Sum of big O notation [duplicate]

Question

Possible Duplicate:
Big O when adding together different routines

What does O(n) + O(log(n)) reduce to? My guess is O(n) but can not give a rigorous reasoning.

I understand O(n) + O(1) should reduce to O(n) since O(1) is just a constant.

Answer 1

Well since O( f(n) ) + O( g(n) ) = O ( f(n) + g(n) ) We are simply trying to calculate an f(n) such that f(n) > n + log(n)

Since as n grows sufficiently log(n) < n we can say that f(n) > 2n > n + log(n)

Therefore O(f(n)) = O(2n) = O(n)

In a more general sense, O( f(n) ) + O( g(n) ) = O( f(n) ) if c*f(n)>g(n) for some constant c. Why? Because in this case f(n) will "dominate" our algorithm and dictate its time complexity.

Answer 2

Order is always reduced to higher order terms. I can give you intuitive reasoning. Suppose you have O(n + n^2). Then which part would play more important role in run time? n or n^2. Obviously n^2. Because where there n^2 you won't notice effect of n when n is increased or decreased.

As example,

let n = 100, then n^2 = 10000
means n is 0.99% and n^2 is 99.01% of total running time.
What would you consider for runtime?
if n is increased then this difference is clearer.

I think you understand now,

Answer 3

the answer is O(n). O(log n) is less than O(n). so their addition sums the maximum value that is O(n).