Solving a recurrence of Time complexity involving Theta notation

Question

T(n) = 4T(n/2) + Θ(n^2 /logn)

How do I solve this recurrence? I cannot use the Master theorem here.

Answer 1

Without using Master theorem, just simply keep expanding the recurrence relation until you see a pattern.

T(n) = 4T(n/2) + Θ(n^2 /log(n))
= 4*(4t(n/4) + theta((n/2)^2 / log(n/2))) + theta(n^2/log(n)) 
= 4^2 * (t(n/4) + theta((n/2)^2 / log(n/2)) / 4) + theta(n^2/log(n)) 
theta((n/2)^2 / log(n/2)) / 4 can be simplified to theta(n^2/log(n))
= 4^2 * (t(n/4)) + 2theta(n^2/log(n)) 
= 4^2 * (4t(n/8) + theta((n/4)^2 / log(n/4))) + 2theta(n^2/log(n)) 
= 4^3 * (t(n/8) + theta((n/4)^2 / log(n/4)) / 4) + 2theta(n^2/log(n)) 
theta((n/4)^2 / log(n/4)) / 4 can be simplified to theta(n^2/log(n))
= 4^3 * (t(n/8)) + 3theta(n^2/log(n)) 

So we can further simplify this to

4^k * (t(n/k)) + k*theta(n^2/log(n)) 

This will keep going until k = n, and assuming T(1) = 1, we get

4^n + n*theta(n^2/log(n))

And since 4^n is greater than n*theta(n^2/log(n)), the answer is O(4^n).