hw2_solution_SEJ

# hw2_solution_SEJ - EE103 Spring 2010 HW2 Sol. Prof. S.E....

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EE103 Spring 2010 HW2 Sol. Prof. S.E. Jacobsen 1 Applied Numerical Computing EE103 Spring 2010 Instructor: Prof. S. E. Jacobsen HW2 Solution Problem 1 Machine epsilon 1 2 n M  is defined as the upper bound of the relative round-off error for a (,) n finite precision machine, and its interpretation is that 1+ M is the smallest number larger than one that the machine can distinguish from 1. Therefore, for the IEEE single precision format ( 2, 24 n ), machine epsilon M is 24 2 , and for the IEEE double precision format ( 2, 53 n ) , machine epsilon M is 53 2 . (a) For the 8 (,) ( 8 , ) nn machine, machine epsilon is 8 1 82 n . To achieve the relative error accuracy that an IEEE single precision machine possesses, we should have 8 8 1 33 24 23 8 6 22 2 . 23 n n n   That is, 8 n should at least be 9. Similarly, to achieve the relative error accuracy that an IEEE double precision machine possesses, we should have 8 8 1 53 52 8 85 5 2 . n n n That is, 8 n should at least be 19. (b) For the 10 1 0 machine, machine epsilon is 10 1 10 2 n . To achieve the relative error accuracy that an IEEE single precision machine possesses, we should have 10 10 1 1 24 23 10 10 10 10 2 10 2 1 23log 2 7.92. 2 n n That is, 8 n should at least be 8. Similarly, to achieve the relative error accuracy that an IEEE double precision machine possesses, we should have 10 10 1 1 53 52 10 10 10 10 2 10 2 1 52log 2 16.65. 2 n n That is, 8 n should at least be 17.

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EE103 Spring 2010 HW2 Sol. Prof. S.E. Jacobsen 2 Problem 2 (a) In this problem, the roots of 2 1 x x are   152 x  and the derivative of 2 () 1 gx x  is '( ) 2 g xx . However, 2 5 1 1 x , which means the fixed point method will not converge to either of the roots. Now consider the other function 1/2 () ( 1 ) x  . Notice that x has to be greater than -1 for g x to be valid, so x can only be   . The derivative is  1/2 1 2 1 x , and 1/2 11 1 2 1 1 2 51 x x  . Therefore, the fixed point method will converge to the root   at a linear (geometric) rate (since 0 and 1 ). From the figure below, it is also clear that for any initial point > -1, the fixed point method will converge to the root   . -1 -0.5 0 0.5 1 1.5 2 -1 -0.5 0 0.5 1 1.5 2 x sqrt(1+x) x (b) In the fixed point method, if k x is in the interval where () g x is greater than x , the next point 1 k x must be greater than k x . This can be easily shown by 1 kk k x x  . On the other hand, if k x g x is less than x , the next point 1 k x must be less than k x . Hence, in this problem if the initial point is chosen to be any number greater than   (root1 in the figure), k x will diverge to infinity. This can be easily shown by the following
EE103 Spring 2010 HW2 Sol. Prof. S.E. Jacobsen 3 Matlab code: g = @(x)x^2-1; xk = 2; % initial point x0 xk = g(xk) xk = 3 xk = g(xk) xk = 8 xk = g(xk) xk = 63 xk = g(xk) xk = 3968 xk = g(xk) xk =15745023 diary off; From part (a), we know that even for the initial points less than   152 , k x will not converge to any roots. However,

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## This note was uploaded on 06/01/2010 for the course EE EE 103 taught by Professor Jacobsen during the Spring '09 term at UCLA.

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hw2_solution_SEJ - EE103 Spring 2010 HW2 Sol. Prof. S.E....

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