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Let $f(x) = \sum_{i =0}^\infty a_i x^i$ be a power series which converges for all real $x$. Assume that $f(x)$ is not identically zero. I'm interested in the density of the zeros of $f(x)$. Let $Z$ be the set of zeros of $f(x)$. Which of the following claims about density of $Z$ are true?

Claim 1: $Z$ is nowhere dense.

Claim 2: $Z$ is countable.

Claim 3: For any $a,b \in \mathbb{R}$ , $Z \cap [a,b]$ is finite.

I believe (correct me if I'm wrong) that claim 3 implies the other two. I suspect all three claims are true.

I suspect that the answers to these questions are well-known, though I was not able to find an obvious reference. Can anyone suggest a reference with a nice treatment of these questions?

2 Answers 2

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Claim 3 is true, and it does imply the other 2. It is enough to consider analytic functions, i.e. functions that have a power series expansion in some interval centered at each real number, which in particular holds if you have an everywhere convergent power series. If $[a,b]$ had infinitely many zeros of $f$, then there would be a limit point $c$ of these zeros. By continuity $f(c)=0$. Let $n\gt0$ be the smallest positive integer such that $f^{(n)}(c)\neq0$ (using the assumption that $f$ is not identically $0$). Then $f$ has power series expansion

$\begin{align*} f(x)&=\sum_{k=0}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^k =\sum_{k=n}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^k\\ &=(x-c)^n\sum_{k=n}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^{k-n}=(x-c)^ng(x), \end{align*}$

where $g(x)$ is a continuous function such that $g(c)\neq0$, and hence $g(x)\neq0$ in some open interval $I$ containing $c$. So the only zero of $f$ in $I$ is $c$, contradicting the fact that $c$ is a limit point of the set of zeros. Hence, unless $f$ is identically zero, the limit point $c$ cannot exist.

For reference you can read any good text on complex analysis. If you prefer to stick to the real case, there is the book A primer of real analytic functions by Krantz and Parks.

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    @shaddin: yes, an analytic function on an open interval can have only finitely many zeros on any compact subinterval of $(a,b)$, for the same reason as for the case where $f$ is defined on all of $\mathbb{R}$. A power series with positive radius of convergence defines an analytic function on the interior of its interval of convergence, which in your example will contain at least $(-1,1)$.2010-12-31
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All three claims are true. Claim 3 $\Rightarrow$ Claim 2 and Claim 1. The zeros of an analytic function are isolated. This implies Claim 3.

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    @shaddin: You're welcome. (I didn't mean nonvanishing, I meant not identically zero.)2010-12-31