Finding classes of non-piecewise functions which have graphs that resemble sine curves but contain periodic hole discontinuities

Question

Discussing continuity in my Calc class today got me wondering: Can we define a non-piecewise sine curve with periodic "hole" discontinuities? In this case, when I say sine curve, I am referring to functions of the traditional $f(x)=a\sin{x}$ shape.

So I did a little bit of digging. $f(x)=\frac{\sin{x}}{x}$ did not work. I plotted it on Desmos and the resulting curve was sinusoidal but wonky; the function approached an absolute maximum in the neighborhood of $x=0$.

Next, I tried $f(x)=\frac{\sin{2x}}{\sin{x}}$. This actually worked; the function resembled a "normal" sine curve, the only difference being its discontinuities. I added a photo of it here.

Next, I tried $f(x)=\frac{\sin{3x}}{\sin{x}}$, which worked. $f(x)=\frac{\sin{3x}}{\sin{2x}}$, did not, with the resulting graph resembling a secant function. I didn't stop there. I found that the following functions also worked. $f(x)=\frac{\sin{4x}}{\sin{2x}}$. $f(x)=\frac{\sin{6x}}{\sin{3x}}$, $f(x)=\frac{\sin{8x}}{\sin{4x}}$. Each of these functions produced a curve of the same shape as the previous.

My question then is, is there a class of functions of a certain form that fit this criteria, namely, graphing as a sine curve with periodic "hole" discontinuities?

My investigation would have me think that functions of the form $f(x)=\frac{\sin{mx}}{\sin{nx}}$ where $\frac{m}{n}=2$ satisfy the criteria, but I am not sure if 1, this is always true, or 2, there other classes of functions that also satisfy the criteria.

My testing showed that when $\frac{m}{n}\neq2$, the function did not meet the criteria.

Could anyone provide proof or heuristics regarding the existence of this, or another class of functions that satisfies this criteria? Are there infinitely many classes of functions that work?

A little bit of context. Our teacher had us draw an arbitrary function with 4 discontinuities. We had to then list the left and right hand limits at each discontinuity, and whether or not the limit existed. This got me thinking about drawing a sine curve.

Thank you for reading.

Answer 1

Recall that $\sin (k x) = \sin x \, U_{k-1} (\cos x)$ where $U_{k-1}(x)$ is the $(k-1)$-th Chebyshev polynomial of the second kind. This allows us to write $$f(x) = \frac {\sin (mx)} {\sin (nx)} = \frac {\sin x \, U_{m-1} (\cos x)} {\sin x \, U_{n-1} (\cos x)}$$ So $f$ has removable discontinuities if and only if $U_{n-1}(x)$ divides $U_{m-1}(x)$. Now we apply the following:

Theorem (*). Let $h \ge k$ be two positive integers. $U_h(x)$ is a multiple of $U_k(x)$ if and only if $h = (l+1)k + l$ for some integer $l \ge 0$.

In our case, by letting $h = m-1$ and $k = n-1$ the condition becomes

$$m-1 = (l+1)(n-1) + l \quad\Longleftrightarrow\quad m = (l+1)n \quad\Longleftrightarrow\quad n \mid m$$

Therefore $f$ has removable discontinuities if and only if $n \mid m$.

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(*) Theorem 5 in M. O. Rayes, V. Trevisan, P. S. Wang. Factorization of Chebyshev Polynomials. ICM-199802-0001