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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?

James Warthington
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    Hint: If it were periodic, its derivative would be as well. – lulu Sep 09 '18 at 00:58
  • So I have to prove that $\frac{cos(\sqrt{x})}{2\sqrt{x}}$ is non periodic as well? How could I do that? – James Warthington Sep 09 '18 at 01:00
  • It's really clear that that function is non-periodic. What happens as $x\to \infty$? – lulu Sep 09 '18 at 01:02
  • Worth noting: your attempt doesn't really make sense. Why should $\sin (\sqrt {x+P})=\sin (x)$? Periodicity would mean $\sin (\sqrt {x+P})=\sin (\sqrt x)$. – lulu Sep 09 '18 at 01:05
  • Typo, I am sorry, I am not adept at using Latex yet. – James Warthington Sep 09 '18 at 01:08
  • Could you elaborate more on your answer, lulu? – James Warthington Sep 09 '18 at 01:20
  • Please show some effort. As I said, it is extremely clear that the derivative is non-periodic. It has a limit as $x$ goes to infinity! No non-constant periodic function can take a limit at infinity. – lulu Sep 09 '18 at 01:22
  • Well, note that if $f(x)$ is periodic on the domain $D$ then if $x\in D\iff x+P\in D$. This means that if $x\in D$ then $x-P\in D$, but this is not true in our case. ($x=0$). – Ixion Sep 09 '18 at 01:25
  • Note that when $\sqrt x = 2\pi$ then $\cos {\sqrt x} =1$. This is the greatest value $\cos{\sqrt x}$ can be. And $\frac {\cos{\sqrt x}}{\sqrt{x}} = \frac 1{2\pi}$. For any larger $x$ then $\frac 1{\sqrt{x}} < \frac 1{2\pi}$ and $\cos{\sqrt{x}}{\sqrt x} \le \frac 1{\sqrt{s}} < \frac 1{2\pi}$. So at $x = 4\pi^2$ then $\cos{\sqrt{x}}{\sqrt x}$ a certain size and it will never be that large again. It can't be periodic if it will never be that value again. – fleablood Sep 09 '18 at 03:19

3 Answers3

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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.

fleablood
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  • Shouldn't it be $\cos {\sqrt x} = \pm \sqrt {1 - \sin^2 \sqrt x}$? Btw, a possible duplicate (with an answer using the derivative) was already pointed out. – Martin R Sep 09 '18 at 02:40
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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.

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

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

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 $x$

TonyK
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Mohammad Riazi-Kermani
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