Solving constants in the recursive term with master theorem We are learning how to solve recurrence relations in different ways (Forward Substitution, Backward Substitution, Master Theorem, etc...). I really thought I understood the topic since most of the problems were in the form of $aT(\frac{n}{b}) + f(n)$ where applying the Master Theorem is no sweat, but we're given this recurrence relation wherein there is a constant inside the recursive term, I am not sure anymore if Master Theorem would still work, or is there another way?
\begin{align*}
T(n) = 3T\left(\frac{n}{3}-2\right) + \frac{n}{2}
\end{align*}
This one has a constant 2 inside the term, and I don't know how to show that this recurrence relation is upper bounded by $ O (n \log n) $.
If you are just interested in an upper bound you can notice that $T(n) \le S(n)$ where $S(n) = 3 S(n/3) + \frac{n}{2}$ and has solution $S(n) = O(n \log n)$.
Alternatively there is always induction. You can show that, for $n \ge 2$,  $T(n) \le c n \log n$.
For $2 \le n &lt; 7$, $T(n)$ is a constant and $n \log n \ge 1$. Therefore the claim is true for a sufficiently large (constant) value $c^*$ of $c$.
For $n \ge 7$ you have:
$$
T(n) = 3T\left(\frac{n}{3} - 2\right) + \frac{n}{2} \le 3 c \frac{n}{3} \log \frac{n}{3} + \frac{n}{2}
= cn \log n - cn \log 3 + \frac{n}{2},
$$
which is at most $cn \log n$ when $c n \log 3 \ge \frac{n}{2}$ or, equivalently, $c \ge \frac{1}{2 \log 3}$.
Simply pick $c = \max\{c^*, \frac{1}{2 \log 3} \}$.