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Why is $n^{1000} \in O(5^{\sqrt{n}})$? I thought that since $5^{\sqrt{1000}} < 1000^{1000}$ then it would be the otherway around. Could you please give me the mathematical derivation of this?

Thanks in advance

DariusTheGreat
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    Don't let the notation fool you though. It means $n^{1000}$ is bounded by the exponential function (for large enough $n$), not that they are the same class of complexity in any sense. Exponential complexity and polynomial complexity are often two very different beasts. – mathreadler Aug 03 '17 at 09:44

4 Answers4

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You basically want to show that there exists some $n_o \in \mathbb{N}$, such that for all $n \geq n_o$, we have $n^{1000} \leq 5^{\sqrt{n}}$.

Taking log of both terms, we have, $1000\log(n)$ and $\sqrt{n} \log (5)$. At this step it is clear that $1000\log(n)$ will be dominated by $\sqrt{n} \log (5)$ for some $n_o$, since $\sqrt{n}$ is known to dominate over $\log(n)$ in the long run.

Then this implies the original inequality is true, and hence $n^{1000} \in \mathcal{O}(5^{\sqrt{n}})$.

Code:

sqrt(100000000000000000)*log(5)

ans =

5.0895e+08

log(100000000000000000)*1000

ans =

3.9144e+04

Norman
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Take logs: $$ 1000\,\ln n \le \sqrt{n}\ln 5 $$ for large $n,$ as $\ln x=O(x^\alpha)$ for all $\alpha>0.$

Reiner Martin
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With d'Hospital it is easy to see that $x^{-1/2} \log x \to 0$ for $x \to \infty$

hence there is $N$ such that $n^{-1/2} \log n < \frac{\log 5}{1000}$ for $n >N$. This means

$n^{1000} < 5^{\sqrt n}$ for $n >N$.

Fred
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0

The $\mathcal O$ notation is only relevant for large $n$. For $n=1000$ as in your post, there does not have to be an equality, since $1000$ is not large enough in this context.

For very large $n$, the inequality $$ n^{1000} < 5^{\sqrt n} $$

is indeed true.

supinf
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