Can anyone provide an example of a real-valued function f with a convergent Taylor series that converges to a function that is not equal to f (not even locally)?

  • $\begingroup$ Can we remove the tag-removed tag? $\endgroup$ Commented Feb 9, 2017 at 22:16
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    $\begingroup$ @AndyPutman's link was great, thus I have upvoted his Answer. Otherwise, I would vote for closing the above Question, which asked about a standard fact provided by many textbooks on Mathematical Analysis and Differential Geometry. $\endgroup$ Commented Feb 10, 2017 at 20:26
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    $\begingroup$ @WłodzimierzHolsztyński You are certainly correct that this answer is not currently appropriate for this site. It was asked in the early weeks of the MO community, when the scope had not been well defined. Ironically, the only reason that I asked the question was in a half-hearted attempt to contribute a little content to the site. $\endgroup$ Commented Feb 13, 2017 at 19:57
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    $\begingroup$ Eric, this was nice of you, I appreciate your sacrifice. Thus let this thread stay open for the sake of the MO's history. (Only now I have paid attention to the 2009-10-02 date). $\endgroup$ Commented Feb 13, 2017 at 23:31

3 Answers 3


If you take the classic non-analytic smooth function: $e^{-1/t}$ for $t \gt 0$ and $0$ for $t \le 0$ then this has a Taylor series at $0$ which is, err, $0$. However, the function is non-zero for any positive number so it does not agree with its Taylor series in any neighbourhood of $0$.

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    $\begingroup$ $e^{-1/t^2}$ is a bit nicer, as one can use a single formula for all real numbers. $\endgroup$ Commented Mar 24, 2010 at 20:17
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    $\begingroup$ All real numbers except zero! $\endgroup$ Commented Mar 25, 2010 at 1:21

Another thing to note is that there are smooth functions whose Taylor series do not converge to the function in a neighborhood of ANY point! An easy example of this can be found here:


The example looks as follows:

$$F(x) = \sum_{n=0}^{\infty} \frac {\exp(2^n i x)} {n!}, \quad F^{(k)}(x) = \sum_{n=0}^{\infty} (2^n i)^k \frac {\exp(2^n i x)} {n!}$$

For every $k$, the above series converges absolutely for real $x$, so the function $F$ is smooth. On the other hand, if $x=a/2^N$ for some integers $a$ and natural number $N$, then for $k\in 4\mathbb{Z}$ we have

$$|F^{(k)}(x)| \geq \frac {2^{Nk}} {N!} - \sum_{n<N} \frac {2^{nk}} {n!} > \frac {2^{Nk-1}} {N!}$$

provided that $2^k > 2N$. Therefore, the Taylor series of $F$ has convergence radius $0$ at $x$. Since the set on which $F$ is analytic is open, this means that $F$ is nowhere analytic on $\mathbb{R}$.


I always thought the classic non-analytic smooth function was $\exp(-1/t^2)$ over the reals. This example is probably more satisfying to students (which is why you see it in texts) because when you look at that expression it's not obvious that anything funny should be happening at 0, whereas that's not so obvious for Andrew's piecewise-defined functions

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    $\begingroup$ To make students happy, though, you'll have to define $\exp(-1/0^2)$ to be 0, and you're back in piecewise-town. $\endgroup$ Commented Nov 26, 2009 at 21:58
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    $\begingroup$ @Kevin O'Bryant: But that follows from demanding that the function be continuous, right? And I think most students are okay with extending a function in the unique way that makes it continuous; many of them probably do it without even realizing it. $\endgroup$
    – Vectornaut
    Commented Nov 29, 2009 at 21:37
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    $\begingroup$ Besides, I'd say that for the students it is more useful to communicate the (historically non-trivial) idea that there is a difference between a function and a representation of it by means of a formula; and that the same function may need several representations in various pieces of its domain (for instance this leads to the definition of analytic function by power series). $\endgroup$ Commented Feb 9, 2017 at 19:46

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