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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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 Spring '03
 janicki

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