Is there a smooth real function $f$ on some open interval containing $x=a$ such that its power series expansion at $x=a$ has zero radius of convergence? or maybe the series fails to converge even at $a$?

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    $\begingroup$ @SamuelAdrianAntz: the example you gave in the now-deleted answer is actually cool. If ever you rewrite another answer please include it. It is nice to remark that a Taylor series may converge, but to the wrong function. $\endgroup$ Sep 1, 2022 at 13:44
  • $\begingroup$ @GiuseppeNegro Yes, I will definitly do that. $\endgroup$ Sep 1, 2022 at 13:54
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    $\begingroup$ See here a concrete example. The Taylor series at $x=0$ (and at many other points too) has radius of convergence $0$. $\endgroup$
    – plop
    Sep 1, 2022 at 13:54
  • $\begingroup$ @user85667: great link. It also discusses the theorem of Borel which both David C. Ullrich and I mention without proof. I guess that in the end the function constructed via the Borel theorem is not so different than the one you directly point to. Indeed, the proof of Borel that is in that Wikipedia page also contains an infinite series. $\endgroup$ Sep 1, 2022 at 13:56

3 Answers 3


As the answers by David C. Ullrich and Giuseppe Negro explained, there is a theorem by Borel that every power series can be realized as a Taylor series. First of all, an example with a radius of convergence arbitrarily small can be given by including poles in the complex plane. For example the radius of convergence of the Taylor series with $a=0$ of: $$f(x)=\frac{1}{x^2+\varepsilon^2}$$ with poles at $\pm i\varepsilon$ is $|\varepsilon|$. There are examples of smooth functions where the radius of convergence of the Taylor series is zero: Two of them are given in this MSE answer taken from A primer of real functions by R. Boas Jr., section 24 on page 168, and Counterexamples in Analysis by by B. Gelbaum and J. Olmsted, counterexample 24 on page 68. This forum also mentions counterexample 22 on page 68, a series which converges only for $x=0$: $$f(x)=\sum_{n=0}^\infty n!x^n$$ but is indeed the Taylor series of a function. (This counterexample connects to the answer of Giuseppe Negro by choosing a function $f\colon\mathbb{R}\rightarrow\mathbb{R}$ with $f^{(k)}(0)=(k!)^2$, but it is not this series here due to its convergence.)

Here is also my previously deleted answer, where I misunderstood the question as finding a Taylor series that converges, but only to the function in a single point. (Ironically, somebody in the forum linked above did as well.) It is counterexample 23 on page 68 of Counterexamples in Analysis:

First of all, the power series always converges at $x=a$ since the term of zeroth order is given by the constant $f(a)$ and all terms of higher order vanish when $x=a$ is put into $f^{(n)}(x)(x-a)^n$. But there is a popular example for a smooth function with a Taylor series that converges everywhere (even uniformly), but not to the function in any neighborhood of the point chosen for expansion, which is: $$f(x) =\left\{\begin{array}{cl} e^{-\frac{1}{x^2}} & ;x\neq 0 \\ 0 & ;x=0 \end{array}\right.$$ for $a=0$. There it is differentiable infinity often (the exponential function beats every single power map in the limit) and all derivatives vanish, so the power series does as well even though the function does not.

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    $\begingroup$ The Taylor series centered at $x=0$ of that function does converge. In fact, it converges everywhere. What it doesn't do is to converge to the function. $\endgroup$
    – plop
    Sep 1, 2022 at 13:34
  • $\begingroup$ Ah, you're right. I was a bit too fast there. I'm going to delete my answer and think about it again. $\endgroup$ Sep 1, 2022 at 13:36

As pointed out, of course the series converges at $a$. But yes, it can have radius of convergence $0$. This is clear from the following exercise, somewhere in Rudin Real and Complex Analysis:

Exercise: Given a sequence $(c_n)$ of real numbers there exists a smooth function $f:\Bbb R\to\Bbb R$ such that $$f^{(n)}(0)=c_n$$for all $n$.

  • $\begingroup$ ok, what's the problem with that? $\endgroup$ Sep 1, 2022 at 13:42
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    $\begingroup$ I think maybe because the question asked for a concrete example if there is one? It's a cool theorem though, +1. $\endgroup$ Sep 1, 2022 at 13:50
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    $\begingroup$ @SamuelAdrianAntz What you suggest may be what was intended, but no it's not what was asked. The question asks only about existence... $\endgroup$ Sep 1, 2022 at 13:54
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    $\begingroup$ Your last remark reminds me of the joke where a mathematician is asked “Can you tell me what time it is?” and answers “Yes.” $\endgroup$
    – Martin R
    Sep 1, 2022 at 14:01
  • $\begingroup$ @DavidC.Ullrich That's indeed just my interpretation of the question. This theorem by Borel is also a way more general and elementary answer than just one example. $\endgroup$ Sep 1, 2022 at 14:37

Yes, there are smooth functions $f\colon \mathbb R\to \mathbb R$ such that the series $$ \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$$ converges only for $x=0$.

There is a theorem, attributed to Borel, that states the following:

for each sequence $(a_n)_{n\in \mathbb N_{\ge 0}}$, there is a smooth function $f\colon \mathbb R\to \mathbb R$ such that $f^{(n)}(0)=a_n$.

The answer to the previous question is a corollary of this theorem. A smooth function $f$ such that $f^{(n)}(0)=(n!)^2$ is an example.

The theorem of Borel is discussed here in considerable detail.


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