It seems to me that there is a contradiction between the Taylor's theorem and the properties of the Taylor series of the function $f(x)=e^{-1/x^2}$ if $x\ne0$ and $f(x)=0$ if $x=0$.

On the one side, with the above extension $f$ is $C^\infty$ at $x=0$ and the Taylor series $TS[f]_0(x)$ of $f(x)$ at $x=0$ is identically null with a radius of convergence $R_c=\infty$, whereas $f(x)\ne 0$ for $x \ne 0$. As a consequence, $TS[f]_0(x)\ne f(x)$, $\forall x\ne0$.

On the other side, Taylor's theorem states that an expansion at the order $n$ gives: $f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+...+\frac{f^{(n)}}{n!}x^n+h_n(x)x^n, \qquad$ with $\lim_{x\rightarrow0}h_n(x)=0$

So this means that we should get better approximations of $f(x)$ when $n$ increases. Which is then not the case in this particular example as increasing $n$ doesn't make a difference in the polynomial expansion.

Now, what is the problem here? Is it that we cannot apply Taylor's theorem in this particular case, and if so, why? Or is it that it is wrong to express the rest as $h_n(x)x^n$ with $\lim_{x\rightarrow0}h_n(x)=0$?

Alternatively, what special properties could have the functions $h_n(x)$ that doesn't make $h_n(x)x^n$ become smaller when $n$ increases?

Or am I missing something else?

  • $\begingroup$ The taylor series only converges to the function if the remainder tends to $0$ in some neighborhood, not just at the point we are looking for. This is the case, for example, if the derivates are all bounded in a neighborhood. In this case, the taylor series converges to the function in this neighborhood. $\endgroup$
    – Peter
    Aug 8, 2016 at 13:16
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    $\begingroup$ $f(x)$ is a non-analytic smooth function, see here en.wikipedia.org/wiki/… $\endgroup$ Aug 8, 2016 at 13:17
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    $\begingroup$ @DavidC.Ullrich I assume the OP is referring to the Peano remainder version of the theorem? $\endgroup$
    – Clement C.
    Aug 8, 2016 at 13:20
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    $\begingroup$ @DavidC.Ullrich Fair enough -- but the OP never explicitly mentioned anything about $\lim_{n\to \infty}h_n(x)$, except the confusing sentence "So this means that we should get better approximations\cdot". Her/his statement of the theorem is correct. $\endgroup$
    – Clement C.
    Aug 8, 2016 at 13:24
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    $\begingroup$ Fine. As has been said a few times already: What you say about $x\to0$ is correct. It simply does not follow from that that we get better approximations as $n\to\infty$. No contradiction. $\endgroup$ Aug 8, 2016 at 13:27

3 Answers 3


You seem to be confused about what the theorem you cite says.

Take $h_n(x) = f(x) x^{-n}= \begin{cases} 0 &\text{ if} x=0 \\ \frac{e^{-\frac{1}{x^2}}}{x^n} &\text{ otherwise.} \end{cases}$

It does satisfy $\lim_{x\to 0} h_n(x) = 0$, and all the properties promised by the theorem. There is no contradiction.

But it is utterly useless in order to approximate $f$ more and more finely by a sequence of polynomials forming the partial sums of a power series, "as $n$ grows." The whole point is exactly that: $f$ cannot be well approximated by such a family of polynomials around $0$. Trying to do so, no matter what degree $n$ you choose, you get a zero polynomial, and the remainder/error is... well, $f$.

  • $\begingroup$ Thanks a lot. Expressing $h_n(x)$ using the expression of $f(x)$ makes it all clearer. I should have thought about it! Nice. $\endgroup$
    – The Quark
    Aug 8, 2016 at 17:51
  • $\begingroup$ You're welcome! Also, this function is a nasty piece of work. (A useful one, but nasty.) $\endgroup$
    – Clement C.
    Aug 8, 2016 at 17:52

Taylor's theorem is perfectly true, even in this case. Simply, the value of the function is entirely in its complementary term. This is one case where the Taylor's series does not converge to the value of the function.

$\mathcal C^\infty$ functions with a Taylor's series which converge to the value of a function are called analytic functions.

Thus you see analytic functions and $\mathcal C^\infty$ *functions are different notions, for functions of a real variable.

For functions of a complex variable, the situations is quite different: a function which is differentiable is ipso facto $\mathcal C^\infty$ and analytic.


I would agree that Taylor's theorem has a weak indication that the approximations should be improving. In particular, we can rephrase the theorem as follows:

Let $f_n$ denote the $n$th-order approximation of $f$. Then the error $e_n(x) = f(x) - f_n(x)$ satisfies $$ \lim_{x \to 0} \frac{e_n(x)}{x^n} = 0 $$ which is to say that $e_n(x) \to 0$ faster than $x^n$ (i.e. $e(x) = o(x^n)$).

In a more "typical" Taylor approximation, this pattern would indicate an improvement of $e_n$ as $n \to \infty$, which is to say that $e_n$ approaches zero progressively more quickly and that $e_n(x) \to 0$ as $n \to \infty$. We could see this with the Taylor series for $f(x) = e^x$, for example.

However, note that the theorem does not actually imply that this is the case. For our problem, we will always have $$ e_n(x) = f(x) $$ and it just so happens that $f(x) \to 0$ faster than $x^n$ for any $n$.

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    $\begingroup$ ??? You say this would typically indicate that. I don't see how the true fact about $x\to0$ says anything at all about the approximation improving as $n\to\infty$. Given that it doesn't actually imply any such thing, could you explain how it would typically "indicate" improvement as $n\to\infty$? (Other forms of the remaineder do typically indicate this - I really don't see at all how this version says anything in this direction, even heuistically) $\endgroup$ Aug 8, 2016 at 13:48
  • $\begingroup$ @David First of all, when I say "typically", I mean "with the usual function one encounters in a calculus class" (as opposed to something about cardinalities or measures). If we knew that at each step the error $e_n(x)$ is as bad as it can possibly be (something like $e_n(x) \sim Cx^{n+1}$), then this theorem does imply an improvement as $n \to \infty$. Moreover, with a "typical sequence of functions", a steady decrease of error indicates that error will tend to zero. Again, none of these "expected behaviors" are actually implied by the theorem. $\endgroup$ Aug 8, 2016 at 13:59
  • $\begingroup$ @David while it is true that nothing truly surprising happened if you read the theorem closely enough and have zero other expectations, it's important to address why exactly the result is so counterintuitive (without an appeal to complex analysis). $\endgroup$ Aug 8, 2016 at 14:02
  • $\begingroup$ Thanks, but nothing you said answers my question, as far as I can see. Yes, it's important to address why things are counterintuitve, when they are. Seems to me the reason the current situation is countereintuitive is just that we have an inutition based on experience that every function is analytic - I simply don't see what the result about $x\to0$ says about that. $\endgroup$ Aug 8, 2016 at 14:16
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    $\begingroup$ Again, I agree that other forms of Taylor's theorem do "indicate" that the series "typicaly" converges; they give explicit bounds on the remainder, which typically tend to $0$. $\endgroup$ Aug 8, 2016 at 14:38

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