Jeremy Ting edited P1.tex  about 10 years ago

Commit id: 988e1693b1cb3ea4e0783405daba5b1dad5e7172

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$f(n)=2^{7n/2}, g(n)=2nlg(7)$  $f(n)=lg(2^{7n/2}, g(n)=lg(2nlg(7))$     $f(n)=lg(2^{7n/2}, g(n)=lg(2nlg(7))$     $f(n)=7n/2, g(n)=lg(2n) + lg(lg(7))$     $f(n)=7n/2, g(n)=lg(2n)$  $f=\Omega(g)$  \item  $f(n)=2^{nln(n)}, g(n)=n!$