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@kokx
Created February 3, 2011 08:55
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\item $n^{-1}$
\item $16 = 2^{\log 16}$
\item $\log \sqrt{n}$
\item $\log n^3$
\item $n$
\item $n + \log n^4$
\item $n^{\log 4} \ (= n^2)$
\item $n^2 \log n$
\item $4^{\log n}$
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