Why do these functions satisfy that f(n) is not O(g(n)) and g(n) is not O(f(n))?

Question

I don't understand what these function are like and why they satisfy that f(n) is not O(g(n)) and g(n) is not O(f(n)).

Where is x?

\begin{eqnarray} f(x)= \begin{cases} k^{2k}, &x\in(2𝑘,2𝑘+1)& \\ k^{2k+1}, &x\in(2𝑘+1,2𝑘+2)& \end{cases} \end{eqnarray}

\begin{eqnarray} g(x)= \begin{cases} k^{2k-2}, &x\in(2𝑘,2𝑘+1)& \\ k^{2k+2}, &x\in(2𝑘+1,2𝑘+2)& \end{cases} \end{eqnarray}

Also, could you teach me how to write these functions in Grapher on MacOS? Cuz I want to know what they are like.

Answer 1

Let's firstly imagine function $f$: for $k=2$ we have two intervals $(4,5)$ and $(5,6)$ and for $k=3$ we have two intervals $(6,7)$ and $(7,8)$. Accordingly for $f$ we have:

$$\quad\quad\quad x\quad\ \overbrace{(4,5)\quad (5,6)}^{k=2}\quad\overbrace{(6,7)\quad (7,8)}^{k=3} \cdots\\ f(x)\quad 2^4\quad\quad 2^5\quad\quad\quad 3^6 \quad\quad3^7$$ for function $g$ we have: $$\quad\quad\quad x\quad\ \overbrace{(4,5)\quad (5,6)}^{k=2}\quad\overbrace{(6,7)\quad (7,8)}^{k=3} \cdots\\ g(x)\quad 2^2\quad\quad 2^6\quad\quad\quad 3^4 \quad\quad3^8$$ So, as you see both $f$ and $g$ are increasing, but alternately one is more then other with more then constant factor, so they cannot be big-$O$ of each other.

To get an idea of an example you can take a look at DNA molecule chain and imagine, that $f$ is one line and $g$ another.