How to prove that $\sin(\sqrt{x})$ is not periodic?

Question

How to prove that $\sin(\sqrt{x})$ is not periodic? THe definition of a periodic function is $f(x+P)=f(x)$.

So I assume that $\sin(\sqrt{x+P})=\sin(\sqrt{x})$. This is equivalent to $\sin(\sqrt{x+P})-\sin(\sqrt{x})=0$. This implies $2cos(\frac{\sqrt{x+P}+\sqrt{x}}{2})\sin(\frac{\sqrt{x+P}-\sqrt{x}}{2})$. What should I do next?

Answer 1

$$\sin(\sqrt{x+P})=\sin(\sqrt{x})$$

$$\implies \sqrt{x+P}=\sqrt{x}+2k\pi \text { or }\sqrt {x+P}=2k\pi+ \pi - \sqrt{x} $$

Upon squaring we get $$ x+P = x+4 k^2\pi ^2 +4k\pi \sqrt x $$

or $$x+P =((2k+1) \pi) ^2 + x-2(2k+1)\pi \sqrt x$$

Note that neither of the above holds for a constant $P$ and every $x$

Answer 2

Note that the domain $D$ of a $P$-periodic function $f$ must be "invariant by translations of $P$", i.e.: $D+P=D$. In this case $\forall P>0, \ D+P$ is a proprer subset of $D$ hence $f(x)$ can not be a periodic function.

Answer 3

If $\sin {\sqrt x}$ is period with period $P$ then so is $\cos {\sqrt x}$ because $\cos {\sqrt x} = \sqrt {1 - \sin^2 \sqrt x}=\sqrt {1 - \sin^2 \sqrt {x+P}}=\cos (\sqrt{x+P})$. Likewis so is the derivative of $\sin {\sqrt x}$ because $ \lim \frac {\sin (\sqrt {x + h}) -\sin(\sqrt{x})}h= \lim \frac {\sin (\sqrt {x +P + h}) -\sin(\sqrt{x+P})}h$.

But the derivative of $\sin (\sqrt{x})$ is $\frac{cos(\sqrt{x})}{2\sqrt{x}}$ and if $\cos (\sqrt{x + P}) = \cos (\sqrt{x})$ then $\frac{cos(\sqrt{x+P})}{2\sqrt{x+P}}=\frac{cos(\sqrt{x})}{2\sqrt{x+P}} \ne \frac{cos(\sqrt{x})}{2\sqrt{x}}$

This is a contradiction.