Rationality of Euler-type Generalised Continued Fraction

Question

My question concerns the following:

If we have a convergent series $S$ (in some field) equivalent to a Euler-type GCF:

$$S = a_0 + a_0 a_1 + a_0 a_1 a_2 + \cdots = \cfrac{a_0}{1-\cfrac{a_1}{1+a_1 - \cfrac{a_2}{1 + a_2 - ...}}}$$ where $a \in \mathbb{Q}$

And $S \in \mathbb{Q}$ or $S \notin \mathbb{Q}$

And now we take a second series:

$$T = a_0 + a_0 a_1 x + a_0 a_1 a_2 x^2 + \cdots = \cfrac{a_0}{1-\cfrac{a_1 x}{1+a_1 x - \cfrac{a_2x}{1 + a_2x - ...}}}$$

for some $ x \in \mathbb{Q}$

Is there any relation between the rationality of S and that of T, especially in a p-adic field?

Thank you

Answer 1

Basically, the question asks whether the rationality of $f(1)$ is related to the rationality of $f(x)$ for some rational $x$, where $f(x)$ is any power series in $x$ with nonzero rational coefficients and $f(1)$ convergent.

If that's correct, then it has nothing to do with continued fractions, and the answer is negative as I see it.

Consider $f(x)=\exp\big(ax(1-x)\big)$ with $a\neq 0$ rational (and $|a|_p$ small enough in the $p$-adic case). Then $f(1)=1$, but $f(x)$ is irrational (even transcendental as known since Hermite in the real case and since Mahler in the $p$-adic case) for rational $x\notin\{0,1\}$. To ensure $f(x)$ has nonzero coefficients, it suffices to add a suitable rational function if needed.

This is easily reversed to have $f(x)$ rational but not $f(1)$ [replace $f(z)$ by $f(xz)$ and $x$ by $1/x$].