I need to find a function $f$ which is in $o(\log^{k} n)$ for fixed value of $k$ with $f = \omega(1)$. I know that for little $o$ the function should be strictly less than $c\log^k n$ for all $c$ and large enough $n$; and for little $\omega$ it should be strictly greater than $c\cdot 1$ for all $c$ and large enough $n$, but I am stuck here. How does one usually solve such type of problems?
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@PålGD: That doesn't go to infinity. $\log^{k/2} n$ works. – Louis Apr 06 '14 at 22:26
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@Louis sorry for deleting my comment, what I meant was $\log^k(n) / \log^{k-1}(n)$. – Pål GD Apr 06 '14 at 22:28
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@Louis I am still not sure how can the value be log^(k/2)n be greater than c.1 – Rahul Mehrotra Apr 06 '14 at 23:22
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@PålGD wouldnt that turn out to be log(n) ? – Rahul Mehrotra Apr 06 '14 at 23:47
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Your understanding of $o$ and $\omega$ is flawed; see here. – Raphael Apr 07 '14 at 06:13
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Hint: Try $f(n) = \log\log n$.
Yuval Filmus
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well i dont think it will satisfy ω(1). the value will be less than 1. we wont it strictly greater than it. – Rahul Mehrotra Apr 06 '14 at 23:24
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@Rahul As it happens, this $f$ is a solution to your question. Go over the definitions again. The condition $f = \omega(1)$ is the same as $\lim_{n\to\infty} f(n) = \infty$, which is certainly satisfied by my $f$. – Yuval Filmus Apr 07 '14 at 01:52