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128a_su09_final_sol

128a_su09_final_sol - UCB MATH 128A-2 SUMMER 2009 FINAL...

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UCB MATH 128A-2, SUMMER 2009: FINAL EXAM SOLUTIONS JUSTIN BLANCHARD 1. (a) Prove that the bisection method on [1 , 2] will converge to a root of x 3 - x - 1. (b) Find a bound on the absolute error in approximating this root after n iterations. (c) How many iterations are necessary to ensure the relative error is at most 2 - 5 ? (Your answer does not have to be exact.) Solution: (a) This function is continuous. Its values at the endpoints are - 1 < 0 and 5 > 0. Therefore, the interval brackets a root and the bisection method will converge. (b) On the n th iteration, the root is known to lie in an interval of width 2 - 1 2 n - 1 . Its distance from this interval’s midpoint can be at most 1 / 2 n . (The answer 1 / 2 n +1 is also acceptable, depending on how the phrase “after n iterations” is interpreted.) (c) Since the root lies in [1 , 2], the relative error is at most equal to the absolute error. By (b), we must use n = 5 iterations. 2. (a) Find an algebraic expression for the unique root (in R ) of f ( x ) = x 2 - 2 /x . (b) Newton’s method searches for this root using the iteration p n +1 = g ( p n ). Find g ( x ). Simplify your answer. (c) Is the convergence linear? Is it at least quadratic? (d) Find the exact order of convergence. Solution: (a) x 2 - 2 /x = 0 ⇐⇒ x 3 - 2 = 0 ⇐⇒ x = 2 1 / 3 . (b) g ( x ) = x - f ( x ) f 0 ( x ) = x - x 2 - 2 /x 2 x +2 /x 2 = x - x x 3 - 2 2 x 3 +2 = x ( x 3 + 4) 2( x 3 + 1) (c) The convergence is at least quadratic, for either of the following reasons: 2 1 / 3 is a simple root of f ( x ), as f 0 (2 1 / 3 ) = 2 · 2 1 / 3 + 2 / 2 2 / 3 6 = 0. Alternatively, g 0 (2 1 / 3 ) = 0 after the calculation g 0 ( x ) = (4 x 3 +4)2( x 3 +1) - x ( x 3 +4)(6 x 2 ) 2 2 ( x 3 +1) 2 = 8( x 3 +1) 2 - 6 x 3 ( x 3 +4) 2 2 ( x 3 +1) 2 . (d) (This was intended to be a harder problem.) By definition, the order of convergence is α iff lim n →∞ p n +1 - 2 1 / 3 p n - 2 1 / 3 = λ 6 = 0. If p n = x , this fraction’s numerator is g ( x ) - 2 1 / 3 = x 4 - 2 4 / 3 x 3 +4 x - 2 2( x 3 +1) . Factor out ( x - 2 1 / 3 ) as many times as possible: ( x - 2 1 / 3 ) 3 ( x +2 1 / 3 ) 2( x 3 +1) . (Alternatively, Taylor-expand x 4 - 2 4 / 3 x 3 + 4 x - 2 with center x = 2

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128a_su09_final_sol - UCB MATH 128A-2 SUMMER 2009 FINAL...

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