Proofs for $g(x)=e^{-1/x^2}$ when $x\neq0$, and $g(x)=0$ when $x=0$

Question

Sorry for the non-descriptive title - the question is a bit long. I have $g(x)$ as in the title, and we proved previously that $g'(0)=0$ using L'Hôpital's rule.

Now I must show by induction that there exists a $P_{n}(x)$ such that $g^{n}(x)=P_{n}(1/x)e^{-1/x^2}$ when $x\neq0$.

I find the $P_{n}(1/x)$ notation hard to keep straight. But for the base case, I found that $g'(x)=\frac{2}{x^{3}}e^{-1/x^2}$. How can I find $P_{1}(1/x)$ that fits the requirements? This question has me quite flustered.

EDIT: Thanks for the help on the base case. How can I show it for $n=k+1$? Is there a general formula for the $n^{th}$ derivatives of $g(x)$?

Answer 1

With $P_1(x)=2x^3$ we have $$P_1\left(\frac1x\right)e^{-1/x^2}=\frac2{x^3}e^{-1/x^2}=g'(x)$$

Edit We have $$g^{(n+1)}(x)=\left(g^{(n)}\right)'(x)=\frac{d}{dx}\left(P_n\left(\frac1x\right)e^{-1/x^2}\right)=\left(-\frac1{x^2}P'_n\left(\frac1x\right)+\frac2{x^3}P_n\left(\frac1x\right)\right)e^{-1/x^2}$$ hence we deduce that $$P_{n+1}=-x^2P'_n+2x^3P_n$$

Answer 2

Extending the below answer, we are looking for $P_n(\frac{1}{x})$ that is iterable, indeed as below:

$P_1(\frac{1}{x})e^{-\frac{1}{x^2}}=\frac{2}{x^3}e^{-\frac{1}{x^2}}=g'(x)$

So a good inference for $P_n(\frac{1}{x})$ is:

$P_{n+1}(\frac{1}{x})=(-1)^n(n+1)\frac{P_n(\frac{1}{x})}{x}+P_n(\frac{1}{x})^2$

I think this is right, give it ago, but starting with the $P_1$ already stated.