Question

Given the growth rate function T(n) = 4n3 + 3n2 - 2n for an arbitrary algorithm, show that T(n) = O(n3). Show as much work as necessary.

Answer 1

T(n) = O(n3)

To prove that T(n) = O(n3), we need to find positive constants c and n0 such that T(n) ≤ c * n3 for all n ≥ n0. We can start by simplifying T(n) as follows: T(n) = 4n3 + 3n2 - 2n T(n) = n3(4 + 3/n - 2/n2) Since the term inside the parentheses approaches 4 as n gets larger, we can say that for all n ≥ 1, 4 + 3/n - 2/n2 ≤ 5. Therefore, we can choose c = 5 and n0 = 1, so that T(n) ≤ 5n3 for all n ≥ 1. This shows that T(n) = O(n3), as required.