Is the value of the power series at 0.1 transcendental?

Question

Let $$f(x)= \sum_{n=1}^{\infty}a_n x^n$$ where $a_n\in \{0,1\}$, and the $f(x)$ has a natural boundary. By the way, $$f(x)= \sum_{n=1}^{\infty}a_n x^n$$ is rational function or transcendental one on $\mathbb{C}$。

Is $f(x)$ at $\frac{1}{10}$ transcendental? Or is the value of such a transcendental function of power series at $\frac{1}{10}$ a transcendental number?

UPDATE: in addition, let $$f(x)= \sum_{n=1}^{\infty}a_n x^n$$ where $a_n\in \mathbb{N},\exists M\in \mathbb{N}, a_n < M$, and the $f(x)$ has a natural boundary. Is $f(x)$ at $\frac{1}{10}$ transcendental?

Answer 1

This question is likely open.

We can tell whether $f(1/10)$ is rational or irrational (by asking whether $a_n$ is eventually periodic); in this case, definitely irrational.

Can we have an algebraic irrational whose expansion base $10$ consists only of $0$s and $1$s? It is believed that is impossible, but has not been proved. Indeed, it is believed that algebraic irrational numbers are normal in all bases.

Similar question: is there an algebraic irrational point in the middle-thirds Cantor set?

Answer 2

For the UPDATE, allowing coefficients $a_n < M$ for a fixed $M$. Then there are examples with $f(1/10)$ rational.

Let's do this. Define a sequence $(a_n)$ of coefficients as follows: Start with a rapidly increasing sequence $(t_k)$ of natural numbers with $t_{k+1} > t_k+3$. Let $$ a_n = \begin{cases} 2 \quad & n=t_k\text{ for some }k \\ 13\quad & n=t_k+1\text{ for some }k \\ 3 \quad &\text{otherwise} \end{cases} $$ Now our series is $f(x) = \sum_{n=1}^\infty a_n x^n$. Most of the digits are $3$s, but occasionally there is a $2$ followed by a $13$. Because the sequence $(t_k)$ is lacunary, $f(x)$ has a natural boundary at its radius of convergence. But of course $f(1/10) = 1/3$.