I am stuck with a problem where I need to prove
$2n^3 − 3n^2 + 17n = \Theta(n^3)$
I am a novice student of asymptotic notations. Can you please push me towards the right direction here?
Theta notation for polynomials
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$\begingroup$
proof-writing
asymptotics
2 Answers
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read some explanations
e.g.
https://en.wikipedia.org/wiki/Big_O_notation
in your problem: $$ \frac{2n^3 − 3n^2 + 17n}{n^3}=2-\frac{3}{n}+\frac{17}{n^2} $$ since $\frac{3}{n}\to 0$ and $\frac{17}{n^2}\to 0$ as $n\to\infty$, these expressions are bounded. Hence $\frac{2n^3 − 3n^2 + 17n}{n^3}$ is bounded, what completes the proof.
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DEFINITION A function $f(n)$ is said to be $O(g(n))$ if there exist $n_0 \in\mathbb{N},c \in \mathbb{R}_+$ such that $$f(n) \leq cg(n) \quad \forall n \geq n_0.$$
For your function we get the following. If $n \geq 3$ then $3n^2 \leq n^3$, so $$ 2n^3 − 3n^2 + 17n \geq 2n^3 -n^3 = n^3,$$ that is, choose $n_0 = 3$ and $c = 1$.