Roots lie outside the circle of radius
ProvedMetodosNumericos.root_lower_boundLet have real coefficients with and , and put . Then every complex root satisfies . This is Proposição 4.2.2, obtained in the source from the outer bound applied to the reversed polynomial.
import Mathlib import Definitions.Def_MetodosNumericos_polinomiosDefs
namespace MetodosNumericos
theorem root_lower_bound (a : ℕ → ℝ) (n : ℕ) (hn : 1 ≤ n) (han : a n ≠ 0)
(B : ℝ) (hB : B = Finset.sup' (Finset.range n) (by simp [Finset.nonempty_range_iff]; omega)
(fun i => |a i|))
(z : ℂ) (hz : polyValC a n z = 0) :
1 / (1 + B / |a n|) ≤ ‖z‖ := by sorry
end MetodosNumericosRead-back
What the Lean code literally says, in plain math · self-authored-by-drafting-agent (non-blind)
Disclosure: this read-back is not blind. It was written by the same agent that drafted the Lean statement, at the explicit instruction of the mission's human owner, and not by an independent auditor with fresh context.
The statement fixes , a natural number with , and a real number , under the hypotheses and
the maximum being taken over the nonempty finite index set . For a complex with
the conclusion is
where is the modulus of . The inequality is non-strict. Since and , the left-hand side is a positive real number, so the statement in particular forbids from being a root.
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.