ECE3040 Chapter 3.pdf - Chapter 3 The lecture notes are selected from the public Powerpoint lecture documents provided by Authors Autar Kaw Sri Harsha

ECE3040 Chapter 3.pdf - Chapter 3 The lecture notes are...

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Chapter 3 The lecture notes are selected from the public Powerpoint lecture documents provided by Authors: Autar Kaw, Sri Harsha Garapati
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Bisection Method
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Basis of Bisection Method Theorem x f(x) x u x An equation f(x)=0, where f(x) is a real continuous function, has at least one root between x l and x u if f(x l ) f(x u ) < 0. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign.
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x f(x) x u x Basis of Bisection Method Figure 2 If function does not change sign between two points, roots of the equation may still exist between the two points.   x f   0 x f
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x f(x) x u x Basis of Bisection Method Figure 3 If the function does not change sign between two points, there may not be any roots for the equation between the two points. x f(x) x u x   x f   0 x f
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x f(x) x u x Basis of Bisection Method Figure 4 If the function changes sign between two points, more than one root for the equation may exist between the two points.   x f   0 x f
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Algorithm for Bisection Method
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Step 1 Choose x and x u as two guesses for the root such that f(x ) f(x u ) < 0, or in other words, f(x) changes sign between x and x u . This was demonstrated in Figure 1. x f(x) x u x Figure 1
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x f(x) x u x x m Step 2 Estimate the root, x m of the equation f (x) = 0 as the mid point between x and x u as x x m = x u 2 Figure 5 Estimate of x m
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Step 3 Now check the following a) If , then the root lies between x and x m ; then x = x ; x u = x m . b) If , then the root lies between x m and x u ; then x = x m ; x u = x u . c) If ; then the root is x m. Stop the algorithm if this is true. 0 m l x f x f 0 m l x f x f 0 m l x f x f
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Step 4 x x m = x u 2 100 new m old m new a x x x m root of estimate current new m x root of estimate previous old m x Find the new estimate of the root Find the absolute relative approximate error where
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Step 5 Is ? Yes No Go to Step 2 using new upper and lower guesses. Stop the algorithm Compare the absolute relative approximate error with the pre-specified error tolerance . a s s a  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.
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Example 1 You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The floating ball has a specific gravity of 0.6 and has a radius of 5.5 cm. You are asked to find the depth to which the ball is submerged when floating in water. Figure 6 Diagram of the floating ball
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Example 1 Cont.
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