Consider a function $f(t)=\exp\left(-\dfrac{1}{t}\right)$, $t>0$. For each positive integer $n$, let $P_n$ be the polynomial such that$$ \frac{\mathrm{d}^n}{\mathrm{d}t^n}f(t)=P_n\left(\frac{1}{t}\right)\exp\left(-\frac{1}{t}\right). \quad \forall t>0 $$
Well to start with this is a long problem, I have seen that for $n=1$. it holds that$$ P_{n+1}(x)=x^2\left(P_n(x)-\frac{\mathrm{d}}{\mathrm{d}x}P_n(x)\right). $$
Now I wanted to prove this using induction.
Assume that for $n=m$, this is true. Thus
$$P_{m+1}(x)=x^2\left(P_m(x)-\frac{\mathrm{d}}{\mathrm{d}x}P_m(x)\right). $$
Now we are required to prove that:
$$P_{m+2}(x)=x^2\left(P_{m+1}(x)-\frac{\mathrm{d}}{\mathrm{d}x}P_{m+1}(x)\right).$$
While trying to simplify the above expression, I found that$$ P_{m+2}(x)=x^2\left(\left(-\frac{\mathrm{d}}{\mathrm{d}x}P_m(x)\right)(x^2-2x+1)+P_m(x)\cdot (x^2-2x)+x^2\frac{\mathrm{d}^2}{\mathrm{d}x^2}P_m(x)\right). $$
But I am stuck with this $P_m(x)$. How do I get a substantial polynomial to start working with.
Also while finding $\dfrac{\mathrm{d}^n}{\mathrm{d}t^n}f(t)$, I get no pattern! I tried it by putting $t=\dfrac{1}{z}$ but that was a stupid decision anyway!
Solution:
$\dfrac{d}{dt}\cdot (\dfrac{d^n}{dt^n}f(t))=\dfrac{d}{dt} (P_n(\dfrac{1}{t})e^{-\dfrac{1}{t}})=e^{-\dfrac{1}{t}}\dfrac{d}{dt}P_n(\dfrac{1}{t})+P_n(\dfrac{1}{t})\cdot \dfrac{e^{-\dfrac{1}{t}}}{t^2} $
$\implies \dfrac{d^{n+1}}{dt^{n+1}}f(t)=e^{-\dfrac{1}{t}}\dfrac{d}{dt}P_n(\dfrac{1}{t})+P_n(\dfrac{1}{t})\cdot \dfrac{e^{-\dfrac{1}{t}}}{t^2}$
$\implies P_{n+1}(\dfrac{1}{t})\cdot e^{-\dfrac{1}{t}}=e^{-\dfrac{1}{t}}\dfrac{d}{dt}P_n(t)+P_n(t)\cdot \dfrac{e^{-\dfrac{1}{t}}}{t^2}$
Cancelling $e^{-\dfrac{1}{t}} $ from both sides, we get,
$\implies P_{n+1}(\dfrac{1}{t})=\dfrac{d}{dt}P_n(\dfrac{1}{t})+P_n(\dfrac{1}{t})\cdot \dfrac{1}{t^2}$
$P_{n+1}(\dfrac{1}{t})=\dfrac{d}{d(\dfrac{1}{t})}P_n(\dfrac{1}{t})+P_n(\dfrac{1}{t})\cdot \dfrac{1}{(\dfrac{1}{t})^2}$
Again $\dfrac{d}{d(\dfrac{1}{t})}P_n(\dfrac{1}{t})=-t^2\dfrac{d}{dt}P_n(\dfrac{1}{t})$
$\implies P_{n+1}(\dfrac{1}{t})=t^2 P_n(\dfrac{1}{t})-t^2\dfrac{d}{dt} P_n(\dfrac{1}{t})$
$\implies P_{n+1}(t)=t^2 P_n(t)-t^2\dfrac{d}{dt} P_n(t)$
Hence proved.
If you are looking for an explicit formula note that $$P_n(x)=\sum_{k=1}^{n}\binom{n}{k}\frac{(n-1)!}{(k-1)!}(-x)^{n+k}.$$ The above formula verifies the given recurrence $$\begin{align}x^2\left(P_n(x)-\frac{\mathrm{d}}{\mathrm{d}x}P_n(x)\right)&= \sum_{k=1}^{n}\binom{n}{k}\frac{(n-1)!}{(k-1)!}(-x)^{n+k+2}+\sum_{k=1}^{n}\binom{n}{k}\frac{(n-1)!}{(k-1)!}(n+k)(-x)^{n+k+1}\\ &=\sum_{k=2}^{n+1}\binom{n}{k-1}\frac{(n-1)!}{(k-2)!}(-x)^{n+1+k}+\sum_{k=1}^{n}\binom{n}{k}\frac{(n-1)!}{(k-1)!}(n+k)(-x)^{n+1+k}\\ &=\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{n!}{(k-1)!}(-x)^{n+1+k}=P_{n+1}(x).\end{align}$$