# Showing that this function is infinitely differentiable

Show, that the function

$$\mathcal E: \mathbb R \to \mathbb R: x \mapsto \begin{cases} \exp(-\frac{1}{x^2}), & \text{if x \neq 0}, \\ 0, & \text{otherwise}, \end{cases}$$

is infinitely differentiable and that $\frac{d^k\mathcal E}{dx}(0) = 0$ for all $k \in \mathbb N$.

We just introduced differentiation, so the solution should not contain very advanced techniques to solve this. We also got the tip it should be done by induction.

@fgp I found that it is $$f^{(n)}(x) = P_n \left(\frac1x\right)e^{-\frac1{x^2}}$$ where $P_n$ is a polynomial with integer coefficients.

I could also do the initial step for the induction:

For $n=1:$ $$f'(x) = \frac2{x^2}e^{-\frac1{x^2}} = P_1 \left(\frac1t\right) e^{-\frac1{x^2}}$$ where $P_1(x)=2x^2.$

I am stuck at the induction step: $f^{(n+1)}(x) =$.....

• Compute the first few derivatives and see if you can find a pattern. Then prove your hypothesis using induction... – fgp Jan 6 '14 at 15:45
• @fgp I answered you in my post (edited) – sj134 Jan 6 '14 at 16:11

Since $x \mapsto {1 \over x}$ is smooth for $x \neq 0$ and $x \mapsto e^x$ is smooth, it is clear that $\cal E$ is smooth for $x \neq 0$.

Suppose $x \ne 0$, then ${\cal E}^{(k)}$ has the form ${\cal E}^{(k)}(x) = e^{-{1 \over x^2}} p_k({1 \over x})$ for some polynomial $p_k$. This is clearly true for $k=0$, so suppose it is true for $k=0,...,n$. Then ${\cal E}^{(n)}(x) = e^{-{1 \over x^2}} p_n({1 \over x})$ and the chain rule gives ${\cal E}^{(n+1))}(x) = {\cal E}^{(1)}(x) p_n({1 \over x}) - {\cal E}^{(0)}(x) p_n'({1 \over x}) ({1 \over x^2}) = e^{-{1 \over x^2}} ({2 \over x^3}p_n({1 \over x})-p_n'({1 \over x}) ({1 \over x^2}) )$. If $p_{n+1}(y) = 2 y^3p_n(y)-p_n'(y) y^2$, then ${\cal E}^{(n+1))}(x) = e^{-{1 \over x^2}} p_{n+1}( {1 \over x} )$, and so the result is true for all $n$.

If $x \neq 0$, we have $e^{-{1 \over x^2}} = {1 \over {e^{1 \over x^2}}}$, and since $e^{1 \over x^2} \ge \sum_{k=0}^n {1 \over k!} {1 \over x^{2k}}$, we have $e^{-{1 \over x^2}} \le {x^{2n} \over \sum_{k=0}^n {1 \over k!} {x^{2(n-k)}}} \le {x^{2n} \over n!}$.

Suppose $p$ is a polynomial of degree $d$. Then for any $n$ we see that there is some constant $K$ such that $|e^{-{1 \over x^2}} p({1\over x})| \le K |x|^{2n-d}$ whenever $0 <|x| \le 1$. In particular, there is some $K$ such that $|e^{-{1 \over x^2}} p({1\over x})| \le K x^2$ for all $0 < |x| \le 1$.

We have ${\cal E}^{(0)}(x) \le x^2$ for all $x$, and so ${\cal E}$ is continuous at $x=0$. Since $|{\cal E}^{(0)}(x) - {\cal E}^{(0)}(0) -0| \le x^2$, we see that ${\cal E}^{(0)}$ is differentiable at $x=0$, and ${\cal E}^{(1)}(0) = 0$.

Now suppose ${\cal E}^{(k)}$ is differentiable at $x=0$ and ${\cal E}^{(k)}(0) = 0$ for $k=0,...,n$. Then $|{\cal E}^{(n)}(x) - {\cal E}^{(n)}(0) -0| \le K x^2$ for some $K$ and $|x| \le 1$. Hence ${\cal E}^{(n)}$ is differentiable at $x=0$, and ${\cal E}^{(n+1)}(0) = 0$.

• In the expression $e^{-{1 \over x^2}} p_n({1 \over x})$ is $({1 \over x})$ a factor or is it part of the expression for the polynomial? – Sam Nov 24 '14 at 20:08
• It is a parameter to the (polynomial) function $p_n$. Unfortunate ambiguity (but I would have written $p_n(x) {1 \over x}$ otherwise, but that is not always common practice). – copper.hat Nov 24 '14 at 20:10
• Thanks. When you computed ${\cal E }^{(n+1)}(x)$, you used the product rule as well as the chain rule, correct? In the first term there, you have not taken the derivative of either $e^{-1 \over x^2}$ or $p_n({1 \over x })$ and in the second term it appears you took the derivative of both. Is this a typo or am I missing something? – Sam Nov 24 '14 at 20:23
• @user151852: Thanks for catching that. I had transposed the $0,1$ derivatives of $\cal E$ and this mistake rippled a little. I have fixed it, I believe. – copper.hat Nov 24 '14 at 20:34

(This is more a hint than a full answer) Note that

\begin{align*} \frac{d\mathcal{E}}{dx}(0) &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{\mathcal{E}(x)-\mathcal{E}(0)}{x} \\ &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{1}{x} \exp \left( -\frac{1}{x^2} \right) \\[2mm] &= 0 \end{align*}

So, $\mathcal{E}'(0)=0$. As well, we have :

\begin{align*} \frac{d^{2}\mathcal{E}}{dx^{2}}(0) &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{\mathcal{E}'(x)-\mathcal{E}'(0)}{x} \\ &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{2}{x^{2}} \exp \left( -\frac{1}{x^2} \right) \\[2mm] &= 0 \end{align*}

So, $\mathcal{E}''(0)=0$. I think you got the idea : you can prove (by induction, for example) that, for all $n \in \mathbb{N}^{\ast}$ there exists a polynomial $P_{n}$ such that

\begin{align*} \frac{d^{n}\mathcal{E}}{dx^{n}}(0) &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{\mathcal{E}^{(n-1)}(x)-\mathcal{E}^{(n-1)}(0)}{x} \\ &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} P_{n} \left( \frac{1}{x} \right) \exp \left( -\frac{1}{x^2} \right) \\[2mm] &= 0 \end{align*}

Let $S=\left\{x\mapsto P\left(\frac{1}{x}\right)\mathcal E(x)\mid P\in \Bbb R\left[X\right]\right\}$

Can you prove that $f\in S \implies f'\in S$?

What can you say about $\lim\limits_{x\to 0}f(x)$ for $f \in S$?