mid1review_some_solns_MORE - CMPS 101 Midterm 1 Review...

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Unformatted text preview: CMPS 101 Midterm 1 Review Solutions to selected problems Problem 2 State whether the following assertions are true or false. If any statements are false, give a related statement which is true. a. )) ( ( ) ( n g O n f = implies )) ( ( ) ( n g o n f = . False )) ( ( ) ( n g o n f = implies )) ( ( ) ( n g O n f = b. )) ( ( ) ( n g O n f = if and only if )) ( ( ) ( n f n g Ω = . True c. )) ( ( ) ( n g n f Θ = if and only if L n g n f n = ∞ → )) ( / ) ( ( lim , where ∞ < < L . False ∞ < < L and L n g n f n = ∞ → )) ( / ) ( ( lim implies )) ( ( ) ( n g n f Θ = Problem 3 Prove that )) ( ) ( ( )) ( ( )) ( ( n g n f n g n f ⋅ Θ = Θ ⋅ Θ . In other words, if )) ( ( ) ( 1 n f n h Θ = and )) ( ( ) ( 2 n g n h Θ = , then )) ( ) ( ( ) ( ) ( 2 1 n g n f n h n h ⋅ Θ = ⋅ . Proof: By hypothesis there exist positive constants 1 n , 2 n , 1 a , 1 b , 2 a , and 2 b such that ) ( ) ( ) ( : 1 1 1 1 n f b n h n f a n n ≤ ≤ ≤ ≥ ∀ and ) ( ) ( ) ( : 2 2 2 2 n g b n h n g a n n ≤ ≤ ≤ ≥ ∀ If ) , max( 2 1 n n n n = ≥ , then both inequalities hold. Let 2 1 a a c = , and 2 1 b b d = . Since everything in sight is non-negative, we can multiply these two inequalities to get ) ( ) ( ) ( ) ( ) ( ) ( : 2 1 n g n f d n h n h n g n f c n n ≤ ≤ ≤ ≥ ∀ , and hence )) ( ) ( ( ) ( ) ( 2 1 n g n f n h n h ⋅ Θ = ⋅ as required. /// Problem 4 Let ) ( n f and ) ( n g be asymptotically positive functions (i.e. ) ( > n f and ) ( > n g for all sufficiently large n ), and suppose that )) ( ( ) ( n g n f Θ = . Does it necessarily follow that         Θ = ) ( 1 ) ( 1 n g n f ? Either prove this statement, or give a counter-example. Solution: The statement is true , as we now prove. By hypothesis there exist positive numbers 1 c , 2 c , and n such that for all n n ≥ : ) ( ) ( ) ( 2 1 n g c n f n g c ≤ ≤ < . (Note: the strict inequality < on the left follows from the fact that ) ( n f and ) ( n g are asymptotically positive.) Taking the reciprocals of all the positive terms in this inequality gives: ) ( 1 1 ) ( 1 ) ( 1 1 1 2 n g c n f n g c ⋅ ≤ ≤ ⋅ < for all n n ≥ . Observe that both 1 2 > c and 1 1 > c , whence         Θ = ) ( 1 ) ( 1 n g n f . /// Problem 5 Let ) ( n g be an asymptotically non-negative function. Prove that...
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This note was uploaded on 11/16/2009 for the course CMPS 101 taught by Professor Tantalo,p during the Fall '08 term at University of California, Santa Cruz.

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mid1review_some_solns_MORE - CMPS 101 Midterm 1 Review...

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