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This question already has an answer here:

The Dirichlet function is defined as the indicator function of rational numbers. I have also seen this function described by: $$f(x)=\lim_{k\to\infty}\lim_{j\to\infty}\cos^{2j}k!\pi x$$ How does this limit act as the indicator, and how does it yield an answer if cosine is limited to Infinity?

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marked as duplicate by Paul Frost, Brian Borchers, Cesareo, Xander Henderson, Don Thousand Oct 28 '18 at 23:42

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Suppose $x$ is rational and $x=\frac pq$ for $\gcd(p,q)=1$. If $k\ge q$, $k!x$ will be an integer, so $\cos k!\pi x=\pm1$ and $\cos^{2j}k!\pi x=1$. Thus the function is 1 for sufficiently large $k,j$, so it evaluates to 1.

Suppose $x$ is irrational, then $k!x$ will never be an integer, so $|\cos k!\pi x|<1$ for all $k$. As $j\to\infty$, this variable being in the exponent yields $\cos^{2j}k!\pi x\to0$, so the function evaluates to 0.

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  • $\begingroup$ Why is it k! instead of k? $\endgroup$ – mtung Oct 28 '18 at 12:18
  • $\begingroup$ @mtung If it were $k$, then $kx$ would not be an integer for sufficiently large $q$, and we would not be able to derive the result for rational $x$. $\endgroup$ – Parcly Taxel Oct 28 '18 at 12:25
  • $\begingroup$ +1. Actually, for $k$ large enough, $k!x$ is an even integer, so we could say that $\cos(k!\pi x)=1$. But this, of course, does not add much to the proof. $\endgroup$ – Taladris Oct 28 '18 at 13:21

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