The finite-difference stability conjecture for stable polynomials

About 7 years old · traced to 2019

Let $p$ be a stable polynomial, meaning that all of its zeros lie in the open left half-plane. For parameters $\theta$ and $h$ as in the operator $]\Delta_{\theta,h}]$ , and for integers $m=1,\ldots,\deg p-1$ , consider the iterates

\Delta_{\theta,h}^m(p)(z).

Finite-difference stability conjecture. For any stable polynomial $p$ , the polynomials

\Delta_{\theta,h}^m(p)(z),\qquad m=1,\ldots,\deg p-1,

have no nonreal zeroes in the closed right half-plane. The conjecture concerns the action of the complex zero-decreasing operator on stable polynomials; calculations suggest it, but the source gives no resolution.

References

Primary source

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

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

An unverified posted calculation claims a simple counterexample, but no independent source confirms that the conjecture is false.

Katkova, Tyaglov, and Vishnyakova stated this as Conjecture 1 in a 2019 paper: for stable $p$ , every permitted iterate $\Delta_{\theta,h}^{m}(p)$ should have no nonreal zero in the closed right half-plane. Their paper reports supporting calculations but no resolution.

Known results

The operator $\Delta_{\theta,h}$ is complex zero-decreasing, and the paper proves several related root-preservation and root-location results, but not this conjecture (Katkova, Tyaglov, and Vishnyakova, 2019).

Posted attempt

An explicit quadratic calculation claims a counterexample: $p(z)=z^2+2z+10$ has zeros $-1\pm3i$ , while with $h=2$ and $\theta=3\pi/4$ , $\Delta_{\theta,h}p(z)=(z^2-2z+2)/\sqrt{2}$ has zeros $1\pm i$ . This is a claimed complete disproof, but it has not been independently verified.

Current status (as of August 2026): The conjecture has an explicit but unverified counterexample claim; absent independent confirmation, its resolution remains unsettled.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

Use the finite-difference operator in the normalization of the conjecture:

\Delta_{\theta,h}p(z) =\frac{e^{i\theta}p(z+ih)-e^{-i\theta}p(z-ih)}{2i}.

Take

p(z)=z^2+2z+10=(z+1)^2+9,\qquad h=2,\qquad \theta=\frac{3\pi}{4}.

The zeros of $p$ are $-1\pm3i$ , so $p$ is strictly Hurwitz stable.

Since

p(z+2i)=z^2+2z+6+i(4z+4),

direct substitution gives

\Delta_{3\pi/4,2}p(z) =\frac{z^2-2z+2}{\sqrt2}.

The zeros of the resulting polynomial are

z=1+i,\qquad z=1-i,

both in the open right half-plane.

Thus the first permitted iterate $m=1=\deg(p)-1$ already violates the conjectured stability. This provides a counterexample of the minimum possible degree.