# Why is n log n dominated by n log^2 n?

Does the rule of $$n ^ a$$ dominate $$n ^ b$$ if $$a > b$$ apply here as well?

My understanding is that $$n \log n$$ will be dominated by $$n \log ^2 n$$ because of $$\log$$ being raised to the power of $$2$$.

• Can you please try and use LaTeX/MathJax to present formulas? – greybeard Jan 12 at 23:07

little-oh proof

An equivalent but more straightforward question would be why $$\lg n$$ is dominated by $$\lg^2 n$$, that is why $$\lg n \in o(\lg^2 n)$$.

Then based on the definition of little-oh we need to show that for any choice of constant $$c > 0$$, we can find a constant $$n_0$$ such that the inequality $$\lg n < c \lg^2n$$ holds for all $$n > n_0$$.

We prove that if we pick $$n_0 = \sqrt[c]{b}$$ where $$b$$ is the base of $$\lg$$, then the definition above holds.

If $$n_0 = \sqrt[c]{b}$$ then we have $$n > \sqrt[c]{b}$$ or $$n > b^{\frac{1}{c}}$$. Now since $$\lg n$$ with a base $$b > 1$$ is an increasing function, then $$\lg n > \lg b^\frac{1}{c}$$ or $$\lg n > \frac{1}{c}$$. If we multiply both sides by $$c > 0$$, we have $$c \lg n > 1$$. Now we can multiply both sides by $$\lg n > 0$$ (Note that for $$\lg n > 0$$ to be true, we must have $$n_0 \geqslant 1$$ which leads to $$\sqrt[c]{b} \geqslant 1$$ or $$b^\frac{1}{c} \geqslant 1$$ which results in $$\lg b^{\frac{1}{c}} \geqslant \lg 1$$ or $$\frac{1}{c} \geqslant 0$$ or $$c > 0$$; which is already guaranteed.) to have $$c \lg^2 n > \lg n .$$ This is what we needed to show.

Big-Oh proof sketch

A slightly different but related question could be why $$\lg n$$ is bounded above by $$\lg^2 n$$, that is why $$\lg n \in O(\lg^2 n)$$.

The answer is that since $$\lg n > 1$$ for $$n$$ larger than the base of $$\lg$$, then if we raise $$\lg n$$ to any power greater than $$1$$ (including power of $$2$$), it will be larger than $$\lg n$$ itself.

(We could rather argue that since we proved that $$\lg n \in o( \lg^2 n)$$, it follows from $$o(f) \subset O(f)$$ that $$\lg n \in O( \lg^2 n)$$.)

Note

So yes, the rule of $$n^a > n^b$$ if $$a > b$$ and $$n > 1$$ does apply here. Note that this rule is equivalent to the statement that any exponential function with a base greater than $$1$$ is an increasing function.

• (I seem to remember $f(n) \in o(g(n))$ for dominated, $f(n) \in O(g(n))$ for bounded above.) – greybeard Jan 12 at 23:00
• Consider that $O(f) \supset o(f)$. – greybeard Jan 13 at 5:43
• @greybeard I updated my answer. – Pooya Jan 13 at 16:22

Assume $$O(n \log n) = O(n \log ^2 n)$$. This implies that for any function $$f(n) \in O(n \log ^2 n)$$, there exist a positive constant $$k$$ and a constant $$n'$$ such that $$f(n) < k (n \log n)$$ for all $$n \geq n'$$.

Take $$f(n) = n \log ^2 n$$. Clearly $$f(n) \in O(n \log ^2 n)$$.

So we have that there exists a value for $$k$$ and $$n_0$$ such that $$n \log ^2 n < k (n \log n)$$ for all $$n \geq n'$$.

Noting that $$n \log ^2 n = n \log n \log n$$, we arrive at a contradiction: whatever constant value we choose for $$k$$, there exists a value $$n'$$ such that $$\log n > k$$ for all $$n \geq n'$$.

In simpler terms, were $$O(n \log n)$$ to be the same as $$O(n \log ^2 n)$$, it would mean that $$n \log ^2 n$$ does not grow faster than some constant factor of $$n \log n$$. But this is obviously not correct as $$n \log ^2 n$$ grows at a rate of $$\log n$$ times $$n \log n$$ (and $$\log n$$ grows faster than a constant).