Conditions for Moving Function Outside Sine Argument

Question

Are there multiple discrete functions $f[n]$ such that \begin{align} \sin \left( f[n] x[n] \right) = f[n] \sin \left( x[n] \right) \end{align} I know that the above equation holds for $f[n] = 0$, $f[n] = 1$, $f[n] = u[n]$ (where $u[n] = 0$ for $n < 0$ and 1 otherwise) and $f[n] = \delta[n]$ (where $\delta[n] = 1$ for $n = 0$ and $0$ otherwise). If there are others, are there any special conditions for $f[n]$ and $x[n]$ that allow for this type of "commuting"? Is it basically just variations of functions that modulate between 0 and 1?

Answer 1

One has $\sin(f\cdot x)=f\cdot\sin(x)$ for every $x$ if and only if $f=-1$ or $0$ or $1$.

The if part is obvious. For the only if part, consider the limit of the identity $\sin(f\cdot x)=f\cdot\sin(x)$ when $x\to0$. Then $\sin(x)=x-x^3/6+o(x^3)$ hence $$ f\cdot\sin(x)=f\cdot x-f\cdot x^3/6+o(x^3). $$ Furthermore $f\cdot x\to0$ hence $$ \sin(f\cdot x)=f\cdot x-f^3\cdot x^3/6+o(x^3). $$ The expansion along powers of $x$ is unique hence $f/6=f^3/6$, that is, $f^3-f=0$, that is, $f\cdot(f-1)\cdot(f+1)=0$. Finally, $f=-1$ or $0$ or $1$.

Answer 2

If $x[n]=pi/2$, $sin(f[n]*\pi/2)=f[n]$. Sine can only take on values between -1 and 1 so $f[n]\in [-1,1]$

\begin{align} \sin \left( f[n] x[n] \right) - f[n] \sin \left( x[n] \right) =0 \end{align} This equation has to hold for all functions $x[n]$ so let's assume $x[n]=\sqrt{2}$ so that we can find some functions $f[n]$ that may work. A plot of $f[n]$ vs $sin(f[n]\sqrt{2})-f[n]sin(\sqrt{2})$ for $f[n]\in [-1,1]$ shows that the only values that $f[n]$ can take are -1, 0 and 1. (plot)

$f[n]={-1, 0, 1}$

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Alternate (analytic) solution: \begin{align} &fsin(x)=sin(fx)\\ \\ \text{apply $\frac{d^2}{dx^2}$ to both sides}\\ &-fsin(x)=-f^2sin(fx)\\ \text{divide both sides by $-f$}\\ &sin(x)=fsin(fx)\\ \text{substitute into first equation}\\ &f^2sin(fx)=sin(fx)&\\ \text{this equation can be true iff $sin(fx)=0$ or $f^2=1$}\\ \\ &f=-1,0,1 \end{align}