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# Comparing asymptotic complexity of functions $\log{n}$, $(\log{n})^c$ and $\sqrt{n}$

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Problem Detail:

I usually follow approach of taking logs and putting arbitrary large powers of $2$ for $n$ and reducing the given function to some constant value for large value of $n$. So in this case I did it as follows:

• $\log{n}$

Putting $n=2^{2^{2^{16}}}$ and taking $\log$

$\log{\log(2^{2^{2^{16}}})}={2^{16}}$

• $(\log{n})^c$

Putting $n=2^{2^{2^{16}}}$, $c=2$ and taking $\log$

$\log({(\log{2^{2^{2^{16}}}})^2})=\log({2\times(\log{2^{2^{2^{16}}}})})=\log({2\times(2^{2^{16}})})=\log{2}\times\log{2^{2^{16}}}$

$=1+2^{16}$

• $\sqrt{n}$

Putting $n=2^{2^{2^{16}}}$ and taking $\log$

$\log{\sqrt{2^{2^{2^{16}}}}}=\log{2^{2^{2^{{16}^{\frac{1}{2}}}}}}=\log{2^{2^{2^8}}}=2^{2^8}=2^{16}$

So it looks like:

$\log{n}=\sqrt{n}<(\log {n})^c$

However I checked that $\log{8}>\sqrt{8}$. But, $\log{(2^{2^{2^{16}}})}=64$, whereas $\sqrt{2^{2^{2^{16}}}}$ is much larger number making equality between the two unlikely.

1. So where I am making mistakes in above calculations?
2. How above three functions compare asymptotically?
###### Answered By : Ran G.

for any $c>1,\epsilon>0$, $$\lim_{n\to\infty} \frac{(\log n)^c}{n^\epsilon} = 0$$

This means that $\log n=o( (\log n)^c)$ and $(\log n)^c = o(n^\epsilon)$. In particular, for $\epsilon =0.5$, $(\log n)^c = o(\sqrt{n})$.