This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 counterexample 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 counterexample, 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 > 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 counterexample, 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 thirtytwo. 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) 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 ....
View
Full
Document
This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at University of California, Berkeley.
 Fall '07
 Simon

Click to edit the document details