Recurrence - Master Theorem - Asymptotic Question 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.
 A: 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 $nlogn=\Omega(n^{\log_43+\epsilon})$
A: Well, actually, the fact you state is not quite s interesting.


*

*If $G(n)$ satisfies $G(n)  = 3G(n/4)$, then clearly, $G(n) < T(n)$ (for the same boundary condition). As you tell from the master theorem, or easily check yourself, $G(n) = A \, n^{\log_4{3}} + B$.

*$T(n)$ is surely $> n\, \log n$. Because $\log_4{3} < 1$, $n^{\log_4{3}} = o(n) = o(n\, \log n)$. So $T(n)$ is $\omega(n) = \omega(n^{\log_4{3}+\epsilon})$ for some small positive epsilon.
A: Master's theorem deals with recurrences of the form
$$T(n) = aT(n/b) + f(n)$$
The theorem is divided into three cases, depending on the behaviour of $f(n)$ "relative" to $n^{\log_b a}$
Case 2) $f(n) = \Theta(n^{\log_b a})$
Case 1) $f(n) = \mathcal{O}(n^{\log_b a - \epsilon})$
Case 3) $f(n) = \Omega(n^{\log_b a + \epsilon})$
In your case, $f(n) = n\log n$, $a = 3$, $b=4$.
In order to apply the Master we need $f(n)$ to fit in one of those cases.
Now $\log_4(3) \lt \log_4 4 = 1$.
Thus we can eliminate Cases 1) and 2).
What remains is Case 3), and we can apply it, by picking $\epsilon = 1 - \log_4 3$, because $n \log n = \Omega(n) = \Omega(n^{\log_4 3 + (1- \log_4 3)})$.
A: Take $c=\lg\left(n_{0}\right)$ in the following definition for the
$\Omega$-notation:
$$
\Omega\left(g\left(n\right)\right)=\left\{ f\left(n\right):\exists\left(c>0\wedge n_{0}>0\right)\backepsilon0\leq cg\left(n\right)\leq f\left(n\right)\forall n\geq n_{0}\right\} 
$$
which appears on page 48 of the third edition of the book. Thus, with
$g\left(n\right)=n$, for $n\geq n_{0}$ we have that
$$
n\lg\left(n_{0}\right)\leq n\lg\left(n\right)
$$
This shows that $f\left(n\right)=\Omega\left(n\right)$, which answers
the question.
