I've got two function $f(x)=\sin(x)$ and $g(x)=\sin\left(\frac{x}{4}\right)$.
I'd like to construct a function $h$, which would equal $0$ when $\sin(x)=0$ except when $\sin\left(\frac{x}{4}\right)=0$.
So
for $f(\pi)=0$ and $g(\pi)\ne0$ we have $h(\pi)=0$;
for $f(2\pi)=0$ and $g(2\pi)\ne0$ we have $h(2\pi)=0$;
for $f(3\pi)=0$ and $g(3\pi)\ne0$ we have $h(3\pi)=0$;
for $f(4\pi)=0$ and $g(4\pi)=0$ we have $h(4\pi)\ne0$ (exact value doesn't matter to me).
Any ideas?
Thanks in advance.
What about $$f(x) = \begin{cases} \frac{\sin(x)}{\sin\left(\tfrac x4\right)} & ; x\ne 4k\pi \\ 4(-1)^{k} & ; x = 4k\pi \end{cases}$$ Its continuous at all $x=4k\pi$ ($k\in\mathbb Z$) because of $$\lim_{x\rightarrow 4k\pi} \frac{\sin(x)}{\sin\left(\tfrac x4\right)} = \lim_{x\rightarrow 4k\pi} \frac{\cos(x)}{\tfrac 14 \cos\left(\tfrac x4\right)} = \frac {(-1)^{4k}}{\tfrac 14 (-1)^{k}} = 4(-1)^{k}$$ So all $x=4k\pi$ are removable singularities and thus $f$ is an analytic function and also $f|_{\mathbb R}$ is $C^\infty$.