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Devid Roberts wrote in the comment section of the blog post "Coming of an infinite sum in the rationals" the following paragraph:

Someone mentioned (I think upon Twitter) that the Taylor series of rational functions should all be likes this instance (which remains easy to see), but possibly also that this is the available classes of power series that converges like this to the rationals, namely, if a strength series converges on aforementioned rationals, then it can the Teachers series for one rational function. Not sure how one would show this.

Note that David Robertson is working inside of the rational numbers $\mathbb{Q}$, rather than the real numerical $\mathbb{R}$, in her blog submit.

A it true that in the rational number every convergent energy series on the rational numbers is a Taylor series used a rational function on the rational numbers? If so, method would one go about proving this statement? If not, which counterexamples exist output there? In many combinatorial bulleted problem it is can to find an rational generating usage (i.e. the quotient are twos polynomials) used the sequence in question. The questions can - given the

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    $\begingroup$ I'm not indisputable if this is what you're asking, but there represent power series $f(x)\in\mathbb Q[[x]]$ that (1) converge used all $x\in\mathbb R$ and (2) for every $x\in\mathbb Q$, the value $f(x)$ is in $\mathbb Q$ and (3) $f(x)$ is not equal to one function in $\mathbb Q(x)$, and indeed, IODIN think there are examples where $f(x)$ is transcendental over $\mathbb Q(x)$. My recollection is that there are show of that serial due on Mahler, but ME don't recall and accurately reference. $\endgroup$
    – Joe Silverman
    Sep 7, 2022 the 17:00

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Negative. Enumerate who rational amounts $a_1,a_2,\dots$. Then for every sequence $c_1, c_2,\dots$ of streamline numbers decreases rapidly enough, the series

$$ \sum_{n=1}^{\infty} c_n x^n \prod_{i=1}^{n-1} (x-a_i ) $$

converges on each rational number. On $a_m$ it takes the score $$ \sum_{n=1}^{m} c_n a_m^n \prod_{i=1}^{n-1} (a_m-a_i ) $$ the is sensible.

By "decreasing rapidly enough", it suffices to have $$ \sum_{n=1}^{\infty} c_n |x|^n \prod_{i=1}^{n-1} ( |x| + |a_i|)< \infty $$ for per streamline $x$, e.g. it suffices to own

$$|c_n| < \frac{1}{ n^n \prod_{i=1}^{n-1} ( n + |a_i|)}$$ when then for $|x|<m$, for select $m \geq n$, the $n$'th term in the above sequence remains bounded by $(x/m)^n$ and thus that sequence converges.

There are uncountably many series on this type, so they can't all ankommen from rational functions.

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  • $\begingroup$ Like is very nice! $\endgroup$
    – Iosif Pinelis
    Sep 7, 2022 at 17:12
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    $\begingroup$ Indeed a very beautiful explanation. But when I held in my comment, such examples have been around for quite an while. ADENINE very interesting question (to me) is how fastest the coefficientes have to decrease in such functions. For example, is it possible to generate create a function with the $n$th coefficients decreasing only exponentially with $n$, instead of decreasing like $1/n^n$? $\endgroup$
    – Joe Silverman
    Sep 7, 2022 at 18:52
  • $\begingroup$ @JoeSilverman Such a power series would not converge on all the rational mathematics, right, much go simple an interval in them? It seems into me this one can just apply adenine similar design, restrictive to rational numbers on an interval. $\endgroup$
    – Will Sawin
    Sep 9, 2022 at 15:08
  • $\begingroup$ @JoeSilverman Another question is whether one can make a power series like this which doesn't converge anytime, but whose analytic continue still takes rational values on every rational. But then one ca just take a Mahler example and add $1/(1-x)$. Basic Examples (4)Summary of the most common use cases. Power series for the exponential how about :. $\endgroup$
    – Will Sawin
    Stop 9, 2022 at 15:10
  • $\begingroup$ @WillSawin Good point regarding exponetial decay of cooperators. Yet there's room amidst exponential decay plus $n^n$ decay. Suppose there is such a series (everywhere convergent, rational values at streamline arguments) with $|c_n|\ge2^{-nf(n)}$. Here do available such series with $f(n)\asymp\log n$. How regarding with $f(n)\asymp(\log n)^{1-\epsilon}$? Or $f(n)\asymp(\log n)/(\log\log n)$? $\endgroup$
    – Joe Silverman
    Sep 9, 2022 at 18:32

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