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Lecture 4

Numerical Analysis. Lecture 4. Chapter 2. Solution of Non-Linear Equations. Introduction Bisection Method Regula-Falsi Method Method of iteration Newton - Raphson Method Muller’s Method Graeffe’s Root Squaring Method. Bisection Method (Bolzano). Example

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Lecture 4

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  1. Numerical Analysis Lecture 4

  2. Chapter 2 Solution of Non-Linear Equations

  3. IntroductionBisection MethodRegula-Falsi MethodMethod of iterationNewton - Raphson MethodMuller’s MethodGraeffe’s Root Squaring Method

  4. Bisection Method (Bolzano)

  5. Example Solve x3 – 9x + 1 = 0 for the root between x = 2 and x = 4 by the bisection method.

  6. Solution Given f (x) = x3 – 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 x0 = 2, x1 = 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 (x3) = f (2.5) < 0, so that f (2.5) f (3) < 0. Therefore, we define the mid-point, Similarly, x5 = 2 . 875 and x6 = 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.

  9. Regula-Falsi Method Method of false position

  10. Here, we choose two points xn and xn -1 such that f (xn) and f (xn-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[xn, f(xn)] and B[xn-1, f (xn-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 (xn-1) and f (xn+1) are of opposite sign. Thus, it is possible to apply to above procedure, to determine the line through B and A1 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 x3 – 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) = x3 – 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 x1 = 2, x2 = 4 and using Regula-Falsi method, the first approximation is given by

  17. Now f (x3) = –6.12644. Since f (x2) and f (x3) are of opposite signs, the root lies between x2 and x3. The second approximation to the root is given as

  18. Therefore f (x4) = – 3. 090707. Now, since f (x2) and f (x4) are of opposite signs, the third approximation is obtained from and f (x5) = – 1.32686.

  19. This procedure can be continued till we get the desired result. The first three iterations are shown as in the table.

  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 x1 = 2, x2 = 3 and using Regula-Falsi method, the first approximation is given by

  21. Since f (x2) and f (x3) are of opposite signs, the root lies between x2 and x3. The second approximation to the root is given as

  22. Now, we observe that f (x2) and f (x4) 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.

  24. We observe that the value of the root as a third approximation is evidently different in both the cases, while the value of x5, when the interval considered is (2, 3), is closer to the root. Important observation: The initial interval (x1, x2) 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 x1 = 1, x2 = 2.

  27. The first approximation is obtained from

  28. Now, since f (x1) and f (x3) are of opposite signs, the second approximation is obtained as

  29. Similarly, we observe that f (x1) and f (x4) are of opposite signs, so, the third approximation is given by The required real root is 1.3030.

  30. Numerical Analysis Lecture 4

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