$f(x)=\exp(-x^{-1})$ infinitely differentiable, induction?

Question

$f:\mathbb{R}\rightarrow{\mathbb{R}}; f(x)=\exp(-x^{-1})$ if $x>0$ and $f(x)=0$ if $x\leq 0$.

Show that you can differentiate $f$ on $\mathbb{R}_{>0}$ as often as you want. And that for every $n \in \mathbb{N}$ polynomals exist so that $f^{(n)}(x)=\frac{p_n(x)}{q_n(x)}f(x)$.

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$f'(x)=\frac{e^{-\frac{1}{x}}}{x^2}$

$f''(x)=\frac{e^{-\frac{1}{x}}-2e^{-\frac{1}{x}}x}{x^4}$

$f'''(x)=\frac{6e^{-\frac{1}{x}}x^2-6e^{-\frac{1}{x}}x+e^{-\frac{1}{x}}x}{x^6}$

Induction sounds very good to me but I don't really know how.

Answer 1

First show the formula holds for $n=1$.

If $f^{(n)}(x)=\frac{p_n(x)}{q_n(x)}f(x)$ then $f^{(n+1)}(x)=\frac{p_n(x)}{q_n(x)}f'(x) + {p'_n(x) q_n(x)-p_n(x)q'_n(x) \over q_n^2(x)}f(x)$, and you are given that $f'(x) = \frac{p_1(x)}{q_1(x)}f(x)$, so $f^{(n+1)}(x)= (\frac{p_1(x)p_n(x)}{q_1(x) q_n(x)} + {p'_n(x) q_n(x)-p_n(x)q'_n(x) \over q_n^2(x)} ) f(x)$.

Now figure out a suitable $p_{n+1}, q_{n+1}$ to show the formula holds.

Answer 2

Note $f^{(n)}(x)$ is $0$ if $x\le0$ and $g_n(x)e^{-1/x}$ if $x>0$ with $g_0(x):=1,\,g_{n+1}:=x^{-2}g_n+g_n^\prime$ so, by induction, $g$ is a degree-$2n$ polynomial in $1/x$.