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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

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  • $\begingroup$ What's given about $a_0,a_1,a_2,\ldots$? And what are CFs doing here at all? $\endgroup$
    – metamorphy
    Jul 5, 2020 at 15:36
  • $\begingroup$ Its a reference to this en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula. The '$a$'s should be rational numbers. $\endgroup$
    – Niklas
    Jul 5, 2020 at 19:19
  • $\begingroup$ The reason for referencing it, is because CFs have a very clear way of indicating rationality, which I suspect might be relevant. $\endgroup$
    – Niklas
    Jul 5, 2020 at 19:26

1 Answer 1

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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 $1/x$].

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  • $\begingroup$ How about if the '$a$'s are integer prime powers, and the field is $Q_p$? $\endgroup$
    – Niklas
    Jul 6, 2020 at 7:12
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    $\begingroup$ I think it deserves a dedicated question (at least to attract someone else's attention). $\endgroup$
    – metamorphy
    Jul 7, 2020 at 13:43

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