# Ubiquity/scarcity of non-analytically continuable functions

Suppose f(z) is a power series with positive integer coefficients centered at zero and positive radius of convergence. What is the likelihood that f has a dense set of singularities on its circle of convergence (or any larger circle for that matter).

For instance, a classical theorem shown in Polya & Szego assumes f has integer coefficients and radius of convergence one. In this case, f is either rational or has a dense set of singularities on the unit circle. There are only countably many of the former and uncountably many of the latter, so the latter case is ubiquitous. (The supplied proof depends heavily on integer coefficients and radius 1.)

What can be said for more general radii? This question arises from combinatorial investigations: such f are generating functions for counting problems and several that I have run into seem to have dense sets of singularities on a circle (they also have a dominant real singularity, per Pringsheim's theorem, with slightly smaller modulus).

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An old an famous result of Borel states that this happens most of the time. Here is the precise statement $\newcommand{\bsP}{\boldsymbol{P}}$

Suppose that $(X_n)_{n\geq 0}$ are independent complex valued random variables on the same probability space $(\Omega, \mathscr{A},\bsP)$. Assume additionally that each $X_n$ is symmetric, i.e., $X_n$ and $-X_n$ have identical distributions. Form the random Taylor series

$$f_\omega(z)= \sum_{n\geq 0} X_n(\omega) z^n.$$

Then the probability that $F_\omega$ has a dense set of singularities on its circle of convergence is $1$.

For details see Chapter 4 of the book Some random series of functions by Jean Pierre Kahane.

Here is a simple application of this result.

Suppose that

$$f(z)=\sum_{n\geq 0} a_nz^n$$

is a series with positive radius of convergence. Consider its modifications $\newcommand{\ve}{\varepsilon}$

$$f_{\ve}(z):= \sum_{n\geq 0} \ve_n a_n z^n,$$

where $\ve_n=\pm 1$. If the sign changes $\ve_n$ are chosen randomly, independently, each sign with probability $1/2$, then with probability $1$ the series $f_\ve$ will have a dense set of singularities on its circle of convergence.

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