# computational mathematic

Post on 03-Dec-2014

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• 1. Summary of LecturesJust click to see next animation Prepared by: Dr. Suhaila Mohamad Yusuf suhailamy@utm.my Prepared by Dr. Suhaila Mohamad Yusuf

2. CHAPTER 6NON-LINEAR EQUATIONSPrepared by Dr. Suhaila Mohamad Yusuf 3. Centre Limit Theorem Given an equation of f(x) with an interval of[a,b], you need to determine whether thereexist at least a real root in that interval CLT said that if f(a) and f(b) have opposite sign(one is ve and another is +ve) then thereexist at least a real root in that intervala bf(a) +ve f(b) -ve Prepared by Dr. Suhaila Mohamad Yusuf 4. Bisection Methodf(x) = x3 3x2 + 8x - 5 c = (a + b) / 2 [0,1] =0.005 iab f(a)f(b)c f(c) 001-5 10.5 -1.625 1These f(c) > then,0.5are from 1the-1.625 given Calculated from this stop!1Is this < ? Yes,0.75 -0.266 2new interval! 0.75 interval 1-0.266No, next iteration!1 equation0.8750.373 3 0.750.875-0.266 0.3730.8125 0.056 4 0.750.813How to choose 0.056new 0.7815 -0.266 the-0.103 5 0.782 0.813 interval?-0.103 0.0560.7975 -0.021 6 0.798 0.813-0.021 0.0560.8055 0.020 7 0.798 0.806-0.021 0.02 0.8020.002Make sure these parts 0Repeat all the steps0.51 This is theCAREFULLY until f(c) < have opposite sign!-ve f(x)-ve f(x)+ve f(x) root!! We need to take this c value. CLT said that f(a) and f(b) How about another one? should have opposite sign! Prepared by Dr. Suhaila Mohamad Yusuf 5. False Position Methodf(x) = x3 3x2 + 8x - 5 c = [af(b) - bf(a)] / [f(b) f(a)] [0,1] =0.005 i a bf(a)f(b)cf(c) 0 0 1-51 0.833 -1.62510 0.8333 -5 These are from the given0.162Is this < ? Yes, 0.807 Calculated from this0.029stop!200.807 interval-50.029 No, next iteration! 0.802 equation 0.004This is theRemember how to choose theroot!!interval value? What does the CLT said about interval?Prepared by Dr. Suhaila Mohamad Yusuf 6. Secant Methodf(x) = sin (x) + 3x e3 xi+2 = [xif(xi+1) xi+1f(xi)] / [f(xi+1) f(xi)]x0 = 1 , x1 = 0 =0.0005 ixixi+1 xi+2f(xi+2) 010 0.47100.26521 0 0.4710 ? Yes, stop! These are from the given Is this

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