VIEWS: 19 PAGES: 4 CATEGORY: College POSTED ON: 5/11/2012 Public Domain
www.VUsolutions.com Assignment # 1 spring 2012 Question 1 Find the real root of the equation correct to three decimal places by Bisection method. Solution We will take to points in which the root can lie x0 1 x1 2 Root may lie between 0 and 1 so we will put it as f (0) 2(0)3 0 2 0 f (0) 0 0 2 0 = -2 Now f (1) 2(1)3 1 2 0 f (1) 2(1)3 1 2 0 2 1 2 1 So both have opposite signs 1 and -2 so root lies between these two points Now x0 2 And x1 1 As we know the formula of Bisection method x x1 x2 0 2 Putting the values in the equation we get 2 1 1 x2 0.5 2 2 Now to perform next approximation we will calculate f (0.5) 2(0.5)3 (0.5) 2 0 2(0.125) (0.5) 2 0.25 (0.5) 2 0.25 0.5 2 2.25 - Now we will apply formula as (2) (0.5) x3 3.5 2 Now putting values in equation we get (2) (0.5) x3 3.5 2 Now the same as we have done before f (3.5) 2(3.5)3 (3.5) 2 0 85.75 (5.5) 85.75 5.5 91.25 Now for the last 3rd iteration Follow the same above procedure do it yourself Question 2 Use Regula-Fasli method to find the real root of the equation e x 3x 0 Correct to four decimal places after three successive approximations in 0 x 1 Solution As it is an algebric equation so we can write this as in the form of f ( x) e x 3 x = 0 Now as we know the condition is 0 x 1 So first approximation will be 0 and 1 as root must lie between these two points 0 and 1 So f ( x) e x 3 x f (0) e0 3(0) 1 f (1) e1 3(1) 2.71828 3 0.2817 Since f (0) f (1) 0,Soliesbetween and 1 1st iteration x2 x1 x3 x2 f ( x2 ). f ( x2 ) f ( x1 ) letsay x1 0and x2 1 x2 x1 1 0 1 x3 x2 f ( x2 ) (0.2817) 1 (0.2817) f ( x2 ) f ( x1 ) ( 0.2817) (1) (1.2817) 0.7802 f ( x3 ) f (0.7802) e0.7802 3(0.7802) 0.1587 sin ce f ( x3 ) 0,soweneed tocalculatex4 Now f ( x3 ) f ( x1 ) 0Thereforesolutionliesbetween 0.7802and 2nd iteration x3 x1 (0.7802) 0 x4 x3 f ( x3 ) (0.7802) (0.1587) f ( x3 ) f ( x1 ) (0.1587) (1) 0.6733 f ( x4 ) f (.6733) e.6733 3(.6733) 0.0592 sin ce f ( x4 ) 0,soweneed tocalculate x Now f ( x4 ) f ( x2 ) 0Thereforesolutionliesbetween.6733and 3rd iteration x4 x1 x5 x4 f ( x4 ) f ( x4 ) f ( x1 ) .6733 0 x5 .6733 (0.0592) (0.0592) (1) x5 0.6357 f ( x5 ) f (0.6357) e0.6357 3(0.6357) 0.0188 f ( x5 ) 0.0188 sin ce f ( x5 ) 0,soweneed tocalculatex Now f ( x5 ) f ( x1 ) 0Thereforesolutionliesbetween0.6357and 4th iteration x5 x1 x6 x5 f ( x5 ) f ( x5 ) f ( x1 ) 0.6357 0 x5 0.6357 (0.0188) (0.0188) (1) x6 0.6240 f ( x6 ) f (0.6240) e0.6240 3(0.6240) 0.0056 f ( x6 ) 0.0056 sin ce f ( x6 ) 0,soweneed tocalculatex Now f ( x7 ) f ( x1 ) 0Thereforesolutionliesbetween0.6240and 5th iteration x6 x1 x7 x6 f ( x6 ) f ( x6 ) f ( x1 ) 0.6240 0 x7 0.6240 ( 0.0056) (0.0056) (1) x7 0.6205 f ( x7 ) f (0.6205) e0.6205 3(0.6205) 0.0016 similarly 6th iteration x8 0.6195 The required answer is up to 4th iteration so take only 4 iterations that’s enough Other Question will upload soon stay blessed