How could one show that:

3n³ – 2n² + 1 ∈ O(n³).

3n³ – 2n² + 1 ∈ Θ(n³).

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- September 21st 2009, 11:06 AMvexikedBig oh and theta calculation
How could one show that:

3n³ – 2n² + 1 ∈ O(n³).

3n³ – 2n² + 1 ∈ Θ(n³).

- September 21st 2009, 06:58 PMSkerven
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.

- September 21st 2009, 09:40 PMCaptainBlack