# Solving Recurrence relation with master method?

I have to solve the following recurrence equation and I thought to solve it with case #3 of the master theorem. Can I do that?

$$T(1) = c>0$$ $$T(n) = 9T(n/3) + f(n)$$ $$f(n) = n^2\cdot lg^3 (n) + n^3 \cdot lg(n)$$

I think that $f(n) = \Omega(n^{log_ba})$ where $a=9, b=3$ so $n^{log_ba}=n^2$.

Since $f(n) = n^3 lg(n)$, we can prove that $$a\cdot f(n/b) \leq c\cdot f(n)$$.

Is it a correct solution?

The only piece I am afraid could be tricky is proving $a\cdot f(n/b) \leq c\cdot f(n)$.

Thank you, Alan

• We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. – Raphael Feb 3 '18 at 11:13

You are claiming that "we can prove that $af(n/b) \leq cf(n)$". There are two problems with this claim:
1. This is now what you actually need in order to apply the master theorem. The master theorem needs you to show that there exists $c < 1$ for which $af(n/b) \leq c f(n)$ for all large enough $n$.
Reiterating, to prove the claim you need to come up with a value of $c < 1$ such that the following inequality holds for large enough $n$: $$9 (n/3)^3 \log(n/3) \leq cn^3 \log n.$$ This holds for $c = 1/3$, since $$9(n/3)^3 \log(n/3) < 9(n/3)^3 \log n = \frac{1}{3} n^3 \log n.$$
• Fantastic! thank you. That's exactly what I was missing. I though initially to go with $c=\frac{1}{2}$ but I couldn't be sure I got it right. – Alan Feb 3 '18 at 12:30