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midterm_2006Ans

# midterm_2006Ans - Fall 2006 Midterm Exam Answer key ARE211...

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Fall 2006 ARE211 Midterm Exam - Answer key Problem 1 : Metrics and continuity . (10 points) Let X be an arbitrary universe, let d be an arbitrary metric. For arbitrary y X , consider the function f y : X R n defined by f y ( · ) = d ( · , y ). Hint: To provide a counter-example to either of the conjectures below, you will need to specify an X , a metric d , an element y X , a sequence { x n } and a ¯ x , and show that for these variables, the selected conjecture is false. a) Prove, or provide a counter-example, to the following conjecture. Conjecture: The function f y is continuous when d is the metric on the domain and the metric on R n is the Pythagorian metric. Ans: Fix a sequence { x n } that converges to ¯ x X , w.r.t. the metric d . We need to show that { f y ( x n ) } converges to f y x ) . Fix epsilon1 > 0 arbitrarily and pick N N such that for all n > N d ( x n , ¯ x ) < epsilon1 . From the triangle inequality, we have d ( x n , y ) d ( x n , ¯ x ) + d x, y ) , so that d ( x n , y ) - d x, y ) d ( x n , ¯ x ) < epsilon1 . Hence | f y ( x n ) - f y x ) | = d ( x n , y ) - d x, y ) < epsilon1 . b) Prove, or provide a counter-example, to the following conjecture: Conjecture: ( { f y ( x n ) } converges to f y x ) in the Pythagorian metric) implies ( { x n } converges to ¯ x in the metric d ). Ans: Let X = R , d be the Pythagorian metric and y = 0 . Let x n = ( - 1) n and let ¯ x = 1 . For each n , f y ( x n ) = d ( x n , 0) = 1 . Also f y x ) = d x, 0) = 1 . Therefore, the sequence { f y ( x n ) } trivially converges to f y x ) . However, the sequence { x n } does not converge to 1. Problem 2 : Extrema . (20 points) Let F denote the set of all continuous functions from R to R and consider the following five properties: P : f is strictly quasiconcave Q : f is concave R : f attains a global maximum S : f attains a global minimum T : f is bounded (that is, the image f ( R ) is bounded above and bounded below.) There are 2 5 = 32 ways of combining these five properties (i.e., each can be satisfied or not). We have selected ten of these thirty-two. For each one of these ten, say whether or not there is an element of F that satisfies the combination. If your answer is “yes,” provide an example; if it is “no,” prove that the combination is internally inconsistent. (Note that you may not need to use all five properties in order to show internal inconsistency.) For example, for the combination ( P Q ¬ R ¬ S ¬ T) there does exist f F satisfying these conditions: f ( x ) = x . The ten combinations are a) ( ¬ P ¬ Q ¬ R ¬ S ¬ T) Ans: Yes. f ( x ) = x ( x + 1)( x - 1) . b) ( ¬ P ¬ Q R S ¬ T)

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2 Ans: No. R and S implies T. To see this, let f be maximized at ¯ x and minimized at x ¯ and pick N N such that N > max( | f x ) | , | f ( x ¯ ) | ) . But this means that f is bounded above by N and below by - N . c) ( ¬ P ¬ Q R S T) Ans: Yes. f ( x ) = braceleftBigg 0 if | x | > 1 x ( x + 1)( x - 1) if | x | ≤ 1 . d) ( ¬ P Q ¬ R ¬ S T) Ans: No. ¬ P and Q implies R. Specifically, the condition ¬ P implies that there exists x * , y * , z * and t * such that z * = t * x * +(1 - t * ) y * and f ( z * ) min { f ( x * ) , f ( y * ) } . The condition Q implies that for all x , y s.t. x negationslash = y and all t (0 , 1) , f ( tx + (1 - t ) y ) tf ( x ) + (1 - t ) f ( y ) . First suppose f ( x * ) negationslash = f ( y * ) and, without loss of generality, let f ( x * ) < f ( y * )
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