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If we have the function $f : \mathbb{N}_0 \rightarrow \mathbb{N}$ with $f(n) = n^2$ and we look at the following representations of the sets $\mathcal{o}(f),\mathcal{O}(f),\Theta(f),\Omega(f),\omega(f) $: Asymptotic Notation

Now I have to list all necessary statements which are proving that the illustration is correct.

The necessary statements are in my opinion:

\begin{align} \Theta (f)&\subseteq \mathcal{O}(f) \\ \Theta (f)&\subseteq \Omega (f) \\ \Theta (f)&= \mathcal{O}(f) \cap \Omega (f) \\ \omega (f)&\subseteq \Omega (f) \\ \hbox{o}(f)&\subseteq\mathcal{O}(f) \\ \, \emptyset &=\, \omega (f) \cap \hbox{o}(f) \end{align} Now I have to proof two statements of my choice. But how to proof that in a formal correct way. (I understand the intuitive proofs but I don't know how to do it formally correctly.)

Hope somebody can help.

Yuval Filmus
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fragant
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1 Answers1

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Here is an explanation by example. Let $\Theta'(f) = \{ g : g(n)/f(n) \text{ tends to a positive limit} \}$. I will show that $\Theta'(f) \subseteq \Theta(f)$. The first step is to write the definitions. There are several variants of the definitions, and I picked one arbitrarily; you should use the definitions that were stated in class.

Let $f\colon \mathbb{N} \to \mathbb{N}$.

  1. $\Theta'(f)$ consists of all functions $g\colon \mathbb{N} \to \mathbb{N}$ such that for some positive $x \in \mathbb{R}_+$, $\lim_{n\to\infty} g(n)/f(n) = x$.
  2. $\Theta(f)$ consists of all functions $g\colon \mathbb{N} \to \mathbb{N}$ such that for some positive $a,b \in \mathbb{R}_+$ and $n_0 \in \mathbb{N}$, for all $n \geq n_0$ it holds that $a f(n) \leq g(n) \leq b f(n)$.

The next step is to take a function $g \in \Theta'(f)$.

Let $g \in \Theta'(f)$. Then there exists a positive $x \in \mathbb{R}_+$ such that $$ \lim_{n\to\infty} \frac{g(n)}{f(n)} = x.$$

Finally, we have to show that $g \in \Theta(f)$. For that we have to come up with positive $a,b \in \mathbb{R}_+$ and $n_0 \in \mathbb{N}$ such that $af(n) \leq g(n) \leq bf(n)$ for all $n \geq n_0$.

By the definition of limit, there exists $n_0$ such that for $n \geq n_0$, $$ \left|\frac{g(n)}{f(n)}-x\right| \leq \frac{x}{2}. $$ Therefore for $n \geq n_0$, $$ \frac{x}{2} \leq \frac{g(n)}{f(n)} \leq \frac{3x}{2}, $$ implying $$ \frac{x}{2} f(n) \leq g(n) \leq \frac{3x}{2} f(n). $$ Taking $a = x/2$ and $b = 3x/2$, we see that $g \in \Theta(f)$.

Only the latter two steps really belong to the proof. The first one states the claim. It is a very important step — without knowing what you're trying to prove, you won't be able to prove it.

Yuval Filmus
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