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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?

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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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  • $\begingroup$ Actually, what if $f$ is not everywhere convergent? In particular, what if $f$ converges only for $x \in [0,1]$? Are there finitely many zeros of $f$ in $[0,1]$? It seems to me that the answer is yes,and nothing really changes. Right? $\endgroup$ – sd234 Dec 31 '10 at 7:40
  • $\begingroup$ It is impossible for a power series centered at $0$ to converge only on $[0,1]$. If it converges on $[0,1]$, then its radius of convergence is at least $1$ and it converges on at least $(-1,1]$. Nothing changes on the interior of the interval, where $f$ is analytic, but you have to be more careful at $1$. It is possible that $f$ cannot be extended to an analytic function in an interval at $1$, so the argument above doesn't apply with $c=1$. That means you have to consider the possibility that the function has a sequence of zeros converging to $1$. $\endgroup$ – Jonas Meyer Dec 31 '10 at 8:20
  • $\begingroup$ There are functions that are analytic on (at least) $(-1,1)$ and continuous on $(-1,1]$ with an infinite sequence of zeros converging to $1$, like $f(x)=(1-x)\sin(1/(1-x))$, $f(1)=0$. That may not precisely answer your question, because I'm not sure whether the series expansion for $f$ centered at $0$ converges at $1$. $\endgroup$ – Jonas Meyer Dec 31 '10 at 8:27
  • $\begingroup$ I meant to say "converges on at least $x \in [0,1]$". I think what you say answers my question. I interpret it as follows. If $f$ converges on at least $[0,1]$, then for any $a,b \in (0,1)$ we know $Z \cap [a,b]$ is finite. $\endgroup$ – sd234 Dec 31 '10 at 8:53
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    $\begingroup$ @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 $(-1,1)$. $\endgroup$ – Jonas Meyer Dec 31 '10 at 8:57
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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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  • $\begingroup$ I'm not sure I see why the fact that the zeros are isolated implies claim 3. Suppose for example the zeros were $Z=\{1/2^i\}_{i \in \mathbb{N}}$. This set of points is infinite. However, it is isolated, since given any $x\in Z$ I can present you an open interval containing $x$ and no other point in $Z$. Am I misinterpreting the definition of "isolated"? $\endgroup$ – sd234 Dec 31 '10 at 3:31
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    $\begingroup$ @shaddin: That cannot be the set of all zeros of a continuous function on $\mathbb{R}$; it contains $0$ as a limit point, so $0$ must also be a zero, and it is not isolated. Hence, because analytic functions have isolated zeros, no nonvanishing analytic function can be zero at $1/2^i$ for all $i\in \mathbb{N}$. $\endgroup$ – Jonas Meyer Dec 31 '10 at 3:41
  • $\begingroup$ @Jonas Meyer: I see, thanks! $\endgroup$ – sd234 Dec 31 '10 at 3:45
  • $\begingroup$ @shaddin: You're welcome. (I didn't mean nonvanishing, I meant not identically zero.) $\endgroup$ – Jonas Meyer Dec 31 '10 at 3:49

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