# Calculus conjecture

When I was a senior in high school in 2001, as I took calculus, I made the following conjecture that proves resistive to attack. It goes like this: For every positive integer $n$, there are exactly $n$ positive real zeroes of $$\frac {d^n}{dx^n}x^{1/x}$$ and no two derivatives has a common root.

A few things to keep in mind are that it is not hard to show that $$\lim_{x\rightarrow 0^+}\frac {d^n}{dx^n}x^{1/x}=\lim_{x\rightarrow \infty}\frac {d^n}{dx^n}x^{1/x}=0$$ for all $n\geq 1$ and that $x^{1/x}$ is smooth and then to use these to show that the $n$th derivative has at least $n$ positive real zeroes. The proof of smoothness uses induction on $n$.

• I assume you want the limits to be as $x\to 0^+$ and $x\to\infty$, correct? – J. Loreaux Aug 30 '12 at 2:38
• Or pointwise convergence? – Tunococ Aug 30 '12 at 2:41
• I wonder if it's easier to work with the logarithm, $(\log x)/x$. – Gerry Myerson Aug 30 '12 at 2:42
• Ok, what I said didn't make sense since $n \to 0^+$ doesn't make sense. – Tunococ Aug 30 '12 at 2:56
• The $n^{th}$ derivative will be an exponential mutiplied by a two variable polynomial in $x$ and $\log x$. The highest power of $\log x$ will be $n$, so we can look at this as a polynomial in $\log x$ of degree $n$, but there are these extra other $x$'s hanging around. If we could ignore these other $x$'s since they don't move two much around zero, which might be possible using some variable changes and complex analysis stuff, we could show that this polynomial has exactly $n$ zeros, and none outside of some disk. However, to show everything is distinct seems extremely difficult to me. – Eric Naslund Aug 30 '12 at 3:29

## 3 Answers

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.

• Good Job on proving the weaker statement. For the stronger statement, the idea I always had was to view the problem as a a system of two equations, $p(x,y)=0$ and $y=\ln x$. – Roman Chokler Aug 30 '12 at 16:29
• Using this interpretation, case $n=1$ is trivial that the only root is $x=e$ and for $n=2$ case is not hard to prove that there are exactly two roots, for higher $n$ it gets increasingly tough though. One nice property of my conjecture is the above proof of the weaker version implies $D^{n+1}(x^{1/x})$ has more roots than $D^n(x^{1/x})$ so that if the strong version is true for $n=k$, then it is true for all $n\leq k$. – Roman Chokler Aug 30 '12 at 16:40

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 $\exp(H_n)$: Try logarithmic differentiation:

$$\frac{d}{dx} \left[ x^{1/x}\right] = \frac{d}{dx} \exp\left[ \frac{\ln x}{x}\right] = \frac{\frac{1}{x^2} - \frac{\ln x}{x^2} }{\exp\left[ \frac{\ln x}{x}\right] } = x^{-2}\cdot (1 - \ln x) \cdot x^{-1/x}$$

What about with a general $f$? $$\frac{d}{dx} \left[ x^f\right] = \frac{d}{dx} e^{\ln f}=(\log f)'x^f$$ or the 2nd derivative: $$\frac{d^2}{dx^2} \left[ x^f\right] = \frac{d^2}{dx^2} \exp\left[ \ln f\right] = \left[(\log f)'' + [(\log f)']^2\right]e^{\ln f}$$ let's try the 3rd derivative $$\frac{d^3}{dx^3} \left[ x^f\right] = \frac{d^3}{dx^3} \exp\left[ \ln f\right] = \left[(\log f)'''+ 3(\log f)''[(\log f)']^2 + [(\log f)']^3\right]e^{\ln f}$$ So there is some sort of Pascal's triangle in the coefficients: \begin{array}{ccc} 1 \\\\ 1 & 1 \\\\ 1 & 3 & 1 \\\\ \end{array} Maybe it's possible to compute the general rows? This might be Sloane sequence A208510 related to the "(2,1) Pascal triangle" or Lucas Triangle

Then the various derivatives of $\log f$ as encoded by the Taylor expansion, may have our answer.

• There are some errors/typos with how you took the log. $x^f\neq e^{\log f}$. – Eric Naslund Aug 30 '12 at 7:35