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The self-interlacing conjecture for the geometric-sum polynomial

From papers

Let

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

Let Δθ,h\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)p(x) strictly interlace those of p(−x)p(-x).

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

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

Progress summary

Open

The conjecture remains open: its original paper records it as a calculation-based expectation, with no publicly reported proof or verified counterexample.

The conjecture concerns whether applying the finite-difference operator Δθ,h\Delta_{\theta,h} to the geometric-sum polynomial produces a self-interlacing polynomial, possibly after multiplication by xx. The directly relevant paper presents this as a conjecture motivated by calculations, not as a theorem, and gives no resolution.

Current status (as of August 2026): The conjecture is unsettled; the retrieved source contains only its formulation and motivation, with no verified proof, counterexample, or claimed resolution.

Sources
Sources & referencesView supporting material

Primary source

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

finite-difference-operatorsinterlacing-polynomialsreal-rootedness

Solutions 1

Counterexample

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

p(z)=1+z+z2+z3.p(z)=1+z+z^2+z^3.

Choose the admissible parameters

θ=0,h=12.\theta=0,\qquad h=\frac12.

By the definition of the finite-difference operator,

Δ0,hp(z)=p(z+ih)−p(z−ih)2i=h(3z2+2z+1−h2).\Delta_{0,h}p(z) =\frac{p(z+ih)-p(z-ih)}{2i} =h(3z^2+2z+1-h^2).

Consequently,

Δ0,1/2p(z)=12z2+8z+38.\Delta_{0,1/2}p(z)=\frac{12z^2+8z+3}{8}.

The discriminant of its numerator is

82−4⋅12⋅3=−80<0.8^2-4\cdot12\cdot3=-80<0.

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

More generally, every

0<h<230<h<\sqrt{\frac23}

gives a counterexample, because the discriminant of 3z2+2z+1−h23z^2+2z+1-h^2 is

4(3h2−2)<0.4(3h^2-2)<0.

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

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Shivam Patel ·