Final answer:
The solution to the recurrence T(n) = 2T(n/2) + n² using the Master theorem falls under Case 3, where the non-recursive part f(n) dominates, leading to an overall time complexity of O(n²).
Step-by-step explanation:
Solution to the Recurrence Using the Master Theorem
The student's question involves solving the recurrence relation T(n) = 2T(n/2) + n² using the Master theorem. According to the Master theorem, we can categorize the recurrence relation into three cases based on the comparison between the amount of work done at each level of recursion and the work done by the recursive calls.
For the recurrence T(n) = 2T(n/2) + f(n), we compare f(n) with n⁰ (where log base 2 of 2 is 1). In this case, f(n) is n², which is larger than n¹. This means we are dealing with Master theorem's Case 3, where f(n) = O(n¹⁴) for any ε > 0.
Because the polynomial n² grows faster than n¹, we can conclude that f(n) dominates the recursive term. Therefore, according to Master theorem's third case, the solution to the recurrence relation is the same order as f(n), which is O(n²).