The origin inequality #
This file closes the origin channel, deriving (origin-exact) — the hypothesis horigin of
Sendov.polar_origin_incompatible — from the triangle inequality (tri) of
Sendov.Analytic.Origin, the two origin identities, and the defect estimate of
Sendov.Analytic.Jsum.
The chain is:
- (f1aq) The two origin identities express
F(1) + (n-1) a (x+iy) ∫₀¹Fexactly, in terms ofJ = (∏ qⱼ)(∏ zⱼ)and∑ⱼ 1/zⱼ. Feeding inJsum_estimatecollapses this toJ·W/nplus an error of sizea(1-|J|²)/(n-1), whereW = a²(n-1) + 1 - a(x+iy). - (grow)
‖W‖ ≥ 2an/(n-1), which makes|J|·‖W‖/n + a(1-|J|²)/(n-1)non-decreasing in|J|on[0,1], so|J|may be replaced by1. This reduces to the quadratic(n-1)²a² - (3n-1)a + (n-1) > 0, whose discriminant(3n-1)² - 4(n-1)³is negative exactly fromn ≥ 5on. This is the only place the hypothesisn ≥ 5is used. - Eliminating
y.|s + it| ≤ s + t²/(2s)fors > 0, applied toW, whose real parta²(n-1) + 1 - axis at leasta²(n-1), and whose imaginary part is-aywithy² ≤ 1 - x².
Main statements #
Sendov.origin_exact:(origin-exact), in exactly the formpolar_origin_incompatiblewants;Sendov.grow: the lower bound‖W‖ ≥ 2an/(n-1).
Scalar lemmas #
The quantity W, the growth bound (grow), and the two real-arithmetic steps #
theorem
Sendov.xy_bounds
{n : ℕ}
(hn : 2 ≤ n)
{x y : ℝ}
{q : Multiset ℂ}
(hqcard : q.card = n - 1)
(hq1 : ∀ v ∈ q, ‖v‖ ≤ 1)
(hxy : q.sum = (↑n - 1) * (↑x + ↑y * Complex.I))
:
x and y are determined by q.sum, and inherit two bounds from ‖qⱼ‖ ≤ 1: the real
part is what the polar and origin estimates need, and x² + y² ≤ 1 is what eliminates y.
The origin inequality #
theorem
Sendov.origin_exact
{n : ℕ}
(hn : 5 ≤ n)
{a x y α : ℝ}
{z q : Multiset ℂ}
(ha0 : 0 < a)
(ha1 : a < 1)
(hqcard : q.card = n - 1)
(hq0 : q ≠ 0)
(hz1 : ∀ w ∈ z, ‖w‖ ≤ 1)
(hq1 : ∀ v ∈ q, ‖v‖ ≤ 1)
(hxy : q.sum = (↑n - 1) * (↑x + ↑y * Complex.I))
(hcent : (↑n - 1) * (↑a + z.sum) = ↑n * (Multiset.map (fun (v : ℂ) => ↑a - v⁻¹) q).sum)
(hfirst : ↑n * ∫ (t : ℝ) in 0..1, Fprod a q t = (-1) ^ (n - 1) * q.prod * z.prod)
(hsecond :
↑n * (Multiset.map (fun (v : ℂ) => 1 - ↑a * v) q).prod = (-1) ^ (n - 1) * q.prod * (z.prod + ↑a * sumEraseProd z))
(hα : α = M n * (1 - a ^ 2) / 2)
:
(origin-exact). The origin channel, in the form the reduction consumes.