Proving that $f^{(n)}(t) = p_n(1/t) e^{-1/t}, t\in(0, +\infty)$ where $p_n(x)$ is a polynomial of degree $\ge 2$.

Question

I want to prove that the n-th derivation of $f(t)=e^\frac{-1}{t}, t\in \mathbb{R}, t \gt 0$ is

$f^{(n)}(t) = p_n(1/t) e^{-1/t}, t\in(0, +\infty)$ where $p_n(x)$ is a polynomial of degree $\ge 2$ by induction.

Assuming it's true for $n$, I want to show that it's also true for $n+1$, which means that $f^{(n+1)}(t) = p_{n+1}(1/t) e^{-1/t}$. But I can't get any further.

$f^{(n+1)}(t)={f^{(n)}(t)}'={p_n(1/t)e^{-1/t}}'= p_{n-1}(1/t)e^{-1/t}+p_n(1/t)1/t^2e^{-1/t}=e^{-1/t}(p_{n-1}(1/t)+p_n(1/t)1/t^2) = ?)$

Answer 1

By induction $P_0=1$ has degree $0$ since $f(t) =e^{1/t}$

Now Set $X= 1/t$ , and assume that $P_n$ is of degree $2n$ then after computation we have $$f^{(n+1)}(t)= e^{-1/t}(p_{n-1}(1/t)+p_n(1/t)1/t^2 \\=e^{-1/t}(p_{n-1}(X)+p_n(X)X^2 )= e^{-1/t}p_{n+1}(X) $$

clearly $$p_{n+1}(X)=p_{n-1}(X)+p_n(X)X^2 $$ is a polynomial degree $2n+2$ and $$f^{(n+1)}(t)= e^{-1/t} p_{n+1}(1/t)$$

Answer 2

First we show that if $f(t)=e^{-\dfrac{1}{t}}$ for $t>0$ and if $(\dfrac{d^n}{dt^n}f(t))= P_{n}(\dfrac{1}{t})\cdot e^{-\dfrac{1}{t}} $

Then:

$\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,

$\implies P_{n+1}(\dfrac{1}{t})=\dfrac{d}{dt}P_n(\dfrac{1}{t})+P_n(\dfrac{1}{t})\cdot \dfrac{1}{t^2}$

$P_{n+1}(t)=\dfrac{d}{d(\dfrac{1}{t})}P_n(t)+P_n(t)\cdot \dfrac{1}{(\dfrac{1}{t})^2}$(by replacing $t$ with $\dfrac{1}{t}$)

Again $\dfrac{d}{d(\dfrac{1}{t})}P_n(t)=-t^2\dfrac{d}{dt}P_n(t)$

$\implies P_{n+1}(t)=t^2 P_n(t)-t^2\dfrac{d}{dt} P_n(t)$

Now as we know that the above is true iff $f(t)=e^{-\dfrac{1}{t}}$ for $t>0$ and $(\dfrac{d^n}{dt^n}f(t))= P_{n}(\dfrac{1}{t})\cdot e^{-\dfrac{1}{t}} $ is true.

If somehow you can prove by induction $P_{n+1}(t)=t^2 P_n(t)-t^2\dfrac{d}{dt} P_n(t)$, you can arrive at $(\dfrac{d^n}{dt^n}f(t))= P_{n}(\dfrac{1}{t})\cdot e^{-\dfrac{1}{t}} $