Final answer:
The solution of T(n) = 2T(n/2) + n/logn does not fit the standard form of the Master theorem due to the presence of a logarithmic term in the denominator. More advanced methods may be required to solve this recurrence.
Step-by-step explanation:
The solution of T(n) = 2T(n/2) + n/logn cannot be directly obtained using the standard form of the Master theorem, as the theorem's typical forms cover the additions of terms like n, n^2, or n^logb(a), without functions of n in the denominator like log n. However, we can analyze the given recurrence relation by comparing it to the standard form of the Master theorem. In the Master theorem, a recurrence relation of the form T(n) = aT(n/b) + f(n) is considered, where 'a' is the number of subproblems, 'n/b' is the size of each subproblem, and f(n) is the outside work done at each level. The solution involves comparing f(n) with n^logb(a) to determine the growth rate.
In this case, 'a' is 2, 'b' is 2, and f(n) would be n/logn. However, because the function involves a logarithm in the denominator, a direct application of the theorem is not possible. An approximation or a more advanced method such as Akra-Bazzi theorem or iterative substitution may need to be used to solve this recurrence.The given recurrence relation T(n) = 2T(n/2) + n/logn can be solved using the Master theorem. In this case, the equation falls under Case 2 of the Master theorem.Case 2 states that if the recurrence relation is of the form T(n) = aT(n/b) + f(n), where a >= 1 and b > 1, and if f(n) is Θ(n^klog^pn) with k >= 0 and p >= 0, then the solution to the recurrence relation is Θ(n^klog^(p+1)n).In this case, a = 2, b = 2, and f(n) = n/logn. Since f(n) is Θ(n^1log^0n), we can see that k = 1 and p = 0. Therefore, the solution to the recurrence relation is Θ(n^1log^(0+1)n) = Θ(nlogn).