18
$\begingroup$

I stumbled upon this "relation" (is the name correct?):

$$ \lim_{m \to \infty} \lim_{n \to \infty} \cos^{2n}(m! \pi x) = \begin{cases} 1,&x\text{ is rational}\\ 0,&x\text{ is irrational}\end{cases} $$

How is it called and why is it so? I'm really not asking for a proof since I fear it would be too complicated for me to understand, but rather for an "intuition".

$\endgroup$
18
$\begingroup$

The intuition is already almost the proof.

  • The cosine takes values between $-1$ and $1$, inclusive
  • Taking an even power ensures that the values are between $0$ and $1$.
  • Taking higher and higher powers will not affect those $x$ where the cosine squared is $1$, but all other values will converge to $0$

Thus $f(x):=\lim_{n\to\infty}\cos^{2n}(x)$ is a function that has value $0$ for all $x$, except that $f(x)=1$ if $\cos(x)=\pm1$, i.e. if $x$ is a multiple of $\pi$. After this, we see that multiplying the argument with $\pi$ is done to have the value $1$ at integers instead of multiples of $\pi$. Then the multiplication with $m!$ serves the purpose to obtain $1$ for any integer multiple of $\frac1{m!}$. As $m\to \infty$, this exhausts the rationals (and only them).

$\endgroup$
  • 2
    $\begingroup$ A totally random question, are the limits interchangeable here? Or does the order matter? $\endgroup$ – Fixed Point Dec 25 '12 at 12:21
  • 2
    $\begingroup$ @PrinceAli $\lim_{m\to\infty}\cos^{2n}(m!\pi x)$ is $1$ for rational $x$, but does not exist for most irrational $x$. $\endgroup$ – Hagen von Eitzen Dec 25 '12 at 21:21
14
$\begingroup$

First notice that $-1\le\cos m!\pi x\le 1$ for all $m$ and $x$, so $0\le\cos^2m!\pi x\le 1$. Suppose that $0\le y\le 1$; then

$$\lim_{n\to\infty}y^n=\begin{cases} 0,&\text{if }y\ne 1\\ 1,&\text{if }y=1\;. \end{cases}\tag{1}$$

Now $\cos^2 m!\pi x=1$ if and only if $m!x$ is an integer. Say $m!x=a$ for some integer $a$; then $x=\frac{a}{m!}$, a rational number. Thus, $m!x$ is never an integer when $x$ is irrational, and therefore for all $m$ we have $\cos^2m!\pi x\ne 1$. $(1)$ then implies that $$\lim_{n\to\infty}\cos^{2n}m!\pi x=0$$ for all $m$, and of course in that case $$\lim_{m\to\infty}\lim_{n\to\infty}\cos^{2n}m!\pi x=\lim_{m\to\infty}0=0\;.$$

This explains why the limit is $0$ when $x$ is irrational.

If $x$ is rational, on the other hand, we can write $x=\frac{a}b$ for some integers $a,b$ with $b>0$. Then for all $m\ge b$ we can be sure that $m!x$ is an integer and hence that $\cos^2m!\pi x=1$. But then the sequence

$$\left\langle\cos^{2n}m!\pi x:n\in\Bbb N\right\rangle$$

is constantly $1$ for $m\ge b!$, so for $m\ge b!$ we have

$$\lim_{n\to\infty}\cos^{2n}m!\pi x=1\;.$$

But then the sequence $$\left\langle\lim_{n\to\infty}\cos^{2n}m!\pi x:m\in\Bbb N\right\rangle$$

is constantly $1$ for $m\ge b!$, and

$$\lim_{m\to\infty}\lim_{n\to\infty}\cos^{2n}m!\pi x=1\;.$$

$\endgroup$
  • $\begingroup$ Why limit for $m$ is before the other one? Shouldn't be the opposite, since the cosine argument is evaluated before the exponentiation? $\endgroup$ – rubik Dec 26 '12 at 17:46
  • $\begingroup$ @rubik: The order in which the $\cos$ and exponent are evaluated is irrelevant. What matters is that we hold $m$ constant and evaluate the limit as $n\to\infty$; call this $\ell_m$. Then we take the limit of the $\ell_m$ as $m\to\infty$. $\endgroup$ – Brian M. Scott Dec 26 '12 at 17:50
  • $\begingroup$ ok, got it! Thanks! $\endgroup$ – rubik Dec 27 '12 at 19:57
  • $\begingroup$ @rubik: You’re welcome. $\endgroup$ – Brian M. Scott Dec 27 '12 at 20:27
8
$\begingroup$

I'm not sure it has a name, but here's a fairly simple proof:

If $m!x$ is an integer, then $\cos^{2n}(m!\pi x) = 1$.

If $x$ is rational $\frac{p}{q}$, then, eventually, for large enough $m$, $m!$ will be divisible by $q$ (and stay so for all larger m), so that $m!x$ will be an integer, and we have what we want.

If $x$ is irrational, $m!x$ will never be an integer, and $|\cos(m!\pi x)| \lt 1$, so that $\lim_{n \to \infty} \cos^{2n}(m!\pi x) = 0 $ for all $m>0$.

(This may need some cleaning up to make rigorous...)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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