Function such that zeros$=$order of the derivative

Question

Does there exist a function $f\in C^n(\mathbb{R},\mathbb{R})$ for $n\ge2$ such that $f^{(n)}$ has exactly $n$ zeros, $f^{(n-1)}$ has exactly $n-1$ zeros and so on ? Where $f^{(n)}$ is the nth derivative of $f$

This question arose from lot of question asked here on MSE.

Answer 1

Yes, the function $f(x)=e^{-x^2}$ has this property; the $n$th derivative is $f^{(n)}(x)=(-1)^n H_n(x) \, e^{-x^2}$, where $H_n$ is the Hermite polynomial of degree $n$, which because of orthogonality has exactly $n$ real zeros (they have to interlace with the zeros of $H_{n-1}$).

Answer 2

The function $f(x)=2-\frac{1}{1+x^2}$ has the desired property for at least $0 \le n \le 7.$ The derivatives $f^{(n)}$ (for $n>0$) each have a denominator a power of $1+x^2$, and a constant in front of a polynomial $p_n(x)$ which for odd $n$ hazs a factor of $x$ and the rest is in powers of $x^2$, whereas for even $n$ there is no factor of $x$ in $p_n$ and the rest is in even powers of $x$.

$p_1=x$

$p_2=3x^2-1$

$p_3=x(x^2-1)$

$p_4=5x^4-10x^2+1$

$p_5=x(3x^4-10x^2+3)$

$p_6=7x^6=35x^4+21x^2-1$

$p_7=x(x^6-7x^4+7x^2-1)$

That's where I stopped; to check for example that $p_6$ has $6$ zeros I rewrote $x^2$ as $t$ and looked at the cubic, which at $t=0,0.1,1,0,5,0$ has values respectively $-1,+0.7,-8,+104$ so that there are three positive zeros for the $t$ polynomial, hence $6$ zeros for $p_6$ And $p_7$ was easy sonce the cubic here factors as $(t-1)(t^2-6t+1)$, which has $3$ positive zeros, so that (recalling the factor $x$) $p_7$ ends up with seven zeros.

The pattern here makes it seem possible that this function $f(x)$ has simultaneously the desired numbers of zeros for each derivative; however I don't see any good way to show that, by induction or otherwise.