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Please help me on following two functions, I need to simplify them.
O(nlogn + n^1.01)
O(log (n^2))
My current idea is
O(nlogn + n^1.01) = O(nlogn)
O(log (n^2)) = O (log (n^2))
Please kindly help me on these two simplification problems and briefly give an explanation, thanks.
For the second, you have O(lg(n²)) = O(2lg(n)) = O(lg(n)).
For the first, you have O(nlg(n) + n^(1.01)) = O(n(lg(n) + n^(0.01)), you’ve to decide whatever lg(n) or n^(0.01) grows larger.
For that purpose, you can take the derivative of n^0.01 – lg(n) and see if, at the limit for n -> infinity, it is positive or negative: 0.01/x^(0.99) – 1/x; at the limit, x is bigger than x^0.99, so the difference is positive and thus n^0.01 grows asymptotically faster than log(n), so the complexity is O(n^1.01).