How to draw the graph of f(x)= [sin x] + [sin 2x]

Question

I needed to solve this question:

Let f(x) = [sin x] + [sin 2x] such that x belongs to (0,10) ,where [.] is the greatest integer function, then find the number of points where f(x) is discontinuous.

For solving such type of questions, I usually draw their graphs and find the points of discontinuity of the graph. However, since the trigonometric functions are inside the greatest integer function, I am unable to apply any trigonometric identity and convert them into one single function. I tried defining piece-wise functions separately for each term, and then adding the two, but it gets very tedious and lengthy.

Answer 1

Let’s see…

Function $f$ is the sum of two $2\pi$-periodic functions, hence it suffice to draw the graph in $]-\pi , \pi]$. We got:

$$ \begin{split} \sin x = 1 \quad &\Leftrightarrow \quad x = \frac{\pi}{2} \\ 0 \leq \sin x < 1 \quad &\Leftrightarrow \quad 0\leq x \leq \pi \land x \neq \frac{\pi}{2} \\ -1 \leq \sin x < 0 \quad &\Leftrightarrow \quad -\pi < x < 0 \\ \sin 2x = 1 \quad &\Leftrightarrow \quad x = \frac{\pi}{4}, - \frac{3\pi}{4} \\ 0 \leq \sin 2x < 1 \quad &\Leftrightarrow \quad 0\leq x \leq \frac{\pi}{2} \land x \neq \frac{\pi}{4} \\ &\phantom{\Leftrightarrow \quad} \text{ or } -\pi < x \leq -\frac{\pi}{2} \land x \neq -\frac{3\pi}{4} \\ &\phantom{\Leftrightarrow \quad} \text{ or } x = \pi \\ -1 \leq \sin x < 0 \quad &\Leftrightarrow \quad -\frac{\pi}{2} < x < 0 \text{ or } \frac{\pi}{2} < x < \pi \end{split} $$ therefore: $$\begin{split} [\sin x] &= \begin{cases} 1 &\text{, if } x = \frac{\pi}{2} \\ 0 &\text{, if } 0 \leq x \leq \pi \land x \neq \frac{\pi}{2} \\ -1 &\text{, otherwise} \end{cases} \\ [\sin 2x ] &= \begin{cases} 1 &\text{, if } x = \frac{\pi}{4}, -\frac{3\pi}{4} \\ 0 &\text{, if } 0\leq x \leq \frac{\pi}{2} \land x \neq \frac{\pi}{4} \\ & \text{ or } -\pi < x \leq -\frac{\pi}{2} \land x \neq -\frac{3\pi}{4} \\ & \text{ or } x = \pi \\ -1 &\text{, otherwise} \end{cases} \end{split} $$ When we put everything together we find: $$ f(x) := \begin{cases} -2 &\text{, if } -\frac{\pi}{2} < x < 0 \\ -1 &\text{, if } -\pi < x < -\frac{3\pi}{4}, -\frac{3\pi}{4} < x \leq -\frac{\pi}{2}, \frac{\pi}{2} < x < \pi \\ 0 &\text{, if } x= -\frac{3\pi}{4}, 0\leq x < \frac{\pi}{4}, \frac{\pi}{4} < x < \frac{\pi}{2}, x = \pi \\ 1 &\text{, if } x = \frac{\pi}{4}, \frac{\pi}{2} \end{cases} $$

Answer 2

For a continuous function $g(x)$, the function $[g(x)]$ is discontinuos at most at those points where $g(x)$ is an integer. It happens precisely at those points among these where $g(x+h)<g(x)$ for arbitrarily small $|h|$. For $g(x)=\sin x$, the possible integer values $-1,0,1$ are achieved for $x\in\frac\pi2\Bbb Z$. However, at points where $\sin x=-1$, $[\sin x]$ is still continuous because there $\sin(x+h)\ge \sin x$ holds for all (in particular all small) $|h|$. Thus $[\sin x]$ is discontinuous at $x=\frac{k\pi}2$ with integer $k\not\equiv 3\pmod 4$, so for $x\in(0,10)$, this is for $$x\in\left\{\frac\pi2,\pi,2\pi,\frac52\pi,3\pi\right\}.$$ Similarly, $[\sin 2x]$ is discontinuous at $x=\frac{j\pi}4$ with integer $j\not\equiv 3\pmod 4$, so for $$x\in\left\{\frac\pi4,\frac\pi2,\pi,\frac54\pi,\frac32\pi,2\pi,\frac94\pi,\frac52\pi,3\pi\right\}.$$

The sum of two floored continuous functions, $[g_1(x)]+[g_2(x)]$ is clearly continuous where both summands are and discontinuous where exactly one of the summands is. What happens at places $x$ where both summands have a discontinuity? Consider a sequence $x_n\to x$ such that $g_1(x_n)<g_1(x)\in\Bbb Z$. Then for almost all $n$, $g_2(x_n)<g_2(x)+1$, hence for these $n$, $[g_1(x_n)]+[g_2(x_n)]\le [g_1(x)]-1+[g_2(x)]$, i.e., the sum is discontinuous at $x$. We conclude that $[\sin x]+[\sin 2x]$ is discontinuous precisely for $$x\in\left\{\frac\pi4,\frac\pi2,\pi,\frac54\pi,\frac32\pi,2\pi,\frac94\pi,\frac52\pi,3\pi\right\}.$$