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I am looking at this questions and the proof for it and wondering how this works.Can anyone explain the answer to me or do you have any other way to answer this question.I am new to asymptotic notations

$$2n^3 + 5n + 12 = O(n^3)$$

True Proof: $$\lim_{n\to\infty}\frac{2n^3 + 5n + 12}{n^3}=2$$

This implies $$2n^3 + 5n + 12 = O(n^3)$$

Tim Raczkowski
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user4203815
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A function $f(n)$ is $O(g(n))$ if there is some constant $k$ such that $f(n)\leq k\cdot g(n)$. I prefer directly prove that $2n^3+5n+12=O(n^3)$ as $2n^3+5n+12\leq 2n^3+5n^3+12n^3=19n^3$.

Salomo
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  • I don't understand your proof here . In $2n^3+5n+12\leq 2n^3+5n^3+12n^3=19n^3$ where is the $n^3$ on 5 and 12 in the left hand side coming from and where is the 19 on the right hand side coming from ? – user4203815 Mar 30 '15 at 00:03
  • Here I assume $n$ is in $\mathbb{N}$, then we can have $n\leq n^3$ and $1\leq n^3$. However, the big-O notation can also be used in the domain of real numbers. I assume we are talking about big-O defined for $\mathbb{N}$ as you use variable $n$ but not $x$. Are you doing analysis or discrete math? – Salomo Mar 30 '15 at 00:06
  • I am trying to prove or disprove the asymptotic equation,.I am a beginner to asymptotics. I don't understand what do you mean by n is in N.What is the difference between n and N? – user4203815 Mar 30 '15 at 00:12
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    $\mathbb{N}$ refers to the positive integers (or "natural" numbers, which is where the "N" comes from). It is written "\mathbb{N}". – marty cohen Mar 30 '15 at 00:43