Why $\sin(x)+\sin(\pi x)$ is not periodic?
There is an answer here which tries to explain it, but I somehow do not get it.
If we assume that $T>0$ is a period of $\sin(x)+\sin(\pi x)$, then
$$\sin(x)+\sin(\pi x)=\sin(x+T)+\sin(\pi (x +T))$$
Apparently one needs to differentiate the equation above two times to get:
$$\sin(x)+\pi^2 \sin(\pi x)=\sin(x+T)+ \pi^2 \sin(\pi (x +T))$$
and then what?
Best Answer
Then you subtract the equations to get $\sin \pi x = \sin ( \pi x + \pi T)$. Putting that in your first equation you get $\sin x = \sin (x + T)$. Therefore, $T = 2n \pi$ for some integer $n$. On the otherhand, $\sin \pi x = \sin ( \pi x + \pi T)$ gives you $\sin x = \sin (x + \pi T)$ (replace $x$ by $\frac x \pi$ ). This means that $\pi T = 2k \pi$ for some integer $k$. So, $T = 2k$, an integer. But from before we had that $T = 2n \pi$, which is an irrational number. So, this is a contradiction.