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Assume we have $f(x) = \sum a_n x^n, a_n \in \mathbb{Q}$, and convergent $f(1) \notin \mathbb{Q}$. Assuming $f(x)$ converges at some $f(x \in \mathbb{Q})$, is it possible for $f(x \in \mathbb{Q}) \in \mathbb{Q}$ for some non-zero $x$?

Thanks!

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  • $\begingroup$ You should use "$\mathbb Q$" for the rationals. The symbol "$\mathbb R$" pretty universally already means "real numbers". $\endgroup$
    – MPW
    Aug 13, 2019 at 19:45
  • $\begingroup$ @JoséCarlosSantos : I think OP is using $\mathbb R$ for "rational numbers" -- yikes! $\endgroup$
    – MPW
    Aug 13, 2019 at 19:46
  • $\begingroup$ yep my bad, that is embarrassing >.> $\endgroup$
    – Niklas
    Aug 13, 2019 at 19:47

1 Answer 1

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Consider $$ \sqrt{1+\frac{x}{2}}=\sum_{k\ge0}\binom{1/2}{k}\frac{1}{2^k}x^k $$ that converges to $\sqrt{3/2}$ for $x=1$.

Take $x=-3/2$; then the series converges to $\sqrt{1/4}=1/2$.

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