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# Numerical Analysis (Solution of Non-Linear Equations)

Bisection Method Regula-Falsi Method Method of iteration Newton - Raphson Method Muller’s Method Graeffe’s Root Squaring Method
Published on: Mar 3, 2016
Published in: Education
Source: www.slideshare.net

#### Transcripts - Numerical Analysis (Solution of Non-Linear Equations)

• 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. ExampleExample SolveSolve xx33 – 9– 9xx + 1 = 0+ 1 = 0 for the root betweenfor the root between xx = 2 and= 2 and xx = 4 by the= 4 by the bisection method.bisection method.
• 6. Solution GivenSolution Given ff ((xx) =) = xx33 – 9– 9xx + 1.+ 1. HereHere ff (2) = -9,(2) = -9, ff (4) = 29.(4) = 29. Therefore,Therefore, ff (2)(2) ff (4) < 0 and hence(4) < 0 and hence the root lies between 2 and 4.the root lies between 2 and 4. LetLet xx00 = 2,= 2, xx11 = 4. Now, we define= 4. Now, we define as a first approximation to a root ofas a first approximation to a root of ff ((xx) = 0 and note that) = 0 and note that ff (3) = 1, so(3) = 1, so thatthat ff (2)(2) ff (3) < 0.(3) < 0. Thus the root liesThus the root lies between 2 and 3between 2 and 3 0 1 2 2 4 3 2 2 x x x + + = = =
• 7. We further define,We further define, and note thatand note that ff ((xx33) =) = ff (2.5) < 0, so that(2.5) < 0, so that ff (2.5)(2.5) ff (3) < 0. Therefore, we define the(3) < 0. Therefore, we define the mid-point,mid-point, Similarly,Similarly, xx55 = 2 . 875 and= 2 . 875 and xx66 = 2.9375= 2.9375 and the process can be continued untiland the process can be continued until the root is obtained to the desiredthe root is obtained to the desired accuracy.accuracy. 0 2 3 2 3 2.5 2 2 x x x + + = = = 3 2 4 2.5 3 2.75, etc. 2 2 x x x + + = = =
• 8. These results are presented in the table.These results are presented in the table. nn xxnn f ( xf ( xnn )) 22 33 1.01.0 33 2.52.5 -5.875-5.875 44 2.752.75 -2.9531-2.9531 55 2.8752.875 -1.1113-1.1113 66 2.93752.9375 -0.0901-0.0901
• 9. Regula-Falsi Method Method of falseMethod of false positionposition Method of falseMethod of false positionposition
• 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 1 1 ( ) ( ) ( ) n n n nn n y f x x x f x f x x x− − − − = − − 1 1 ( ) ( ) ( ) n n n n n n x x x x f x f x f x − − − = − −
• 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). 1 1 1 ( ) ( ) ( ) n n n n n n n x x x x f x f x f x − + − − = − −
• 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 2 1 3 2 2 2 1 ( ) ( ) ( ) 2 29 4 2.47368 38 x x x x f x f x f x − = − − × = − =
• 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 3 2 4 3 3 3 2 ( ) ( ) ( ) 2.73989 x x x x f x f x f x − = − − =
• 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. 4 2 5 4 4 4 2 ( ) 2.86125 ( ) ( ) x x x x f x f x f x − = − = −
• 19. This procedure can be continued till we get the desired result. The first three iterations are shown as in the table. n xn+1 f (xn+1) 2 2.47368 -6.12644 3 2.73989 -3.090707 4 2.86125 -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 x1 = 2, x2 = 3 and using Regula-Falsi method, the first approximation is given by 3 2 1 3 2 2 2 1 ( ) ( ) ( ) 1 3 2.9 10 ( ) 0.711 x x x x f x f x f x f x − = − − = − = =−
• 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 3 2 4 3 3 3 2 4 ( ) 2.94156 ( ) ( ) ( ) 0.0207 x x x x f x f x f x f x − = − = − = −
• 22. Now, we observe that f (x2) and f (x4) are of opposite signs, the third approximation is obtained from 4 2 5 4 4 4 2 ( ) ( ) ( ) 2.94275 ( ) 0.0011896 x x x x f x f x f x f x − = − − = =−
• 23. This procedure can be continued till we get the desired result. The first three iterations are shown as in the table. n xn+1 f (xn+1) 2 2.9 -0.711 3 2.94156 -0.0207 4 2.94275 -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 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 2 1 3 2 2 2 1 3 ( ) ( ) ( ) 1.1093 2 1.3275 1.6496 ( ) 02833 0.2409 0.0424 x x x x f x f x f x f x − = − − = − = = − =
• 28. Now, since f (x1) and f (x3) are of opposite signs, the second approximation is obtained as 4 3 4 (.3275)(.0424) 1.3275 0.0424 0.5403 1.3037 ( ) 1.24816 10 x f x − = − + = = ×
• 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. 5 4 5 (0.3037)(0.001248) 1.3037 0.001248 0.5403 1.3030 ( ) 0.6245 10 x f x − = − + = = ×
• 30. Lecture 4 Numerical Analysis