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Show that: The continuous functions $f_{n,k}(x):=(\cos(k!\pi x))^{2n},0\leq x \leq 1$ satisfy the relation

$\lim_{k\to \infty}(\lim_{n\to \infty}f_{n,k}(x))=\begin{cases} 1, & \textit{if x is rational}\\0,&\textit{if x is irrational}\end{cases}$

on the interval $[0,1]$.

Usually I start doing this type of problem by plotting $f$ for some $n$ and $k$ but I can't seem to see a correlation between the plots and the limit of the function.

Anyways, here were my general ideas:

I was thinking that it is $1$ only if the argument of the cosine is an even multiple of $\pi$, meaning that, let's say $x=\frac{a}{b}$ for the rational case, then I was thinking that if $k$ is large enough a number in $k!$ cancels with $b$ leaving only $j!\pi a$ in the argument (where $j!$ is $k!$ without the $b$).

Similarly I was thinking that when $x$ is irrational then it wouldn't cancel with a number in $k!$, right? Meaning that the argument of the cosine is uneven, meaning not $1$.

And since all of it is to the power of $2n$, that would mean that for $\cos(k!\pi x)\neq 1$ the limit of $n\to \infty$ would be $0$, right?

Now, even if my argumentation were correct I don't know how to write it down mathematically correct.

Could anyone help me out here?

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Fabian Henry
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  • You have all the right ideas. Just write them down clearly and you should be fine. Also check http://math.stackexchange.com/questions/264889/double-limit-of-cos2nm-pi-x-at-rationals-and-irrationals?rq=1 –  Jul 05 '15 at 07:02

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I don't really understand your problem...

If $x$ is rational and $k\geq 2$ big enough, $f_{n,k}(x)=1$ because $\cos (k!\pi x)=\cos(2\pi k')=1$. If $x$ is irrational, $\cos(k!\pi x)<1$ and thus $(\cos(k!\pi x))^{2n}\to 0$ if $n\to \infty $, therefore you got your result.

idm
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