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

Polynomial and roots

The perturbation of z7−1 is small on the unit circle but has a nontrivial shape at z=εw: the terms ε4az3, ε5bz2 and ε6c0z all become order ε7 there. The paired conjugate terms give the antisymmetry used to locate the roots. These two features let us control roots near the unit circle and critical points near the origin by different tests.

All constants are rational, so both sets of tests can be made without rounding. Set

s=10−6,ε=s2=10−12,ρ=1−s16=1−10−96,
A0=−3295071600,B0=551827800,C0=−2301381332000,

and define

a=A0−is,b=iB0+95s,c0=−C0−16225is.

The polynomial is

(1)F(z)=z7−1+ε4(az3−a¯z4)+ε5(bz2−b¯z5)+ε6(c0z−c¯0z6),(2)f(z)=ρ7F(z/ρ).

The identity F(z)=−z7F(1/z¯)― is useful, but does not by itself place the zeros on the unit circle. For that purpose put χ(x)=(1+ix)/(1−ix). Direct expansion gives (1−ix)7F(χ(x))=2iH(x), where

H(x)=7x−35x3+21x5−x7−ε4(1+x2)3(s+A0x)+ε5(1+x2)2[B0(1−3x2)+(9s/5)(x3−3x)]+ε6(1+x2)[C0(5x−10x3+x5)−(162s/25)(1−10x2+5x4)].

The signs at the following rational endpoints are obtained by substitution. Both endpoints in each row are divided by 104.

j Left endpoint Right endpoint Left sign Right sign
0 −1 1 − +
1 4815 4816 + −
2 12539 12540 − +
3 43812 43813 + −
4 −43813 −43812 + −
5 −12540 −12539 − +
6 −4816 −4815 + −

The seven intervals give distinct real zeros xj of H, so ζj=χ(xj) are all seven zeros of F. They have modulus one. Hence the zeros bj=ρζj of f are distinct and lie in the open unit disc.

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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