# HW_5 - HOMEWORK#5 SOLUTION 1 f x = e x 4 x 2 a Bisection...

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HOMEWORK#5 SOLUTION 1) 2 ( ) 4 x f x e x = - a) Bisection method xL = 0, xU = 1 Iteration1 0 ( ) (0) 4*0 f xL f e = = - = 1 1 ( ) (1) 4*1 f xU f e = = - = -1.2817 ( )* ( ) 1.2817 f xL f xU = - < 0 ( ) (0 1) 0.5 2 2 xL xU xR + + = = = 0.5 2 ( ) (0.5) 4*0.5 f xR f e = = - = 0.6487 ( )* ( ) 0 f xR f xU < Hence xL=xR=0.5 ( ) 0.6487 f xL = Iteration2 0.5 2 ( ) (0.5) 4*0.5 f xL f e = = - = 0.6487 1 ( ) (1) 4*1 f xU f e = = - = -1.2817 ( ) (0.5 1) 0.75 2 2 xL xU xR + + = = = Thus after 2 iterations, xR=0.75 0.75 2 ( ) (0.75) 4*0.75 f xR f e = = - = - 0. 1329 b) False position method xL = 0, xU = 1 Iteration1 0 ( ) (0) 4*0 f xL f e = = - = 1

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( ) (1) 4*1 f xU f e = = - = -1.2817 ( )* ( ) 1.2817 f xL f xU = - < 0 ( )*( ) ( ) ( ) ( ) f xU xL xU xR xU f xL f xU - = - - 1.2817*(0 1) 1 ( ) 1 ( 1.2817) xR - - = - - - =0.4382 0.4382 2 ( ) (0.4382) 4*0.4382 f xR f e = = - = 0.7818 ( )* ( ) 0 f xR f xU < Hence xL=xR=0.4382 ( ) 0.7818 f xL = Iteration2 0.4382 2 ( ) (0.4382) 4*0.4382 f xL f e = = - = 0.7818 1 ( ) (1) 4*1 f xU f e = = - = -1.2817 ( )*( ) ( ) ( ) ( ) f xU xL xU xR xU f xL f xU - = - - 1.2817*(0.4382 1) 1 ( ) 0.7818 ( 1.2817) xR - - = - - - =0.651 Thus after 2 iterations, xR=0.651 fR = 0.222 c) Fixed point iteration Iteration1
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HW_5 - HOMEWORK#5 SOLUTION 1 f x = e x 4 x 2 a Bisection...

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