Lecture3

# Lecture3 - Lecture 3 Solution of Lecture 3 Solution of...

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ecture 3: Solution of Lecture 3: Solution of equations by iteration 1

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Basis of Bisection Method Theorem 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. f(x) x x u x igure 1 t least one root exists between the two points if the function is 3 Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign.

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Basis of Bisection Method f(x) x x u x igure 2 function does not change sign between two   x f 4 Figure 2 If function does not change sign between two points, roots of the equation may still exist between the two points.   0 x f
Basis of Bisection Method f(x) f(x) x x u x x x u x 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   0 x f 5

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Basis of Bisection Method f(x) x x u x Figure 4 If the function changes sign between two points, ore than one root for the equation may exist between the two   x f 6 more than one root for the equation may exist between the two points.   0 x f
lgorithm for Bisection Method Algorithm for Bisection Method 7

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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 etween x nd x his was demonstrated in Figure 1. between x and x u . This was demonstrated in Figure 1. f(x) x x u x 8 Figure 1
Step 2 x) Estimate the root, x m of the equation f (x) = 0 as the mid point between x and x u as f(x) x x m = x u 2 x x u x x m 9 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 .     0 m l x f x f b) If , then the root lies between x m and x u ; then x = x m ; x u = x u .     0 m l x f x f c) If ; then the root is x m. Stop the algorithm if this is true.  0 m l x f x f 10
Step 4 Find the new estimate of the root x x m = x u 2 Find the absolute relative approximate error 100 new m old m new a x x x m root of estimate previous old m x where 11 root of estimate current new m x

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Step 5 Compare the absolute relative approximate error with the pre-specified error tolerance . a s Is ? Yes Go to Step 2 using new upper and lower guesses.  No Stop the algorithm s a Note one should also check whether the number of iterations is more than the maximum number of iterations llowed If so one needs to terminate the algorithm and 12 allowed. If so, one needs to terminate the algorithm and notify the user about it.
Example 1 You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The floating ball has a ecific gravity of 0.6 and has a radius of 5.5 cm. You 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.

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## This note was uploaded on 08/30/2011 for the course BUSN 1003 taught by Professor Carr,r during the Spring '08 term at Arkansas State.

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Lecture3 - Lecture 3 Solution of Lecture 3 Solution of...

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