Title says it all. It is clear that the set of all power series with integer coefficients is not countable while the set of algebraic numbers is countable, but I do not know whether the set of roots for power series with integer roots is uncountable.
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3 Even ignoring the all-zeroes power series (of which every number is a root), it's still definitely uncountable. Given any $a\in (-1, 1)$ we can construct a nonzero power series $f_a$ with $a$ as a root. Basically, higher powers of $a$ yield smaller numbers, so we build $f_a$ as follows. First, suppose WLOG that $a\not=0$ (otherwise just make the constant term zero). Then note that (i) for all positive reals $\epsilon$, there is some even positive integer $n$ such that $0<a^{n}<\epsilon$, but (ii) for any positive $N$ and any positive even $n$ there is some positive integer $m$ such that $ma^n>N$.
This lets us recursively define a sequence of coefficients $a_i$ such that for each $n$, the sum $a_0+a_1a^2+a_2a^4+...+a_na^{2n}$ is between $0$ and ${1\over n}$; hence the whole power series goes to zero, and so $a$ is a root of $$\sum_{n\in\mathbb{N}}a_nx^{2n}.$$ Since there are uncountably many $x$ in $(-1, 1)$, this finishes the proof.
- 1$\begingroup$ That's what I get for typing on a cell phone. I didn't realize it was taking that long, and I didn't see your answer. $\endgroup$Matt Samuel– Matt Samuel2017-06-03 20:31:15 +00:00Commented Jun 3, 2017 at 20:31
- $\begingroup$ This is interesting, thanks. I wonder of there is a growth condition that can be placed on the coefficients. Similarly, can you get the same result if you bpundits the coefficients. If the answer is not trivial, I can ask separately. $\endgroup$Paul– Paul2017-06-03 22:03:53 +00:00Commented Jun 3, 2017 at 22:03
- $\begingroup$ @MattSamuel Well, your answer goes into much more detail than mine. $\endgroup$Noah Schweber– Noah Schweber2017-06-03 22:11:05 +00:00Commented Jun 3, 2017 at 22:11
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Let $0<a<1$ be irrational. I claim that there is a power series $$\sum{a_nx^n}$$ with integer coefficients that vanishes at $a$.
If $a>\frac12$, let $a_0=1$ and $a_1=-1$, otherwise let $a_0=0$ and $a_1=1$. Suppose recursively that $k\ge 1$ we have defined $a_0,\ldots,a_N$ such that $$0<\sum_{n=1}^N{a_na^n}<\frac1{k+1}$$ We choose $N'$ and $a_{N+1},\ldots,a_{N'}$ such that $$0<\sum_{n=1}^{N'}{a_na^n}<\frac1{k+2}$$ If the sum (let's call it $A$ for brevity) already satisfies this let $N'=N$. Otherwise choose $N'$ large enough that $$a^{N'}<\frac1{2(k+2)}$$ Let $a_i=0$ if $N<i<N'$. Then let $$a_{N'}=-\left\lceil\frac{(A-\frac1{k+2})}{a^{N'}}\right\rceil$$ Then $$0<A+a_{N'}a^{N'}<\frac1{k+2}$$ Letting $k\to\infty$, we end up with a power series with integer coefficients that vanishes at $a$, which should answer your question.
