Show that n^2 is not O(n*log(n))?

Question

Using only the definition of O()?

Should this perhaps go on the Mathematica site? – E.T. Feb 04 '13 at 08:48 — E.T., Feb 04 '13 at 08:48

score 1 · Answer 1 · answered Feb 04 '13 at 08:50

1

You want to calculate the limit of

  (n * log(n)) / (n ^ 2) =
= log(n) / n =
= 0 if n approaches infinity.

because log(n) grows slower than n.

answered Feb 04 '13 at 08:50

I would give this a +1, except that the question really deserves a -1. – Philip Sheard Feb 04 '13 at 08:56
this is not a big-o proof... – UmNyobe Feb 04 '13 at 08:58
@UmNyobe Then what is this? – Feb 04 '13 at 08:59
1

I mean this is not a formal proof, ie it doesnt use the definition of Big-O as the OP requested. – UmNyobe Feb 04 '13 at 09:08

score 1 · Accepted Answer · answered Feb 04 '13 at 09:06

You need to prove by contradiction. Assume that n^2 is O(n*log(n)). Which means by definition there is a finite and non variable real number c such that

n^2 <= c * n * log(n)

for every n bigger than some finite number n0.

Then you arrive to the point when c >= n /log(n), and you derive that as n -> INF, c >= INF which is obviously impossible.

And you conclude n^2 is not O(n*log(n))

Show that n^2 is not O(n*log(n))?

2 Answers2