{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 * )
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}