1. Lecture 4 Numerical Analysis

2. Solution of Non-Linear Equations Chapter 2

3. Introduction Bisection Method Regula-Falsi Method Method of iteration Newton - Raphson Method Muller’s Method Graeffe’s Root Squaring Method

4. Bisection Method (Bolzano)

5. Example Solve x 3 – 9x + 1 = 0 for the root between x = 2 and x = 4 by the bisection method.

6. Solution Given f (x) = x 3 – 9x + 1. Here f (2) = -9, f (4) = 29. Therefore, f (2) f (4) < 0 and hence the root lies between 2 and 4. Let x 0 = 2, x 1 = 4. Now, we define as a first approximation to a root of f (x) = 0 and note that f (3) = 1, so that f (2) f (3) < 0. Thus the root lies between 2 and 3

7. We further define, and note that f (x 3 ) = f (2.5) < 0, so that f (2.5) f (3) < 0. Therefore, we define the mid-point, Similarly, x 5 = 2. 875 and x 6 = 2.9375 and the process can be continued until the root is obtained to the desired accuracy.

8. These results are presented in the table. n xnxnxnxn f ( x n ) 231.0 32.5-5.875 42.75-2.9531 52.875-1.1113 62.9375-0.0901

9. Regula-Falsi Method Method of false position

10. Here, we choose two points x n and x n -1 such that f (x n ) and f (x n-1 ) are of opposite signs. Intermediate value property suggests that the graph of y = f (x) crosses the x-axis between these two points and therefore, a root say lies between these two points.

11. Thus, to find a real root of f (x) = 0 using Regula-Falsi method, we replace the part of the curve between the points A[x n, f(x n )] and B[x n-1, f (x n-1 )] by a chord in that interval and we take the point of intersection of this chord with the x-axis as a first approximation to the root.

12. Now, the equation of the chord joining the points A and B is (2.1) Setting y = 0 in Eq. (2.1), we get

13. Hence, the first approximation to the root of f (x) = 0 is given by (2.2) we observe that f (x n-1 ) and f (x n+1 ) are of opposite sign. Thus, it is possible to apply to above procedure, to determine the line through B and A 1 and so on. Hence, the successive approximations to the root of f (x) = 0 is given by Eq. (2.2).

14. Example Use the Regula-Falsi method to compute a real root of the equation x 3 – 9x + 1 = 0, (i) if the root lies between 2 and 4 (ii) if the root lies between 2 and 3. Comment on the results.

15. Solution Let f (x) = x 3 – 9x + 1. f (2) = – 9 and f (4) = 29. Since f (2) and f (4) are of opposite signs, the root of f (x) = 0 lies between 2 and 4.

16. Taking x 1 = 2, x 2 = 4 and using Regula-Falsi method, the first approximation is given by

17. Now f (x 3 ) = –6.12644. Since f (x 2 ) and f (x 3 ) are of opposite signs, the root lies between x 2 and x 3. The second approximation to the root is given as

18. Therefore f (x 4 ) = – 3. 090707. Now, since f (x 2 ) and f (x 4 ) are of opposite signs, the third approximation is obtained from and f (x 5 ) = – 1.32686.

19. This procedure can be continued till we get the desired result. The first three iterations are shown as in the table. nx n+1 f (x n+1 ) 22.47368-6.12644 32.73989-3.090707 42.86125-1.32686

20. (ii) f (2) = – 9 and f (3) = 1. Since f (2) and f (3) are of opposite signs, the root of f (x) = 0 lies between 2 and 3. Taking x 1 = 2, x 2 = 3 and using Regula-Falsi method, the first approximation is given by

21. Since f (x 2 ) and f (x 3 ) are of opposite signs, the root lies between x 2 and x 3. The second approximation to the root is given as

22. Now, we observe that f (x 2 ) and f (x 4 ) are of opposite signs, the third approximation is obtained from

23. This procedure can be continued till we get the desired result. The first three iterations are shown as in the table. nx n+1 f (x n+1 ) 22.9-0.711 32.94156-0.0207 42.94275-0.0011896

24. We observe that the value of the root as a third approximation is evidently different in both the cases, while the value of x 5, when the interval considered is (2, 3), is closer to the root. Important observation: The initial interval (x 1, x 2 ) in which the root of the equation lies should be sufficiently small.

25. Example Use Regula-Falsi method to find a real root of the equation log x – cos x = 0 accurate to four decimal places after three successive approximations.

26. Solution Given f (x) = log x – cos x. We observe that f (1) = 0-0.5403,and f (2)=0.69315+0.41615=1.1093 Since f (1) and f (2) are of opposite signs, the root lies between x 1 = 1, x 2 = 2.

27. The first approximation is obtained from

28. Now, since f (x 1 ) and f (x 3 ) are of opposite signs, the second approximation is obtained as

29. Similarly, we observe that f (x 1 ) and f (x 4 ) are of opposite signs, so, the third approximation is given by The required real root is 1.3030.

30. Lecture 4 Numerical Analysis