How can I solve the recurrence $f(n) = 3f(\frac{n}{4}) + \log(n)$?

Question

The master theorem didn't work here. I tried to do the substitution method but I ended up with an additional term: $2Σ(i \cdot 3^i)$. Also I should find the solution $g(n)$ such as $f=\Theta(g)$.

Answer 1

The given recurrence is: $T(n) = 3T(n/4) + \log n$

Using the substitution method, we get:

\begin{align} T(n) &= 3^2T \left(n/4^2\right) + \log n + 3\log (n/4) \\ &= 3^3T \left(n/4^3\right) + \log n + 3\log(n/4) + 9 \log(n/4^2)\\ \end{align} and so on.

Can you see a pattern now?

Answer 2

I use a simpler master theorem for scholar recurrences. Consider the general form $T(n) = aT(n/b) + f(n)$. Call $c=log_ba$. Now you have 3 cases:

1. $\theta(n^c) = f(n)$ then $T(n) = \theta(f(n)log^k(n)) \text{ where }k\geq1$
2. $\theta(n^c) \gt f(n)$ then $T(n) = \theta(n^c)$
3. $\theta(n^c) \lt f(n)$ then $T(n) = \theta(f(n))$

In your exercise you have the 2nd case, so $T(n) = \theta(n^{log_43})$.
However this method works only with simple recurrences.