TITLE: Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
QUESTION [50 upvotes]: Set $f(x) = \exp(-x-x^{-1})$. An easy induction shows that
$$\frac{d^n}{(dx)^n} f(x) = \phi_n(x^{-1}) f(x)$$
for $\phi_n$ a polynomial of degree $2n$. Clearly, the roots of $\phi_n(x^{-1})$ are the same as the roots of $f^{(n)}(x)$ and, by repeated use of Rolle's theorem, $f^{(n)}(x)$ has at least $n$ roots in $(0, \infty)$. Since $\phi_n$ has degree $2n$, there is room for many more positive real roots than that.
However, for $n \leq 50$, computer computations show that $\phi_n$ has exactly $n$ positive real roots! Why?
Motivation: Nothing really, I was just thinking about this question and fooling around.
Data: Here are the first $10$ values of $\phi_n(y)$:
1, -1 + y^2, 1 - 2 y^2 - 2 y^3 + y^4, -1 + 3 y^2 + 6 y^3 + 3 y^4 - 6 y^5 + y^6, 1 - 4 y^2 - 12 y^3 - 18 y^4 + 32 y^6 - 12 y^7 + y^8, -1 + 5 y^2 + 20 y^3 + 50 y^4 + 60 y^5 - 50 y^6 - 180 y^7 + 115 y^8 - 20 y^9 + y^10, 1 - 6 y^2 - 30 y^3 - 105 y^4 - 240 y^5 - 200 y^6 + 540 y^7 + 1095 y^8 - 1080 y^9 + 294 y^10 - 30 y^11 + y^12, -1 + 7 y^2 + 42 y^3 + 189 y^4 + 630 y^5 + 1295 y^6 + 420 y^7 - 5075 y^8 - 7140 y^9 + 10521 y^10 - 3990 y^11 + 623 y^12 - 42 y^13 + y^14, 1 - 8 y^2 - 56 y^3 - 308 y^4 - 1344 y^5 - 4256 y^6 - 7560 y^7 + 3430 y^8 + 48160 y^9 + 48664 y^10 - 108360 y^11 + 53788 y^12 - 11424 y^13 + 1168 y^14 - 56 y^15 + y^16, -1 + 9 y^2 + 72 y^3 + 468 y^4 + 2520 y^5 + 10668 y^6 + 31752 y^7 + 45234 y^8 - 83160 y^9 - 478674 y^10 - 330120 y^11 + 1186836 y^12 - 742392 y^13 + 201132 y^14 - 27720 y^15 + 2007 y^16 - 72 y^17 + y^18, 1 - 10 y^2 - 90 y^3 - 675 y^4 - 4320 y^5 - 22800 y^6 - 93240 y^7 - 256830 y^8 - 246960 y^9 + 1272348 y^10 + 5033700 y^11 + 1965810 y^12 - 13829760 y^13 + 10636800 y^14 - 3530520 y^15 + 614925 y^16 - 59760 y^17 + 3230 y^18 - 90 y^19 + y^20

The other roots are shrinking towards $0$ like $1/n$ while accumulating on some sort of curve. Here is a picture for $n=50$:

REPLY [5 votes]: Another solution, inspired by fedja's. This one also establishes Peter Mueller's stronger statement that the coefficients of $\phi_n(x)$ have $n$ sign changes (in the sense of Descartes' Rule of Signs).
For any Laurent polynomial $f(x) \in \mathbb{R}[x, x^{-1}]$, let $D(f)$ denote the number of sign changes in the coefficients of $f$.
Claim 1 For any integers $0 \leq r \leq m$,
$$D \left( \left( \frac{d}{dx} \right)^r (1-x/m)^m (1-x^{-1}/m)^m \right) \leq 2m-r.$$
Proof The only monomials with nonzero coefficient are $x^{-m-r}$, $x^{-m-r+1}$, ..., $x^{-r-1}$, $1$, $x$, ..., $x^{m-r}$, a total of $2m-r+1$ monomials. $\square$
Claim 2 For any Laurent polynomial $h(x)$ and any $a>0$, we have $D((x-a) h(x)) \geq D(h(x))+1$.
Proof This is a standard Lemma established in the course of proving Descartes' Rule of Signs. See, for example, this proof. $\square$
Claim 3
$$D \left(\frac{ \left( \frac{d}{dx} \right)^r (1-x/m)^m (1-x^{-1}/m)^m }{(1-x/m)^{m-r} (1-x^{-1}/m)^{m-r}} \right) \leq r.$$
This follows by $2m-2r$ uses of Claim 2.
Now, sending $m \to \infty$, we obtain
$$D\left( \frac{\left( \frac{d}{dx} \right)^r \exp(-x-x^{-1})}{\exp(-x-x^{-1})}  \right)\leq r$$
as desired.