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?
$\endgroup$ 42 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$Suppose $f(x)=\sin(x)+\sin(\pi x)$ is periodic with period $T$. Then its derivative $f'(x)=\cos(x)+\pi\cos(\pi x)$ has period $T$ as well and the same for $f''(x)=-\sin(x)-\pi^2\sin(\pi x)$
Thus you have, evaluating $f(0)=f(T)$ and $f''(0)=f''(T)$, $$ \begin{cases} \sin(T)+\sin(\pi T)=0\\[4px] \sin(T)+\pi^2\sin(\pi T)=0 \end{cases} $$ which entails $\sin(\pi T)=0$ and $\sin(T)=0$. Since $\pi$ is irrational, this is impossible.
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