Paths in Polynomial Lemniscates: A Degree-Seven Counterexample and Radial Connections

Will Cook

Long paper

Paths in Polynomial Lemniscates: A Degree-Seven Counterexample and Radial Connections

Erdős #1041 By Will Cook 29 September 2026 Authorship, AI use and citation

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Summary. The record gives the rational estimates and geometric proof for ani's degree-seven polynomial: every connected set in its strict unit lemniscate joining two roots has Hausdorff measure greater than two. It develops contained paths for trinomials, collinear roots and two sparse families. Ordinary analytic arguments for critical-value criteria retain their unformalised inputs beside the statements. Critical-value moments, counterexamples to prescribed paths and questions about flow-strip geometry complete the account. The one-polynomial formal statement and correspondence with the historical wording remain distinct.

This paper covers the exact geometry of ani's one-polynomial counterexample, proved positive families and critical-value arguments with their unformalised inputs identified; formal-statement status and historical correspondence remain owned by the claims registry.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or the current status of Erdős #1041, including the later one-polynomial Formal Conjectures path-image Hausdorff-length check and its unreviewed historical curve-length correspondence; those distinctions belong to docs/claims.json in the mathematics repository.

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In this paper

Trinomials

At a zero of a trinomial, the root equation eliminates the middle coefficient. Along the segment from the origin to that zero, the remaining expression is a nonnegative linear combination of two values of modulus less than one. The sum of the coefficients is at most one, which gives the required containment.

Theorem 3.1 (trinomial root connections). Let $n, m$ be integers with $1 \leq m < n$ , and let $f (z) = z^{n} + a z^{m} + b$ have every zero in $D$ . For every zero $ζ$ , the segment $[0, ζ]$ lies in $E_{f}$ . Consequently any two zeros $ζ_{1}, ζ_{2}$ are joined in $E_{f}$ by the broken line $ζ_{1} \to 0 \to ζ_{2}$ , of length $| ζ_{1} | + | ζ_{2} | < 2$ .

Proof. Vieta’s formula gives $| b | < 1$ . At a zero $ζ$ the root equation $ζ^{n} + a ζ^{m} + b = 0$ eliminates the middle coefficient:

\begin{matrix} (5) & f (t ζ) = b (1 - t^{m}) + ζ^{n} (t^{n} - t^{m}) . \end{matrix}

For $0 \leq t < 1$ the two scalar weights $1 - t^{m}$ and $t^{m} - t^{n}$ are nonnegative and sum to $1 - t^{n}$ , so

| f (t ζ) | \leq | b | (1 - t^{m}) + | ζ |^{n} (t^{m} - t^{n}) < 1 - t^{n} \leq 1 .

At $t = 1$ the value is zero. Concatenating two such segments gives the length assertion. ◻

No separate bound on $a$ is needed, since the root equation eliminates it before we take absolute values. The conclusion concerns segments to zeros and makes no assertion that $E_{f}$ is star-shaped.

Summation by parts.

For $f (z) = \sum_{k = 0}^{n} c_{k} z^{k}$ and a zero $ζ$ , put $S_{j} = \sum_{k = 0}^{j} c_{k} ζ^{k}$ . Finite summation by parts gives

\begin{matrix} (6) & f (t ζ) = \sum_{j = 0}^{n - 1} (t^{j} - t^{j + 1}) S_{j} (0 \leq t \leq 1), \end{matrix}

since the coefficient of $c_{k} ζ^{k}$ on the right is $t^{k} - t^{n}$ and $\sum_{k < n} c_{k} ζ^{k} = - c_{n} ζ^{n}$ . The weights in (6) are nonnegative and sum to $1 - t^{n}$ , so the whole radial segment lies in the closed unit sublevel set whenever every partial sum lies in the closed unit disc. For a trinomial, $S_{j} = b$ for $j < m$ and $S_{j} = - ζ^{n}$ for $m \leq j < n$ ; these two values may coincide. This is exactly the estimate above. With two intermediate coefficients, the assumption that all zeros lie in the unit disc need not bound every partial sum by one. The next example shows that a prescribed radial segment can then escape.

Example 3.2 (an escaping root spoke). Let $0 < r < 1$ satisfy $r^{6} > 320 / 327$ , and set

f_{r} (z) = z^{6} + \frac{1}{5} r^{2} z^{4} - \frac{1}{5} r^{4} z^{2} - r^{6} = (z^{2} - r^{2}) (z^{4} + \frac{6}{5} r^{2} z^{2} + r^{4}) .

Every zero has modulus $r$ : for the quartic factor put $z = r w$ , so that the squared roots solve $v^{2} + (6 / 5) v + 1 = 0$ , whose two roots are complex conjugates of product one. Nevertheless

f_{r} (r / 2) = - \frac{327}{320} r^{6},

so the segment $[0, r]$ leaves $E_{f_{r}}$ . This rules out a universal assertion about every origin-to-zero segment. It exhibits no counterexample to the existence of some short connection.

Sources and scope.

Lean checks the identity (6), the constant-term bound, the radial estimate, their combination with the length bound and the evaluation in Example 3.2. No novelty claim is made for this elementary trinomial argument.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

Cite and contact. Cite this paper by its title, author and date above, with its PDF; cite the earlier sources it uses for a mathematical result. For software, use the release citation and give the commit used. Contact Will with questions or corrections.

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