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

To solve this recurrence relation, we can use the Master Theorem.

The recurrence relation can be written in the form:

T(n) = aT(n/b) + f(n)

where a = 2, b = 2, and f(n) = f(n).

We can use the following steps to apply the Master Theorem:

1. Calculate the value of logb(a):

log2(2) = 1

2. Compare f(n) with nlogb(a):

If f(n) = O(nlogb(a)-ε) for some ε > 0, then T(n) = Θ(nlogb(a)).

If f(n) = Θ(nlogb(a)), then T(n) = Θ(nlogb(a) log n).

If f(n) = Ω(nlogb(a)+ε) for some ε > 0, and if af(n/b) ≤ cf(n) for some constant c < 1 and sufficiently large n, then T(n) = Θ(f(n)).

In this case, we have f(n) = f(n), which is not comparable to nlogb(a) = n. Therefore, we cannot use the Master Theorem to determine the asymptotic growth rate of T(n).

However, we can still solve the recurrence relation by using the substitution method.

Let T(n) = 2T(n/2) + f(n)

Assume that T(n) = O(nlog n).

Then, we have:

T(n) ≤ 2T(n/2) + f(n)

≤ 2(c(n/2)log(n/2)) + f(n)

= cnlogn - cn + f(n)

Since f(n) is a non-negative function, we can further simplify:

T(n) ≤ cnlogn + f(n)

Therefore, T(n) = O(nlogn) is a valid upper bound.

Now, assume that T(n) = Ω(nlogn).

Then, we have:

T(n) ≥ 2T(n/2) + f(n)

≥ 2(c(n/2)log(n/2)) + f(n)

= cnlogn - cn + f(n)

Since f(n) is a non-negative function, we can further simplify:

T(n) ≥ cnlogn + f(n)

Therefore, T(n) = Ω(nlogn) is a valid lower bound.

Combining the upper and lower bounds, we get:

T(n) = Θ(nlogn)

Therefore, the solution to the recurrence relation is T(n) = Θ(nlogn).