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How can we see that Dirichlet's function

$$D(x):=\lim_{m\to\infty} \lim_{n\to\infty} \cos^{2n}(m!\pi x)= \begin{cases} 1 & x\in\mathbb Q\\ 0 & x\notin\mathbb Q\\ \end{cases}$$

Martin Sleziak
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superAnnoyingUser
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  • First compute the inner limit. Which values can it have, and where does it attain each. Then take the outer limit. – Daniel Fischer Jun 26 '14 at 14:57
  • See http://math.stackexchange.com/questions/275974/how-to-prove-that-lim-undersetk-rightarrow-infty-lim-cosn-pi-x2k and http://math.stackexchange.com/questions/264889/how-is-this-called-rationals-and-irrationals and other posts shown there among linked questions. – Martin Sleziak Oct 16 '14 at 12:04

1 Answers1

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For $x$ rational there is such (large enough) $m$, that $m!$ contains as divisors all divisors of the $x$ denominator, so $m!x$ is integer. (EDIT Actually for $x=\tfrac pq$ having $m\ge q$ is enough, since $q$ becomes explicitly one of the terms in $m!$ product.) This guarantees cosine being $+1$ or $-1$. When raised to any positive even power ($2n$) it results in $1$.

OTOH for $x$ irrational, there is no such natural $m!$ which makes $m!x$ integer, so cosine will be less than $1$ (in absolute value). Thus raised to a power big enough it approaches zero, so the limit is $0$.

CiaPan
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  • That's false if cos(n!x$\pi$) is dense in ]-1,1[, which you will have to prove or disprove. So your answer is uncomplete, if not false. What's more, the only existing limit is the one where m goes to infinity before n, and switching limits doesn't work here, so it requires to be careful – mvggz Oct 16 '14 at 12:50
  • There is $m!$, not $n!$ in $\cos(m!\pi x)$. 2. For given $x$, for any natural $m$ cosine is either $\pm 1$ or not, the limit as $n\to\infty$ is $1$ or $0$, respectively. As $m$ grows, the cosine will become $\pm 1$ at some point for $x\in\Bbb Q$, for irrationals it won't. So the sequence (indexed by $m$) of inner limits is $(0,0\ldots,0,1,1,1\ldots)$ or zeros forever. Finally the outer limit is $1$ for rational $x$ or $0$ for irrational. Density has nothing to do here.
    • – CiaPan Oct 18 '14 at 06:09