How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.

Question

How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.

I know that by definition $(\lfloor{x}\rfloor=m) ≡ (m≤x<m+1)$, so it follows that $(\lfloor{x^2}\rfloor=m ≡ (m≤x^2<m+1) ≡ (\sqrt{m}≤x^2<\sqrt{m+1})$; from this I can do

$\lfloor{x^2}\rfloor= \begin{cases} ...\\ -3, & -3≤x^2<-2 \\ -2, & -2≤x^2<-1 \\ -1, & -1≤x^2<0 \\ 0, & 0≤x^2<1 \\ 1, & 1≤x^2<2 \\ 2, & 2≤x^2<3 \\ 3, & 3≤x^2<4 \\ ... \end{cases} $

= $\begin{cases} ...\\ -3, & -3≤x^2<-2 \\ -2, & -2≤x^2<-1 \\ -1, & -1≤x^2<0 \\ 0, & 0≤x<1 \\ 1, & 1≤x<\sqrt{2} \\ 2, & \sqrt{2}≤x<\sqrt{3} \\ 3, & \sqrt{3}≤x<2 \\ ... \end{cases} $

Everything is normal until I "solve" for x in the negative intervals of the domain, which results in this

$\lfloor{x^2}\rfloor= \begin{cases} ...\\ -3, & 3i≤x<2i \\ -2, & 2i≤x<i \\ -1, & i≤x^2<0 \\ 0, & 0≤x<1 \\ 1, & 1≤x<\sqrt{2} \\ 2, & \sqrt{2}≤x<\sqrt{3} \\ 3, & \sqrt{3}≤x<2 \\ ... \end{cases} $

I don't know how to graph between those intervals.

Nonetheless when i see a graph of this function it seems that the function is symmetrical to the y axis.

Could you explain me, please, how can I graph the function in those negative intervals?

Thanks in advance.

Answer 1

Guide:

• There are no real $x$ which satisfies $-1 \le x^2 < 0$.

• $0 \le x^2 < 1$ is equivalent to $0 \le |x|^2 < 1$ and the solution is $|x| < 1$, that is $-1 < x < 1$.

• $1 \le x^2 < 2$ is equivalent to $1 \le |x|^2 < 2$ and the solution is $1 \le |x| < \sqrt{2}$, that is $\{ x: -\sqrt{2} < x \le -1\} \cup \{ x: 1 \le x < \sqrt2\}$.

Note that the function is an even function.

Answer 2

Well, by the trivial inequality, we have that $x^2$ is nonnegative. Note that you are solving for the floor of the $y$ outputs, not the $x$ inputs. Also, $\text{floor}(-x)^2=\text{floor}(x^2),$ so both sides are symmetrical.