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Re^2: Big O notationby jettero (Monsignor) |
| on Oct 14, 2011 at 16:37 UTC ( [id://931553]=note: print w/replies, xml ) | Need Help?? |
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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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