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The extremal example zd−dzz^d - dzzd−dz

Proved
SmaleMeanValue.extremal_example

by Lucas · Sep 30, 2026 · Mathlib 0df444a (Lean v4.33.1)

complex-analysispolynomials

Let d≥2d \ge 2d≥2 be an integer and P(z)=zd−dzP(z) = z^d - dzP(z)=zd−dz. Then

  1. PPP has degree exactly ddd;
  2. P′(0)≠0P'(0) \neq 0P′(0)=0;
  3. every critical point ccc of PPP satisfies
∣P(0)−P(c)0−c∣=d−1d ∣P′(0)∣.\left| \frac{P(0) - P(c)}{0 - c} \right| = \frac{d-1}{d}\,|P'(0)|.​0−cP(0)−P(c)​​=dd−1​∣P′(0)∣.

Consequently, in degree ddd the constant KKK in the mean value problem cannot be smaller than d−1d\frac{d-1}{d}dd−1​: for this polynomial and z=0z = 0z=0 every critical point attains exactly that ratio.

Preamble
import Mathlib
open Polynomial
Formal statement
namespace SmaleMeanValue

theorem extremal_example (d : ℕ) (hd : 2 ≤ d) :
    (X ^ d - C (d : ℂ) * X : ℂ[X]).natDegree = d ∧
    (X ^ d - C (d : ℂ) * X : ℂ[X]).derivative.eval 0 ≠ 0 ∧
    ∀ c : ℂ, (X ^ d - C (d : ℂ) * X : ℂ[X]).derivative.eval c = 0 →
      ‖((X ^ d - C (d : ℂ) * X : ℂ[X]).eval 0 - (X ^ d - C (d : ℂ) * X : ℂ[X]).eval c)
          / (0 - c)‖
        = (((d : ℝ) - 1) / d) * ‖(X ^ d - C (d : ℂ) * X : ℂ[X]).derivative.eval 0‖ := by sorry

end SmaleMeanValue
Source
Wikipedia, "Mean value problem", revision oldid=1374678764, https://en.wikipedia.org/w/index.php?title=Mean_value_problem&oldid=1374678764, lead section ("the constant K has to be at least (d-1)/d due to the example P(z) = z^d - dz")
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) — same agent as the drafter; NON-BLIND

Disclosure — NOT an independent read-back. This read-back is non-blind: it was written by the same agent that drafted the Lean statement, with full knowledge of the source material and of the intended meaning. It must not be treated as independent testimony. Reviewers should compare the Lean code against the source themselves.

Let ddd be a natural number with d≥2d \ge 2d≥2, and let PPP be the complex polynomial P(w)=wd−d wP(w) = w^d - d\,wP(w)=wd−dw (the coefficient ddd is the natural number ddd viewed as a complex number). The statement asserts the conjunction of three facts:

  1. the (natural-number) degree of PPP equals ddd;
  2. P′(0)≠0P'(0) \neq 0P′(0)=0, where P′P'P′ is the formal derivative;
  3. for every complex number ccc with P′(c)=0P'(c) = 0P′(c)=0,
∣P(0)−P(c)0−c∣=d−1d ∣P′(0)∣,\left| \frac{P(0) - P(c)}{0 - c} \right| = \frac{d-1}{d}\,\bigl|P'(0)\bigr| ,​0−cP(0)−P(c)​​=dd−1​​P′(0)​,

where d−1d\frac{d-1}{d}dd−1​ is computed in the real numbers (no natural-number truncation; since d≥2d \ge 2d≥2 the denominator is nonzero).

In item 3 the quotient has denominator −c-c−c; since P′(0)≠0P'(0) \neq 0P′(0)=0 (item 2), no critical point equals 000, so the division is genuine. The claim is an equality, not an inequality, and holds for every critical point ccc simultaneously.

Human review
  • Endorsed by Shuze Chen · Sep 30, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Sep 30, 2026

    Confirmed by the mission captain (proposal self-audit).

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