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How to show that:

$f(x)= \lim_{k \to \infty} \Bigl(\lim_{j \to \infty} \cos^{2j}(k!\cdot \pi \cdot x)\Bigl)$

is the Dirichlet's function.

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    @prometheus: Yes there is thanks for the correction.2011-01-04

2 Answers 2

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Suppose $x = \frac {p}{q} \in \mathbb{Q}$. If $k>q$ then $k! \cdot x \in \mathbb {Z}$ so $ \cos(k!\cdot \pi \cdot x) = \pm 1$ $ k>q,\; \; \lim_{j \to \infty} \cos(k!\cdot \pi \cdot x)^{2j} = \lim_{j \to \infty} (\pm 1)^{2j} = 1$

so taking the limit of $k \rightarrow \infty$ you get also 1.

If $x \notin \mathbb{Q}$ then for every k we have $k! \cdot x \notin \mathbb {Z}$ so $ |\cos(k!\cdot \pi \cdot x)| < 1$ therefore

$ \lim_{k \to \infty} \lim_{j \to \infty} \cos(k!\cdot \pi \cdot x)^{2j} = \lim_{k \to \infty} 0 = 0 $

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If x is rational, once k gets large enough the cosine is 1 independent of j. If x is irrational, the cosine is less than one so the j limit will be zero. This is true for all k.