you are comletely wrong you hve to show that [lg n]^2 is small-Oh of n. that is done by showing that lim [lg n]^2/n is zero as n goes to infinity.
you need now some mathematics to show this trivial result. 2009/5/14 console kid <[email protected]> > > hi all ,I am talking about the big-Oh notation , > I had to prove that > [lg n] ^2 is better than the n . > > how can I do that . I have tried to show that n is a big-oh of ([lg n] > ^2) ,but > how can I do that ? > > -- thanks in advance -- > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
