What should be the graph of $[y]=[\sin x]$?

Question

What should be the graph of $$[y]=[\sin x]?$$(here $[a]$ reffers to the floor function.)

In my opinion if we just consider the square bounded by $(0,0) ,(0,1),(\pi,0)$ and $(\pi,1)$ then any point inside the square will satisfy the equation as $[y]=0$ and so is $[\sin x]$.

But as I looked up to desmos the graph shows something else.It just covers the border and not the points inside the square. Why is it so? I am unable to upload the screenshot of the graph here.

Answer 1

I understand that you're looking for the set of pairs $$ \{ (x,y) \in \mathbb R^2: [y] = [\sin x] \}$$ Since this is not in the form $"y=f(x)"$, its not a graph, but a more general collection of points. In this specific case, you end up with filled in regions of the graph paper.

For $0\le x<\pi/2$, $\sin x\in [0,1)$ so $[\sin x]=0$. Then the set of points $y$ such that $[y] = 0$ is not a single point, but in fact a whole interval $[0,1)$.

The same is true for $\pi/2 < x \le \pi$, but at $x=\pi/2$, $[\sin \pi/2] = 1$, so then $y$ can take any value in $[1,2)$.

For the other half of the period, $\sin(x) \in [-1,0)$, and then $[\sin x] = -1$. In the end, you end up with a drawing something like:

     |        |
...[) (]    [) (]    ...
        (..)     (..)

Not sure what you put into Desmos, but here is how I coaxed Desmos into plotting the correct thing: I had to manually add certain features of the graph that I wanted to emphasize, but if you zoom in then my graph becomes "wrong".