Jeremy Ting edited P1.tex  about 10 years ago

Commit id: ef6efedad1272e02476c06dd9ed96499067c9f6a

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