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$$ (n-7)^2 \, \text{is} \, \Theta(n^2) $$

Is this correct?

So far I have:

$ (n-7)^2 \, \text{is} \, O(n^2) \\ n^2 -14n +49 \, \text{is} \, O(n^2) \\ \begin{align} n^2 -14n +49 & \le \, C \cdot n^2 \, , \, n \ge 1 \\ & \le 50n^2 \end{align} \\ $

$ (n-7)^2 \, \text{is} \, \Omega(n^2) \\ n^2 -14n +49 \, \text{is} \, \Omega(n^2) \\ \begin{align} n^2 -14n +49 \, & \ge \, C \cdot (n^2) \; ,n \ge 1 \\ & \ge -14 n^2 \end{align}$

This proves it true right? Must both proofs have the same constant? I'm very new with Asymptotics.

share|improve this question
    
It must be with a positive constant. –  Tobias Kildetoft Mar 26 '13 at 14:35
    
Can both constants in both proofs be different? So would the second proof $ is \Omega $ be okay with a constant of $ \ge 1 \cdot n^2 $ alone? –  user1766555 Mar 26 '13 at 14:39
1  
Yes, the constants do not need to be the same, but they must both be positive. –  Tobias Kildetoft Mar 26 '13 at 14:41
    
The constants need to be positive, but the inequality only has to hold above some $N$, so you don't have to worry about irregular behavior in the low ranges. (for example, for $n>28$, you can assume that $n^2-14>\frac{1}{2}n^2$) –  Alfonso Fernandez Mar 26 '13 at 15:17

1 Answer 1

up vote 3 down vote accepted

Easier: $(n-7)^2 > (\frac{n}{2})^2=\Omega(n^2), \ (n-7)^2 < (2n)^2=O(n^2)$ for $n$ large enough, hence $(n-7)^2 = \Theta(n^2)$

share|improve this answer
    
Was the choice of a constant of $ \frac12 $ purely arbitrary for the $ \Omega $ proof? –  user1766555 Mar 26 '13 at 15:05
1  
Yes. You can get sharper bound of course. –  Alex Mar 26 '13 at 15:09

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