## § Flat functions

Define
$f(x) \equiv
\begin{cases}
0 & x <= 0 \\
e^{-1/x} & x > 0
\end{cases}$

This is smooth, but is badly non-analytic. Any taylor expansion around $x=0$
is going to be identically zero. So we're going to prove that it possesses
all derivatives. This implies that the derivative at zero is equal to zero,
because the left derivative is always equal to zero.
- $f(x) \equiv e^{-1/x}$. Differentiate to get $f'(x) = e^{-1/x}/x^2$. Change $y \equiv 1/x$ to get $y^2 e^{-y}$. As $y \mapsto \infty$, $e^{-y}$decays more rapidly than $y^2$ increases, thus the limit is zero. Hence, $f'(x) = 0$.
- For higher derivatives, let $f^{(n)}(x) \equiv p_n(1/x) e^{-1/x}$ for some polynomial $p_n$. See that $f^{(n+1)}(x) = d/dx [p_n(1/x) e^{-1/x})]$. To compute this, set $y \equiv 1/x$and compute $d/dy [p_n(y) e^{-y}] dy/dx$ which is:

$\begin{aligned}
& = d/dy [p_n(y) e^{-y}] dy/dx \\
& = p_n'(y) e^{-y} + p_n(y) (- e^{-y}) \cdot 1/x^2 \\
& = e^{-1/x} (p_n'(1/x) - p_n(1/x)) 1/x^2 \\
& = e^{-1/x} (q_n'(1/x) - q_n(1/x)) \\
& \text{let $r_{n+1}(x) (\equiv q_n'(t) - q_n(t))t^2$} \\
& = r_{n+1}(1/x) e^{-1/x}
\end{aligned}$

- So we can write higher derivatives too as $poly(1/x)$ times $exp(-1/x)$ which also decays rapidly to $0$.

- Philosophically, what's going on is that a non-zero polynomial can only have a finite number of zeroes. Since this function has an infinite number of zeroes around it's neighbourhood at $(x = 0)$, any polynomial that agrees with this function in any neighbourhood must be identically zero everywhere.