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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)?

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up vote 14 down vote accepted

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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Typo: you mean $e^{-1/t}$ for $t>0$. –  Kevin O'Bryant Nov 26 '09 at 21:59
    
Whoops! Thanks for pointing that out. I've corrected it (and converted to jsMath). –  Andrew Stacey Nov 29 '09 at 19:30
    
$e^{-1/t^2}$ is a bit nicer, as one can use a single formula for all real numbers. –  Zoran Skoda Mar 24 '10 at 20:17
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All real numbers except zero! –  Andrew Stacey Mar 25 '10 at 1:21
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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://www.math.niu.edu/~rusin/known-math/99/nowhere_analy

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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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To make students happy, though, you'll have to define $\exp(-1/0^2)$ to be 0, and you're back in piecewise-town. –  Kevin O'Bryant Nov 26 '09 at 21:58
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@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. –  Vectornaut Nov 29 '09 at 21:37
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