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Proving $2^{\log^{1-\varepsilon}n}\in \omega(\operatorname{polylog}(n))$
Let $\varepsilon \in (0,1)$. I wish to show that $2^{\log^{1-\varepsilon}(n)}\in \omega(\operatorname{polylog}(n))$. I attempted to turn this into a function and use L'Hospital's rule but that got me ...
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interpretation of logarithmic in the dimension
I'm reading a paper that says
"$\ Cn^{6/5}r \log n $ is not logarithmic or polylogarithmic in the dimension and one would like to know if results closer to the $\ nr \log n$ limit are possible."
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