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I am proposing you a self-posed question, but I do not know whether it makes sense or not.

Consider an analytic function $f\colon \mathbb R \to \mathbb R$. For every $x$ define $R(x)$ as the radius of convergence of the Taylor series with center in $x$. What can we say about the map $x \mapsto R(x)$?

For example, for $e^x, \sin{x}, \cos{x}$ we can take $R \equiv +\infty$. Are there any functions for which the map $x \mapsto R(x)$ is constant? What about the continuity? Is there any characterization of the functions $f$ for which $x \mapsto R(x)$ is continuous?

I hope the question is meaningful.

Romeo
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3 Answers3

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Instead of real analytic functions, let us start with some specific class of complex functions which is analytic over some neighborhood of real axis.

There is a theorem by Pólya–Carlson which states that for any power series with integer coefficients

$$f(z) = \sum_{n=0}^{\infty} c_n z^n\quad\text{ where }\quad c_n \in \mathbb{Z}$$

If it is analytic over the open unit disk, then it is either a rational function or have the unit circle $S^1$ as its natural boundary of analyticity. What this mean is the singularities of $f(z)$ is dense over $S^1$ and one cannot analytic continue $f(z)$ outside the open unit disk.

An example of this is the power series $\displaystyle f(z) = \sum_{n=0}^{\infty} z^{n!}$. Consider the function

$$g(z) = f(\frac{e^{-iz}}{2}) = \sum_{n=0}^{\infty} 2^{-n!} e^{-in!z}$$

It is clear $g(z)$ is analytic over $\{z \in \mathbb{C} : \Im z < \log 2 \}$ and has the line $\Im z = \log 2$ as a natural boundary of analyticity. So for every $x \in \mathbb{R}$, the power series expansion of $g(z)$ with respect to $x$ has radius of convergence equal to $\log 2$.

Adding $g(z)$ and $g(-z)$ together, we find the function

$$h(z) = \sum_{n=0}^{\infty} 2^{-n!} \cos( n! z) = \frac12 (g(z) + g(-z))$$

is real over the real axis, complex analytic over the strip $\{ z \in \mathbb{C} : |\Im z| < \log 2 \}$ and has the two lines $\Im z = \pm\log 2$ as natural boundaries.

This give us an example of real analytic function whose radius of convergence is finite and constant over the whole real axis.

achille hui
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If you consider a larger class of analytic functions, that is $f: \mathbb{C} \to \mathbb{C}$ we can say that the radius of convergence of the power series computed at the point $z_0$ is the distance from $z_0$ to the closest point where $f$ is not analytic.

For instance consider the function $f(z) = \frac{1}{1+z^2} = \sum_{n=0}^\infty (-1)^n z^{2n}$. The radius of convergence of this power series is 1, the distance from $0$ to $\pm i$. This is also why it is important to consider complex values. For if we take $f:\mathbb{R}\to\mathbb{R}$ then this function is analytic for all real numbers, but would not have an infinite radius of convergence. In this sense the real analytic function "knows" about the complex poles.

If your power series has an infinite radius of convergence (i.e. $f$ is entire) then your function $R(x)$ would be constant and equal to $\infty$. Otherwise, I don't believe it would be constant.

In response to the question in your comment, suppose we take a function $f(x) = \sum a_n x^n$ where the radius of convergence of the power series is infinite. This means that $$\lim \frac{|a_{n+1}|}{|a_n|} = 0$$ which in turn means the series $F(x) = \sum |a_n| x^n$ has an infinite radius of convergence.

Now let $g: \mathbb{C} \to \mathbb{C}$ be a complex analytic function defined by the power series $g(z) = \sum a_n z^n$. Now with the triangle inequality we have $$|g(z)| = |\sum a_n z^n | \le \sum |a_n| |z|^n < \infty$$ where the last inequality comes from $F(x)$ having an infinite radius of convergence. Hence g(z) has a power series expansion with an infinite radius of convergence, and is entire.

Joel
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    Thank for your answer. Indeed, I was asking another thing in the question but then I did not write it. It is about your last sentence: is it true that if $f: \mathbb R \to \mathbb R$ is analytic with $R \equiv +\infty$ then $f$ comes from an entire function, in the sense that it is the restriction of some entire $g \colon \mathbb C \to \mathbb C$? Thanks. – Romeo Oct 24 '13 at 19:52
  • I appended a response to your comment. :) – Joel Oct 25 '13 at 13:43
  • Can I get an example of real analytic function on $\mathbb R$ for which radius of convergent is finite? @joel – neelkanth Nov 08 '22 at 16:15
  • @Joel is examples $\frac{1}{x^2+1}$ is suitable ? – neelkanth Nov 08 '22 at 16:22
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At the end of this answer, I state

A corollary of Cauchy's Integral Formula is that the radius of convergence of a complex analytic function is the distance from the center of the power series expansion to the nearest singularity.

From this, the function you mention is simply the distance to the nearest singularity of the given analytic function. This is a continuous function on the domain of the analytic function, except when the analytic function in question is entire, in which case $R(x)$ is infinite everywhere.

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robjohn
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