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Formally, given a sequence of real numbers $a_0,a_1,\cdots$ which are not all zero, is it possible that there're an uncountable number of real numbers $x$ such that the sequence $\Big\{S_n=\sum\limits_{i=0}^n a_ix^i\Big\}$ converges to zero?

What I already know about this:

  1. Such $x$ can never constitute an interval, for otherwise we could prove $a_i=0$ via successive differentiation.
  2. The answer is no (i.e. there are at most a countable number of $x$) if the sequence of functions $\Big\{f_n(x)=\sum\limits_{i=0}^n a_ix^i\Big\}$ converges uniformly on any closed subinterval $[a,b]$ of $f_{\infty}(x)$'s domain.
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    It cannot. According to this: https://math.stackexchange.com/questions/310113/accumulation-points-of-uncountable-sets , an uncountable subset of $\mathbb R$ contains at least one of its limit points. But then according to the identity theorem, your power series would have to be identically $0$. – Vercassivelaunos Dec 21 '20 at 11:59
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    Interesting extension: complex power series. The case of interest would be a nonzero power series $\sum a_n z^n$ that converges to $0$ for uncountably many points at the radius of convergence. – GEdgar Dec 21 '20 at 12:34
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    @GEdgar, such a function can't vanish identically on an arc because of the reflexion principle but indeed I can't see any obstruction if these points are, say, a bunch of points with irrational arguments. – Ruy Dec 21 '20 at 17:00
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    OK no arc, but what about vanishing on a Cantor set? – GEdgar Dec 21 '20 at 17:28
  • Have you found an answer @GEdgar and @Ruy? The question has come up again. – Olius Nov 24 '23 at 15:45
  • Conrad's comment to the new question provides the textbook answer. – GEdgar Nov 24 '23 at 17:06

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