Can anyone provide an example of a realvalued 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 tagremoved tag? $\endgroup$ – Gerry Myerson Feb 9 '17 at 22:16

1$\begingroup$ @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. $\endgroup$ – Włodzimierz Holsztyński Feb 10 '17 at 20:26

2$\begingroup$ @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 halfhearted attempt to contribute a little content to the site. $\endgroup$ – Eric Wilson Feb 13 '17 at 19:57

1$\begingroup$ 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 20091002 date). $\endgroup$ – Włodzimierz Holsztyński Feb 13 '17 at 23:31
If you take the classic nonanalytic 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 nonzero for any positive number so it does not agree with its Taylor series in any neighbourhood of $0$.

4$\begingroup$ $e^{1/t^2}$ is a bit nicer, as one can use a single formula for all real numbers. $\endgroup$ – Zoran Skoda Mar 24 '10 at 20:17

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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:
http://web.archive.org/web/20141230224759/http://www.math.niu.edu/~rusin/knownmath/99/nowhere_analy
I always thought the classic nonanalytic 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 piecewisedefined functions

3$\begingroup$ To make students happy, though, you'll have to define $\exp(1/0^2)$ to be 0, and you're back in piecewisetown. $\endgroup$ – Kevin O'Bryant Nov 26 '09 at 21:58

7$\begingroup$ @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 Nov 29 '09 at 21:37

3$\begingroup$ Besides, I'd say that for the students it is more useful to communicate the (historically nontrivial) 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). $\endgroup$ – Pietro Majer Feb 9 '17 at 19:46