Sorry if this question has been asked before, but I am trying to figure this out. I am using the CLRS text, Introduction to Algorithms. In the Recurrences chapter, in the Master Theorem section, the following example is given with the solution:
$$T(n) = 3T\left(\frac{n}4\right) + n\log n $$
Here,
$$ a = 3, b =4, f(n) = n\log n $$
Using the Master method, it says
$$f(n) = \Omega(n^{\log_4 3 + \epsilon}) $$
I don't understand how $f(n)$ was "assumed" to have this
lower bound ($\Omega$). I would be extremely greatful to anyone who points out to me why this is so.
If more details are needed to understand the problem, please let me know. I will add more details to this question.
TIA,
Jake Clawson
EDIT: I thank all those who have responded below. I have replied to each below individually. But my question still stands. The book directly goes on to say that $f(n)$ is $\Omega$. I don't understand this and this is the key to applying the Master Theorem because once $f(n)$ is known that we know which case to apply.
Best Answer
So you want to understand why $n\log n=\Omega(n^{\log_4 3+\epsilon})$
The easiest way is to look at the limit
$\lim_{n->\infty}\frac{n\log n}{n^c}$
which is evaluated to
$\lim_{n->\infty}\frac{\log n+1}{cn^{c-1}}$ where $c=log_4 3+\epsilon$
since $\log_4 3\lt1$, there exists $\epsilon\gt0$ such that $\log_4 3+\epsilon=c\lt1$. Therefore, the limit would tends to $\infty$ and hence $nlogn=\Omega(n^{\log_43+\epsilon})$