Foundation Jcost Hessian C7 Jcost Hessian Coefficient Eq One

The local shape of cost

The cost function J(x) = (x + 1/x)/2 - 1 measures how far a recognition event is from a perfect match, where x = 1 means no cost. The question is what happens when x is close to 1, say x = 1 + ε for a small number ε. The answer, proved in the framework's machine-checked library of formal theorems, is that J(1 + ε) = ε² / (2(1 + ε)). For tiny ε, this behaves like ε²/2: the cost rises quadratically, not linearly, as you move away from equilibrium. A linear term would mean small errors cost proportionally; the quadratic term means they cost almost nothing at first but accelerate.

In the standard Taylor expansion convention, the coefficient of ε² is 1/2, and the Hessian (the second derivative) is twice that, so the Hessian coefficient is exactly 1. The declaration jcostHessianCoefficient_eq_one records precisely this: the framework defines the Hessian coefficient as 2 times the Taylor quadratic coefficient, and proves it equals 1. This is not an approximation or a numerical coincidence; it is an exact algebraic identity, holding for all ε except ε = -1, where the original cost function is undefined. The proof is a direct computation from the definition of J, with no calculus machinery invoked.

In Recognition Science, this local result matters because it pins down the stiffness of the ledger: the discrete record of recognition events. The framework models the cost of a recognition event with J, and the fact that the Hessian is exactly 1 means the cost curve has a universal, unit curvature at equilibrium. This is a structural fact about the forced cost function, not a fitted parameter. It tells a reader that the framework's central object has a clean, canonical local geometry, which is what one would want before building larger structures on top of it.

The declaration does not claim that the cost function is globally quadratic, nor that the Taylor expansion converges everywhere. Away from ε = -1, the exact formula J(1 + ε) = ε² / (2(1 + ε)) holds, and for large ε the cost grows roughly linearly, not quadratically. The Hessian coefficient being 1 is a statement about the local behavior at the equilibrium point x = 1 only. It also does not assert anything about the physical meaning of the cost; it is a pure mathematical fact about the defined function J.

theorem jcost_one_plus_eq (eps : ℝ) (h : eps ≠ -1) :
    Jcost (1 + eps) = eps ^ 2 / (2 * (1 + eps)) := by
  have hx : 1 + eps ≠ 0 := by
    intro hz
    apply h
    linarith
  rw [Jcost_eq_sq hx]
  ring_nf

theorem jcost_one_plus_eq (eps : ℝ) (h : eps ≠ -1) :
    Jcost (1 + eps) = eps ^ 2 / (2 * (1 + eps)) := by
  have hx : 1 + eps ≠ 0 := by
    intro hz
    apply h
    linarith
  rw [Jcost_eq_sq hx]
  ring_nf

theorem jcostHessianCoefficient_eq_one :
    jcostHessianCoefficient = 1 := by
  unfold jcostHessianCoefficient jcostTaylorQuadraticCoefficient
  norm_num

theorem jcost_one_plus_eq (eps : ℝ) (h : eps ≠ -1) :
    Jcost (1 + eps) = eps ^ 2 / (2 * (1 + eps)) := by
  have hx : 1 + eps ≠ 0 := by
    intro hz
    apply h
    linarith
  rw [Jcost_eq_sq hx]
  ring_nf

What this page does not claim

The cost function is globally quadratic; away from equilibrium it grows roughly linearly. The Taylor expansion converges for all real inputs; the exact formula holds only for ε ≠ -1. The Hessian coefficient being one has any physical interpretation beyond being a property of the defined function J.

Verify this page

Every tagged claim above names its theorem. To check one yourself rather than trust this page, elaborate the source module with Lean 4 and audit its axiom basis:

$ lake env lean IndisputableMonolith/Foundation/JCostHessianC7.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)

A page whose claims cannot be reproduced this way does not ship. In production, every anchor links to the exact declaration in the public source release, and this block carries the build receipt for the page itself.

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