Is the function $f:\Bbb R \rightarrow \Bbb R$ defined as $f(x)=\sin(x^2)$, for all $x\in\Bbb R$, periodic?
Here's my attempt to solve this:
Let's assume that it is periodic. For a function to be periodic, it must satisfy $f(x)=f(T+x)$ for all $x\in\Bbb R$, so it must satisfy the relation for $x=0$ as well. So we get that $T^2=k\pi \iff T=\sqrt{k\pi}$, $k\in\Bbb N$ (since $T$ must be positive, we remove the $-\sqrt{k\pi}$ solution).
So what now? I tried taking $x=\sqrt\pi$ and using the $T$ I found, and I get this: $$ \sin\pi=\sin(T+\sqrt\pi)\iff-1=\sin(\pi(\sqrt k+1)^2)\iff k+2\sqrt k+1=3/2+l $$
Is this enough for contradiction? The left side of equation is sometimes irrational and gets rational only when $k$ is perfect square, which doesn't happen periodic, while the right hand side is always rational. Or I'm still missing some steps?
Thanks.
Best Answer
Let $f : \mathbb{R} \to \mathbb{R}$ be periodic with period $T$.
Note that $f(x) = \sin(x^2)$ is differentiable and $f'(x) = 2x\cos(x^2)$ which is unbounded. Therefore, $f'$ cannot be periodic by the first point, and hence $f$ cannot be periodic by the second point.