224 Solutions to Exercises 12. [BB] We have an < b n for n 2: 1 since a < b, so an = O(b n ). We must prove b n -# O(a n ). If b n = O(a n ), then there would exist a constant c such that b n S; can for all sufficiently large n. Dividing by a, we obtain (~)n S; c, for all sufficiently large n, but this is not true because ~ > 1 implies that (~)n grows without bound as n increases. 13. (a) [BB] We established 2 n < n! for n 2: 4 in Problem 6 of Chapter 5. Thus, 2 n = O(n!) (c = 1, no = 4). On the other hand, n! -# 0(2 n ), for consider n! 1234 n 2n = 2222···"2· The product of the first three factors on the right is ~ and each of the remaining n -3 terms is bigger than 2. So ;l > ~(2n-3) for n > 3. If n! = 0(2n), then n! < c2 n for some constant c and all sufficiently large n and so ;l < c. We have shown this is not possible. Since 2 n = O(n!) but n! -# 0(2n), we have 2 n -< n! as required. (b) Since
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.