Explain the graph of $\lfloor \arctan x\rfloor$ I recently started studying continuity and I tried to figure out continuity of $\lfloor \arctan x\rfloor$ using graphs.
I got a similar graph to the one which is in the Desmos graph plotter.
To check the graph of $\lfloor \arctan x\rfloor$ in desmos graph plotter just plug in the function floor(arc(tanx)) in the link given below:
https://www.desmos.com/calculator
My doubt is that at $x=\tan(1)=1.557$ the value of $\lfloor \arctan x\rfloor$ has to be one, but in the graph of Desmos website the value given is $0$.
How is that possible? Could anyone explain the graph and solve my doubt?
Any little help is appreciable.
Moreover, you can check and share about the Desmos website, it is a very good website and can plot even complex graphs. 
 A: $$\tan(1)\approx 1.557407724654902230506974807458360173087250772381520038>1.557\implies \arctan 1.557<1$$(See http://m.wolframalpha.com/input/?i=tan%281%29 )
Note that $\arctan 1.557$ is very close to $1$, but the floor function is not continuous hence unless it is greater or equal how close it is doesn't matter: $\arctan 1.557\approx 0.99981$( see http://m.wolframalpha.com/input/?i=arctan%281.557%29 )
A: We have
$$
\begin{cases}
-{\large{\frac{\pi}{2}}} < \arctan(x) < -1&\;\;\;\text{if}\;\,x < -\tan(1)\\[4pt]
-1\le  \arctan(x) < 0&\;\;\;\text{if}\;\,-\tan(1) \le x < 0\\[4pt]
0\le  \arctan(x) < 1&\;\;\;\text{if}\;\,0 \le x < \tan(1)\\[4pt]
1 \le \arctan(x) < {\large{\frac{\pi}{2}}} &\;\;\;\text{if}\;\,x \ge \tan(1)\\[4pt]
\end{cases}
$$
Noting that $1 < {\large{\frac{\pi}{2}}} < 2$, we get
$$
\lfloor{\arctan(x)}\rfloor=
\begin{cases}
-2&\;\;\;\text{if}\;\,x < -\tan(1)\\[4pt]
-1&\;\;\;\text{if}\;\,-\tan(1) \le x < 0\\[4pt]
0&\;\;\;\text{if}\;\,0 \le x < \tan(1)\\[4pt]
1&\;\;\;\text{if}\;\,x \ge \tan(1)\\[4pt]
\end{cases}
$$
Thus, the graph of the function 
$$f(x)=\lfloor{\arctan(x)}\rfloor$$
is piecewise constant, with $4$ horizontal pieces, one at each of the $y$-values $-2,-1,0,1$.

Thus, based on the piecewise constant form of $f$, we see that if $x=\tan(1)$, then $f(x)=1$, but if $x$ is slightly less than $\tan(1)$, then $f(x)=0$.

Hence, when evaluating $f(x)$ for decimal values of $x$ near $\tan(1)$, you need to worry about the possibility that $x$ might be slightly less than $\tan(1)$.

In particular, since $1.557$ is slightly less than $\tan(1)$, we get $f(1.557)=0$.
