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Why is $\log_{2}n = O(n^{0.00001})$ true?

This is obvious to me when the exponent is $> 1$ but i'm having trouble understanding the cases where the exponent is very close to $0$. I would have to find some constants $c$ and $n_0$ where $\log_{2}n \le cn^{0.00001}$ for all $n \gt n_0$.

Where I'm stumped is that $n^{0.00001} \approx 1$ and $\log_{2}n$ approaches infinity as $n$ gets larger. It feels like regardless of whatever $c$ and $n_0$ I choose, if $n$ was large enough, I could show that $\log_{2}n \ge c$.

Raphael
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nvi
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    Try what happens if n = 2^10,000,000. Now n^0.00001 = 2^100, while log n = 10,000,000. 2^100 is a lot larger than 10,000,000. – gnasher729 Mar 14 '20 at 08:09

1 Answers1

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$n^{0.00001}$ is not approximately $1$. $n^{0.00001}$ goes to infinity as $n \to \infty$.

You can see that $\log_2 n = o( n^{0.00001} )$ by taking the limit of their ratio: $$ \lim_{n \to \infty} \frac{\log_2 n}{n^{0.00001}} = \lim_{n \to \infty} \frac{n^{-1}}{0.00001 \cdot n^{0.00001} \cdot n^{-1}} = \lim_{n \to \infty} \frac{100000}{n^{0.00001}} = 0. $$

This tells you that you can pick any value of $c>0$, for example $c=1$. Now you just need a value $n_0$ such that $n^{0.00001} - \log_2 n \ge 0 \; \forall n \ge n_0$.

The derivative of $n^{0.00001} - \log_2 n$ is $n^{-1}( 0.00001 \cdot n^{0.00001} - 1)$ which is non-negative as soon as $ n^{0.00001} \ge 100000$, i.e., for $n \ge 10^{5 \cdot 10^5}$.

You can then pick any value of $n_0$ such that $(n_0)^{0.00001} - \log_2 n_0$ is non-negative and $n_0 \ge 10^{5 \cdot 10^5}$. For example $n_0 = 2^{10^{7}}$. Indeed:

  • $(n_0)^{0.00001} = 2^{100} = 1024^{10} > 1000^{10} = 10^{30} > 10^7 = \log_2 n_0$; and
  • $n_0 = 1024^{10^6} > 1000^{10^6} = 10^{3 \cdot 10^6} > 10^{5 \cdot 10^5}$.
Steven
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