Is it possible to put some sort of bounds on the smallest difference between elements for a limited range of integers $\mod 2\pi$?
$$ A =\bigl\{n\pmod{2\pi} : n \in \{\ 0, 1, 2, ... 2^x\ \} \bigl\} $$ (or similarly $ A = \bigl\{n\cdot \frac{1}{2\pi} \bmod 1 : n \in \{\ 0, 1, 2, ... 2^x\ \}\} $)
Where the smallest difference is defined as:
$$d_{min} = \min(\ \{\ |a - b|\ :\ a,b\in \ A,\ a \ne b\}\ )$$
Can I show that some sort of bounds exist for $d_{min}$ such as (intuitively/pessimistically via Dirichlet approximation theorem): $$ 2^{-2x} < d_{min} < 2^{-x} $$
If not, what are the tightest bounds we can place on $d_{min}$?
I'm comfortable with the answer here that shows $\{n\pmod{2 \pi}: n\in\Bbb N\}$ is dense in $[0,{2 \pi}]$ since $\pi$ is irrational, but the proof depends on arbitrarily large $n$.
In my concrete (software) case, I need to represent 44-bit integers as an angle in radians i.e. $(x=44)$, and would like to know the smallest number of fixed point precision bits that would be needed to uniquely represent these values $\pmod{2 \pi}$ to ensure no collisions.
It's easy to show that some of the smallest distances where $ b = 0$ are likely to occur for values of $a$ that are numerators in convergents of continued fraction approximations of $2\pi$, e.g. for:
$$a = 44 ,\ b = 0,\ \ (2\pi \approx \frac{2\times22}{7})$$
or $$a = 710,\ b = 0,\ \ (2\pi \approx \frac{2\times355}{113})$$
or even the semiconvergents - the sum of numerators/denominators such that the denominator becomes even, e.g. $$a = 377,\ b = 0, \ \ (2\pi \approx \frac{355+22}{\frac{113+7}{2}})$$
So one might suppose that the smallest distances occur with numerators of the most accurate continued fraction approximations of $2\pi$, but pairs like $$a = 103993, \ b = 0, \ \ (2\pi \approx \frac{103993}{\frac{33102}{2}}) $$ (ostensibly $\pi$ accurate to 9.2 decimal places) produce smaller gaps than the next (more accurate) approximation of $\pi$ (9.5 decimal places)
$$a = 208696,\ b = 0, \ \ (2\pi \approx \frac{2\times104348}{33215} \approx \frac{208341 + 355}{\frac{66317 + 113}{2}})$$
(This is probably because the smaller numerator already has an even denominator, so there's no need to double the numerator, and thus double the error).
Edit: as Eric points out in his answer below, I should be looking at convergents of $2\pi$ directly, since I was using double the convergents of $\pi$, the above musing can be ignored
For the $b = 0$ case, is it sufficient to only look at numerators of convergents/semiconvergents which have an even denominator?
Would another approach to this be better?
Edit, having thought about it a bit more:
If distance is defined $$ d(m, n) = | (m\pmod{2\pi}) - (n\pmod{2\pi}) |\ \ \text{for}\ m,n \in \Bbb{N}$$
I think we can say that the distance $d(a + b, b) = d(a,0)$ for $b \in \Bbb N$, so all the minima from $a$ being a convergent numerator where $n = 0$ will be trivially replicated at $ m = a+b $ for $n = b$.
The same applies for $d(a, b - a) = d(0,b)$ for $a \in \Bbb N$, and due to the symmetry of $a$ and $b$, this allows the reduction of any pair of values into a single measure of accuracy of an approximation of $2\pi$.
Could there be other minima where $ n > 0 $ that don't correspond to a trivial $ m = a+b $ case, i.e. could there be a $b$ such that $d(a, b) < d(a,0)$?
I think all members of $A$ must be positive, and so finding a smaller minimum where $b > 0$ is a matter of looking for $n$ such that $n\pmod{2\pi}$ is as close as possible to but not equal to $ m\pmod{2\pi}$. In the case where
$$n\pmod{2\pi} < m\pmod{2\pi}$$ this is effectively trying to find a tighter approximation of ${2\pi}$, and can likely only happen with a "better" continued fraction convergent numerator.
In the case of $$(m\ \text{mod}\ {2\pi}) < (n\ \text{mod}\ {2\pi}) < 2(m\ \text{mod}\ {2\pi}) $$ due to the symmetry in the definition of distance, $m$ and $n$ can be swapped, and the problem again becomes one of trying to find a "better" continued fraction convergent numerator.
The relationship for a convergent $ \frac{p}{q}$ is: $$p\pmod{2\pi} = (2\pi - \frac{p}{q})\times q$$
Via the mediant inequality, I think we can always find a midpoint between 2 convergents by adding the numerators together (and denominators), and I think this applies whether the 2 convergents are overestimates or underestimates... If you have one overestimate and one underestimate, then the new estimate for $2\pi$ becomes better than either.
$$\frac{p}{q} < \frac{p+a}{q+b} < \frac{a}{b}$$
It follows that you can do this an arbitrary number of times to get successively closer distances between pairs of convergents, but how quickly does that grow the $d_{min}$?
This implies a sort of fractal scale which evenly distributes integer values across $[0,2\pi]$ at a granularity roughly the "scale" of the preceding convergent numerator, then as soon as the next convergent numerator is reached, the scale gets smaller, and so on, so the $d_{min}$ for any number $n$ can never be smaller than the value of the largest convergent numerator smaller than n $\pmod{2\pi}$.
So my question rearranges to
$$\log(p\pmod{2\pi}) = \log(\lvert2\pi - \frac{p}{q}\rvert) + \log(q)$$
Which is almost the same question as this
Does that mean $2^{−(x+\epsilon)}$ is enough, where $\epsilon$ is how much better the convergent is than the size of the denominator (the negation of the right hand term above)?
If so, does that mean for $\mu(\pi) =2$ that $\epsilon \leq x$?
