Solving this recurrence relation: T(n) = 2 T(n/2) + f(n).

To solve this recurrence relation, we can use the master theorem.

The master theorem states that for a recurrence relation of the form T(n) = a T(n/b) + f(n), where a ≥ 1 and b > 1 are constants and f(n) is a function, we can determine the asymptotic behavior of T(n) as follows:

1. If f(n) = O(n^c) for some constant c < logb a, then T(n) = Θ(nlogb a).
2. If f(n) = Θ(n^c logk n) for some constants c ≥ logb a and k ≥ 0, then T(n) = Θ(n^c logk+1 n).
3. If f(n) = Ω(n^c) for some constant c > logb a, and if a f(n/b) ≤ c f(n) for sufficiently large n, then T(n) = Θ(f(n)).

In this case, we have a = 2, b = 2, and f(n) = f(n), so we need to determine the asymptotic behavior of f(n) to apply the master theorem.

Without additional information about f(n), we cannot determine its asymptotic behavior, so we cannot directly apply the master theorem. However, we can make an educated guess based on the form of the recurrence relation.

Since T(n) = 2 T(n/2), we can interpret this as a divide-and-conquer algorithm that splits the input into two equal-sized subproblems, solves each subproblem recursively, and then combines the solutions in linear time (since f(n) is a function of n).

If we assume that f(n) = Θ(n^d) for some constant d ≥ 0, then we can apply the master theorem to obtain the following result:

T(n) = Θ(n^d log n)

To see why, note that we have a = 2, b = 2, and c = d. Since d < log2 2 = 1, we are in case 1 of the master theorem, so T(n) = Θ(nlog2 2) = Θ(n). However, this only accounts for the recursive part of the algorithm, and we need to add the linear time for combining the solutions, which gives us T(n) = Θ(n) + Θ(n^d) = Θ(n^d) for d ≥ 0.

Therefore, we can conclude that T(n) = Θ(n^d log n) for f(n) = Θ(n^d) and d ≥ 0.