About Dirichlet function

Question

caozeyuan · Answer

How are the two def's of Dirichlet function equivalent? one is D(x)=1 if x belongs to Q and 0 if not, the other is the double limit definition

caozeyuan · Answer

$$\lim_{k \rightarrow \infty}(\lim_{j \rightarrow \infty}(\cos ^{2j}(k!\pi x)))$$

caozeyuan · Answer

I cant understand why this function is 1 if x is any rational and 0 if not

amorfide · Answer

I personally can not give help due to now having experience with this but https://en.wikipedia.org/wiki/Nowhere_continuous_function read this incase noone else can help you either

caozeyuan · Answer

Actually I posted this question after reading this page, but thx anyway

amorfide · Answer

Sorry

amorfide · Answer

@mathmale

holsteremission · Answer

You can get some intuition for why the definition work by considering its parts. There are only two major ones.

First, recall that you have $|\cos x|\le1$ for all $x$, so for large $j$ you have $\cos^jx\to0$ whenever $\cos x\neq1$. Graphically, a large $j$ would correspond to something like this:
|dw:1479759851413:dw|
The even power makes it so the function is non-negative, e.g.
|dw:1479760084068:dw|
Note the plots above aren't actually discontinuous. You can think of them more as periodic spikes from values near $0$ jumping instantaneously to $1$.

The second important feature is that coefficient of $x$. Obviously, $\cos^{2j} \pi x=1$ whenever $x\in\mathbb Z$. As you increase $k$, the period of $\cos^{2j} k\pi x$ decreases, so you get a higher frequency of oscillation.

Now for the factorial. We want $D(x)=1$ whenever $x$ is rational, so when $x=\dfrac{p}{q}$ for integers $p,q$.

When $k$ is large, it's guaranteed that $k!$ will contain a factor of $q$ so that $k!\pi x=mp\pi$ for some integer $m$, and thus $\cos^{2j}k!\pi x$ will be equal to $1$.

Meanwhile, when $x$ is irrational, $\cos k!\pi x$ will always fall somewhere between $-1$ and $1$, which means as $j\to\infty$ the cosine function approaches $0$.