Big Omega of $n \log n$

Question

While studying master method at recurrences topic I'm stacked at a point. It is written in the book as:

$T(n) = 3T(n/4) + n \log n$,

we have $a = 3, b = 4$,

$f(n) = n \log n$, and

$n^{\log_b(a)} = n^{\log_4 3} = O(n^{0.793})$.

Since $f(n) = \Omega(n^{\log_4( 3)+\varepsilon} )$, where $\varepsilon \approx0.2$ ....

The authors means that the $n\log n = \Omega(n)$. How will we know this? Is $n \log n = \Omega(n)$ true? Or something is wrong?

It's better to say $n\log n \in \Omega(n)$ instead of $n\log n = \Omega(n)$, I think if you think so, you will have a less problem on understanding this. — , Nov 09 '12 at 09:41

score 6 · Answer 1 · edited Nov 09 '12 at 08:28

6

Yes, $ n\log n = \Omega(n) $ is true.

To see why, note that for this to be true, we have to show that $ n\log n \ge cn$ for all $n \gt n_0 $ and $ \exists c \gt 1 $. If we take $ c = 1 $ then, we get $n\log n \ge n$ for all $n \ge 2$. This shows that $ n\log n = \Omega(n) $ is true.

edited Nov 09 '12 at 08:28

A.Schulz

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answered Nov 08 '12 at 19:54

Arani

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Big Omega of $n \log n$

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