Is $$n\ll\omega(n)\ll n^{1+\epsilon}$$ true at every $\epsilon>0$ where $\omega$ is from Landau notation?
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What makes you think $\omega(n)\ll n^{1+\epsilon}$? – Henry Oct 12 '15 at 07:18
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I dont know I am asking... when I wrote on the board to a prof $n^{1+\epsilon}$ he asked me to write $\omega(n)$ (this is with regard to superlinear functions) – Turbo Oct 12 '15 at 07:21
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1$\exp(n) = \omega(n)$ ... – user251257 Oct 12 '15 at 10:46
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@user251257 how about $n^2$ or $2^{2^n}$. Is $n^2=\omega(n)$ or $2^{2^n}=\omega(n)$? – Turbo Oct 12 '15 at 17:38
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@Turbo both. You know, $\omega(n)$ is not a function but a symbol for any function growing asymptotically 'faster' than $n$ – user251257 Oct 12 '15 at 23:02
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@user251257 What does it mean when my prof said superlinear and I wrote $n^{1+\epsilon}$ and he corrected me and said $\omega(n)$? – Turbo Oct 13 '15 at 07:54
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@Turbo no idea. Sorry. – user251257 Oct 13 '15 at 10:51