I have to prove that f(n)= 5n+2=O(n^2) and I know that it is true for O(n) so obviously, it will be true for a higher degree of n but how to prove this?
OK. Here is a simple way of proving this. I'm including it here in the hope that it could be useful for others with similar problems
5n + 2 <= 5n + 2n ; n >= 1
= 7n ; always
<= n*n ; n >= 7
= n^2 ; always
Therefore, there exists a constant c
, in this case c=1
, and an integer N
, in this case N=7
, such that
5n + 2 <= c*n^2 for all n >= N
Then, by definition
5n + 2 = O(n^2).
Note also that the first two lines
5n + 2 <= 5n + 2n ; n >= 1
= 7n ; always
are enough to show that 5n + 2 = O(n)
. In this case, c=7
and N=1
.