Jeremy Ting edited P1.tex  about 10 years ago

Commit id: 3aaefffec4859ec724e06593a6454ac7bb6a042e

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$f(n)=2^{7n/2}, g(n)=2nlg(7)$  $f=\theta(g)$ $f=\Omega(g)$  \item  $f(n)=2^{nln(n)}, g(n)=n!$  $f(n)=ln(2^{nln(n)}), g(n) = ln(n!)$     $f(n)=nln(n), g(n) = nlg(n) (via previously proved identity)$   $f=\theta(g)$  \end{enumerate}