# Math Help - Big oh and theta calculation

1. ## Big oh and theta calculation

How could one show that:

3n³ – 2n² + 1 O(n³).

3n³ – 2n² + 1 Θ(n³).

2. by referring to their definitions.

3n³ – 2n² + 1 O(n³).
in the definition, just let the constant be 4

3n³ – 2n² + 1 Θ(n³).
in the definition, let the constant be 2 in omega and let the constant be 4 in big-oh.

3. Originally Posted by vexiked
How could one show that:

$3n^3 – 2n^2 + 1 \in O(n^3)$
For $n>1$:

$n^3<3n^3-2n^3+1<3n^3-2n^2+1<3n^3+2n^3+n^3=6n^3$

CB