# Big oh and theta calculation

• Sep 21st 2009, 10:06 AM
vexiked
Big oh and theta calculation
How could one show that:

3n³ – 2n² + 1 O(n³).

3n³ – 2n² + 1 Θ(n³).
• Sep 21st 2009, 05:58 PM
Skerven
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.

• Sep 21st 2009, 08:40 PM
CaptainBlack
Quote:

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