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First task of numerical method lecture

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- 1. Bisection and Fixed-Point Method Maria Priscillya Pasaribu 4103312018 Bilingual Mathematics Education
- 2. Bisection Method Bisection method is one of the closed methods (bracketing method) to determine the root of a nonlinear equation f(x) = 0, with the following main principles: •Using two initial values to confine one or more roots of non-linear equations. •Root value is estimated by the midpoint between two existing initial values
- 3. Bisection Method
- 4. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign
- 5. If , then there may be more than one root between and
- 6. Algorithm for The Bisection Method
- 7. Step 1 Choose x and xu as two guesses for the root such that f(x) f(xu) < 0, or in other words, f(x) changes sign between x and xu. This was demonstrated in Figure 1. f(x) x xu x Figure 1 http://numericalmethods.eng.usf.edu 9
- 8. Step 2 Estimate the root, xm of the equation f (x) = 0 as the mid point between x and xu as f(x) x + xu xm = 2 x xm xu x Figure 5 Estimate of xm http://numericalmethods.eng.usf.edu 10
- 9. Step 3 Now check the following a) If f ( xl ) f ( xm ) < ,0 then the root lies between x and xm; then x = x ; xu = xm. f ( xl ) f ( xm ) > 0 b) If , then the root lies between x m and xu; then x = xm; xu = xu. f ( xl ) f ( xm ) = 0 c) If ; then the root is x m. Stop the algorithm if this is true. http://numericalmethods.eng.usf.edu 11
- 10. Step 4 Find the new estimate of the root x + xu xm = 2 Find the absolute relative approximate error ∈a = old x new − xm m x new m ×100 where old xm = previous estimate of root new xm = current estimate of root http://numericalmethods.eng.usf.edu 12
- 11. Step 5 Compare the absolute relative approximate error ∈a with the pre-specified error tolerance ∈s . Yes Go to Step 2 using new upper and lower guesses. No Stop the algorithm Is ∈a >∈s? Note one should also check whether the number of iterations is more than the maximum number of iterations allowed. If so, one needs to terminate the algorithm and notify the user about it. 13 http://numericalmethods.eng.usf.edu
- 12. Example 1 Find the root of X3 – 9X2 + 18X – 6 = 0 has a unique root in [2, 2.5] with accuracy 10 -3.
- 13. Graph
- 14. Step 1: Assume that x = 2 and x = 2,5 as u two guesses for the root such that f(x) f(xu) < 0, or in other words, f(x) changes sign between x and xu from the equation X3 – 9X2 + 18X – 6 = 0
- 15. Step 2
- 16. Step 3
- 17. Step 4: Iteration 1
- 18. Cont.
- 19. Root of f(x)=0 as function of number of iterations for bisection method Iteration xl xu xm f(xl) f(xm) Error (xu- xm) 0 2 2,5 2,25 2 0,328 0,25 1 2,25 2,5 2,375 0,328 -2,86 0,125 2 2,25 2,375 2,3125 0,328 -0,137451 3 2,25 2,3125 2,28125 0,328 0,09744 0,03125 4 2,28125 2,3125 2,296875 0,09744 -0,019489 0,015625 5 2,28125 2,296875 2,2890625 0,09744 0,039107 6 2,2890625 2,296875 2,29296875 0,039107 0,00984 7 2,29296875 2,296875 2,29492188 0,00984 -0,004816 8 2,29296875 2,29492188 2,29394532 0,00984 0,00251476 9 2,29394532 2,29492188 2,2944336 0,00251476 -0,0011502 0,0625 0,0078125 0,00390625 0,00195312 0,000977 0,000488
- 20. The iteration is stopped because the error is approximate to 10-3. So, the estimated root is 2,294436.
- 21. Fixed-Point Method Fixed-point method is one of the opened methods that is finding approximate solutions of the equation f(x)=0
- 22. Algorithm of Fixed-point Method • Given an equation f(x)=0 • Rewrite the equation f(x)=0 in the form of x=g(x) • Let the initial guess be x0 and consider the recursive process • xn+1=g(xn), n= 0, 1, 2, ...
- 23. Example Find the root of X3 – 9X2 + 18X – 6 = 0 has a unique root in [2, 2.5] with accuracy 10 -3.
- 24. Solution
- 25. With x0 = 2.25, this is the result of the fixedpoint method for all five choices of g.
- 26. The root of the equation we got is 2,2944336, as was noted in example of Bisection Method. Comparing the results to the Bisection method given in that example, it can be seen that the same result at least have been obtained for choice d.

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