In the first paragraph you talk about functions $2^\omega \to \{0,1\}$ that can be computed with oracle access to the input $x \in 2^\omega$. In my model of computation, the input is given as a Turing machine that computes $x$. I don't know how to prove the equivalence of these two models of computation and that is the crux of my question.

In general, if the most efficient algorithm for computing a few digits after the n'th digit requires $\Theta (f (n))$ "computational resources" then I expect a proof which constrains those digits should have length $\Theta (f (n))$, where the exact encoding method for proofs affects how the cost of a computation is calculated. A more efficient algorithm should lead us to expect weaker lower bounds for the proof length. Considering specifically $\alpha = \pi$ and $b = 2$, I believe the BPP algorithm requires $O (n \log n)$ time and $O (\log n)$ space, so I expect my conjecture to still be true.

The $d=1$ case has two interesting behaviors: All the singularities are of the form $Q \to \infty$, none are $Q \to 0$, and these singularities are entirely on the imaginary axis. However, I don't see a reason to expect these two facts to generalize to higher $d$.

Operads are essentially the same thing universal algebras, and the former term is more commonly used in category-theoretical/topological/geometric settings, so searching for "algebraic operads" may be more fruitful. For example, here is the first result I found: irma.math.unistra.fr/~loday/PAPERS/LodayVallette.pdf

@Speiser Good point; I forgot about the fast polynomial evaluation algorithm. In fact I believe I see a way to apply this arbitrary sums of rational functions -- I might write this up later. Perhaps I should have asked whether an algorithm could be faster than a general algorithm for summing $n$ terms applied to $n = p$.