# All smooth functions have Taylor expansion?

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.

• 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. Nov 5, 2020 at 10:35
• The Taylor expansion of $e^{-x^2}$ is indeed $0$. It does exist. The remainder is exactly $e^{-x^2}$.
– user65203
Nov 5, 2020 at 10:37

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 $$\forall n \in \mathbb{N}$$, $$||f^{(n)}||_{\infty|[a,b]} \leq Mn!\alpha^n$$

• 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. Nov 6, 2020 at 2:01
• 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? Nov 6, 2020 at 6:23
• 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 $0$. Nov 6, 2020 at 7:34
• Thanks！Then Peano remainder is not supposed to be used to compute whether a function has Taylor expansion. Nov 6, 2020 at 10:03
• You mean $e^{-1/x^2}$, rather than $e^{-x^2}$, which is obviously analytic. 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?