I want to use the Master theorem to solve the following recurrence.
$$T(n)=9T(n/3) + \Theta(n^2/\operatorname{lg}(n))$$
We can easily see that $a=9$ and $b=3$ and $f(n) = n^2/\operatorname{lg}(n)$.
I am very new to using this theorem, and am hoping for some help or pointers at this stage.
Is there a way to eyeball which 'case' to use or should we go through each of them (that actually takes me a pretty long time).
Thanks in advance.
By way of enrichment we solve another closely related recurrence that admits an exact solution. Suppose we have $T(0)=0$ and $T(1)=T(2)=1$ and for $n\ge 3$
$$T(n) = 9 T(\lfloor n/3 \rfloor) + \frac{n^2}{\lfloor \log_3 n \rfloor}.$$
Furthermore let the base three representation of $n$ be $$n = \sum_{k=0}^{\lfloor \log_3 n \rfloor} d_k 3^k.$$
Then we can unroll the recurrence to obtain the following exact formula for $n\ge 3$ $$T(n) = 9^{\lfloor \log_3 n \rfloor} + \sum_{j=0}^{\lfloor \log_3 n \rfloor-1} 9^j \frac{1}{\lfloor \log_3 n \rfloor-j} \left( \sum_{k=j}^{\lfloor \log_3 n \rfloor} d_k 3^{k-j} \right)^2 \\ = 9^{\lfloor \log_3 n \rfloor} + \sum_{j=0}^{\lfloor \log_3 n \rfloor-1} \frac{1}{\lfloor \log_3 n \rfloor-j} \left( \sum_{k=j}^{\lfloor \log_3 n \rfloor} d_k 3^k\right)^2.$$
Now to get an upper bound consider a string of two digits which yields $$T(n) \le 9^{\lfloor \log_3 n \rfloor} + \sum_{j=0}^{\lfloor \log_3 n \rfloor-1} \frac{1}{\lfloor \log_3 n \rfloor-j} \left( 2 \times \sum_{k=j}^{\lfloor \log_3 n \rfloor} 3^k\right)^2 \\ = 9^{\lfloor \log_3 n \rfloor} + \sum_{j=0}^{\lfloor \log_3 n \rfloor-1} \frac{1}{\lfloor \log_3 n \rfloor-j} \left( 3^{\lfloor \log_3 n \rfloor+1} - 3^j\right)^2 \\ = 9^{\lfloor \log_3 n \rfloor} + 9\times 3^{2\lfloor \log_3 n \rfloor} \sum_{j=0}^{\lfloor \log_3 n \rfloor-1} \frac{1}{\lfloor \log_3 n \rfloor-j} \left(1 - 3^{j-(\lfloor \log_3 n \rfloor+1)}\right)^2 \\ = 9^{\lfloor \log_3 n \rfloor} + 9\times 3^{2\lfloor \log_3 n \rfloor} \sum_{j=1}^{\lfloor \log_3 n \rfloor} \frac{1}{j} \left(1 - 3^{-j-1}\right)^2 \\ = 9^{\lfloor \log_3 n \rfloor} + 9\times 3^{2\lfloor \log_3 n \rfloor} \sum_{j=1}^{\lfloor \log_3 n \rfloor} \frac{1}{j}\left( 1 - \frac{2}{3} 3^{-j} + \frac{1}{9} 3^{-2j}\right).$$
Note that this bound is attained and cannot be improved. It is asymptotic to $$9^{\lfloor \log_3 n \rfloor} + 9\times 3^{2\lfloor \log_3 n \rfloor} \left(H_{\lfloor \log_3 n \rfloor} + \frac{2}{3}\log(2/3) -\frac{1}{9} \log(8/9)\right) \\ = 9^{\lfloor \log_3 n \rfloor} + 9\times 3^{2\lfloor \log_3 n \rfloor} \left(H_{\lfloor \log_3 n \rfloor} + \frac{1}{3}\log 2 - \frac{4}{9}\log 3\right).$$
The lower bound is for the case of a one digit followed by a string of zeros and yields $$T(n) \ge 9^{\lfloor \log_3 n \rfloor} + \sum_{j=0}^{\lfloor \log_3 n \rfloor-1} \frac{1}{\lfloor \log_3 n \rfloor-j} \left( 3^{\lfloor \log_3 n \rfloor}\right)^2 \\ = 9^{\lfloor \log_3 n \rfloor} + 3^{2\lfloor \log_3 n \rfloor} \sum_{j=0}^{\lfloor \log_3 n \rfloor-1} \frac{1}{\lfloor \log_3 n \rfloor-j} = 9^{\lfloor \log_3 n \rfloor} + 3^{2\lfloor \log_3 n \rfloor} H_{\lfloor \log_3 n \rfloor}.$$
The lower bound too is attained.
Joining the dominant terms of the upper and the lower bound we obtain the asymptotics $$\color{#0A0}{3^{2\lfloor \log_3 n \rfloor} H_{\lfloor \log_3 n \rfloor} \in \Theta\left(\log\log_3(n) \times 3^{2 \log_3 n}\right) = \Theta\left(\log\log n \times n^2\right)}.$$
These are in agreement with what the Master theorem would produce.
Addendum. Wikipedia lists the above as an inadmissible case for the Master Theorem. However Akra-Bazzi applies and the reader is invited to do this calculation.
master theorem– Oct 13 '14 at 22:31