For functions f(n) and g(n), we say that "f(n) is Big-O of g(n)" if

There exists a constant c > 0 there exists a constant n_{0}such that

for all n ≥ n_{0}, 0 ≤ f(n) ≤ c * g(n)

We write f(n) = O(g(n)), but the "=" is not the usual meaning.

The intention is to allow us to say

- 3*n
^{2}+ 14*n - 7 = O(n^{2}) - 0.0007 * n
^{3}is*not*O(n^{2})

Technically, 10^{1,000,000}*n is O(n), but programs with that
running time is still very slow.