Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume that $g: [0, \infty) \rightarrow \mathbb{R}$ is $\mathbb{R}$-analytic on $[0, \infty)$ (i.e. for every point of $a \in [0,\infty)$ there exist $R_a>0$ and a real sequence $(a_n)_{n=0}^\infty$ (depending on $a$) such that the series $\sum_{n=0}^\infty a_n (x-a)^n$ is convergent for all $x \in(a-R_a, a+R_a)$ and $\sum_{n=0}^\infty a_n (x-a)^n=f(x)$ for all $x \in(a-R_a, a+R_a)$. Does $g$ can be extended to analytic function on the whole $\mathbb{R}$ ?

P.S. Please without complex analysis, if possible.


share|cite|improve this question
Wouldn't a function like $g(x)=e^{-1/(x+1)^2}, g(-1)=0$ give the usual headache with all its derivatives vanishing at $x=-1$? – Jyrki Lahtonen Nov 21 '11 at 12:55
up vote 1 down vote accepted

Since the Taylor series converges in a neighbourhood of 0, the function can be extended to $(-\varepsilon,\infty)$ for a small $\varepsilon>0$, but one cannot go beyond that; simply take $$f(x):=\frac{1}{x+\varepsilon}$$ as a counterexample; it has a pole at $-\varepsilon$.

share|cite|improve this answer
Thanks for answer. – arc Nov 21 '11 at 14:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.