# Prove that $\sin(\sqrt x)$ not periodic

$\sin\sqrt x$ is not a periodic function. How can one prove this?

• Look at its zeroes. – Pedro Tamaroff Aug 8 '14 at 15:30

An other way to prove it is to remark that if this function is $T-$ periodic, the derivate would be $T-$ periodic too. Indeed,

$$f'(x+T)=\lim_{h\to 0}\frac{f(x+T+h)-f(x+T)}{h}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f'(x).$$

Therefore if $f(x)=\sin(\sqrt{x})$, then $f'(x)=\frac{\cos(\sqrt x)}{2\sqrt x}$ would be periodic, and this is impossible because $$\lim_{x\to \infty }f'(x)=0.$$

• @idm Sir can you please explain how after this part $\lim_{x\to\infty}f^\prime(x)=0$ you are concluding that the function is aperiodic,Please reply – Rohit Oct 30 '17 at 12:59
• Because if $f$ is $T-$periodic and not constant, there is $x,y$ s.t. $f(x)\neq f(y)$. In particular, if you set $x_n=x+T$ and $y_n=y+T$, you have that $$\lim_{n\to \infty }f(x_n)=f(x)\neq f(y)=\lim_{n\to \infty }f(y_n),$$ and thus $\lim_{x\to \infty }f(x)$ can't exist (in particular, it can't be $0$). @Rohit – idm Oct 30 '17 at 15:24
• @idm sir that means for any aperiodic signal this will follow? means as $t \to \infty$ it will give particular value(constant) then we can conclude that its a aperiodic signal right sir? – Rohit Oct 30 '17 at 15:29
• @idm Thanks for the reply – Rohit Oct 30 '17 at 15:31

Assume that it is. Then $\sin{\sqrt{x}}=\sin{\sqrt{x+p}}=$ for all $x$ and some $p$. Solve for $p$.

Take $f(x)=\sin \sqrt{x}$. If $f(x+T)=f(x)$ for all $x$, and $T \neq 0$, then $f(T)=f(0)=0$. So $T$ is a root of $f$. So $T=n^2 \pi^2$ for some integer $n$.

If $f(x+T)=f(x)$ for all $x$. Then $f(2T)=f(T)=0$. Recall we assumed $T \neq 0$. That implies that there exists two distinct roots of $f$, $x_1$ and $x_2$ such that $\frac{x_2}{x_1}=2$. That means,

$$\frac{m^2 \pi^2}{n^2 \pi^2}=\frac{m^2}{n^2}=\left(\frac{m}{n} \right)^2=2$$

For some integers $n$ and $m$. This is impossible since $\sqrt{2}$ is irrational.