Given a polynomial $P(x)=\sum_{n=0}^{d}a_nx^n\in\mathbb{R}[x]$ with all roots on the unit circle.

Question: Is it true that all the roots of $Q(x)=\sum_{n=0}^{d}a_n{{x+d-n}\choose{d}}$ lie on a straight line? If it is not true,is it possible to give a counterexample?

  • $\begingroup$ Very interesting. It appears they always lie on the line $x=-1/2$. Here's a plot. They do not seem to be bounded vertically. $\endgroup$ – Antonio Vargas May 1 '12 at 15:57
  • 1
    $\begingroup$ What makes you think the roots lie on a straight line? $\endgroup$ – Did May 1 '12 at 18:47
  • 2
    $\begingroup$ How did this question originate? $\endgroup$ – Olivier Bégassat May 1 '12 at 21:38
  • 2
    $\begingroup$ @Didier @ Olivier Bégassat: In fact,the original question is: Given a polynomial $P(x)$ that all the roots lie on the unit circle. Consider $P(x)/(1-x)^{d+1}=\sum_{n=0}^{\infty}c_nx^n$(d is the degree of $P(x)$). It is quite easy to find an interpolating polynomial $Q(x)$ that $c_n=Q(n)$. Then it is natural to ask about the location of the zeros. I find this question on a math forum, and this question was entitled with "Easy Riemann Hypothesis". This question seems very interesting, but I don't know how to work it out. (It seems that $P(x)\in\mathbb{C}[x]$ is OK) $\endgroup$ – zy_ May 2 '12 at 1:41

With the help of Polya and Szego's book "Problems and Theorems in Analysis(II,Chapter 3,Problem 196.1)",I found this paper:


It is just what I want.


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.