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.

I want to find the period of $\sin(t) \cos(\pi t)$.

I started off by transforming that into $\frac{1}{2}\left [ \sin((\pi +1)t) - \sin((\pi - 1)t\right ]$, but then I get stuck. How do I find the least common multiple of $\pi + 1$ and $\pi - 1$? Is that what I need to do to find the period of the whole thing?

share|improve this question
There is no nonzero common multiple of $\pi-1$ and $\pi+1$. By the way, what makes you think this function is periodic? –  Did Jan 23 '12 at 6:54
Seeing as $\pi$ and $1$ are independent over the rationals, I doubt that this function is periodic. –  Alex Becker Jan 23 '12 at 6:58
So then the function $\sin(2t)cos(2\pi t)$ is not periodic? It has no period? –  stackedAE Jan 23 '12 at 7:11
@stackedAE: Finally typed out a proof that your function is not periodic. –  André Nicolas Jan 25 '12 at 6:12
show 1 more comment

3 Answers

up vote 9 down vote accepted

We show that $\sin(t)\cos(\pi t)$ is not periodic. Suppose to the contrary that it is periodic. Let $f(t)=|\sin(t)\cos(\pi t)|$. Then $f(t)$ is periodic. Let $p$ be a period of $f(t)$.

Let $m$ be the maximum value of $f(t)$ in the interval $[0,p]$. If $f(t)$ is periodic, then $m$ is the maximum value of $f(t)$ as $t$ ranges over all the reals. We will show that this is not the case, by showing that there is a $t$ such that $f(t)>m$.

Note first that $m\ne 1$. For if $f(t)$ ever takes on the value $1$, then $|\sin(t)|$ and $|\cos(\pi t)|$ must be simultaneously equal to $1$. So $t$ is an odd multiple of $\pi/2$, say $t=q \pi/2$. Also, $\pi t$ is a multiple of $\pi$, so $t$ is an integer. It follows that $\pi=2t/q$. This is impossible, since $\pi$ is irrational.

We now show that there is a $t$ such that $f(t)>m$. This is easy, but uses some machinery.

The sequence $(\sin(n))$ is dense in the interval $[-1,1]$. Thus there is an integer $t$ such that $\sin(t)>m$. Since $|\cos(\pi n)|=1$, it follows that $f(t) >m$.

Comment: A quick search shows that there are many proofs of the fact that the sequence $(\sin(n))$ is dense in $[-1,1]$. Indeed the problem has been posed and solved on MSE. The most intuitive argument shows that the points $(\cos(n), \sin(n))$ are dense on the unit circle. The result for $(\sin(n))$ then follows by projecting on the $y$-axis. In general, if $\theta$ is not a rational multiple of $\pi$, then the points $(\cos(n\theta), \sin(n\theta))$ are dense on the unit circle.

share|improve this answer
Nice. $ $ $ $ $ $ –  Did Jan 25 '12 at 6:14
@Didier Piau: But not as nice as dozens of yours. –  André Nicolas Jan 25 '12 at 6:19
$\langle$ Blushes $\rangle$. –  Did Jan 25 '12 at 6:39
add comment

Here's a more self-contained proof that $f(t) = \sin(t) \cos(\pi t)$ is not periodic, using only the fact that $\pi$ is irrational. If $f(t)$ had period $p$, then we'd also have $f(p) = f(0) = 0$. Now this implies either $\sin(p) = 0$, i.e. $p = n \pi$ for some nonzero integer $n$, or $\cos(\pi p) = 0$, i.e. $p = n+1/2$ for some integer $n$. But $$f(1/2 + n \pi) - f(1/2) = (-1)^{n+1} \sin(1/2) \sin(n \pi^2) \ne 0$$ since neither $1/2$ nor $n \pi^2$ is an integer multiple of $\pi$ (if it were $m \pi$, then $\pi = m/n$ would be rational). Similarly, $$f(\pi + (n+1/2)) - f(\pi)= (-1)^n \sin(\pi^2) \sin(n+1/2) \ne 0$$ since neither $\pi^2$ nor $n+1/2$ is an integer multiple of $\pi$.

share|improve this answer
Nice solution. But $f(\pi/2)\ne0$, hence I guess one should use $\pi$ instead of $\pi/2$ to disqualify $p=n+1/2$. –  Did Jan 25 '12 at 8:13
I have taken the liberty of correcting this answer to meet Didier's objection. (I had to do this so I could upvote it...) –  TonyK Jan 25 '12 at 18:45
@TonyK: Thanks. Upvoting too, then. –  Did Jan 25 '12 at 18:49
add comment

Suppose there is a common multiple $p$ of $\pi+1$ and $\pi-1$. Then $$ \begin{align} p & = n(\pi+1) \\ p & = m(\pi-1) \end{align} $$

(Later note: The context of the problem should make it clear that this means $n$ and $m$ are positive integers.)

It follows via a bit of algebra that $$ \pi=\frac{n+m}{n-m}. $$ Therefore $\pi$ is rational. But in this article it is proved that $\pi$ is irrational. At least two of the proofs given there can be understood by someone who knows nothing beyond first-year calculus.

The function is therefore not periodic, but it is almost periodic.

share|improve this answer
you probably mean a «common non-zero integer multiple» or something, but both $0$ and $(\pi+1)(\pi-1)$ are surely common multiples :) –  Mariano Suárez-Alvarez Jan 25 '12 at 18:49
I meant "multiple" in the sense that makes sense in the context of the problem. –  Michael Hardy Jan 25 '12 at 20:58
@mariano: I was pretty heavily abused for making this joke earlier math.stackexchange.com/a/74730/18005 ('integer' has since been edited into the title). –  opt Jan 25 '12 at 21:40
@opt: in that case, you used a whole answer to make the joke, and did not take the chance to actually —well...— answer the question! –  Mariano Suárez-Alvarez Jan 25 '12 at 21:43
add comment

Your Answer


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.