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Re^2: Big O notation

by jettero (Monsignor)
on Oct 14, 2011 at 16:37 UTC ( #931553=note: print w/replies, xml ) Need Help??


in reply to Re: Big O notation
in thread Big O notation

You could use a graphing calculator to show that's probably not true. n^2 log(n) will grow faster than log(n), as will n^2, so surely O(log(n)) can't be an upper bound for something that has n^2 log(n) in it. Maybe it was supposed to be n^(2 * log(n)) ... can't really tell. Probably homework anyway.

update: lol, figured

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[Eily]: muthusathish ok, the fact that you can't tell what perl has to do with it proves it: this is not perl

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