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 $n$.

  • 2
    $\begingroup$ I assume you want the limits to be as $x\to 0^+$ and $x\to\infty$, correct? $\endgroup$ – J. Loreaux Aug 30 '12 at 2:38
  • $\begingroup$ Or pointwise convergence? $\endgroup$ – Tunococ Aug 30 '12 at 2:41
  • 1
    $\begingroup$ I wonder if it's easier to work with the logarithm, $(\log x)/x$. $\endgroup$ – Gerry Myerson Aug 30 '12 at 2:42
  • $\begingroup$ Ok, what I said didn't make sense since $n \to 0^+$ doesn't make sense. $\endgroup$ – Tunococ Aug 30 '12 at 2:56
  • 1
    $\begingroup$ 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. $\endgroup$ – Eric Naslund Aug 30 '12 at 3:29

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.

  • $\begingroup$ 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 $y=\ln x$. $\endgroup$ – Roman Chokler Aug 30 '12 at 16:29
  • $\begingroup$ 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 $n\leq k$. $\endgroup$ – 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[
      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 $\exp(H_n)$:

enter image description here


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.

  • 6
    $\begingroup$ There are some errors/typos with how you took the log. $x^f\neq e^{\log f}$. $\endgroup$ – Eric Naslund Aug 30 '12 at 7:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.