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

In reply to Re^2: Big O notation by jettero
in thread Big O notation by rolandomantilla

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