O（n）+ o（nlogn）是什么？

Question

What would o(nlogn) + O(n) be ?

My guess is O(nlogn) ?

f1(n) = o(nlogn) means for each constant c, there is an n0 such that 0 <= f1(n) < cnlogn .

And f2(n) = O(n) means there exists some constant c1, such that for n > n1, 0 <= f2(n) <= c1n .

so , all I can get from this is that, there is some constant c1, such that for n > max(n0,n1), 0 <= f1(n) + f2(n) <= c1(nlogn) .

Answer 1

If f is o(n log n) and g is O(n), then g is also o(n log n) so f+g is o(n log n) .

You do not get a positive lower bound. You do not get to say f+g can't be o(n) or even o(1). The function 0 is both O(n) and o(n log n). So is sqrt(n).

Big-O and little-o notations are asymptotic upper bounds for the magnitude. They are not lower bounds. Big-Omega and little-omega are lower bounds. Theta means you have both an upper bound and a lower bound of the same form. When you add together two upper bounds, you get an upper bound. You do not get a positive lower bound. In particular, you can't claim that the sum is not o(n).

Answer 2

Lets look at asymptotical limits of those functions and their sum.

First, upper limit:

F1 = o(n*log(n)), F2 = O(n) = o(n*log(n)), so
F1 + F2 = o (n* log(n))

Second, lower limit: we know F2 is O(N), but F1 can be zero for all we know, or anywhere between zero and n * log(n). So all we can say for sure is that F1+F2 is not o(N).

I will let you put all that in terms of 'for each constant C there exists an index n' yourself.

EDIT: Douglas pointed out that big-Oh and little-Oh are upper bounds only, and strictly speaking, you can't say anything about lower bounds at all. This is absolutely correct.

However, in programming most of the time what we actually care about are worst-case scenarios, and nobody says that sqrt(n) = O(n) – which would technically be correct. So yes, we can talk about lower bounds, in a sense, if we mean worst-case.