MidI2008Sol - Midterm Solutions Problem 2(a Our function is f(x =(3x 1)1\/3 At x = 1 f(x x = 41\/3 1 This is positive since 13 < 4 At x = 2 f(x 2 = 71\/3 2

# MidI2008Sol - Midterm Solutions Problem 2(a Our function is...

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Midterm Solutions Problem 2: (a) Our function is f ( x ) = (3 x + 1) 1 / 3 . At x = 1, f ( x ) - x = 4 1 / 3 - 1. This is positive, since 1 3 < 4. At x = 2, f ( x ) - 2 = 7 1 / 3 - 2. This is negative, since 2 3 > 7. By the Intermediate Value Theorem, f ( x ) - x has a root in the interval [1 , 2]. Equivalently, f ( x ) has a fixed point in this interval. (b) Fixed point iteration for f ( x ) converges in an interval [ a, b ] if there is a root in this interval and | f 0 ( x ) | < λ < 1 for all x [ a, b ]. From part (a), we know there is a root. We compute f 0 ( x ) = 1 3 3(3 x + 1) - 2 / 3 = (3 x + 1) - 2 / 3 . Since 3 x + 1 4 for x in [1 , 2], we conclude that (3 x + 1) - 2 / 3 4 - 2 / 3 < 1 for all x in the interval. This satisfies the conditions of the theorem, so the fixed point iteration p k +1 = (3 p k + 1) 1 / 3 convergers for any starting point p 0 [1 , 2]. Problem 3: We want to find a polynomial P ( x ) of degree at most 2 such that P (0) = P (1) = P (2) = 1. We compute the Lagrange polynomials L 0 = ( x - 1)( x - 2) (0 - 1)(0 - 2) , L 1 = ( x - 0)( x - 2) (1 - 0)(1 - 2) , L 2 = ( x - 0)( x - 1) (2 - 0)(2 - 1) . Our polynomial P should be 1 L 0 +1 L 1 +1 L 2 = ( x 2 - 3 x + 2) / 2 - ( x 2 - 2 x ) + ( x 2 - x ) / 2 = 0 x 2 + 0 x + 1. So P ( x ) = 1. We verify P (1) = P (2) = P (3) = 1, so this is the correct polynomial. Problem 4: Relative error is | p - p * | | p | , and this should be at most 10
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