Discrete Mathematics with Graph Theory (3rd Edition) 245

# Discrete Mathematics with Graph Theory (3rd Edition) 245 -...

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Chapter 8 243 (b) S · 12 22 2 n(n + 1)(2n + 1) mce + + ... + n = --'------'-6-'-------'- see Problem 4 of Section 5 . I-the result follows because a polynomial has the same order as its highest power, by Proposition 8.2.7. 8. (a) Letf(n) = 3n 3 -5n 2 +2n+1. Bythetriangleinequality,forn ~ l,lf(n)1 ~ 3n 3 +5n 2 +2n+ 1 ~ 3n 3 +5n 3 +2n 3 +n 3 = 11n 3 , so f(n) = O(n 3 ) with no = 1 andc = 11 in Definition 8.2.1. Conversely (and here we borrow an idea from the proof of Proposition 8.2.7), 3[ 5 2 1] 3[ 5 2 1] f(n) = n 3 - - + - + - ~ n 3 - - - - - - . n n 2 n 3 n n 2 n 3 Choose no such that !, ~ and ;;ta are all less than ~ for n ~ no, for example, no = 16. Then 5 1 -- >-- n - 3' 2 1 -->-- n 2 - 3 and 1 1 -->-- n 3 - 3 so that f(n) ~ n 3 [3 - ~ - ~ - ~] = 2n 3 ; hence n 3 ~ ~f(n) for n ~ no. So n 3 = O(f(n)) with c = ~ in Definition 8.2.1. (b) Let f(n) = n 4 - 2n 3 +3n 2 - 5n+7. By the triangle inequality, forn ~ l,lf(n)1 ~ n 4 +2n 3 + 3n 2 +5n+7 ~ n 4 +2n4 +3n 4 +5n 4 +7n 4 = 18n 4 , so f(n) = O(n4)
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