Prove that the function $x\mapsto\sin(x^{2})$ is not periodic

Question

Prove that the function $f:\Bbb R \rightarrow \Bbb R ;x\mapsto\sin(x^{2})$ is not periodic.

Let's assume the opposite, i.e that $f$ is periodic. Then for all $\tau \in\Bbb R$, $\sin(\tau +x)^{2}=\sin(x^{2})$.

How do I continue from here?

Answer 1

You have $\sin (x+\tau)^2 = \sin x^2$, then $$\cos\left(\frac{(x+\tau)^2-x^2}{2}\right)\sin\left(\frac{(x+\tau)^2-x^2}{2}\right) = 0$$

Then, you can find the condition of $\tau$.

Answer 2

Hint:

The function $f(x)$ has period $T$ if $f(x)=f(x+T)$ for some constant $T$, independent from $x$. In your case this means: $$ \sin x^2=\sin(x+T)^2 $$ but we have: $$ x^2+2k\pi=(x+T)^2 \iff T^2+2xT-2k\pi=0 $$ and the solutions of this equation in $T$ are dependent from $x$.

Answer 3

As per your approach, assume that $\sin(x^2)$ is periodic .

Then since every continuous periodic function is uniformly continuous so will be $\sin (x^2)$.

But $\sin (x^2)$ is not uniformly continuous.

Take $x_n=\sqrt{2n\pi+\frac{\pi}{2}}$ and $y_n=\sqrt {2n\pi}$

$|x_n-y_n|$ can be made sufficiently small but $|f(x_n)-f(y_n)|=1$ can't be made so.

Answer 4

If $\sin(x^2)$ were periodic, then its derivative $2x\cos (x^2)$ would also be periodic. But a continuous periodic function is bounded, and $2x\cos (x^2)$ is not. The conclusion follows.

Answer 5

There should exist $T>0$ such that, if $z$ is a zero of $\sin(x^2)$, then also $z+T$ is a zero of $\sin(x^2)$.

Since $0$ is a zero, also $T$ should be, which means $$ T^2=h\pi $$ for some integer $h$.

Since $\sqrt\pi$ is a zero, also $\sqrt{\pi}+T$ should be, which means $$ (\sqrt{\pi}+T)^2=(k+1)\pi $$ for some integer $k$.

Also $\sqrt{2\pi}$ is a zero, so we get $$ (\sqrt{2\pi}+T)^2=(l+2)\pi $$ for some integer $l$. Therefore $$ \begin{cases} T^2=h\pi \\[4px] T^2=k\pi-2T\sqrt{\pi} \\[4px] T^2=l\pi-2T\sqrt{2\pi} \end{cases} $$ From the first two equations we get $2T=(k-h)\sqrt{\pi}$ and, substituting in the third equation, $$ h\pi=l\pi-(k-h)\sqrt{2}\pi $$ so $$ \sqrt{2}=\frac{l-h}{k-h} $$ would be rational.

Answer 6

The function $f(x)=\sin(x^2)$ has a finite number of zeroes in any finite interval. If it were periodic with period $T$, then there would be a uniform upper bound on the number of zeroes in any interval of the form $(x,x+T)$. But the spacing between consecutive zeroes of $\sin(x^2)$ tends to $0$ as $x\to\infty$. Hence there is no upper bound on the number of zeroes in intervals of the form $(x,x+T)$, hence $f(x)=\sin(x^2)$ is not periodic.