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Macroeconomics Exam Review 87

Macroeconomics Exam Review 87 - Solutions for Foundations...

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𝑋 is closed and hence complete (Exercise 1.107). Therefore, 𝑓 has a fixed point. That is, there exists 𝑥 0 𝑋 such that 𝑥 0 = 𝑓 ( 𝑥 0 ) = 1 2 ( 𝑥 0 + 2 𝑥 0 ) Rearranging 2 𝑥 2 0 = 𝑥 2 0 + 2 = 𝑥 2 0 = 2 so that 𝑥 0 = 2. Letting 𝑥 0 = 2 𝑥 1 = 1 2 (2 + 1) = 3 2 Using the error bounds in Corollary 2.5.1, 𝜌 ( 𝑥 𝑛 , 2) 𝛽 𝑛 1 𝛽 𝜌 ( 𝑥 0 , 𝑥 1 ) = (1 / 2) 𝑛 1 / 2 1 / 2 = 1 2 𝑛 = 1 1024 < 0 . 001 when 𝑛 = 10. Therefore, we conclude that 10 iterations are ample to reduce the error below 0 . 001. Actually, with experience, we can refine this a priori estimate. In Example 1.64, we calculated the first five terms of the sequence to be (2 , 1 . 5 , 1 . 416666666666667 , 1 . 41421568627451 , 1 . 41421356237469) We observe that 𝜌 ( 𝑥 3 , 𝑥 4 ) = 1 . 41421568627451 1 . 41421356237469) = 0 . 0000212389982 so that using the second inequality of Corollary 2.5.1 𝜌 ( 𝑥 4 , 2) 1 / 2 1 / 2 0 . 0000212389982 < 0 . 001 𝑥 4 = 1 . 41421356237469 is the desired approximation. 2.121 Choose any 𝑥 0 𝑆 . Define the sequence 𝑥 𝑛 = 𝑓 ( 𝑥 𝑛 ) = 𝑓 𝑛 ( 𝑥 0 ). Then ( 𝑥 𝑛 ) is a Cauchy sequence in
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