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Unformatted text preview: ARE211 FINAL EXAM DECEMBER 12, 2003 This is the final exam for ARE211. As announced earlier, this is an open-book exam. Try to allocate your 180 minutes in this exam wisely, and keep in mind that leaving any questions unanswered is not a good strategy. Make sure that you do all the easy questions, and easy parts of hard questions, before you move onto the hard questions. Problem 1 (15 points) (1) In Euclidean space, under the Pythagorean metric, assume that a sequence ( x n ) satisfies | x n- x n +1 | ≤ α | x n- x n- 1 | for each n = 2 , 3 , ... for some fixed 0 < α < 1. Show that ( x n ) is a convergent sequence. Solution : Let c = | x 2- x 1 | . An easy inductive argument shows that for each n we have | x n +1- x n | ≤ cα n- 1 . Thus, | x n + p- x n | ≤ p X i =1 | x n + i- x n + i- 1 | ≤ c p X i =1 α n + i- 2 ≤ c 1- α α n- 1 holds for all n and all p . Since lim α n = 0 , it follows that ( x n ) is a Cauchy sequence, and hence a convergent sequence, under this complete metric space. (2) Let ( X, d ) be a complete metric space. A function f : X → X is called a contraction if there exists some < α < 1 such that ∀ x, y ∈ X, d ( f ( x ) , f ( y )) ≤ αd ( x, y ) α is called a contraction constant. Show that for every contraction f on a complete metric space ( X, d ), there exists a unique point x ∈ X such that f ( x ) = x . (Such a point x ∈ X is called a fixed point .) (Hint: To prove this, construct a sequence that has the property defined in (1). An understanding of com- pleteness will also be helpful.) Solution : Note first that if f ( x ) = x and f ( y ) = y hold, then the inequality d ( f ( x ) , f ( y )) ≤ αd ( x, y ) easily implies that d ( x, y ) = 0 , and so x = y . That is, f has at most one fixed point. To see that f has a fixed point, choose some a ∈ X , and then define the sequence ( x n ) inductively by x 1 = a and x n +1 = f ( x n ) for n = 1 , 2 , ... From our condition, it follows that d ( x n +1 , x n ) = d ( f ( x n ) , f ( x n- 1 )) ≤ αd ( x n , x n- 1 ) holds for n = 2 , 3 , ... Thus, as in part (1), we have shown that ( x n ) is a convergent sequence. Let x = lim x n ....
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This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at Berkeley.
- Fall '07