# Thread: is n^0.01 Omega log(n) or BigO(log(n))

1. ## is n^0.01 Omega log(n) or BigO(log(n))

I was plugging in huge values for n and noticed n^0.01 grows incredibly slowly, much slower then log(n) it seems. So my initial thought was that $$n^{0.01}~is~BigO(log(n))$$.
However I came across the rule in my book that,
$$log^x(n)~is~BigO(n^y)$$ for any fixed constant x > 0 and y > 0,
which made me think my initial thought was wrong.

2. ## Re: is n^0.01 Omega log(n) or BigO(log(n))

Consider that $\displaystyle n^{0.01} = e^{0.01 ln(n)}$

You can use the Maclaurin series expansion of e^x in order to show that by the Formal Definition of Big O notation, that it's not the case your initial thought was true.
https://en.wikipedia.org/wiki/Big_O_...mal_definition

ie. Show that for a constant M, you can choose an $\displaystyle n_0$ such that for $\displaystyle n \geq n_0 , |e^{0.01 ln(n)}| \geq M |ln(n)|$

3. ## Re: is n^0.01 Omega log(n) or BigO(log(n))

that is an incredibly creative way to solve this. I never in a million years would have thought to go about it this way.
I ended up just proving that $$lim_{x\to\infty}\frac{n^{0.01}}{\log{n}} = \infty$$

4. ## Re: is n^0.01 Omega log(n) or BigO(log(n))

Your method is quicker. Great catch.