Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 6 down vote accepted

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

  • The range of $f$ is precisely $f([0, T])$, in particular it 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 second point, and hence $f$ cannot be periodic by the first point.

share|improve this answer
    
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. –  Lazar Ljubenović Jan 20 '13 at 11:40
1  
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. –  Michael Albanese Jan 20 '13 at 11:56
1  
@Lazar Ljubenović Note the following: $f'(x+T)=\frac{f(x+T+\Delta x)-f(x+T)}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}=f'(x)$. –  Michael Li Jan 20 '13 at 12:17
add comment

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!

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.