Showing that $f^{(n)}(0)=0$ for $f(x)=e^{-1/x^2}$ if $x\neq 0$ and $f(0)=0$.

Question

Define $f$ as follows: $f(x)=e^{-1/x^2}$ if $x\neq 0$ and $f(0)=0$.

Show that $f^{(n)}(0)$ is continuous for all $x$ and $f^{(n)}(0)=0$. $n=1,2,\dots$.

To show this, I have shown that for $x\neq 0$, we have $f^{(n)}(x)=e^{-1/x^2}P_{3n}(1/x),$ where $P_{3n}(t)$ is a real polynomial of degree $3n$.

Hence, to show that $f^{(n)}(0)=0,$ I will use induction. $n=1$ case is trivial. So suppose that $n=k$ holds. Then as $n=k+1$, we have

$\frac{f^{(k)}(x)-f^{(k)}(0)}{x-0}=\frac{f^{(k)}(x)}{x}=\frac{e^{-1/x^2}P_{3k}(1/x)}{x}$.

Now I need to show that the limit of the above fraction as $x\to 0$ tends to $0$. For the case $x\to 0^{+}$, I can argue as follows.

Replace $t=1/x$ in the above fraction, then we get

$\frac{e^{-1/x^2}P_{3k}(1/x)}{x}=\frac{tP_{3k}(t)}{e^{t^2}}=(\frac{tP_{3k}(t)}{e^t})(\frac{e^t}{e^{t^2}})\to 0$ as $t\to \infty$, since we have $\lim_{x\to \infty}\frac{P(x)}{e^x}=0$.

However, I have trouble showing the case for $x\to 0^{-}$, since $\lim_{\to -\infty}\frac{P(x)}{e^x}$ does not exist for all polynomials.

How can I solve this problem? I would greatly appreciate any help.

(added)

My attempt: $\lim_{x\to 0^{-}}\frac{e^{-1/x^2}P_{3k}(1/x)}{x}=\lim_{x\to 0^{-}}e^{-1/x^2}P_{3k+1}(1/x)=\lim_{x\to 0^{+}}e^{-1/x^2}P_{3k+1}(-1/x)=\lim_{t\to \infty}e^{-t^2}P_{3k+1}(-t)=\lim_{t\to\infty}\frac{P_{3k+1}(-t)}{e^t}\cdot \frac{e^t}{e^{t^2}}=0\cdot 0=0$

$\lim_{t\to\infty}\frac{P_{3k+1}(-t)}{e^t}=0$, since

$|\frac{P_{3k+1}(-t)}{e^t}|\le\frac{P_{3k+1}(|-t|)}{e^t} \to 0$ as $t\to \infty$.

You might want to try taking the $n-\ th $ derivative of the Taylor expansion of your function. — Victor, Sep 15 '15 at 18:52
The case $x\to0-$ is not different from the case $x\to0+$ since polynomials are continuous at $0$ and $x\mapsto e^{-1/x^2}$ is even. — Christian Blatter, Sep 15 '15 at 19:09
Taylor expansion hasn't appeared yet in the text where I got this problem. I added my latest attempt. Is it correct? — nomadicmathematician, Sep 15 '15 at 19:50
For the $n$-variable result, see Smoothness of a real-valued function on $\mathbb{R}^n$. For a sketch in the one-variable case, see my answer here. — Dave L. Renfro, Sep 15 '15 at 19:57

score 1 · Answer 1 · answered Sep 15 '15 at 20:16

1

$\lim_{x\to -\infty}\frac{P(x)}{e^x} $ does not exist but $\lim_{x\to -\infty}\frac{P(x)}{e^{x^2}} =0 $ since $e^{(-x)^2} =e^{x^2} $.

answered Sep 15 '15 at 20:16

marty cohen

110,450

So my attempt is correct? – nomadicmathematician Sep 16 '15 at 02:18
You don't need the $P(t)/e^{t^2} = (P(t)/e^t)(e^t/e^{t^2})$. Work directly with $e^{t^2}$. – marty cohen Sep 16 '15 at 05:15

Showing that $f^{(n)}(0)=0$ for $f(x)=e^{-1/x^2}$ if $x\neq 0$ and $f(0)=0$.

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