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Dragonfall
Jun15-09, 02:53 AM
(\lg n)!\in\mathcal{O}((\lg n)^{\lg n}) right?
CRGreathouse
Jun15-09, 03:53 AM
Right, you can get that from Stirling's formula. (Assuming that either n is a power of 2 or you use gamma(lg(n) + 1), of course.)
Dragonfall
Jun15-09, 11:53 AM
Excellent, thanks.