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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?

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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
    
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3 Answers 3

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.

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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
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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 $\pi$.

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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
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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.

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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
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