Is $f(x)=\sin(x^2)$ periodic?

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.

Let $f : \mathbb{R} \to \mathbb{R}$ be periodic with period $T$.

• The range of $f$ is precisely $f([0, T])$; in particular, if $f$ is continuous, the range of $f$ is bounded.
• If $f$ is differentiable, then $f'$ is 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.

• Could you be a bit more specific or give me some links for this theorem about connection between periodicity and derivatives? We didn't mention that in our classes and it seems very useful. Jan 20, 2013 at 11:40
• Well, if you differentiate both sides of $f(x) = f(x + T)$, you get $f'(x) = f'(x + T)$, so $f'$ must be periodic with period at most $T$. I'm not sure how to prove that the period must be $T$. Either way, all you need is that $f'$ is periodic. Jan 20, 2013 at 11:56
• @Lazar Ljubenović Note the following: $f'(x+T)=\frac{f(x+T+\Delta x)-f(x+T)}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}=f'(x)$. Jan 20, 2013 at 12:17

Your approach simply forces $k$ to be a square. Observing that the LHS will be irrational sometimes is a crucial idea but it cannot apply when you are using only two intervals $[0, T]$ and $[\sqrt{\pi}, \sqrt\pi + T]$ to form equations. Here is an approach that uses more.

Assume that $f$ is periodic with period $T$. Note that the solution set for $f(x) = 0$ is $\{\pm\sqrt{n\pi}\ |\ n = 0, 1, 2, \dotsc\}$. For any nonnegative integer $m$, $f(\sqrt{m\pi}) = 0$ thus $f(\sqrt{m\pi} + T) = 0$. Hence, there exists some integer $k_m$ such that $\sqrt{m\pi} + T = \sqrt{k_m\pi}$. Note that $T = \sqrt{k_0\pi}$, which gives us \begin{align} \sqrt m + \sqrt k_0 &= \sqrt k_m\\ m + k_0 + 2\sqrt{mk_0} &= k_m. \end{align}

And now we have a number-theoretic question. Let $k_0 = a^2b$, where $a^2$ is the greatest square that divides $k_0$.

1. If $b = 1$, choose $m = 2$ to have an irrational LHS and an integer RHS.
2. If $b \neq 1$, choose $m = 1$ to have an irrational LHS and an integer RHS.