2nd Three-point formula: 1 { f ( x0 + h) − f ( x0 − h) } f ′( x0 ) ≈ 2h 1 [ f ( 2.1) − f (1.9)] f ′( 2 ) ≈ 2 × 0.1 1 [ 17.148957 − 12.703199 = 0.2 = 22.228790 The exact value of f ′( 2 ) is : 22.167168 ]
26. 26. Comparison of the results with h = 0.1 The exact value of f ′ ( 2 ) is 22.167168 Formula f ′ ( 2) Error Forward Difference 23.708450 1.541282 1st Three-point 22.032310 0.134858 2nd Three-point 22.228790 0.061622
27. 27. Second-order Derivative h2 ( 2 ) h3 ( 3 ) f ( x + h) = f ( x ) + hf ′( x ) + f ( x) + f ( x) +  2 3! 2 3 h ( 2) h ( 3) f ( x − h) = f ( x ) − hf ′( x ) + f ( x) − f ( x) +  2 3! Add these two equations. 2h 2 ( 2 ) 2h4 ( 4 ) f ( x + h) + f ( x − h) = 2 f ( x ) + f ( x) + f ( x) +  2 4! 2h2 ( 2 ) 2 h4 ( 4 ) f ( x + h) − 2 f ( x ) + f ( x − h) = f ( x) + f ( x) + 2 4!
28. 28. f ( x + h) − 2 f ( x ) + f ( x − h) 2h ( 4 ) ( 2) = f ( x) + f ( x) +  2 h 4! 2 f ( x + h ) − 2 f ( x ) + f ( x − h ) 2h 2 ( 4 ) f ( 2) ( x ) = − f ( x) +  2 h 4! f ( 2) f ( x + h) − 2 f ( x ) + f ( x − h) ( x) ≈ h2
29. 29. NUMERICAL INTEGRATION b ∫ f ( x )dx = a area under the curve f(x) between x = a to x = b. In many cases a mathematical expression for f(x) is unknown and in some cases even if f(x) is known its complex form makes it difficult to perform the integration.
30. 30. y f(b) f(x) f(a) x 0 =a x 0 =b x
31. 31. y f(b) f(x) f(a) x 0 =a x 0 =b x
32. 32. Area of the trapezoid The length of the two parallel sides of the trapezoid are: f(a) and f(b) The height is b-a b ∫ a b−a [ f ( a ) + f ( b )] f ( x )dx ≈ 2 h = [ f ( a ) + f ( b )] 2
33. 33. Simpson’s Rule: ∫ x2 x0 x2 f ( x )dx ≈ ∫ P ( x )dx x0 x1 = x0 + h and x 2 = x0 + 2h
34. 34. ∫ x2 x0 P ( x ) dx = ∫ + ∫ + x2 x0 x2 x0 ∫ x2 x0 ( x − x1 )( x − x2 ) f ( x0 ) dx ( x0 − x1 )( x0 − x2 ) ( x − x0 )( x − x2 ) f ( x1 ) dx ( x1 − x0 )( x1 − x2 ) ( x − x0 )( x − x1 ) f ( x2 ) dx ( x2 − x0 )( x2 − x1 )
35. 35. ∫ x2 x0 x2 f ( x )dx ≈ ∫ P ( x )dx x0 h = [ f ( x0 ) + 4 f ( x1 ) + f ( x 2 )] 3
36. 36. y f(b) f(x) f(a) x 0 =a x 0 =b x
37. 37. y f(b) f(x) f(a) x 0 =a x 0 =b x
38. 38. y f(b) f(x) f(a) x 0 =a x 0 =b x
39. 39. y f(b) f(x) f(a) x 0 =a x 0 =b x
40. 40. Composite Numerical Integration
41. 41. Riemann Sum The area under the curve is subdivided into n subintervals. Each subinterval is treated as a rectangle. The area of all subintervals are added to determine the area under the curve. There are several variations of Riemann sum as applied to composite integration.
42. 42. ∆ xIn Left Riemann = ( b − a) / n sum, x1 = a the left- x2 x3 side sample of = afunction is the + ∆ x used as the = a + 2∆the x height of individual M rectangle. xi = a + ( i − 1) ∆ x ∫ b a n f ( x ) dx ≈ ∑ f ( xi ) ∆ x i =1
43. 43. ∆x = ( b − a) / n In Right Riemann sum, the right-side x1 = a + ∆ x sample of the x2 = a is used function+ 2∆ x as the height of the x3 = a + 3∆ x individual rectangle. M xi = a + i∆ x n ∫ f ( x ) dx ≈ ∑ f ( x ) ∆ x b a i =1 i
44. 44. In the ) / n ∆ x = ( b − aMidpoint Rule, the sample at x1 = the (middle1of ∆ x / 2 ) a + 2 ×1 − ) ( the x2 =subinterval1is( used 2 ) a + ( 2× 2 − ) ∆x / as the height of the x3 = individual 1) ( ∆ x / 2 ) a + ( 2×3− M rectangle. xi = a + ( 2 × i − 1) ( ∆ x / 2 ) n ∫ f ( x ) dx ≈ ∑ f ( x ) ∆ x b a i =1 i
45. 45. Composite Trapezoidal Rule: Divide the interval into n subintervals and apply Trapezoidal Rule in each subinterval. b ∫ a  h f ( x )dx ≈  f ( a ) + 2∑ f ( x k ) + f (b ) 2 k =1  n −1 where b−a h= and x k = a + kh for k = 0, 1, 2, ... , n n
46. 46. π Find ∫ sin (x)dx 0 by dividing the interval into 20 subintervals. n = 20 b−a π h= = n 20 kπ x k = a + kh = , k = 0, 1, 2, ....., 20 20
47. 47. π  h ∫ sin( x )dx ≈ 2  f ( a ) + 2∑ f ( xk ) + f (b) k =1   0 n −1 19  π   kπ  = sin( 0 ) + 2∑ sin 20  + sin( π )  40    k =1  = 1.995886
48. 48. Composite Simpson’s Rule: Divide the interval into n subintervals and apply Simpson’s Rule on each consecutive pair of subinterval. Note that n must be even. b ∫ a h f ( x )dx ≈  f ( a ) + 2 3 n/ 2 + 4∑ k =1 ( n / 2 ) −1 ∑ f (x k =1 2k )  f ( x 2 k −1 ) + f (b ) 
49. 49. where b−a h= and x k = a + kh for k = 0, 1, 2, ... , n n π Find ∫ sin (x)dx 0 by dividing the interval into 20 subintervals. b−a π n = 20 h= = n 20 kπ x k = a + kh = , k = 0, 1, 2, ....., 20 20
50. 50. π 9 π   2kπ  ∫ sin( x )dx ≈ 60 sin( 0) + 2∑ sin 20    k =1  0   ( 2k − 1)π  + 4∑ sin  + sin( π) 20   k =1  = 2.000006 10