Show that \begin{equation} f(t) = \begin{cases} e^{-1/t} & x > 0 \\ 0 & x \le 0 \end{cases} \end{equation} is in $C^{\infty}$.
To show that this is true, I think we have to show that $f^{(k)}$ is continuous for all $k \in \mathbb{N}$, where $f^{(k)}$ is the $k$-th derivative of $f$ ($f^{(0)} := f$).
Since the only point where these could be discontinuous is at $x = 0$, I think I need to show that \begin{equation} f^{(k)}(0) = 0 \end{equation} for all $k$.
I've calculated the first few derivatives in order to see a pattern emerge and create a closed form, but I can't do it. Here are the derivatives.
\begin{align} f^{(0)} & = e^{-1/t}\\ f^{(1)} & = {\frac {1}{{t}^{2}}{{e}^{-1/t}}}\\ f^{(2)} & = -2\,{\frac {1}{{t}^{3}}{{e}^{-1/t}}}+{\frac {1}{{t}^{4}}{{e}^{-1/t}}}\\ f^{(3)} & = 6\,{\frac {1}{{t}^{4}}{{e}^{-1/t}}}-6\,{\frac {1}{{t}^{5}}{{e}^{-1/t}}}+{\frac {1}{{t}^{6}}{{e}^{-1/t}}}\\ f^{(4)} & = -24\,{\frac {1}{{t}^{5}}{{e}^{-1/t}}}+36\,{\frac {1}{{t}^{6}}{{e}^{-1/t}}}-12\,{\frac {1}{{t}^{7}}{{e}^{-1/t}}}+{\frac {1}{{t}^{8}}{{e}^{-1/t}}}\\ \end{align}
I can see that \begin{equation} f^{(k)}(t) = f^{(0)}(t) \sum_{i = k+1}^{2k} (-1)^{i}t^{-i} c_i \end{equation}
where $c_i \stackrel{?}{=} k!$ or something like that. So in fact, I just need to find an expression for $c_i$. Any ideas?