# Relationship between "analytic" functions and their taylor series

Analytic functions are those that are differentiable infinitely.

let $f$ be a function. My claims are as follows :

C1 : Functions that are analytic are exactly those who have taylor series expansions which converges for all x. (Radius of converges infinite)

C2 : If $f$ is an analytical functions (it's differentiable infinitely) I can create it's taylor series expansion around any point $x=x_0$ such that radius of convergence is infinite.

Are my Claims true? false?

• Analytic functions do converge on an open set about each point in its domain Jul 19, 2017 at 9:25
• Your tag says real analysis but maybe you are talking about complex analytic functions? Then yes, all the things you are saying follow from the fact that the radius of convergence of a power series at a given point is just the distance to the nearest singularity. Jul 19, 2017 at 9:32
• You seem to be talking about "entire functions", a special sort of analytic functions. Jul 19, 2017 at 12:30

First of all, an analytic function is not the same thing as a function which has derivatives of all orders. It is stronger than that: it is a function $f$ such that, for each $x_0$ in its domain, the Taylor series of $f$ converges to $f$ in some interval $(x_0-\varepsilon,x_0+\varepsilon)$.
Besides, the function $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by $f(x)=\frac1{1+x^2}$ is analytic, but the radius of convergence of its Taylor series at $0$ is $1$, not $+\infty$. And, of course, this shows that your second claim is also false.
• The OP tagged the question as real-analysis not as complex-analysis. But, yes, the radius of convergence of the Taylor seres about any point of the domain of an analytic function $f\colon\mathbb C\longrightarrow\mathbb C$ is $\infty$. Aug 24, 2019 at 6:51