# $f_m(x)=\lim_{n→\infty}(\cos m!\pi x)^{2n}$

$f_m(x)=\lim_{n→\infty}(\cos m!\pi x)^{2n}$

Define $f(x)=\lim_{m→\infty}f_m(x)$
For irrational $x$, $f_m(x)=0$ for every $m$ hence $f(x)=0$.
For rational $x$, say $x=p/q$, where $p$ and $q$ are integers,
we see that $m!x$ is an integer if $q \le m$ so that $f(x)=1$.

I can't understand why for irrational $x$, $f_m(x)=0$.

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## 1 Answer

If $x$ is irrational, we have $m!x \not\in \frac 12\mathbb Z$ for every $m \in \mathbb N$, that is $|\cos m!x\pi| < 1$, which gives $(\cos m!x\pi)^{2n} \to 0$ for $n \to \infty$, and hence $f_m(x) = 0$.

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Thanks! By the way, we can say $m!x \not\in \mathbb Z$ too? I think if $m!x \not\in \mathbb Z$ then $m!x$ is not an interger so your explanation can be applied. –  anstein Nov 29 '12 at 6:54
Sure. That suffices, you are right. –  martini Nov 29 '12 at 7:07