Discrete Mathematics with Graph Theory (3rd Edition) 225

Discrete Mathematics with Graph Theory (3rd Edition) 225 -...

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Section 8.2 223 7. (a) [BB] Since 5n :s; n 3 for all n ~ 3, we can take C = 1, no = 3, or C = 5, no = 1. (b) For n ~ 1, we have 17n 4 + 8n 3 + 5n 2 + 6n + 1 :s; 17n 4 + 8n n + 5n 4 + 6n 4 + n 4 = 37n4. So we can take C = 37 and no = 1. (c) [BB] For n ~ 1, we have 8n 3 + 4n 2 + 5n + 1 :s; 9n 4 + 18n 2 + 24n + 6 = 3(3n4 + 6n 2 + 8n + 2), so we can take C = 3, no = 1. (d) For n ~ 1,2 n :s; 3 n , so take C = 1, no = 1. (e) For n ~ 1, we have f(n) = 2n 2 + 3n + 1 :s; 2n2 + 3n 2 + 5 = 5n 2 + 5 = 5g(n), so take no = 1 andc = 5. (f) We have f(n) = 2n 2 - 3n + 5 :s; 2n 2 + 3n + 5 :s; 2n 2 + 3n 2 + 5n 2 = 10n 2 for any n ~ 1. Now n 2 :s; 2(n 2 - 7n) = 2n2 -14n = 2g(n) for n ~ 14 since (2n2 -14n) - n 2 = n 2 -14n = n(n - 14) ~ 0 for n ~ 14. So, for n ~ 14, f(n) :s; 10n 2 :s; 20g(n). So f = O(g) with no = 14 and c = 20. (g) Since 10g2 n 5 = 510g 2 n and since 10g2 n :s; n for any n ~ 1 we have 10g2 n 5 :s; 5n for n ~ 1. So take c
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