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I'm wondering, is the function $f=(\sin{x})(\sin{\pi x})$ is periodic?

My first inclination would be two assume that if the periods of the individual sine expressions, $p_1 \text{and}\space p_2$ have the quality that $p_1 \times a = p_2 \times b$ where $a \space\text{and}\space b$ are integers, then the entire function will eventually repeat after a period of $p_1 \times a$.

If that is true, than I think $f$ might not be periodic due to the fact that two Pi is irrational.

Does anyone know the answer and/or weather my thinking is correct? I've never seen a function like this before, so I'm really curious.

  • What is the definition of periodic? – Ali Caglayan Mar 26 '16 at 23:48
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    Just for your own pleasure, you should go to fooplot.com, and plot the function $7sin(x)sin(\pi x)$, forget about periodicity , it looks gorgeous. – Sarvesh Ravichandran Iyer Mar 26 '16 at 23:52
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    Yes @астон вілла олоф мэллбэрг nice barbed wire – Jean Marie Mar 27 '16 at 00:04
  • It is periodic if and only if $\pi$ is rational. For example, if you think $\pi=22/7$, then you think this function is periodic. See http://math.stackexchange.com/questions/93222/is-22-7-equal-to-the-pi-constant – GEdgar Mar 27 '16 at 01:01
  • no it is not periodic, but it is a perfect example of an almost periodic function : $f(x) = \sum_{n=1}^N c_n e^{i a_n x}$ which is periodic iff there is a $lcm(a_1,\ldots, a_N)$, otherwise there still are approximate $lcm$, i.e. for every $\epsilon$ there is an almost period $T_\epsilon$ such that $|f(x+T_\epsilon)-f(x)| < \epsilon$ – reuns Mar 27 '16 at 01:06
  • If you plot sin(ax)sin(bx), if a is smaller than b, you get sin(ax) as an envelope modulated by sin(bx). – marty cohen Mar 27 '16 at 04:56
  • @ астон вілла олоф мэллбэрг for (x < 0 < 100)! – Narasimham Mar 27 '16 at 05:12
  • Your reasoning is correct, as the two periods are not rationally related, the function can't be periodic. –  Jul 04 '16 at 11:58
  • Note that the relevant keyword here is quasiperiodic. – Wrzlprmft Nov 26 '17 at 16:48

3 Answers3

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Using the product-to-sum formula, $$ \sin x \sin\pi x = \frac{1}{2} \left( \cos ((1-\pi)x) - \cos((1 + \pi)x) \right) $$ but $\frac{1-\pi}{1+\pi} \not\in \mathbb{Q}$ and $\sin x \sin \pi$ is continuous, so this function is not periodic.

Henricus V.
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  • Can we take $LCM(\frac{2π}{1-π},\frac{2π}{1+π})$?what is the value here? Please explain. –  Aug 16 '17 at 11:18
  • @N.Maneesh such a least common multiple doesn't exist. – Henricus V. Aug 16 '17 at 11:20
  • but $\frac{1-\pi}{1+\pi} \not\in \mathbb{Q}$ and $\sin x \sin \pi$ is continuous, so this function is not periodic. How this statement implies function is not periodic? Please explain. –  Aug 16 '17 at 11:23
  • @N.Maneesh Consider the values of $x$s for which the function is maximised. (Maybe the continuity criterion is unnecessary) – Henricus V. Aug 16 '17 at 11:28
  • I understood upto product formula.Can you please explain after that? –  Aug 16 '17 at 11:44
  • @N.Maneesh After that the proof boils down to any period of the function implies $(1-\pi)/(1+\pi)$ being rational. – Henricus V. Aug 16 '17 at 11:46
  • $\frac{1-\pi}{1+\pi} \not\in \mathbb{Q}$ , how do you draw the conclusion? , if there exists a period. Can you give hints? I assumed T be the period. Then tried to solve $f(t+π)=f(t)$. Then I tried to simplify the expression . I could not draw any conclusions like this. Please help me. –  Aug 16 '17 at 15:02
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If $f(x)=\sin(x)\sin(\pi x)$ were periodic, then in particular its set of zeroes would be periodic. But the zero set is $\mathbb Z\cup\pi\mathbb Z$, and $a\mathbb Z\cup b\mathbb Z$ (for nonzero $a$, $b$) is not periodic unless $b/a$ is rational.

(Suppose $a\mathbb Z\cup b\mathbb Z$ is periodic with period $P>0$. Then $0+P\in a\mathbb Z\cup b\mathbb Z$; assume without loss of generality that $P=ka$ for some $k\in Z$. Now $b+P = ka+b$ must be in $a\mathbb Z\cup b\mathbb Z$ too, but if $ka+b=bn$ then $\frac ba=\frac{k}{n-1}$, and if $ka+b=am$, then $\frac ba = m-k$; in both cases $\frac ba$ is rational.)

hmakholm left over Monica
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Assume

$$\sin(x+T)\sin(\pi(x+T))=\sin(x)\sin(\pi x)$$ for all $x$.

Then with $x=0$,

$$\sin(T)\sin(\pi T)=0$$ so that $T=k\pi$ or $T=k$.

But you can find examples of

$$\sin(x+k\pi)\sin(\pi x+k\pi^2)\ne\sin(x)\sin(\pi x),$$ i.e. $$\sin(\pi x+k\pi^2)\ne\sin(\pi x),$$ and

$$\sin(x+k)\sin(\pi x+k\pi)\ne\sin(x)\sin(\pi x),$$i.e. $$\sin(x+k)\ne\sin(x).$$

For solutions to be possible, you would indeed need

$$k\pi^2=2k'\pi$$ or $$k=2k'\pi,$$ requiring $\pi$ to be rational.