How could one show that: 3n³ – 2n² + 1 ∈ O(n³). 3n³ – 2n² + 1 ∈ Θ(n³).
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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.
Originally Posted by vexiked How could one show that: $\displaystyle 3n^3 – 2n^2 + 1 \in O(n^3)$ For $\displaystyle n>1$: $\displaystyle n^3<3n^3-2n^3+1<3n^3-2n^2+1<3n^3+2n^3+n^3=6n^3$ CB
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