Let $f(x)=e^{-1/x}$ for $x>0$ and $f(x)=0$ for $x\leq 0$. Prove that $f^{(n)}(0)=0$ for all $n\in \Bbb N$.

Question

I'm reading the solution, and I understand how to prove that all derivates must be of the form:

$$f^{(n)}(x)=e^{-1/x}\left[\sum_{k=0}^{2n} \frac{a_k}{x^k}\right]$$

After this, the solution manual begins to prove $f^{(n)}(0)=0$ for $n\in \Bbb N$ by induction.

Assume that $f^{(n)}(0)=0$ for some $n\geq 0$. We need to prove $$\lim_{x\to 0} \frac{f^{(n)}(x)-f^{(n)}(0)}{x-0}=0$$

I don't understand why they do this. Isn't easier to just prove that for $n\in \Bbb N$, $f^{(n)}(x) \to 0$. I would think that using l'hopitale this must be possible.

Not sure if this is of interest to you: "$n$th derivative of $e^{1/x}$" — draks ..., Mar 07 '13 at 15:55
I think this is for using induction, that expression is equal to $f^{n+1}(0)$ — S L, Mar 07 '13 at 15:57
I think we can use the value of $f^n(x)$ from the series above and show that $f^{n+1}(0) = \lim_{x \to 0} f^{n}(x)/x $ assuming $f^{n}(0) = 0$ which says that if $f^n(0) = 0$ then $f^{n+1}(0) = 0$ and hence proved by induction. — S L, Mar 07 '13 at 16:02

score 1 · Answer 1 · answered Mar 07 '13 at 16:02

1

The formula you give for $f^{(n)}(x)$ is valid for $x > 0$. So the problem is that it is not necessarily true that $f^{(n)}(0) = \lim_{x\to 0^+} f^{(n)}(x)$, as this would be assuming that the derivative was continuous.

answered Mar 07 '13 at 16:02

Morgan Sherman

1,246

score 1 · Answer 2 · answered Sep 28 '21 at 12:41

In the theory of differentiable manifolds, the function \begin{equation}\label{part-unity-function}\tag{#} f(t)= \begin{cases} \textrm{e}^{-1/t}, & t>0\\ 0, & t\le0 \end{cases} \end{equation} plays an indispensable role in the proof of the existence of partitions of unity. In the paper [1] below, among other things, some properties of the function $f(t)$ defined by \eqref{part-unity-function} were investigated.

For $i\in\mathbb{N}=\{1,2,\dotsc\}$, \begin{equation}\label{f(t)-derivative} f^{(i)}(t)= \begin{cases}\displaystyle \frac{\textrm{e}^{-1/t}}{t^{2i}} \sum_{k=0}^{i-1}(-1)^k{k!}\binom{i}{k}\binom{i-1}{k}{t^{k}}, & t>0;\\ 0, & t\le0. \end{cases} \end{equation} See Theorem 2.3 in the paper [1] below.
The function $f(t)$ defined by \eqref{part-unity-function} is infinitely differentiable on $\mathbb{R}$ and \begin{equation}\label{f-der-at0=0} f^{(i)}(0)=0 \end{equation} for all $i\in\{0\}\cup\mathbb{N}$, but it is not analytic at $t=0$. See Theorem 2.4 in the paper [1] below.

By the way, at the site https://math.stackexchange.com/a/4262498/945479, there are discussions on the functions $\textrm{e}^{\pm1/t}$.

References

Xiao-Jing Zhang, Feng Qi, and Wen-Hui Li, Properties of three functions relating to the exponential function and the existence of partitions of unity, International Journal of Open Problems in Computer Science and Mathematics 5 (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.

https://mathoverflow.net/q/405015/147732 – qifeng618 Sep 28 '21 at 13:28 — qifeng618, Sep 28 '21 at 13:28

Let $f(x)=e^{-1/x}$ for $x>0$ and $f(x)=0$ for $x\leq 0$. Prove that $f^{(n)}(0)=0$ for all $n\in \Bbb N$.

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