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

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?

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

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

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