Let g∈widehatmathbfP, and define F_m(x,y) by fracd^mdx^mf(g(x))=∑_k f^(k)(g(x))A_m,k(x),qquad F_m(x,y)=∑_k y^kA_m,k(x). The Faà di Bruno real-rootedness conjecture. For every alpha∈R and m=0,1,…, F_m(alpha,y) has only real roots and F_m+1(alpha,y)lessdot F_m(alpha,y). The source says this is known in the first few cases, for e^x, and for x^d, but leaves the general assertion open.

The Faà di Bruno real-rootedness conjecture

At least 19 years old · documented by 2006

Let $g\in\widehat{\mathbf{P}}$ , and define $F_m(x,y)$ by

\frac{d^m}{dx^m}f(g(x))=\sum_k f^{(k)}(g(x))A_{m,k}(x),\qquad F_m(x,y)=\sum_k y^kA_{m,k}(x).

The Faà di Bruno real-rootedness conjecture. For every $\alpha\in\mathbb{R}$ and $m=0,1,\dots$ , $F_m(\alpha,y)$ has only real roots and

F_{m+1}(\alpha,y)\lessdot F_m(\alpha,y).

The source says this is known in the first few cases, for $e^x$ , and for $x^d$ , but leaves the general assertion open.

References

Primary source

Steve Fisk, “Polynomials, roots, and interlacing”, arXiv:math/0612833 (2008).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Open

An unverified posted calculation claims to disprove the conjecture in order four, so the problem may be settled negatively if the calculation is correct.

The conjecture asserts universal real-rootedness and consecutive interlacing for the associated Faà di Bruno polynomials. The supplied source description records only special cases, not the general assertion.

Posted attempt

A posted calculation claims a counterexample with a real-rooted quartic input at $\alpha=0$ and $m=4$ : the resulting polynomial has a cubic factor with negative discriminant, hence a nonreal conjugate pair. This would disprove the conjecture, but the calculation has not been independently verified.

Current status (as of August 2026): A purported order-four counterexample is the only reported development; the conjecture is not verified false, and no verified proof or disproof is recorded.

CounterexampleThis solution needs a summarySee full solutionHide full solution

The conjecture is false already for $m=4$ , the first case beyond those established in the source.

Take

g(x)=(x-1)(x+2)^3,\qquad \alpha=0.

All zeros of $g$ are real, so $g\in\mathcal P\subseteq\widehat{\mathcal P}$ , exactly as required. Direct differentiation gives

g'(0)=-4,\qquad g''(0)=12,\qquad g'''(0)=30,\qquad g^{(4)}(0)=24.

The Faà di Bruno polynomial at order four is

F_4(\alpha,y) =(g'(\alpha))^4y^4 +6g''(\alpha)(g'(\alpha))^2y^3 +\bigl(4g'''(\alpha)g'(\alpha)+3(g''(\alpha))^2\bigr)y^2 +g^{(4)}(\alpha)y.

Consequently,

F_4(0,y)=8y\bigl(32y^3+144y^2-6y+3\bigr).

For

q(y)=32y^3+144y^2-6y+3

the discriminant is

\operatorname{disc}(q) =-36{,}799{,}488 =-2^{10}\cdot3^3\cdot11^3<0.

A real cubic with negative discriminant has one real zero and a nonreal complex-conjugate pair. Therefore $F_4(0,y)$ is not real-rooted. This contradicts the conjectured real-rootedness and, consequently, also precludes the proposed universal consecutive interlacing.