Nth Derivative of Exponential of Minus One over x

Theorem
Let $n \in \N$ be a natural number.

Let $\map {P_n} x$ be a real polynomial of degree $n$.

Then:


 * $\ds \dfrac {\d^n} {\d x^n} \map \exp {- \frac 1 x} = \frac {\map {P_n} x}{x^{2n}} \map \exp {- \frac 1 x}$

where $\map {P_n} x$ satisfies the following recurrence relation:


 * $\map {P_{n + 1}} x = x^2 \dfrac \d {\d x} \map {P_n} x - \paren {2 n x - 1} \map {P_n} x$

and $\map {P_0} x = 1$.

Proof
Proof by induction:

Basis for the Induction
We have that:


 * $\ds \map \exp {- \frac 1 x} = \frac {\map {P_0} x} {x^{2 \cdot 0}} \map \exp {- \frac 1 x}$

Induction Hypothesis
This is our induction hypothesis:
 * $\ds \dfrac {\d^n} {\d x^n} \map \exp {- \frac 1 x} = \dfrac {\map {P_n} x}{x^{2n}} \map \exp {- \frac 1 x}$

Now we need to show true for $n + 1$:
 * $\ds \dfrac {\d^{n + 1}} {\d x^{n + 1}} \map \exp {- \frac 1 x} = \dfrac {\map {P_{n + 1}} x}{x^{2\paren {n + 1}}} \map \exp {- \frac 1 x}$

Induction Step
This is our induction step:

The result follows by induction.

For all $n \in \N$ $\map {P_n} x$ is a real polynomial because we have $\map {P_0} x = 1$, and every step involves only addition, subtraction and multiplication of polynomials and their derivatives.