The simple-roots conjecture for finite difference operators

About 7 years old · traced to 2019

Let $p$ be a complex polynomial, and let $\Delta_{\theta,h}$ be the finite difference operator from the paper. The exceptional polynomials are those with multiple roots of the form $\pm nih$ , where $n\in\mathbb{N}$ .

Simple-roots conjecture. For almost all polynomials $p$ , the polynomial $\Delta_{\theta,h}(p)$ has only simple roots, apart from the stated exclusion.

The conjecture is motivated by the observation that most polynomials in the image of the operator have simple roots, while the source describes the exceptional multiple-root configurations. No resolution is given.

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 polynomial-roots

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

An unverified posted calculation claims the generic part is true but gives a counterexample to the proposed list of all exceptions, so the conjecture is not settled.

Katkova, Tyaglov, and Vishnyakova posed Conjecture $3$ in 2019: for almost all inputs, the finite-difference output has simple roots outside the stated exceptional configurations. The paper leaves the general claim open.

Known results

Katkova, Tyaglov, and Vishnyakova (2019): all roots are simple when the input lies in the Hermite–Poulain class $\mathcal{HP}$ .

Posted attempt

An unverified partial attempt argues that the generic assertion follows because the output discriminant is a nonzero polynomial in the input coefficients. It also claims the exception list is false, using $\theta=0$ , $h=1$ , and $p(z)=z^3+z$ , for which $\Delta_{0,1}p(z)=3z^2$ despite $p$ having distinct roots; the argument has not been independently verified.

Current status (as of August 2026): The paper’s special case is settled, while the general simple-roots assertion and the claimed counterexample to its exception characterization remain unverified.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

The generic simple-root assertion is true, but the proposed exhaustive characterization of its exceptions is false.

Fix $h\ne0$ , $\theta\in[0,\pi)$ , and the input degree $d$ . The condition that

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

have a multiple zero is the vanishing of the discriminant of the output polynomial, a polynomial in the coefficients of $p$ . This discriminant is not identically zero: for $p(z)=z^d$ , the finite zeros are determined by

\left(\frac{z+ih}{z-ih}\right)^d=e^{-2i\theta},

and are distinct. When $\theta=0$ , the ratio $1$ corresponds to the degree drop and the remaining $d-1$ finite zeros are distinct. Thus the exceptional locus is a proper algebraic hypersurface, establishing the generic assertion.

However, the exceptional locus is not restricted to the multiple-root inputs specified in the conjecture. Take

\theta=0,\qquad h=1,\qquad p(z)=z^3+z=z(z-i)(z+i).

This input has three distinct zeros, so it belongs to none of the claimed exceptional classes. Nevertheless,

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

which has a double zero at $z=0$ .

Therefore the generic statement holds, while the proposed characterization of all exceptional polynomials is disproved.