The $n$-th derivative has $n$ zeros. Can such a function be unbounded? I asked this on Math.SE some days ago, but without any success. For some application I need a formal definition of bell-shaped function. So I had the following idea:

Definition. A $C^\infty$-function $f:\Bbb R\to\Bbb R$ should be called bell-shaped if for all $n\geq 1$ the $n$-th derivative has exactly $n$ zeros (counted with multiplicity).

E.g. this works for $\exp(-x^2)$. As this is my try to formalize an informal term, I have no way to check if this is correct. However, I tried to prove that such functions must be bounded (for me, intuitively bell-shaped functions are always bounded). But I had no success.
Question: Is a bell-shaped function (in the sense above) always bounded?
 A: If I remember correctly, $\sqrt{1 + x^2}$ is bell-shaped, and it is not bounded. (The $n$-th derivative is a polynomial of degree $n$ times $(1 + x^2)^{1/2-n}$, and it has at least $n$ zeroes by the intermediate value theorem).
By the way, is there any progress on the characterisation of bell-shaped functions?
A: As suggested by Mateusz Kwaśnicki, the function $f : x \mapsto (1+x^2)^{s}$ is bell-shaped and unbounded for any $s \in (0,\frac{1}{2})$.
It is easy to see that $f^{(n)}(x) = P_n(x) (1+x^2)^{s-n}$ where $P_n$ is a polynomial of degree $\leq n$. Actually
$$
P_{n+1}(x) = (1+x^2) P_n'(x) - 2(n-s)xP_n(x).
$$
Let $a_n$ be the coefficient of $x^n$ in $P_n$. Then $a_{n+1} = n a_n - 2(n-s)a_n = (2s-n) a_n$. In particular $(-1)^{n-1} a_n > 0$ for $n \geq 1$, so that $P_n$ has degree exactly $n$ (this is where we use $s < \frac{1}{2}$).
One first checks that $f'$ has a single zero.
We then assume that $f^{(n)}$ has exactly $n$ distinct simple zeroes $x_n < x_{n-1} < \dots < x_1$ for some $n \geq 1$. Then
$$
(-1)^{n-1} f^{(n+1)}(x_1) = (-1)^{n-1} P_n'(x_1) (1+x_1^2)^{s-n} > 0
$$
since $(-1)^{n-1} P_n(x) > 0$ for $x > x_1$. But $(-1)^{n-1} f^{(n+1)}(x) < 0$ for large $x$ so $f^{(n+1)}$ has a zero in $(x_1,+\infty)$. Similarly $f^{(n+1)}$ has a zero in $(-\infty ,x_n)$. Together with the zeroes of $f^{(n+1)}$ between the $x_i$'s, we get $\geq n+1$ distinct zeroes (and therefore exactly $n+1$ zeroes).
