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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$ Jan 20, 2013 at 11:40
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    $\begingroup$ 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. $\endgroup$ Jan 20, 2013 at 11:56
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    $\begingroup$ @Lazar Ljubenović Note the following: $f'(x+T)=\frac{f(x+T+\Delta x)-f(x+T)}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}=f'(x)$. $\endgroup$
    – Michael Li
    Jan 20, 2013 at 12:17
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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.

Contradiction!

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