I'm studying elementary mathematical analysis class and I've learned that a function is called real analytic if its taylor expansion at any point in its domain converges to it in some neighborhoods of the point.

And here's my question. The definition above seems to emphasize the importance to have convergent expansions at every point of its domain. Until now all analytic functions I've met can be approximated by its Maclaurin series and I thought that would be another possible equivalent definition. Now I need to show (or disprove) that:

  1. If a function is analytic, then its taylor expansion at any fixed point is convergent to the function in $\mathbb R$

  2. If a taylor expansion at one point pointwise converge to the function in $\mathbb R$, the the function is analytic, which means that its taylor expansions at other points also converge to the function.

Unfortunately, I solved neither of the above two problems, not even able to determine whether they are true or false. So I came here for help. I'm new to this site (and worse, not a native English speaker) so point me out if I've done anything not suitable for this site. Thanks!


1 Answer 1


1) and 2) are both false. The easy one is 2). Just take any nice $f$ and redefine it outside some interval around $0$. The Taylor series converges in that interval but the function need not analytic elsewhere.

For 1) consider $\frac 1 {1+x^{2}}$. This is analytic everywhere. The Taylor series at $0$ converges in $(-1,1)$ but it diverges for $|x| >1$.

  • $\begingroup$ Thanks for the counterexample for 1)! but I still have problems on 2), by "converge to f" I mean that it converges on the whole domain of f, not only locally. I've edited my question to clarify this. $\endgroup$
    – GZZ
    Nov 20, 2019 at 8:29

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