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Wikipedia talks about two groups of functions with asymptotic growth rates between polynomial and exponential – quasi-polynomial functions and sub-exponential functions.

It only gives two examples of such a function, however, and those functions are of the form $2^{(n^{1/3})}$ (the runtime of the prime field sieve) and $2^{(n \log n)^{1/2}}$ (the runtime of the graph isomorphism problem). Both of these functions have the property that $f(p(n))$ (where $p$ is some polynomial) is still a function in $\Omega(2^n)$.

Is there an example of a function that doesn't have this property? Specifically, in terms of growth behaviour, I want the following to be true for $f(n)$:

  • $f(n)$ grows faster than any polynomial function – that is, for any polynomial $p(n)$, $f \notin O(p(n))$.

  • No polynomial of $f(n)$ grows faster than an exponential function – that is, for any polynomial $p(n)$, a new function defined as $g(n) = f(p(n))$ is in $o(2^n)$.

Joe Z.
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  • That got a downvote pretty quickly. What happened? – Joe Z. Jul 08 '14 at 19:37
  • I didn't downvote, but your comment about 'still a function in $O(2^n)$ probably isn't what you mean to say - I think you mean that 'there is some polynomial such that $f(p(n))\in\Omega(2^n)$. ($O(2^n)$ bounds above; $2^{n^{1/3}}$ is still in $O(2^n)$.) – Steven Stadnicki Jul 16 '14 at 03:20
  • Yeah, I meant $\Omega(2^n)$. I'm too used to using big $O$ notation as a misnomer for that. – Joe Z. Jul 16 '14 at 03:23

1 Answers1

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Let $f(n) = n^{\log n}$. Then for any $k$ we have $$ \frac{n^k}{f(n)} = n^{k-\log n} \to 0 $$ that is $n^k \in o(f)$, any $k$.

Moreover, for any $k$ with $p_k(n)=n^k$, we have $$ f(p_k(n)) = f(n^k) = n^{k\log n^k} = n^{k^2\log n} = 2^{k^2\log^2 n}$$ which is $o(a^n)$ for any $a > 1$, as $a^n = 2^{n\log a}$.

martini
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