Unformatted text preview: 3 = 6 × n × n 2 ≥ 60n 2 . Then, max(6n 3 ,10n 2 ) ≤ 6n 3 < 7n 3 hold. d.[5] log n + n 1/2 = O(n 1/2 ). [2] First, we can show (for instance by induction) that, for all m ≥ 7, m 2 ≤ 2 m1 . [3] Let c=2, n =128 =2 7 , for all n ≥ n , 2 m1 < n ≤ 2 m , for some m. And m ≥ 7 Therefore, (log n) 2 ≤ m 2 ≤ 2 m1 < n ⇒ log n ≤ n 1/2 ⇒ log n + n 1/2 ≤ 2n 1/2 e.[3] log n + n 1/2 = O(n) Let c=2, n =1. For all n ≥ 1, log n ≤ n, and n 1/2 ≤ n Then, log n + n 1/2 ≤ 2n hold. 2.[17] Using only definition of O(f(n)) proof that the following statements are false: You should use the following definition: g(n) = O(f(n)) if and only if ∃ C>0 ∃ n >0 ∀ n>n g(n) ≤ Cf(n), which means: g(n) ≠ O(f(n)) if and only if ∀ C>0 ∀ n >0 ∃ n>n g(n) > Cf(n)...
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This note was uploaded on 12/10/2009 for the course CAS 2c03 taught by Professor Janicki during the Spring '03 term at McMaster University.
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