hw2_solutions

hw2_solutions - MAE 107 Computational Methods in...

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Unformatted text preview: MAE 107 Computational Methods in Engineering Spring 2007 Homework #2 Solutions Problem 1: Problem 4.7 on page 110 of the textbook. The function (correcting a typographical error in the text) ( ) 2 3 1 1 ) ( x x f = has the derivative ( ) 2 2 3 1 6 ) ( x x x f = Evaluating this derivative at x = 0.577 using double precision arithmetic, rounded to the nearest integer 911 , 352 , 2 ) 577 . ( = f Since the denominator is close to zero, we expect problems with subtractive cancellation when using limited precision arithmetic. Using three-digit arithmetic with chopping (normalized base 10): ( ) 000 , 216 10 216 . 10 160 . 10 346 . ) 577 . ( 10 160 . 3 1 10 400 . 3 1 10 996 . 3 10 332 . 10 346 . 6 6 4 1 4 2 2 2 2 2 2 1 = = = = = = = = f x x x x x In this case, the relative error is % 8 . 90 or 908 . 911 , 352 , 2 000 , 216 911 , 352 , 2 = = Using four-digit arithmetic with chopping (normalized base 10): ( ) 000 , 048 , 2 10 2048 . 10 1690 . 10 3462 . ) 577 . ( 10 1690 . 3 1 10 1300 . 3 1 10 9987 . 3 10 3329 . 10 3462 . 6 7 4 1 5 2 2 2 2 2 2 1 = = = = = = = = f x x x x x And in this case the relative error is % 96 . 12 or 1296 . 911 , 352 , 2 000 , 048 , 2 911 , 352 , 2 = = Problem 2 : Problem 4.13 on page 110 of the textbook. Equation (4.11) on page 94 of the textbook is 2 1 1 1 ) )( ( ! 2 1 ) )( ( ) ( ) ( i i i i i i i i x x x f x x x f x f x f + + = + + + and for the function c bx ax x f + + = 2 ) ( we have a x f b ax x f 2 ) ( 2 ) ( = + = . Therefore, equation (4.11) becomes c bx ax ax x ax ax bx bx ax x ax c bx ax x x a x x b ax c bx ax x f i i i i i i i i i i i i i i i i i i i i i + + = + + + + + + = + + + + + = + + + + + + + + + 1 2 1 2 1 2 1 1 2 1 2 2 1 1 2 1 2 2 2 ) ( ! 2 2 ) )( 2 ( ) ( MAE 107 Spring 2007 HW 2 Problem 3 Contents bisect.m Define f(x), and plot it Use bisect.m to compute root of f(x) Use a smaller interval to see if number of iterations is reduced bisect.m type bisect.m % Function to find the root of function 'f' near between [x1, x2] % using bisection method function xm = bisect(f, x1, x2) itmax = 100; % set maximum number of iterations accuracy = 1e-4; % set required accuracy prev_xm = NaN; % used for computing estimated error disp(sprintf('Desired accuracy = %e\n', accuracy)); for iter = 1:itmax % calculate midpoint of dcurrent domain xm = (x1 + x2) / 2; disp( sprintf(... 'Iter#=%2d, x-range=(%+6.4f,%+6.4f), xm=%+8.6f, f(xm)=%+8.6e, Estim error=%+6.4e', ......
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hw2_solutions - MAE 107 Computational Methods in...

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