Answer:
f(n) is O(n¹⁴logn) by definition of Big-O.
Explanation:
To prove that f(n) = 5n³ + n² + ylogn is O(n¹⁴logn), we need to show that there exist positive constants c and n₀ such that:
|f(n)| ≤ c|n¹⁴logn| for all n > n₀
Let's start by finding an upper bound for f(n). We can do this by simplifying the expression and getting rid of constants and lower-order terms:
f(n) = 5n³ + n² + ylogn
≤ 5n³ + n³ + n³ (since logn ≤ n³ for all n > 1)
= 7n³
Now, we can use this upper bound to find suitable values for c and n₀:
|f(n)| ≤ 7n³ ≤ 7n¹⁴/n¹¹ (since n¹¹ ≤ n³ for all n > 1)
≤ 7n¹⁴logn/n¹¹ (since logn ≤ n⁰ for all n > 1)
= 7n³logn
So, we can choose c = 7 and n₀ = 1 as our positive constants. Then, for all n > n₀, we have:
|f(n)| ≤ 7n³logn ≤ 7n¹⁴logn
Therefore, f(n) is O(n¹⁴logn) by definition of Big-O.