The self-interlacing conjecture for the geometric-sum polynomial

About 7 years old · traced to 2019

Let

p(x)=x^n+x^{n-1}+\cdots+x+1=\dfrac{x^{n+1}-1}{x-1}.

Let $\Delta_{\theta,h}$ be the finite difference operator from the paper, and let a polynomial be self-interlacing when it has real simple zeros and the zeros of $p(x)$ strictly interlace those of $p(-x)$ .

Self-interlacing conjecture. The polynomial $\Delta_{\theta,h}(p)(x)$ is self-interlacing, or is a self-interlacing polynomial multiplied by $x$ .

The claim is motivated by calculations concerning the action of the operator on subclasses of complex polynomials. The source gives no proof or resolution.

References

Primary source

Olga Katkova, Mikhail Tyaglov and Anna Vishnyakova, “Hermite-Poulain theorems for linear finite difference operators”, arXiv:1901.06398 (2019).

finite-difference-operators interlacing-polynomials real-rootedness

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-written calculation claims the conjecture is false already in three dimensions, but the counterexample has not been independently verified.

Katkova, Tyaglov, and Vishnyakova state this assertion as Conjecture 2 in their 2019 paper: applying $\Delta_{\theta,h}$ to the geometric-sum polynomial should produce a self-interlacing polynomial, possibly after multiplication by $x$ . The paper presents calculations motivating the conjecture but gives no proof or resolution.

Posted attempt

A reader-written calculation claims a counterexample at $n=3$ , $\theta=0$ , and $h=\frac{1}{2}$ . It obtains $\Delta_{0,h}p(z)=h(3z^2+2z+1-h^2)$ , whose discriminant is negative for $0<h<\sqrt{\frac{2}{3}}$ , so the output has nonreal zeros and violates the proposed dichotomy. This is a claimed complete disproof, but it has not been independently verified.

Current status (as of August 2026): the original conjecture has no verified proof, while a reader-written calculation claims a counterexample for $n=3$ ; the conjecture therefore remains unsettled pending verification of that calculation.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

Take $n=3$ , and consider the geometric-sum polynomial

p(z)=1+z+z^2+z^3.

Choose the admissible parameters

\theta=0,\qquad h=\frac12.

By the definition of the finite-difference operator,

\Delta_{0,h}p(z) =\frac{p(z+ih)-p(z-ih)}{2i} =h(3z^2+2z+1-h^2).

Consequently,

\Delta_{0,1/2}p(z)=\frac{12z^2+8z+3}{8}.

The discriminant of its numerator is

8^2-4\cdot12\cdot3=-80<0.

Therefore this polynomial has two nonreal zeros. In particular, it is neither self-interlacing nor $z$ times a self-interlacing polynomial.

More generally, every

0<h<\sqrt{\frac23}

gives a counterexample, because the discriminant of $3z^2+2z+1-h^2$ is

4(3h^2-2)<0.

Thus the proposed dichotomy fails already for $n=3$ and $\theta=0$ .