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

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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?

## marked as duplicate by Math Lover, Martin R, lulu, Henning Makholm, xbhSep 9 '18 at 3:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Hint: If it were periodic, its derivative would be as well. – lulu Sep 9 '18 at 0:58
• 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 9 '18 at 1:00
• It's really clear that that function is non-periodic. What happens as $x\to \infty$? – lulu Sep 9 '18 at 1:02
• 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 9 '18 at 1:05
• Typo, I am sorry, I am not adept at using Latex yet. – James Warthington Sep 9 '18 at 1:08

## 3 Answers

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

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

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 $x$

• I understand your proof except for $\pi-\sqrt{x}$ part. – James Warthington Sep 9 '18 at 1:33
• two angles have the same sines if either the difference is $2k\pi$ or one is $\pi$ minus the other. – Mohammad Riazi-Kermani Sep 9 '18 at 1:39
• $\sin(\sqrt{x+P})=\sin(\sqrt{x})$ implies $\sqrt{x+P}=\sqrt{x}+2k\pi$ or $\sqrt {x+P}= (2k+1)\pi - \sqrt{x}$, for some integer $k$ which may depend on $x$. – Martin R Sep 9 '18 at 1:45

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.

• 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 9 '18 at 2:40

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.