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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.

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    $\begingroup$ 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 '20 at 10:35
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    $\begingroup$ The Taylor expansion of $e^{-x^2}$ is indeed $0$. It does exist. The remainder is exactly $e^{-x^2}$. $\endgroup$
    – user65203
    Nov 5 '20 at 10:37
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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 $\forall n \in \mathbb{N}$, $$||f^{(n)}||_{\infty|[a,b]} \leq Mn!\alpha^n$$

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  • $\begingroup$ 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 '20 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 '20 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 '20 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 '20 at 10:03
  • $\begingroup$ You mean $e^{-1/x^2}$, rather than $e^{-x^2}$, which is obviously analytic. $\endgroup$
    – pyon
    Jul 8 at 18:51
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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?

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