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

  • $\begingroup$ 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


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 $1/x$].

  • $\begingroup$ How about if the '$a$'s are integer prime powers, and the field is $Q_p$? $\endgroup$
    – Niklas
    Jul 6, 2020 at 7:12
  • 1
    $\begingroup$ I think it deserves a dedicated question (at least to attract someone else's attention). $\endgroup$
    – metamorphy
    Jul 7, 2020 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.