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12 x1 t04 07 approximations to roots (2013)

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  • 1. Approximations To Roots(1) Halving The Interval
  • 2. Approximations To Roots(1) Halving The Interval y y  f x x If y = f(x) is a continuous function over the interval a  x  b , and f(a) and f(b) are opposite in sign,
  • 3. Approximations To Roots(1) Halving The Interval y y  f x f a  a b x f b  If y = f(x) is a continuous function over the interval a  x  b , and f(a) and f(b) are opposite in sign,
  • 4. Approximations To Roots(1) Halving The Interval y y  f x f a  a b x f b  If y = f(x) is a continuous function over the interval a  x  b , and f(a) and f(b) are opposite in sign, then at least one root of the equation f(x) = 0 lies in the interval a  x  b
  • 5. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 in the interval 1  x  3
  • 6. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 in the interval 1  x  3 f  x   x 4  2 x  19 f 1  14  2  19 f 3  34  23  19  16  0  68  0
  • 7. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 in the interval 1  x  3 f  x   x 4  2 x  19 f 1  14  2  19 f 3  34  23  19  16  0  68  0 1 3 x1  f 2   2 4  22   19 2 1 0 1 2 3 2
  • 8. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 in the interval 1  x  3 f  x   x 4  2 x  19 f 1  14  2  19 f 3  34  23  19  16  0  68  0 1 3 x1  f 2   2 4  22   19 2 1 0 1 2 3 2  solution lies in interval 1  x  2
  • 9. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 in the interval 1  x  3 f  x   x 4  2 x  19 f 1  14  2  19 f 3  34  23  19  16  0  68  0 1 3 x1  f 2   2 4  22   19 2 1 0 1 2 3 2  solution lies in interval 1  x  2 1 2 x2  f 1.5  1.54  21.5  19 2  10.9  0 1 1.5 2  1 .5
  • 10. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 in the interval 1  x  3 f  x   x 4  2 x  19 f 1  14  2  19 f 3  34  23  19  16  0  68  0 1 3 x1  f 2   2 4  22   19 2 1 0 1 2 3 2  solution lies in interval 1  x  2 1 2 x2  f 1.5  1.54  21.5  19 2  10.9  0 1 1.5 2  1 .5  solution lies in interval 1.5  x  2
  • 11. 1 .5  2x3  f 1.75  1.75 4  21.75  19 2  6.12  0 1.5 1.75 2  1.75
  • 12. 1 .5  2x3  f 1.75  1.75 4  21.75  19 2  6.12  0 1.5 1.75 2  1.75  solution lies in interval 1.75  x  2
  • 13. 1 .5  2x3  f 1.75  1.75 4  21.75  19 2  6.12  0 1.5 1.75 2  1.75  solution lies in interval 1.75  x  2 1.75  2 f 1.88  1.88 4  21.88  19x4  2  2.75  0 1.75 1.88 2  1.88
  • 14. 1 .5  2x3  f 1.75  1.75 4  21.75  19 2  6.12  0 1.5 1.75 2  1.75  solution lies in interval 1.75  x  2 1.75  2 f 1.88  1.88 4  21.88  19x4  2  2.75  0 1.75 1.88 2  1.88  solution lies in interval 1.88  x  2
  • 15. 1 .5  2x3  f 1.75  1.75 4  21.75  19 2  6.12  0 1.5 1.75 2  1.75  solution lies in interval 1.75  x  2 1.75  2 f 1.88  1.88 4  21.88  19x4  2  2.75  0 1.75 1.88 2  1.88  solution lies in interval 1.88  x  2 1.88  2 f 1.94   1.94 4  21.94   19x5  2  0.96  0 1.88 1.94 2  1.94
  • 16. 1 .5  2x3  f 1.75  1.75 4  21.75  19 2  6.12  0 1.5 1.75 2  1.75  solution lies in interval 1.75  x  2 1.75  2 f 1.88  1.88 4  21.88  19x4  2  2.75  0 1.75 1.88 2  1.88  solution lies in interval 1.88  x  2 1.88  2 f 1.94   1.94 4  21.94   19x5  2  0.96  0 1.88 1.94 2  1.94  solution lies in interval 1.94  x  2
  • 17. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97
  • 18. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97  solution lies in interval 1.94  x  1.97
  • 19. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97  solution lies in interval 1.94  x  1.97 1.94  1.97x7  2 f 1.96   1.96 4  21.96   19 1.94 1.96 1.97  1.96  0.32  0
  • 20. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97  solution lies in interval 1.94  x  1.97 1.94  1.97x7  2 f 1.96   1.96 4  21.96   19 1.94 1.96 1.97  1.96  0.32  0  solution lies in interval 1.96  x  1.97
  • 21. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97  solution lies in interval 1.94  x  1.97 1.94  1.97x7  2 f 1.96   1.96 4  21.96   19 1.94 1.96 1.97  1.96  0.32  0  solution lies in interval 1.96  x  1.97 so is the solution closer to 1.96 or 1.97?
  • 22. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97  solution lies in interval 1.94  x  1.97 1.94  1.97x7  2 f 1.96   1.96 4  21.96   19 1.94 1.96 1.97  1.96  0.32  0  solution lies in interval 1.96  x  1.97 so is the solution closer to 1.96 or 1.97? 1.96 1.965 1.97
  • 23. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97  solution lies in interval 1.94  x  1.97 1.94  1.97x7  2 f 1.96   1.96 4  21.96   19 1.94 1.96 1.97  1.96  0.32  0  solution lies in interval 1.96  x  1.97 so is the solution closer to 1.96 or 1.97? f 1.965   1.9654  2 1.965   19  0.16  0 1.96 1.965 1.97
  • 24. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97  solution lies in interval 1.94  x  1.97 1.94  1.97x7  2 f 1.96   1.96 4  21.96   19 1.94 1.96 1.97  1.96  0.32  0  solution lies in interval 1.96  x  1.97 so is the solution closer to 1.96 or 1.97? f 1.965   1.9654  2 1.965   19  0.16  0 1.96 1.965 1.97
  • 25. 1.94  2x6  f 1.97   1.97 4  21.97   19 2  0.001  0 1.94 1.97 2  1.97  solution lies in interval 1.94  x  1.97 1.94  1.97x7  2 f 1.96   1.96 4  21.96   19 1.94 1.96 1.97  1.96  0.32  0  solution lies in interval 1.96  x  1.97 so is the solution closer to 1.96 or 1.97? f 1.965   1.9654  2 1.965   19  0.16  0 1.96 1.965 1.97  an approximation for the root is x  1.97
  • 26. NOTE:(2) Newton’s Method of Approximation y x0 must be a good first approximation y  f x Newton’s method finds where the tangent at x0 cuts the x axis x
  • 27. NOTE:(2) Newton’s Method of Approximation y x0 must be a good first approximation y  f x Newton’s method finds where the tangent at x0 cuts the x axis x0 x
  • 28. NOTE:(2) Newton’s Method of Approximation y x0 must be a good first approximation y  f x Newton’s method finds where the tangent at x0 cuts the x axis x1 x0 x
  • 29. NOTE:(2) Newton’s Method of Approximation y x0 must be a good first approximation y  f x Newton’s method finds where the tangent at x0 cuts the x axis x1 x0 x If f  x0   0 i.e. tangent || x axis the method will fail
  • 30. NOTE:(2) Newton’s Method of Approximation y x0 must be a good first approximation y  f x Newton’s method finds where the tangent at x0 cuts the x axis x1 x0 x If f  x0   0 i.e. tangent || x axis the method will failUsing the tangent at x0 to find x1
  • 31. NOTE:(2) Newton’s Method of Approximation y x0 must be a good first approximation y  f x Newton’s method finds where the tangent at x0 cuts the x axis x1 x0 x If f  x0   0 i.e. tangent || x axis the method will failUsing the tangent at x0 to find x1 f  x0   0 slope of tangent  x0  x1
  • 32. NOTE:(2) Newton’s Method of Approximation y x0 must be a good first approximation y  f x Newton’s method finds where the tangent at x0 cuts the x axis x1 x0 x If f  x0   0 i.e. tangent || x axis the method will failUsing the tangent at x0 to find x1 f  x0   0 slope of tangent  x0  x1 f  x0   0 f   x0   x0  x1
  • 33.  x0  x1  f   x0   f  x0  f  x0  x0  x1  f   x0 
  • 34.  x0  x1  f   x0   f  x0  f  x0  x0  x1  f   x0 If x0 is a good first approximation to a root of the equation f(x) = 0,then a closer approximation is given by; f  x0  x1  x0  f  x0 
  • 35.  x0  x1  f   x0   f  x0  f  x0  x0  x1  f   x0 If x0 is a good first approximation to a root of the equation f(x) = 0,then a closer approximation is given by; f  x0  x1  x0  f  x0 Successive approximations x2 , x3 ,  , xn , xn 1are given by; f  xn  xn 1  xn  f  xn 
  • 36. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0
  • 37. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 f  x   x 4  2 x  19 f  x   4 x 3  2
  • 38. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 f  x   x 4  2 x  19 f  x   4 x 3  2 x0  1.5 f 1.5  1.54  21.5  19 f 1.5  41.5  2 3  10.9375  15.5
  • 39. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 f  x   x 4  2 x  19 f  x   4 x 3  2 x0  1.5 f 1.5  1.54  21.5  19 f 1.5  41.5  2 3  10.9375  15.5 f  x0  x1  x0  f  x0   10.9375  1 .5  15.5  2.21
  • 40. e.g Find an approximation to two decimal places for a root of x 4  2 x  19  0 f  x   x 4  2 x  19 f  x   4 x 3  2 x0  1.5 f 1.5  1.54  21.5  19 f 1.5  41.5  2 3  10.9375  15.5 f  x0  x1  x0  f 2.21  2.214  22.21  19 f  x0   9.2744  10.9375  1 .5  f 2.21  42.21  2 3 15.5  2.21  45.1754
  • 41. 9.2744x2  2.21  45.1754  2.00
  • 42. 9.2744 f 2   2 4  22   19x2  2.21  45.1754 1  2.00 f 2   42   2 3  35
  • 43. 9.2744 f 2   2 4  22   19x2  2.21  45.1754 1  2.00 f 2   42   2 3  35 1 x3  2  35  1.97
  • 44. 9.2744 f 2   2 4  22   19x2  2.21  45.1754 1  2.00 f 2   42   2 3  35 x3  2  1 f 1.97   1.97 4  21.97   19 35  0.001  1.97 f 1.97   41.97   2 3  32.58
  • 45. 9.2744 f 2   2 4  22   19x2  2.21  45.1754 1  2.00 f 2   42   2 3  35 x3  2  1 f 1.97   1.97 4  21.97   19 35  0.001  1.97 f 1.97   41.97   2 3  32.58 0.001 x4  1.97  32.58  1.97
  • 46. 9.2744 f 2   2 4  22   19x2  2.21  45.1754 1  2.00 f 2   42   2 3  35 x3  2  1 f 1.97   1.97 4  21.97   19 35  0.001  1.97 f 1.97   41.97   2 3  32.58 0.001 x4  1.97  32.58  1.97  x  1.97 is a better approximation for the root
  • 47. (ii ) Use Newtons Method to obtain an approximation to 23 correct to two decimal places
  • 48. (ii ) Use Newtons Method to obtain an approximation to 23 correct to two decimal places f  x   x 2  23
  • 49. (ii ) Use Newtons Method to obtain an approximation to 23 correct to two decimal places f  x   x 2  23 f  x  2x
  • 50. (ii ) Use Newtons Method to obtain an approximation to 23 correct to two decimal places xn12  23 f  x   x  23 2 xn  xn1  2 xn1 f  x  2x xn12  23  2 xn1
  • 51. (ii ) Use Newtons Method to obtain an approximation to 23 correct to two decimal places xn12  23 f  x   x  23 2 xn  xn1  2 xn1 f  x  2x xn12  23  2 xn1 x0  5
  • 52. (ii ) Use Newtons Method to obtain an approximation to 23 correct to two decimal places xn12  23 f  x   x  23 2 xn  xn1  2 xn1 f  x  2x xn12  23  2 xn1 x0  5 52  23 x1  2 5 x1  4.8
  • 53. (ii ) Use Newtons Method to obtain an approximation to 23 correct to two decimal places xn12  23 f  x   x  23 2 xn  xn1  2 xn1 f  x  2x xn12  23  2 xn1 x0  5 52  23 4.82  23 x1  x2  2 5 2  4.8  x1  4.8 x2  4.795833333 x2  4.80 (to 2 dp)
  • 54. (ii ) Use Newtons Method to obtain an approximation to 23 correct to two decimal places xn12  23 f  x   x  23 2 xn  xn1  2 xn1 f  x  2x xn12  23  2 xn1 x0  5 52  23 4.82  23 x1  x2  2 5 2  4.8  x1  4.8 x2  4.795833333 x2  4.80 (to 2 dp)  23  4.80 (to 2 dp)
  • 55. Other Possible Problems with Newton’s Method
  • 56. Other Possible Problems with Newton’s Method Approximations oscillate
  • 57. Other Possible Problems with Newton’s Method y Approximations oscillate x
  • 58. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root x
  • 59. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root x1 x2 x
  • 60. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root x1 x2 x
  • 61. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root x1 x2 x wrong side of stationary point converges to wrong root
  • 62. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root x1 x2 x wrong side of stationary point y converges to wrong root x
  • 63. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root x1 x2 x wrong side of stationary point y converges to wrong root x want to find this root
  • 64. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root x1 x2 x wrong side of stationary point y converges to wrong root x1 x2 x want to find this root
  • 65. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root x1 x2 x wrong side of stationary point y converges to wrong root x1 x2 x want to find this root
  • 66. Other Possible Problems with Newton’s Method y Approximations oscillate want to find this root Exercise 6E; 1, 3ac, 6adf, 8a, 10, 12 x1 x2 x wrong side of stationary point y converges to wrong root x1 x2 x want to find this root

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