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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?

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    $\begingroup$ Subtract the two equations you wrote to get $(1-\pi^2)\sin(\pi x)=(1-\pi^2)\sin(\pi(x+T))$. Thus, $\sin(\pi x)=\sin(\pi(x+T))$. So we have also $\sin(x)=\sin(x+T)$. $\endgroup$
    – user72829
    May 8, 2016 at 12:59
  • $\begingroup$ That proof has been recognized as false by the poster. You just need to use the sum formula and you will notice an absurd. $\endgroup$
    – N74
    May 8, 2016 at 13:03
  • $\begingroup$ Then subtract the first equation from the second and divide both sides by $\pi^2-1$ to get $\sin\pi x=\sin\pi(x+T)$. Also multiply the first equation by $\pi^2$ and subtract the second equation, divide by $\pi^2-1$ to get $\sin x=\sin(x+T)$. So $T$ is a period of both $\sin x$ and $\sin\pi x$, which is impossible. $\endgroup$
    – almagest
    May 8, 2016 at 13:04
  • $\begingroup$ @N74 No, it is fine for proving that $\sin x+\sin\pi x$ is not periodic. The problem was that it did not answer the original question asked. $\endgroup$
    – almagest
    May 8, 2016 at 13:06

3 Answers 3

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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.

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    $\begingroup$ Thanks, I got it, I would rather argue that from $\sin (x) = \sin(x +T)$ follows that $\sin(T) = 0 , (x=0)$ and thus $T = \pi k_1 , k_1 \in \mathbb Z$, and from $\sin (\pi x) = \sin(\pi (x +T))$ follows that $\sin(\pi T) = 0, (x=0)$, thus $T = k_2 \in \mathbb Z$, thus $k_2 = \pi k_1$ and $\pi =k_2 / k_1$ a rational number. $\endgroup$
    – zesy
    May 8, 2016 at 13:14
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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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Other answers have explained this using proofs ,contradictions, etc. This post is not an explanation, rather a technique to determine the period length if it exists.

In the question, the period length of sin(πx) is 2, while that of sin(x) is 2π. As we know LCM of rational and irrational numbers do not exist.This is because LCM must be an integral multiple. If LCM existed we could have said that an integral multiple of fundamental period is given by the LCM of the two fundamental periods. Since LCM doesnot exist we can say that the given function is not periodic.

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