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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 openbook 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
 Simon

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