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I'm reading The Algorithm Design Manual and this is one of the excersizes:

Prove or disprove the following statement:

$\Theta(n^2 ) = \Theta(n^2 + 1)$

I think this is untrue because the right side of the equation claims to have a multiple $c$ such that it is smaller than all the multiples of $n^2$, which is untrue. But since the book doesn't have answers and I'm not really sure about this question I had to ask.

Thanks a lot :)

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As we have on both sides of equality sets, then it should be prove correspondingly i.e we need to show, that $\Theta(n^2) \subset \Theta(n^2+1)$ and $ \Theta(n^2+1)\subset \Theta(n^2) $. Let's start with first:

Assume $f \in \Theta(n^2)$. This means, that for some constants $A_1, A_2 \gt 0$ holds

$$A_1 n^2 \leqslant f(n) \leqslant A_2 n^2$$. To obtain inequalities for $\Theta(n^2+1)$ we need $$B_1 (n^2+1) \leqslant f(n) \leqslant B_2 (n^2+1)$$ It's easy to see, that it's enough to take $B_2 = A_2$ for right hand and $0<B_1 \leqslant \frac{A_1}{2}\leqslant \frac{A_1 n^2}{n^2+1}$ for left hand, as $A_2>0$.

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  • $\begingroup$ Thanks a lot! :) $\endgroup$ – kasra Feb 1 at 13:29

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