How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$?

Question

How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$, where $\lfloor \cdot \rfloor$ denotes the greatest integer function?

Please help. I have no idea about this.

Answer 1

First, notice that the domain of $x$ is $$2-x^2\ge0\iff-\sqrt2\le x\le \sqrt 2.$$

And the floor function $\lfloor x\rfloor$ is defined as $$\lfloor x\rfloor=t\iff t\le x\lt t+1.$$

So, in your question, we have $$\lfloor \sqrt{2-x^2}\rfloor =t\iff t\le\sqrt{2-x^2}\lt t+1$$ Now note that $t\ge 0 \in\mathbb Z$ (This is because $\sqrt{2-x^2}\ge0$). So we have $$t^2\le 2-x^2\lt (t+1)^2\iff 2-(t+1)^2\lt x^2\le 2-t^2.$$

The fact that $\lfloor \sqrt{2-0^2}\rfloor=1, \lfloor \sqrt{2-({\sqrt 2})^2}\rfloor=0$ tells us that $t=0,1$.

Hence, we know

$$2-(0+1)^2\lt x^2\le 2-0^2\Rightarrow t=0,$$ $$2-(1+1)^2\lt x^2\le 2-1^2\Rightarrow t=1.$$

Answer 2

Let $$f(x)=\sqrt{2-x^2}$$

$2-x^2\geq 0$ or $ x^2 - 2\leq 0$ this gives $|x|\leq\sqrt 2 $

so find x in which f(x) is lies between 0 and 1 , 1 and 2 , 2 and 3 ....

this will give you a set of solution in that region take Greatest Integer of f(x) and now you can draw graph easily .

Answer 3

Are you looking for this?

This was rather easy to do with Wolfram Alpha:

y = floor(sqrt(2-x^2))