# The radius of convergence of a power series about a point interior to the domain of an analytic function

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a real analytic function with domain an open, non-empty set $(a, b) \subseteq \mathbb{R}$, $-\infty \leq a < b \leq \infty$ and let $c \in (a, b)$. Since $f$ is analytic, there is an interval $(c - r, c + r)$ about $c$ ($r > 0$), inside of which it is possible to represent $f$ as a power series about $c$: $$\forall x \in (c - r, c + r),\hspace{1cm} f(x) = \sum_{n = 0}^\infty h_n (x - c)^n$$ for some uniquely determined coefficients $h_0, h_1, h_2, \dots \in \mathbb{R}$.

Is it true that the radius of convergence $R$ of this power series is at least $$R \geq \min(c - a, b - c)$$

## 1 Answer

No. Consider $\dfrac{1}{1+x^2}$, $c=0$, $a<-1$, $b>1$.

Work in the complex plane to learn that the analogous result would be true there (analyticity on a disk implies the series converges there), and to see why the example above has radius of convergence $1$.

• Thank you. Then what am I missing in my understanding of the following excerpt from a German book on ordinary differential equations? – Evan Aad Feb 13 '15 at 15:55
• Unless my German fails me, the author made a mistake. – Robert Israel Feb 13 '15 at 16:13
• @RobertIsrael: Thanks, Robert. – Evan Aad Feb 13 '15 at 16:28
• @RobertIsrael: Your German is good. These are (trivial) upper bounds for the radius of convergence. – LutzL Feb 14 '15 at 20:09