Plectis

Long paper

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

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.

In this paper

A sufficient critical-value threshold in degree five

The reciprocal balance at a critical point bounds the sum of the two nearest root distances. We begin with this formal geometric statement, then distinguish distance from containment in the degree-five argument.

Theorem 10.5 (a geometric-mean bound for distances to a critical point). Let n≥2, let z1,…,zn,c∈C with c≠zk for every k, and suppose

∑k=1n1c−zk=0.

If r>0 is determined by

rn=∏k=1n|c−zk|,

then there are distinct indices i,j such that

|c−zi|+|c−zj|≤2r.

Proof. Relabel the points so that their distances satisfy d1≤d2≤⋯≤dn. The reciprocal balance and the triangle inequality give

1d1=|∑j=2n1c−zj|≤∑j=2n1dj≤n−1d2.

Thus d2≤(n−1)d1. Also d1d2n−1≤rn, since every remaining distance is at least d2. For t=d2/d1∈[1,n−1],

ddtlog⁡2t(n−1)/n1+t=n−1−tnt(1+t)≥0.

The logarithm vanishes at t=1, so d1+d2≤2(d1d2n−1)1/n≤2r. ◻

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.

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