Timeline for In the rational numbers, is every convergent power series a Taylor series for a rational function?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2022 at 18:46 | comment | added | Joe Silverman | @WillSawin Very nice!! | |
| Sep 9, 2022 at 18:45 | comment | added | Will Sawin | @JoeSilverman I think my argument gives $g(n)^n$ decay for any function $g(n)$ going to $\infty$ with $n$ , by replacing $|x|$ by $\sqrt{ g(n)/2}$ and choosing the sequence $a_i$ so $|a_n | \leq \sqrt{ g(n)/2}$. | |
| Sep 9, 2022 at 18:32 | comment | added | Joe Silverman | @WillSawin Good point regarding exponetial decay of coefficients. But there's room between exponential decay and $n^n$ decay. Suppose there is such a series (everywhere convergent, rational values at rational arguments) with $|c_n|\ge2^{-nf(n)}$. There do exist such series with $f(n)\asymp\log n$. How about with $f(n)\asymp(\log n)^{1-\epsilon}$? Or $f(n)\asymp(\log n)/(\log\log n)$? | |
| Sep 9, 2022 at 15:10 | comment | added | Will Sawin | @JoeSilverman Another question is whether one can make a power series like this that doesn't converge everywhere, but whose analytic continuation still takes rational values on every rational. But then one can just take a Mahler example and add $1/(1-x)$. | |
| Sep 9, 2022 at 15:08 | comment | added | Will Sawin | @JoeSilverman Such a power series would not converge on all the rational numbers, right, rather on just an interval in them? It seems to me that one can just apply a similar construction, restricted to rational numbers on an interval. | |
| Sep 7, 2022 at 19:14 | vote | accept | Madeleine Birchfield | ||
| Sep 7, 2022 at 18:52 | comment | added | Joe Silverman | Indeed a very nice explanation. But as I noted in my comment, such examples have been around for quite a while. A very interesting question (to me) is how fast the coefficients have to decrease for such functions. For example, is it possible to create such a function with the $n$th coefficients decreasing only exponentially with $n$, instead of decreasing like $1/n^n$? | |
| Sep 7, 2022 at 17:12 | comment | added | Iosif Pinelis | This is very nice! | |
| Sep 7, 2022 at 17:02 | history | answered | Will Sawin | CC BY-SA 4.0 |