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XL: I've revised the question substantially; in particular, the bits about 'computable' unbounded coefficients are rather out of place; CF coefficients for any algebraic number are certainly computable (there are known algorithms) but they're also largely 'patternless' so far as anyone knows (again, at least for degree $\gt 2$). AFAIK even the simpler question of unboundedness is open, and as that's the most central question here I've revised your Q to be about that specifically.
@ToddTrimble Much belated, but I like the usage of $O()$ as a set of functions, and the notation $f\in O(g)$. It offers a much clearer intuition as to what's going on, and helps to prevent some of the most egregious mistakes that the equality notation tends to trigger.
@YCor There are in fact similar results to (2) in graph theory; the Robertson-Seymour theorem essentially says that every family of graphs that's closed under the operation of taking minors is the compliment of a union of cones, but there can be no algorithm for taking a description of a minor-closed family and producing the minimal elements of its compliment; see mathoverflow.net/a/48025/7092
Ahh, mea culpa - in my head I actually have a slightly different version of this question; essentially, one with an arbitrary, possibly even $d$-dependent constant of approximation. (I only actually need the $a=1$ case for what I'm trying to prove, but was curious about the general result.)
@DouglasZare Ideally, but it's not essential for my purposes (obviously if $m$ and $n$ don't have to be coprime then the question essentially reduces to the $a=1$ case)
@ChristianRemling More specifically, the convergents of that continued fraction - but it's possible to have 'good' approximations that aren't convergents, and it's certainly not true that convergents have to be in all residue classes. (For instance, the convergents for $\sqrt{2}$ have no $n\equiv 3\bmod 4$)
Incidentally, Alpha gives $\frac{a_1(1/2)}{a_2(1/2)}$ as approximately $4.663632$, and passing that into the ISC doesn't yield anything particularly interesting. I imagine it's probably transcendental as well. Perhaps if the ratio of two automatic reals is also automatic...
This follows from some results about numbers whose binary (actually arbitrary-base, but binary is the relevant case here) representations form automatic sequences ; see en.wikipedia.org/wiki/Automatic_sequence#Automatic_real_number for more pointers. (The automata for recognizing each individual sequence is fairly trivial, just a 'count-to-$k$' FSM. There's no uniform automaton that covers all the sequences, but that's irrelevant here.)
I don't have my original reference (Melzak's Companion to Concrete Mathematics, IIRC) handy, but isn't there a variant of Hadamard convolution that produces the function $f(x) = \sum_n a_nb_nx^n$ from the two functions $g(x)=\sum_n a_nx^n$ and $h(x)=\sum_n b_nx^n$?
One possible way of formulating the 3-dimensional problem: you can define a frame along your curves in a couple of different ways (either 'intrinsically' via parallel transport, or using the orthogonal components of your two surfaces as the two axes); intersecting this foliation with your approximation allows you to define a 2d locus corresponding to the deviation of the approximation. Then maybe you can say something about the approximate circularity of this locus, or at least talk meaningfully about its centroid...
