(15 points) Prove that the following assertion is true for all values of n ≥ n0. Identify both n0 and c.2n + 5∈ O(n2)
Question
(15 points) Prove that the following assertion is true for all values of n ≥ n0. Identify both n0 and c.2n + 5∈ O(n2)
Solution
To prove that 2n + 5 ∈ O(n^2), we need to find constants c and n0 such that 2n + 5 ≤ c*n^2 for all n ≥ n0.
Step 1: Choose c = 7 (or any number greater than 2).
Step 2: Choose n0 = 1 (or any number greater than 0).
Step 3: Prove the assertion.
For all n ≥ n0, we have:
2n + 5 ≤ 7n^2
This is true because for all n ≥ 1, n^2 is always greater than or equal to n. Therefore, 7n^2 is always greater than or equal to 2n.
Also, 7n^2 is always greater than 5 for all n ≥ 1.
Therefore, 2n + 5 is always less than or equal to 7n^2 for all n ≥ 1.
So, we have proved that 2n + 5 ∈ O(n^2) with c = 7 and n0 = 1.
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