If we have a series $F(x)=\sum a_n x^n$ where $a_n$ is in {0,1} for every integer $n$. Is $F$ algebraic over $\mathbb Q$ (set of rational numbers). If it is, under what conditions? Thank you.
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$\begingroup$ I take the question to be, under what conditions on the sequence will the (formal) series be an algebraic function. $\endgroup$– Gerry MyersonCommented Nov 10, 2011 at 11:34
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1$\begingroup$ I tried to improve the title... $\endgroup$– Gjergji ZaimiCommented Nov 10, 2011 at 11:38
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2 Answers
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There is a classical rational-transcendental dichotomy for power series with coefficients which don't grow too fast. Here are two famous results:
P. Fatou, "Series trigonometriques et series de Taylor", Acta Math. (1906), no. 30, 335–400.
Fatou proves that if a power series $F(x)\in \mathbb Z[[x]]$ converges inside the unit disk, then either $F(x)\in \mathbb Q(x)$ or $F(x)$ is transcendental over $\mathbb Q(x)$.
F. Carlson, "Uber Potenzreihen mit ganzzahligen Koeffizienten.", Math. Zeitschr. (1921), no. 9, 1–13.
Carlson proves that if $F(x)\in \mathbb Z[[x]]$ converges inside the unit disk, then either $F(x)$ is rational or $F(x)$ admits the unit circle as a natural boundary.
This tells you that your series if it is algebraic then it must be a rational function with rational coefficients. This happens if and only if the binary number $0.a_1a_2\dots$ is eventually periodic (to see this just plug in $x= \frac{1}{2}$).
answered Nov 10, 2011 at 11:32
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3$\begingroup$ You could wonder why this "corresponding" question is much harder: if a number $a \in (0,1)$ has only $0,1$ in its decimal expansion, then either it is rational or transcendental. That is surely true, but there is no known proof. $\endgroup$ Commented Nov 10, 2011 at 12:42
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$\begingroup$ @GeraldEdgar "if a number a∈(0,1) has only 0,1 in its decimal expansion, then either it is rational or transcendental" is implied from"Carlson proves that if F(x)∈Z[[x]] converges inside the unit disk, then either F(x) is rational or F(x) admits the unit circle as a natural boundary.", right? Since, power series with natural boundary is transcendental function, Is it necessary to give a proof? $\endgroup$ Commented Jul 7, 2017 at 12:06
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$\begingroup$ @XL_at_China ... I don't see how Carlson's result about a function can be applied to my question about a number. (But of course if we can find out how to do this, we will be famous.) $\endgroup$ Commented Jul 7, 2017 at 12:21
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$\begingroup$ @GeraldEdgar Good morning,Gerald. what do you think about a power series with natural boundary? Could it be non-transcendental (which means rational or algebraic)? Or have I misunderstood you? $\endgroup$ Commented Jul 7, 2017 at 12:39
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$\begingroup$ My comment was about a number $a$. Not a function. So what is the connection with Carlson's theorem about a function? $\endgroup$ Commented Jul 7, 2017 at 12:43
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I think the answer to Asymptotics/growth for coefficients of algebraic power series will be helpful.
answered Nov 10, 2011 at 11:22