At the CMFT international conference in Turkey (2009), the following open problem was given:

Show that $$p_n(x):=\sum_{k=0}^n \frac{(n-k)^k}{k!}x^{n-k}$$ has only real simple zeros for every $n$. Assume $p_0(x)=1$.

I have been struggling with this problem for days and as always my lack of knowledge of differential equations seems to get in the way. Letting $f^*(x)=x^{\deg(f)}\cdot f(\frac{1}{x})$, we derive the following $$\bigg(x\frac{d}{dx}p^*_{n+1}(x)\bigg)^*=x\frac{d}{dx}p_n(x) .$$ Following through the calculations we arrive at $$(n+1)p_{n+1}=xp'_{n+1}(x)+xp'_n(x).$$ Letting $F(x,t):=\sum_{k=0}^\infty p_k(x)t^k$, we arrive at the following differential equation $$t F_t = xF_x+xtF_x.$$ Please help me solve for $F$ to get the a generating function for $\{p_n\}$, that will hopefully help to solve the problem.

I think I have found the generating function. After using the formula in here provided by Raymond Manzoni and then guessing for hours. I arrived at $$\frac{1}{1-xte^t}=\sum_{k=0}^\infty p_k t^k,$$ which satisfies the differential equation and has the appropriate initial conditions. Still haven't been able to prove these have only real zeros.

  • $\begingroup$ $f$ is actually $F$? $\endgroup$ – Gerry Myerson Jul 14 '13 at 0:26
  • $\begingroup$ Yes, sorry for the typo. $\endgroup$ – Bobby Ocean Jul 14 '13 at 0:46
  • $\begingroup$ I can't help but notice that all of the zeros seem to lie in the interval $(-e,0]$. $\endgroup$ – Antonio Vargas Jul 14 '13 at 1:21
  • $\begingroup$ I don't see your result $(n+1)p_{n+1}=xp_{n+1}'+xp_n'$. This would imply that $x^{-n-1}p_{n+1}$ has the same (nonzero) extrema as $p_n$. But if the claim to be shown is true, it must have one more. $\endgroup$ – Hagen von Eitzen Jul 14 '13 at 10:00
  • $\begingroup$ I am certainly prone to error, however I checked the statement on Mathematica. Here is my calculation: $$(xDP_{n+1}^*)^*=\left(xD\left(\sum_{k=0}^{n+1} \frac{(n-k+1)^k}{k!}x^{n-k+1}\right)^*\right)^* = \left(xD\sum_{k=0}^{n+1} \frac{(n-k+1)^k}{k!}x^k\right)^* = \left(\sum_{k=1}^{n+1} \frac{(n-k+1)^k}{(k-1)!}x^k\right)^* = \sum_{k=1}^{n+1} \frac{(n-k+1)^k}{(k-1)!}x^{n-k+1}$$ likewise $$xDP_n = \sum_{k=1}^{n+1} \frac{(n-k+1)^k}{(k-1)!}x^{n-k+1}.$$ Next if you replace $p^*_{n+1}$ with $\frac{1}{x^{n+1}}p_{n+1}\left(\frac{1}{x}\right)$ and run through $xD$ and then $*$ again, we get the formula. $\endgroup$ – Bobby Ocean Jul 14 '13 at 19:57

Your Answer

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

Browse other questions tagged or ask your own question.