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  2
x3  f 1.75  1.75 4  21.75  19
2
 6.12  0 1.5 1.75 2
 1.75

12. 1 .5  2
x3  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  2
x3  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  19
x4 
2
 2.75  0 1.75 1.88 2
 1.88

14. 1 .5  2
x3  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  19
x4 
2
 2.75  0 1.75 1.88 2
 1.88
 solution lies in interval 1.88  x  2

15. 1 .5  2
x3  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  19
x4 
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   19
x5 
2
 0.96  0 1.88 1.94 2
 1.94

16. 1 .5  2
x3  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  19
x4 
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   19
x5 
2
 0.96  0 1.88 1.94 2
 1.94
 solution lies in interval 1.94  x  2

17. 1.94  2
x6  f 1.97   1.97 4  21.97   19
2
 0.001  0 1.94 1.97 2
 1.97

18. 1.94  2
x6  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  2
x6  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  1.97
x7 
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  2
x6  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  1.97
x7 
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  2
x6  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  1.97
x7 
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
1.96  1.97
x8 
2
 1.97

22. 1.94  2
x6  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  1.97
x7 
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
1.96  1.97
x8 
2
 1.97
 an approximation for the root is x  1.97

23. (2) Newton’s Method of Approximation

24. (2) Newton’s Method of Approximation
y
x

25. (2) Newton’s Method of Approximation
y
x
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 

26. (2) Newton’s Method of Approximation
y
x
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 

27. NOTE:
(2) Newton’s Method of Approximation
y x0 must be a good first
approximation
Newton’s method finds
where the tangent at x0
cuts the x axis
x
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 

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
x0 x
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 

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 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 

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 fail
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 

31. e.g Find an approximation to two decimal places for a root of
x 4  2 x  19  0

32. 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

33. 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3  2
 10.9375  15.5

34. 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3  2
 10.9375  15.5
f  x0 
x1  x0 
f  x0 
 10.9375
 1 .5 
15.5
 2.21

35. 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3  2
 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

36. 9.2744
x2  2.21 
45.1754
 2.00

37. 9.2744 f 2   2 4  22   19
x2  2.21 
45.1754 1
 2.00
f 2   42   2
3
 35

38. 9.2744 f 2   2 4  22   19
x2  2.21 
45.1754 1
 2.00
f 2   42   2
3
 35
1
x3  2 
35
 1.97

39. 9.2744 f 2   2 4  22   19
x2  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

40. 9.2744 f 2   2 4  22   19
x2  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

41. 9.2744 f 2   2 4  22   19
x2  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

42. (ii ) Use Newton's Method to obtain an approximation to 23
correct to two decimal places

43. (ii ) Use Newton's Method to obtain an approximation to 23
correct to two decimal places
f  x   x 2  23

44. (ii ) Use Newton's Method to obtain an approximation to 23
correct to two decimal places
f  x   x 2  23
f  x  2x

45. (ii ) Use Newton's 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

46. (ii ) Use Newton's 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

47. (ii ) Use Newton's 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

48. (ii ) Use Newton's 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)

49. (ii ) Use Newton's 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)

50. Other Possible Problems with Newton’s Method

51. Other Possible Problems with Newton’s Method
Approximations oscillate

52. Other Possible Problems with Newton’s Method
y Approximations oscillate
x

53. Other Possible Problems with Newton’s Method
y Approximations oscillate
want to find this root
x

54. Other Possible Problems with Newton’s Method
y Approximations oscillate
want to find this root
x1 x2 x

55. Other Possible Problems with Newton’s Method
y Approximations oscillate
want to find this root
x1 x2 x

56. 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

57. 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

58. 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

59. 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

60. 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

61. 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