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definition
IsReciprocalCost
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IndisputableMonolith.Cost.FunctionalEquation on GitHub at line 593.
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590
591/-- **Definition 2.1 (Reciprocal Cost)**
592A function F : ℝ₊ → ℝ is a reciprocal cost if F(x) = F(1/x) for all x > 0. -/
593def IsReciprocalCost (F : ℝ → ℝ) : Prop :=
594 ∀ x : ℝ, 0 < x → F x = F x⁻¹
595
596/-- **Normalized**: F(1) = 0. -/
597def IsNormalized (F : ℝ → ℝ) : Prop := F 1 = 0
598
599/-- **Calibration (Condition 1.2)**:
600lim_{t→0} 2·F(e^t)/t² = 1, equivalently G''(0) = 1 where G(t) = F(e^t). -/
601def IsCalibrated (F : ℝ → ℝ) : Prop :=
602 deriv (deriv (G F)) 0 = 1
603
604/-- **Calibration (limit form)**:
605lim_{t→0} 2·F(e^t)/t² = 1, expressed on H = G + 1. -/
606def IsCalibratedLimit (F : ℝ → ℝ) : Prop :=
607 HasLogCurvature (H F) 1
608
609lemma taylorWithinEval_succ_real (H : ℝ → ℝ) (n : ℕ) (x₀ x : ℝ) :
610 taylorWithinEval H (n + 1) Set.univ x₀ x =
611 taylorWithinEval H n Set.univ x₀ x +
612 (((n + 1 : ℝ) * (Nat.factorial n))⁻¹ * (x - x₀) ^ (n + 1)) *
613 iteratedDerivWithin (n + 1) H Set.univ x₀ := by
614 simpa [smul_eq_mul] using taylorWithinEval_succ H n Set.univ x₀ x
615
616lemma taylorWithinEval_one_univ (H : ℝ → ℝ) (x : ℝ) :
617 taylorWithinEval H 1 Set.univ 0 x = H 0 + deriv H 0 * x := by
618 have h := taylorWithinEval_succ_real H 0 0 x
619 -- simplify the Taylor term at order 1
620 have h' :
621 taylorWithinEval H 1 Set.univ 0 x = H 0 + x * deriv H 0 := by
622 simp [taylor_within_zero_eval, iteratedDerivWithin_univ, iteratedDerivWithin_one,
623 iteratedDeriv_one, derivWithin_univ, sub_eq_add_neg] at h
papers checked against this theorem (showing 2 of 2)
-
Multidimensional cost geometry
"Permutation symmetry plus reduction to 1D when all variables coincide forces αᵢ = 1/n; the cost depends only on the geometric mean (xᵢ-form) or arithmetic mean (tᵢ-form)."
-
A Noble-Gas-Centered Coordinate for Within-Period Atomic Property Trends
"The model's input cost function J is the unique solution on R>0 of the functional equation J(xy) + J(x/y) = 2J(x)J(y) + 2J(x) + 2J(y), under the regularity conditions ... The unique solution is J(x) = ½(x + x⁻¹) − 1 = cosh(ln x) − 1 (Corollary 3.1 of Ref. [5])."