Sendov's conjecture for a real interior zero #
This file assembles everything: if p has degree n ≥ 2, all zeroes in the closed unit disk,
and a zero at a real a with 0 < a < 1, then p' has a zero within distance 1 of a.
The proof is by contradiction. Suppose not; then every critical point w satisfies
‖w - a‖ ≥ 1, so Sendov.exists_crit_multiset writes p' over points qⱼ of the closed unit
disk, and Sendov.exists_root_multiset writes p over the other zeroes zⱼ. Setting
x + iy = (∑ⱼ qⱼ)/(n-1), the two channels are:
- polar: the polar identity and
p'(a)two ways feed the branch point(⋆).
From (⋆) the argument branches on the degree.
- For
n ≥ 5, AM–GM relaxes(⋆)to the raw polar inequality(1Q), the centroid identity and the two origin identities give(origin-exact), andSendov.polar_origin_incompatiblesays the two cannot hold together. - For
n ≤ 5,Sendov.low_degree_contradictionshows(⋆)is already contradictory on its own, with no recourse to the origin channel. The two branches overlap at degree five.
The zero a = 0 is excluded above (the polar identity divides by a) but needs none of this
machinery: p'(a) two ways already says (∏ⱼ qⱼ)(∏ⱼ (a - zⱼ)) = n, and at a = 0 both factors
have norm at most one.
Main statements #
Sendov.sendov_interior: the conjecture for a real zero0 < a < 1, every degreen ≥ 2;Sendov.sendov_center: the conjecture ata = 0;Sendov.sendov_interior_real: the two combined,0 ≤ a < 1.
Sendov's conjecture for a real interior zero. If every zero of p lies in the
closed unit disk and p(a) = 0 for a real a with 0 < a < 1, then some zero of p' lies
within distance 1 of a.
The conjecture at a = 0. p'(a) two ways gives (∏ⱼ qⱼ)(∏ⱼ (a - zⱼ)) = n. At
a = 0 the second factor is ∏ⱼ zⱼ up to sign, so both factors have norm at most one and
n ≤ 1 — impossible for n ≥ 2. No polar or origin estimate is involved.
The conjecture for a real zero in [0,1), every degree n ≥ 2.