Prove that $\sin(\sqrt{x})$ is not periodic, without using derivation

Question

So basically I'm supposed to prove that identity, without using derivations. I've seen this question posted here already, and one answer offered derivations, another said, what I basically tried to do before coming here, to assume f(x) = f(x+T) and solve for T, but I'm not sure how to proceed on $\sin(\sqrt{x})=\sin(\sqrt{x+T})$.

Answer 1

Hint:

You have to solve for $T$ the equation $$ \sin \sqrt{x}=\sin(\sqrt{x+T}) $$ and find a solution that is a constant (i.e. independent from $x$). But we have:

$$ \sqrt{x} +2k\pi=\sqrt{x+T} \iff x+4k^2\pi^2+4k\pi\sqrt{x}=x+T $$ so T cannot be a constant

Answer 2

If $T$ is the period,

$$\sin\sqrt T=\sin0.$$

But then $$\sin\sqrt{2T}=0$$ can't be true. (Solve for $T$ to check.)

Answer 3

just a Hint

Assume

$\forall x\geq 0$

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

$\implies$

$\sqrt{x+T}=\sqrt{x}+2k\pi$

and by squaring

$T=4k^2\pi^2+4k\pi\sqrt{x}$

thus if $x=0$, it gives $T=4\pi^2$

and if $x=\frac{1}{16}$, we get

$T=4\pi^2+\pi$ which is impossible.

Answer 4

Two reasons:

1. $f(x)=\sin\sqrt x$ is not defined for $x\lt0$, hence cannot satisfy an equation of the form $f(x)=f(x+T)$ for all $x$.

2. The zeroes of $f(x)=\sin\sqrt x$ are $x=n^2\pi^2$ for $n=0,1,2,3,\ldots$. The spacing between consecutive zeroes is $(2n+1)\pi^2$, which eventually exceeds any possible period $T\gt0$.

Remark: The first reason might be dismissed if your definition of periodicity as $f(x)=f(x+T)$ for some $T\gt0$ is restricted to $x$ (and $x+T$) in the domain of $f$, or if you use periodicity to extend the domain of $f$ to values of $x$ where it's not defined.