# mid1solns - CMPS 101 Algorithms and Abstract Data Types...

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CMPS 101 Algorithms and Abstract Data Types Summer 2009 Midterm Exam 1 Solutions 1. (20 Points) Prove that )) ( ) ( ( )) ( ( )) ( ( n g n f n g n f Ω = Ω Ω . In other words, if )) ( ( ) ( 1 n f n h Ω = and )) ( ( ) ( 2 n g n h Ω = , then )) ( ) ( ( ) ( ) ( 2 1 n g n f n h n h Ω = . Proof: Assume )) ( ( ) ( 1 n f n h Ω = and )) ( ( ) ( 2 n g n h Ω = . Then there exist positive constants 2 1 2 1 , , , n n c c such that for all 1 n n : ) ( ) ( 0 1 1 n h n f c , and for all 2 n n : ) ( ) ( 0 2 2 n h n g c . Let ) , max( 2 1 0 n n n = . Then for all 0 n n , we can multiply the above inequalities to get ) ( ) ( ) ( ) ( 0 2 1 2 1 n h n h n g n f c c . Now let 2 1 c c c = . Observe that c , and 0 n are positive, and for all 0 n n : ) ( ) ( ) ( ) ( 0 2 1 n h n h n g n cf , showing that )) ( ) ( ( ) ( ) ( 2 1 n g n f n h n h Ω = , as required. /// 2. (20 Points) Use Stirling’s formula to prove that ) log ( ) ! log( n n n Θ = . Proof: Taking ) log( of both sides of Stirling’s formula: ( ) ) / 1 ( 1 2 ! n e n n n n Θ + = π , we get ( ) ) / 1 ( 1 log log log 2 log ) ! log( n e n n n n Θ + + + + = π ( ) ) / 1 ( 1 log log log log ) 2 / 1 ( 2 log n e n n n n Θ + + - + + = π , whence ( ) n n n n e

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