Dirichlet’s approximation theorem

Last summer I was attempting to show that sin(\mathbb{N}) was dense in [-1,1] . After more reflection I realised that this was an interesting corollary of Dirichlet’s approximation theorem which is stated as follows:

Let \alpha \in \mathbb{R_+} and n \in \mathbb{N} . Dirichlet’s approximation theorem says that \exists k, b \in \mathbb{N}, |k\alpha -b | < \frac{1}{n k} .


Each of the n+1 numbers a_i = i \alpha -\left\lfloor i \alpha \right\rfloor, i \in [0,n] lies in 0 \leq a_i < 1 so by the pigeon-hole principle there’s at least one semi-closed interval of the form [\frac{r}{n},\frac{r+1}{n}), 0 \leq r < n which contains two of them.

It follows that there exists m, j \in \mathbb{N} such that:

\begin{aligned} |(m-j)\alpha -(\left\lfloor m \alpha \right\rfloor) -\left\lfloor j \alpha \right\rfloor) | < \frac{1}{n} \end{aligned} 

Now if we set k= m-j  and

\begin{aligned} b = \left\lfloor m \alpha \right\rfloor -\left\lfloor j \alpha \right\rfloor, i \in [0,n] \end{aligned}

we have the result we desire.

The above proof can be found anywhere on the internet but using this I managed to deduce that integer angles are dense in the unit circle. I also tried to show that sin(\mathbb{P}) is dense in [-1,1] where \mathbb{P} is the set of prime numbers but this appears to be a more difficult result.

An interesting further question I wondered about is the following: what’s the necessary and sufficient condition on S \subset \mathbb{N} such that sin(S) is dense in [-1,1] ?


The Rocquiasami function

A month ago it occurred to me to try and figure out how well the rationals approximated the irrationals. I am aware of the irrationality measure but I wondered whether there might be a method for determining how well it was approximated by an infinite number of rationals. The function that resulted from my analysis was the following:

\forall \alpha \in \mathbb{R}_+ \setminus \mathbb{Q} , let \{\alpha_q\}_{q=1}^\infty \subset \mathbb{Q} denote the optimal rational approximants of \alpha:

\forall q \in \mathbb{N}, |\alpha_q -\alpha| = \min_{n \in \mathbb{N}}|\frac{n}{q}-\alpha| \quad (1)

Using the language of functions rather than sequences we may define:

\begin{cases} f: \mathbb{R}_+ \setminus \mathbb{Q} \rightarrow\overline{\mathbb{R}} \\ f(\alpha) = \sum_{q=1}^{\infty} |\alpha_q-\alpha| \end{cases} \quad (2)

It’s not known to me whether f is convergent for any irrational x . But, I believe that the function is not bounded anywhere:

For any open interval I \subset \mathbb{R_+},\forall M \in \mathbb{R_+} \exists \alpha \in I, f(\alpha) > M

While I haven’t had any success with this particular question so far I have managed to show that f(\alpha)>\alpha for uncountably many irrationals:

Let’s define the uncountable sets:

M = [0,1] \setminus {{\mathbb{Q} \cap [0,1]}}  \quad (3)

S = \{\frac{1}{2}+\frac{m}{2^n} : m \in M\}_{n=5}^\infty \quad (4)

It isn’t too difficult to show that:

\forall \alpha \in S, f(\alpha) \geq |1-\alpha|+|\frac{1}{2}-\alpha|+|\frac{2}{3}-\alpha| \quad (5)

Now, using (5) it follows that:

\forall \alpha \in S, f(\alpha) > \alpha \quad (6)

p.s. This function has probably been studied before but after several unsuccessful google searches I decided to give it this name.