Calculus conjecture

Put $$a_n(x):=D^n\bigl(x^{1/x}\bigr)\qquad (n\geq 0, \ x>0)$$ and $$b_n(x)=x^{2n} a_n(x) x^{-1/x}\ .$$ Then $b_n(x)$ is the polynomial in $x$ and $\log x$ Eric Naslund alluded to in his comment. The $b_n$ satisfy the recursion $$b_0(x)\equiv1\ ,\qquad b_{n+1}(x)=(1-\log x - 2nx) b_n(x)+ x^2 b_n'(x)\quad (n\geq0)\ .$$ In 2001 the average high school student didn't have Mathematica or similar at his disposal, but now we do. Graphing the first twenty $b_n$'s, resp. their signs, could corroborate the conjecture, or might provide a counterexample.

Looking at the resulting expressions for $b_n(x)$ and at the corresponding graphs I have come to the following conclusion:

For $x\to0\!+$ the worst term is $(-\log x)^n$; therefore $\lim_{x\to0+} b_n(x)=\infty$ for all $n\geq1$.

For $x\to\infty$ the largest term is of the form $(-1)^n c_n x^{n-1}\log x$ with a positive constant $c_n$ (one should be able to prove this by induction). Therefore $\lim_{x\to\infty} b_n(x)=(-\infty)^n$.

Now ${\rm sgn}\bigl(a_n(x)\bigr)={\rm sgn}\bigl(b_n(x)\bigr)$. Therefore we have $\lim_{x\to 0+}{\rm sgn}\bigl(a_n(x)\bigr)=1$ and $\lim_{x\to \infty}{\rm sgn}\bigl(a_n(x)\bigr)=(-1)^n$.

As $a_{n+1}=a_n'$, by Rolle's theorem between any two zeros of $a_n$ there is at least one zero of $a_{n+1}$. Let $\xi$ be the first of these intermediate zeros. It belongs to a local minimum of $a_n$, so $a_{n+1}=a_n'$ is increasing at $\xi$. This implies that immediately to the left of $\xi$ the function $a_{n+1}$ is negative, and as $\lim_{x\to 0+}{\rm sgn}\bigl(a_{n+1}(x)\bigr)=1$ we are guaranteed at least one zero of $a_{n+1}$ to the left of $\xi$. Similarly we get at least one additional zero at the right end.

What all this means is that $a_n$ has at least $n$ zeros.

The following is simply a comment, grown large.

Since $x^{1/x} = \exp\left(\frac{\log(x)}{x} \right)$, Bell polynomial can be used to expresses the $n$-th order derivative using Faà di Bruno's formula: $$ \frac{\mathrm{d}^n}{\mathrm{d}x^n} f(g(x)) = \sum_{k=1}^n f^{(k)}(g(x)) B_{n,k}\left(g^\prime(x),g^{\prime\prime}(x),\ldots, g^{(n)}(x)\right) $$ Taking $f = \exp$ and $g(x) = \frac{\log(x)}{x}$ and using $f^\prime = f$, as well as a known result: $$ \frac{\mathrm{d}^n}{\mathrm{d}x^n} \frac{\log(x)}{x} = (-1)^n\frac{n!}{x^{n+1}} \left(\log(x) - H_n\right) $$ we have: $$ \frac{\mathrm{d}^n}{\mathrm{d}x^n} x^{1/x} = x^{1/x} \sum_{k=1}^n B_{n,k}\left(-\frac{\log(x)-1}{x},\ldots, (-1)^n\frac{n!}{x^{n+1}} \left(\log(x) - H_n\right) \right) $$ Using well known homogeneity properties of the Bell polynomial: $$\begin{eqnarray} B_{n,k}\left( \lambda y_1, \lambda^2 y_2, \ldots, \lambda^n y_n\right) &=& \lambda^n B_{n,k}\left( y_1, y_2, \ldots, y_n\right) \\ B_{n,k}\left( \lambda y_1, \lambda y_2, \ldots, \lambda y_n\right) &=& \lambda^k B_{n,k}\left( y_1, y_2, \ldots, y_n\right) \end{eqnarray} $$ the expression can be further simplified: $$ \frac{\mathrm{d}^n}{\mathrm{d}x^n} x^{1/x} = (-1)^n x^{1/x-n} \sum_{k=1}^n \frac{1}{x^k} B_{n,k}\left(\log(x)-1,\ldots, n! \left(\log(x) - H_n\right) \right) $$ Bell polynomials are multivariate polynomials with positive coefficients. Notice that this implies that all the real roots are bounded from above by $\exp(H_n)$.

Here are few low order Bell polynomials: $$ \begin{array}{c|ccccc} n\backslash k & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & y_1 & \text{} & \text{} & \text{} & \text{} \\ 2 & y_2 & y_1^2 & \text{} & \text{} & \text{} \\ 3 & y_3 & 3 y_1 y_2 & y_1^3 & \text{} & \text{} \\ 4 & y_4 & 3 y_2^2+4 y_1 y_3 & 6 y_1^2 y_2 & y_1^4 & \text{} \\ 5 & y_5 & 10 y_2 y_3+5 y_1 y_4 & 10 y_3 y_1^2+15 y_2^2 y_1 & 10 y_1^3 y_2 & y_1^5 \end{array} $$ Notice that this implies that derivatives of $x^{1/x}$ evaluated at $x=1$ are all integers. In fact this is known as sequence A008405.

Also, here are few zeros computed explicitly in Mathematica: $$ \begin{array}{c|llllll} n & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ \hline 1 & \mathrm{e} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2 & 0.581933 & 4.36777 & \text{} & \text{} & \text{} & \text{} \\ 3 & 0.342086 & 0.826389 & 6.00406 & \text{} & \text{} & \text{} \\ 4 & 0.246206 & 0.453114 & 1.05675 & 7.63532 & \text{} & \text{} \\ 5 & 0.19403 & 0.312965 & 0.555416 & 1.28152 & 9.26413 & \text{} \\ 6 & 0.16101 & 0.239864 & 0.373541 & 0.654014 & 1.5034 & 10.8915 \\ \end{array} $$ The following was the command used:

zeros = N[
   Table[Exp[
      t] /. {ToRules[
       Reduce[0 == 
         BellY[Table[{1, n! Exp[-t] (t - HarmonicNumber[n])}, {n, 1, 
            m}]], t, Reals]]}, {m, 1, 6}]];

Notice that the largest root is quite close to the upper bound $\exp(H_n)$, here are few values, side-by-side, the upper row being actual larges zeros for $1 \leqslant n \leqslant 12$ and the lower row being approximate values of $\exp(H_n)$: