The theories are:

  1. $f(x)$ has Taylor expansion equals the remainder of Taylor polynomial converge to $0$.

  2. a smooth function such as $f(x) = e^{-1/x^2}$, $x>0$, $f(x) = 0$, $x \leqslant 0$ do not has Taylor expansion near $0$.

However, I think about these propositions and find:

smooth - exist Taylor polynomial - select Peano remainder = $o(x^n$) - the remainder must converge to $0$ - smooth function must have Taylor expansion near $0$

I will be appreciated if someone can point out the error in this infer.

  • 1
    $\begingroup$ All smooth functions $f$ (even your example in $2.$) have a Taylor expansion : it is $\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k$. But it has no need to converge to $f$ itself, has shows your example in $2.$. In fact, it has no need to be convergent on any open interval. $\endgroup$
    – Didier
    Nov 5, 2020 at 10:35
  • 3
    $\begingroup$ The Taylor expansion of $e^{-x^2}$ is indeed $0$. It does exist. The remainder is exactly $e^{-x^2}$. $\endgroup$
    – user65203
    Nov 5, 2020 at 10:37

2 Answers 2


For every smooth function $f$, you can consider the Taylor series around $0$ $$\sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n$$

However, there are two possible obstructions in general to link directly this series to the function :

  1. It is possible that the radius of convergence is $0$. Actually, one can prove that for every sequence $(a_n)$, there exists a smooth function $f$ satisfying $f^{(n)}(0)=a_n$. Therefore the Taylor series can be every series, so it can be a series which converges nowhere.

  2. It is possible that the Taylor series converges everywhere, but is not equal to $f$ in a neighbourhood of $0$. The classical example is the one you mention with $e^{-1/x^2}$, you get a Taylor series which is the null series, and the function is non constant equal to $0$ in any neighbourhood of $0$.

Finally, you have the following condition for a function to be analytic over an interval $I$ : a smooth function $f : I \rightarrow \mathbb{R}$ is analytic over $I$ iff $\forall [a,b]\subset I$, there exists $M \in \mathbb{R}$ and $\alpha > 0$ such that $\forall n \in \mathbb{N}$, $$||f^{(n)}||_{\infty|[a,b]} \leq Mn!\alpha^n$$

  • $\begingroup$ Thanks to your reply! It's proved that not every smooth function has a Taylor expansion. However, I still have doubts. Please see my answer below. $\endgroup$
    – Sarakio
    Nov 6, 2020 at 2:01
  • $\begingroup$ take $f(x) = e^{-x^2}, x \not= 0, f(0) = 0$ as an example. Consider $R_{n}(x) = f(x) - \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k$. On one hand, $f^{(k)}(0) = 0$ so $R_{n}(x) = f(x)$, then $\lim\limits_{n\to+\infty}R_{n}(x) = f(x) \not= 0, if x \not= 0$; on the other hand, smooth function $f(x)$ has $Maclaurin formula$ so $R_{n}(x) = o(x^n) $for every n, then $\lim\limits_{n\to+\infty}R_{n}(x) = \lim\limits_{n\to+\infty}o(x^n) = 0, if |x| < 1$. Why there are two different results? $\endgroup$
    – Sarakio
    Nov 6, 2020 at 6:23
  • $\begingroup$ The $o(x^n)$ is considered when $x$ tends to $0$, not when $n$ tends to $+\infty$. As you said, $R_n(x)=f(x)$ for every $n$, and $f(x)$ is indeed a $o(x^n)$ when $x$ tends to $0$. $\endgroup$ Nov 6, 2020 at 7:34
  • $\begingroup$ Thanks!Then Peano remainder is not supposed to be used to compute whether a function has Taylor expansion. $\endgroup$
    – Sarakio
    Nov 6, 2020 at 10:03
  • $\begingroup$ You mean $e^{-1/x^2}$, rather than $e^{-x^2}$, which is obviously analytic. $\endgroup$
    – isekaijin
    Jul 8, 2021 at 18:51

Assume $f(x)$ as a smooth function and consider its $Maclaurin formula$,

then for every n: $f(x) = \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k + o(x^n)$

select a sufficiently small neighborhood of zero, then: $\lim\limits_{n\to+\infty}R_{n}(x) = \lim\limits_{n\to+\infty}(f(x) - \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k)=\lim\limits_{n\to+\infty}o(x^n)=0$

then smooth function $f(x)=\lim\limits_{n\to+\infty}\sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k=\sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n$.

then smooth function $f(x)$ has Taylor expansion.

the above conclusion is wrong, because $e^{-1/x^2}$ is smooth and do not have Taylor expansion. However, what's the error during my infer?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .