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I can obviously tell by inspection that it is not true but I cannot figure out how to prove it with witnesses (C and k). Thanks!

chase1745
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    Hint: try to prove that $\lim_{n\to \infty}\frac{n^2}{3^n}=0$. – Shaull Sep 14 '16 at 13:32
  • Welcome to Computer Science! What have you tried? Where did you get stuck? We do not want to just hand you the solution; we want you to gain understanding. However, as it is we do not know what your underlying problem is, so we can not begin to help. See here for tips on asking questions about exercise problems. If you are uncertain how to improve your question, why not ask around in [chat]? – Raphael Sep 14 '16 at 16:22
  • "I can obviously tell by inspection that it is not true" -- I don't think you can. – Raphael Sep 14 '16 at 16:22

1 Answers1

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Let's assume that $3^n = O(n^2)$.

This would by definition mean that there are real positive numbers $a$ and $b$ such that for any sufficiently large $n$ (or to say as $lim_{n \to +\infty}$):

$ an^2 \leq 3^n \leq bn^2 $

Dividing everything by $3^n$ we get:

$ a\frac{n^2}{3^n} \leq 1 \leq b\frac{n^2}{3^n}$

And since if we let $lim_{n \to +\infty}$ we get a contradiction:

$ 0 \leq 1 \leq 0 $

Hence inital statement is false and $ 3^n \neq O(n^2) $

tms1337
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    This is wrong. $O$ does not imply such lower bounds. Also, I don't think your double-limit (you use an inequality that only holds sometimes, then manipulate it as if it holds everywhere, then let $n\to\infty$) is very rigorous. – Raphael Sep 14 '16 at 16:23
  • @itsRAWRtime007: $O$ notation only establishes upper bounds (not lower). The reasoning should go as follows: $3^n \neq O(n^2)$ because there exist some constant $c$ (for example, $c=1$) such that $3^n > c \cdot n^2$ for all $n \geq 0$. – Mario Cervera Sep 14 '16 at 19:12
  • Sorry mixed it up with $\theta$. What I gave above is proof for theta> I think you can just use one side of the inequalities for $O$ – tms1337 Sep 18 '16 at 14:30
  • Also it wasn't meant to be truly rigorous. Then we should define what does it mean "to have sufficiently large n". Just wanted to give a general idea of the proof. But if n is chosen such that both sides of inequality are satisfied (e.g taking the max $n_0$) then i think we can work it out in this double inequality notation. – tms1337 Sep 18 '16 at 14:39