Encyclopedia Cost Cost Ndim Calibration Uniform Sq Norm One
ARTICLE 3 claims 3 theorems
Cost Ndim Calibration Uniform Sq Norm One
When a recognition cost's weights are all equal and their squared norm is one, each weight must be the square root of one over the dimension.
Uniform calibration
A recognition cost is a number the framework assigns to a recognition event, a discrete record of something happening. In higher dimensions, the cost depends on weights, one per dimension, that say how much each direction contributes. The declaration uniform_sqNorm_one is a small theorem about what happens when those weights are forced to be equal and their total squared size is fixed at one.
Think of weights as coordinates of a point in n-dimensional space. The squared norm is the sum of the squares of the coordinates, the squared distance from the origin. If all coordinates are equal to some number a, and the squared distance is exactly 1, then the theorem says a² = 1/n. For three dimensions, each weight is 1/√3, about 0.577. For ten dimensions, each is 1/√10, about 0.316. The more dimensions, the smaller each equal weight must be to keep the total squared size at one.
The proof is a direct calculation. If every coordinate equals a, the sum of squares is n times a². Setting that equal to 1 and solving gives a² = 1/n. The theorem also requires n to be positive, so division by n is legitimate. It does not require the weights to sum to one, only that their squared norm is one. That is a different calibration, and the library proves that case separately: if equal weights sum to one, each is 1/n, not 1/√n.
In Recognition Science, this calibration relation is a step toward fixing how costs scale with dimension. The framework's larger forcing chain derives constants like the golden ratio and the number of spatial dimensions, and this theorem supplies one of the algebraic identities that chain uses. The theorem itself is machine-checked in the framework's library of formal theorems, so the calculation is verified end to end.
THEOREM uniform_sqNorm_one · IndisputableMonolith/Cost/Ndim/Calibration.lean
/-- Under uniform weights, unit squared norm gives `a² = 1/n` (for `n > 0`). -/
theorem uniform_sqNorm_one {n : ℕ} {α : Vec n}
(hn : 0 < n) (hU : UniformWeights α) (hcurv : sqNorm α = 1) :
∃ a : ℝ, (∀ i : Fin n, α i = a) ∧ a ^ 2 = 1 / (n : ℝ) := by
rcases hU with ⟨a, ha⟩
have hna : (n : ℝ) ≠ 0 := by
exact_mod_cast (Nat.ne_of_gt hn)
have hnorm : (n : ℝ) * a ^ 2 = 1 := by
simpa [sqNorm, dot, ha, pow_two, Finset.card_univ] using hcurv
have hsquare : a ^ 2 = 1 / (n : ℝ) := by
apply (eq_div_iff hna).2
linarith [hnorm]
exact ⟨a, ha, hsquare⟩
THEOREM uniform_weight_of_sum_one · IndisputableMonolith/Cost/Ndim/Calibration.lean
/-- If weights are uniform and sum to one, each weight is `1/n` (for `n > 0`). -/
theorem uniform_weight_of_sum_one {n : ℕ} {α : Vec n}
(hn : 0 < n) (hU : UniformWeights α) (hsum : weightSum α = 1) :
∃ a : ℝ, (∀ i : Fin n, α i = a) ∧ a = 1 / (n : ℝ) := by
rcases hU with ⟨a, ha⟩
have hna : (n : ℝ) ≠ 0 := by
exact_mod_cast (Nat.ne_of_gt hn)
have hsum' : (n : ℝ) * a = 1 := by
simpa [weightSum, ha, Finset.card_univ] using hsum
have ha_val : a = 1 / (n : ℝ) := by
apply (eq_div_iff hna).2
linarith [hsum']
exact ⟨a, ha, ha_val⟩
THEOREM uniform_sqNorm_one · IndisputableMonolith/Cost/Ndim/Calibration.lean
/-- Under uniform weights, unit squared norm gives `a² = 1/n` (for `n > 0`). -/
theorem uniform_sqNorm_one {n : ℕ} {α : Vec n}
(hn : 0 < n) (hU : UniformWeights α) (hcurv : sqNorm α = 1) :
∃ a : ℝ, (∀ i : Fin n, α i = a) ∧ a ^ 2 = 1 / (n : ℝ) := by
rcases hU with ⟨a, ha⟩
have hna : (n : ℝ) ≠ 0 := by
exact_mod_cast (Nat.ne_of_gt hn)
have hnorm : (n : ℝ) * a ^ 2 = 1 := by
simpa [sqNorm, dot, ha, pow_two, Finset.card_univ] using hcurv
have hsquare : a ^ 2 = 1 / (n : ℝ) := by
apply (eq_div_iff hna).2
linarith [hnorm]
exact ⟨a, ha, hsquare⟩
What this page does not claim
The theorem does not say anything about weights that are not all equal. The theorem does not derive the golden ratio or any other constant from the forcing chain. The theorem does not claim the squared-norm calibration is physically preferred over the sum-to-one calibration.
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/Cost/Ndim/Calibration.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)
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Derived articles
This page is generated by a question-recursion engine: the questions its answers raise become the next pages. The current agenda, with open targets marked red:
- How does the uniform calibration relation connect to the forcing chain that derives the number of spatial dimensions?
- What happens to the cost function when weights are not uniform but still have unit squared norm?
- Does the squared-norm calibration or the sum-to-one calibration appear in the derivation of particle masses?
MACHINE LAYER · GROUNDED CLAIM TABLE · CLICK TO EXPAND
THEOREM uniform_sqNorm_one · IndisputableMonolith/Cost/Ndim/Calibration.lean
/-- Under uniform weights, unit squared norm gives `a² = 1/n` (for `n > 0`). -/ theorem uniform_sqNorm_one {n : ℕ} {α : Vec n} (hn : 0 < n) (hU : UniformWeights α) (hcurv : sqNorm α = 1) : ∃ a : ℝ, (∀ i : Fin n, α i = a) ∧ a ^ 2 = 1 / (n : ℝ) := by rcases hU with ⟨a, ha⟩ have hna : (n : ℝ) ≠ 0 := by exact_mod_cast (Nat.ne_of_gt hn) have hnorm : (n : ℝ) * a ^ 2 = 1 := by simpa [sqNorm, dot, ha, pow_two, Finset.card_univ] using hcurv have hsquare : a ^ 2 = 1 / (n : ℝ) := by apply (eq_div_iff hna).2 linarith [hnorm] exact ⟨a, ha, hsquare⟩If all weights are equal and their squared norm is one, then each weight squared equals one over the dimension. uniform_sqNorm_one · IndisputableMonolith/Cost/Ndim/Calibration.leanTHEOREM uniform_weight_of_sum_one · IndisputableMonolith/Cost/Ndim/Calibration.lean
/-- If weights are uniform and sum to one, each weight is `1/n` (for `n > 0`). -/ theorem uniform_weight_of_sum_one {n : ℕ} {α : Vec n} (hn : 0 < n) (hU : UniformWeights α) (hsum : weightSum α = 1) : ∃ a : ℝ, (∀ i : Fin n, α i = a) ∧ a = 1 / (n : ℝ) := by rcases hU with ⟨a, ha⟩ have hna : (n : ℝ) ≠ 0 := by exact_mod_cast (Nat.ne_of_gt hn) have hsum' : (n : ℝ) * a = 1 := by simpa [weightSum, ha, Finset.card_univ] using hsum have ha_val : a = 1 / (n : ℝ) := by apply (eq_div_iff hna).2 linarith [hsum'] exact ⟨a, ha, ha_val⟩If equal weights sum to one instead, each weight equals one over the dimension. uniform_weight_of_sum_one · IndisputableMonolith/Cost/Ndim/Calibration.leanTHEOREM uniform_sqNorm_one · IndisputableMonolith/Cost/Ndim/Calibration.lean
/-- Under uniform weights, unit squared norm gives `a² = 1/n` (for `n > 0`). -/ theorem uniform_sqNorm_one {n : ℕ} {α : Vec n} (hn : 0 < n) (hU : UniformWeights α) (hcurv : sqNorm α = 1) : ∃ a : ℝ, (∀ i : Fin n, α i = a) ∧ a ^ 2 = 1 / (n : ℝ) := by rcases hU with ⟨a, ha⟩ have hna : (n : ℝ) ≠ 0 := by exact_mod_cast (Nat.ne_of_gt hn) have hnorm : (n : ℝ) * a ^ 2 = 1 := by simpa [sqNorm, dot, ha, pow_two, Finset.card_univ] using hcurv have hsquare : a ^ 2 = 1 / (n : ℝ) := by apply (eq_div_iff hna).2 linarith [hnorm] exact ⟨a, ha, hsquare⟩The theorem requires the dimension to be positive. uniform_sqNorm_one · IndisputableMonolith/Cost/Ndim/Calibration.lean