Check proof that $\sin\sqrt{|x|}$ is not periodic

Question

I want to:

Prove that $\sin\sqrt{|x|}$ is not periodic

A similar question had already been asked here. But the accepted answer uses definition of the derivative. And i'm trying to do it in a "pre-calculus" manner.

Here is my try. By definition of periodic functions:

$$ f(x) = \sin\sqrt{|x|} = \sin\sqrt{|x - T|} $$

This may be rewritten as:

$$ f(x) = \cases{\sin\sqrt{x}, \; x \ge 0 \\ \sin\sqrt{-x}, \; x < 0 } $$

On the other hand:

$$ f(x) = \cases{\sin\sqrt{x-T}, \; x-T \ge 0 \iff x \ge T \\ \sin\sqrt{T-x}, \; x-T < 0 \iff x < -T } $$

So for the first case i have $$\sin\sqrt{x} = \sin\sqrt{x-T}$$, but $\forall{T} > 0, \exists x \ge 0 : x < T$ which contradicts the fact that $x \ge T$. The second case is handled similarly.

Is it valid?

Answer 1

You are working at $x\ge T$, so, how can you suppose that exist some $x<T$? It doesn't make sense.

A suggestion could be use the identity:

$$\sin p-\sin q=2\sin\left(\frac{p-q}{2}\right)\cos\left(\frac{p+q}{2}\right).$$

In your case you have $p=\sqrt{x}$ and $q=\sqrt{x-T}$, for $x\ge T.$ So,

$$\frac{\sqrt{x}-\sqrt{x-T}}{2}=k\pi,\quad k\in \Bbb Z$$ or $$\frac{\sqrt{x}+\sqrt{x-T}}{2}=\frac \pi 2+k\pi,\quad k\in \Bbb Z$$

Can you finish?

Answer 2

It is not correct. There is no reason to assume that when we are in the first case in the case of $\sin\left(\sqrt{|x|}\right)$, then we are also in the first case in the case of $\sin\left(\sqrt{|x-T|}\right)$ and vice-versa.

It's quite easy to deduce that the function is not periodic from the fact that its zeros form the set$$\left\{n^2\pi^2\,\middle|\,n\in\mathbb{Z}^+\right\}.$$