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Assume we have a base $b$ and a power series $f(x)$ with the coefficient of each power $x^n$ being equal to the nth non-zero $b$-ary digit times $b^{-i}$ (where $i$ is the index of the b-ary digit) in the expansion of an irrational number $m$.

For instance, in base three: $\pi = 10.010211012222..._3$, so:

$a_0 = 3^{-1}, a_1 = 2 \cdot 3^{-3}, a_2 = 3^{-4}, ... $

and:

$f(x) = 3^{-1} + 2 \cdot 3^{-3} \cdot x + 3^{-4} \cdot x^2 \cdots$

My question is, can $f(x \in \mathbb{Q}, x \neq 0) \in \mathbb{Q}$?

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Suppose, for example, $b > 3$ and $(b-1) x + 1 = x + 2$, which is true if $x = 1/(b-2)$. Let $m = \sum_{i=1}^\infty c_i b^{-2i}$, where each $c_i$ is either $(b-1)b+1$ or $b + 2$. Thus the base-$b$ expansion of $m$ consists of pairs $((b-1),1)$ and $(1,2)$. There are uncountably many possible $m$ (corresponding to two choices of $c_i$ for each $i \in \mathbb N$), so some of them will be irrational (in fact, those where the sequence of $c_i$ is not eventually periodic). But since $(b-1) x+1 = x+2$, the value of $f(x)$ is the same for all such $m$: in fact it is $$\sum_{i=1}^\infty (x + 2) x^{2i} = \frac{x^3 + 2 x^2}{1-x^2}$$ which is rational.

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  • $\begingroup$ Great answer, and very smart technique! However, what I was after was the case that the coefficients of the power series are the digit times the power of the digit in the base-b expansion (e.g. 0.101 in binary --> 0.5 + x*0.125 + ...) -- sorry for the lack of clarity. So I believe the left sum would have an additional term: ${(\frac{1}{x} + 2)}^{2i}$. $\endgroup$
    – Niklas
    Sep 25, 2019 at 9:06

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