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300 lines
7.8 KiB
Markdown
300 lines
7.8 KiB
Markdown
# WORKSHEET G3 — Eigensolid Fixed Point
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## Verifiable with any calculator. No English inside formulas.
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---
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## PART A: The Crossing Operator C (applied numerically)
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**INPUT:** p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
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**Check:** 0.3+0.1+0.15+0.05+0.2+0.08+0.07+0.05 = 1.0 ✓
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**FORMULA:**
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```
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C(p)₁ = C(p)₂ = (p₁ + p₂) / 2
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C(p)₃ = C(p)₄ = (p₃ + p₄) / 2
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C(p)₅ = C(p)₆ = (p₅ + p₆) / 2
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C(p)₇ = C(p)₈ = (p₇ + p₈) / 2
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```
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**WORK:**
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```
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C(p)₁ = C(p)₂ = (0.3 + 0.1) / 2 = 0.4 / 2 = 0.2
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C(p)₃ = C(p)₄ = (0.15 + 0.05) / 2 = 0.2 / 2 = 0.1
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C(p)₅ = C(p)₆ = (0.2 + 0.08) / 2 = 0.28 / 2 = 0.14
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C(p)₇ = C(p)₈ = (0.07 + 0.05) / 2 = 0.12 / 2 = 0.06
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```
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**OUTPUT:** C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
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**Check:** 0.2+0.2+0.1+0.1+0.14+0.14+0.06+0.06 = 1.0 ✓
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---
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## PART B: Idempotence C∘C = C (verified numerically)
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**INPUT:** C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
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**WORK:**
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```
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C(C(p))₁ = C(C(p))₂ = (0.2 + 0.2) / 2 = 0.4 / 2 = 0.2
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C(C(p))₃ = C(C(p))₄ = (0.1 + 0.1) / 2 = 0.2 / 2 = 0.1
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C(C(p))₅ = C(C(p))₆ = (0.14 + 0.14) / 2 = 0.28 / 2 = 0.14
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C(C(p))₇ = C(C(p))₈ = (0.06 + 0.06) / 2 = 0.12 / 2 = 0.06
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```
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**OUTPUT:** C(C(p)) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
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**VERIFICATION:** C(C(p)) = C(p) ✓
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**RESULT:** C∘C = C. One application reaches the fixed point.
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---
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## PART C: Image M ≅ Δ₃ (verified numerically)
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**Image M:** All vectors with p₁=p₂, p₃=p₄, p₅=p₆, p₇=p₈.
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**Map from Δ₃ to M:**
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```
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(q₁, q₂, q₃, q₄) ↦ (q₁/2, q₁/2, q₂/2, q₂/2, q₃/2, q₃/2, q₄/2, q₄/2)
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```
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**TEST:** (q₁, q₂, q₃, q₄) = (0.4, 0.2, 0.28, 0.12)
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**Check:** 0.4 + 0.2 + 0.28 + 0.12 = 1.0 ✓
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**WORK:**
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```
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φ(q) = (0.4/2, 0.4/2, 0.2/2, 0.2/2, 0.28/2, 0.28/2, 0.12/2, 0.12/2)
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= (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
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```
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**OUTPUT:** φ(0.4, 0.2, 0.28, 0.12) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
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**VERIFICATION:** This equals C(p) from Part A. ✓
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**Inverse map:**
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```
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ψ(p₁, p₂, ..., p₈) = (2p₁, 2p₃, 2p₅, 2p₇)
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```
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**TEST:** ψ(0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
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```
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= (2×0.2, 2×0.1, 2×0.14, 2×0.06)
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= (0.4, 0.2, 0.28, 0.12)
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```
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**VERIFICATION:** ψ(φ(q)) = q ✓ and φ(ψ(C(p))) = C(p) ✓
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---
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## PART D: Contraction (verified numerically)
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**INPUT:** p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
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**INPUT:** q = (0.2, 0.2, 0.1, 0.1, 0.15, 0.1, 0.1, 0.05)
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**From Part A:** C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
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**Compute C(q):**
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```
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C(q)₁ = C(q)₂ = (0.2 + 0.2) / 2 = 0.2
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C(q)₃ = C(q)₄ = (0.1 + 0.1) / 2 = 0.1
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C(q)₅ = C(q)₆ = (0.15 + 0.1) / 2 = 0.125
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C(q)₇ = C(q)₈ = (0.1 + 0.05) / 2 = 0.075
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```
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**OUTPUT:** C(q) = (0.2, 0.2, 0.1, 0.1, 0.125, 0.125, 0.075, 0.075)
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**Check:** 0.2+0.2+0.1+0.1+0.125+0.125+0.075+0.075 = 1.0 ✓
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---
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**STEP 1: Compute d_F(p, q)**
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```
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√(p₁q₁) = √(0.3×0.2) = √0.06 = 0.24495
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√(p₂q₂) = √(0.1×0.2) = √0.02 = 0.14142
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√(p₃q₃) = √(0.15×0.1) = √0.015 = 0.12247
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√(p₄q₄) = √(0.05×0.1) = √0.005 = 0.07071
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√(p₅q₅) = √(0.2×0.15) = √0.03 = 0.17321
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√(p₆q₆) = √(0.08×0.1) = √0.008 = 0.08944
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√(p₇q₇) = √(0.07×0.1) = √0.007 = 0.08367
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√(p₈q₈) = √(0.05×0.05) = √0.0025= 0.05000
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```
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**Sum:**
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```
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S_pq = 0.24495 + 0.14142 + 0.12247 + 0.07071
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+ 0.17321 + 0.08944 + 0.08367 + 0.05000
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= 0.97587
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```
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**d_F(p,q) = 2·arccos(0.97587)**
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Type into calculator: `2 * arccos(0.97587)`
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---
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**STEP 2: Compute d_F(C(p), C(q))**
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```
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√(C(p)₁·C(q)₁) = √(0.2×0.2) = 0.2
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√(C(p)₂·C(q)₂) = √(0.2×0.2) = 0.2
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√(C(p)₃·C(q)₃) = √(0.1×0.1) = 0.1
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√(C(p)₄·C(q)₄) = √(0.1×0.1) = 0.1
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√(C(p)₅·C(q)₅) = √(0.14×0.125) = √0.0175 = 0.13229
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√(C(p)₆·C(q)₆) = √(0.14×0.125) = 0.13229
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√(C(p)₇·C(q)₇) = √(0.06×0.075) = √0.0045 = 0.06708
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√(C(p)₈·C(q)₈) = √(0.06×0.075) = 0.06708
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```
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**Sum:**
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```
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S_CpCq = 0.2 + 0.2 + 0.1 + 0.1 + 0.13229 + 0.13229 + 0.06708 + 0.06708
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= 0.99874
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```
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**d_F(C(p), C(q)) = 2·arccos(0.99874)**
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Type into calculator: `2 * arccos(0.99874)`
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---
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**STEP 3: Compare**
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| Value | Calculator Input | Expected |
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|-------|-----------------|----------|
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| arccos(0.97587) | `arccos(0.97587)` | ~0.2197 |
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| arccos(0.99874) | `arccos(0.99874)` | ~0.0502 |
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| d_F(p,q) | `2*arccos(0.97587)` | ~0.4394 |
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| d_F(C(p),C(q)) | `2*arccos(0.99874)` | ~0.1004 |
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**VERIFICATION:** d_F(C(p), C(q)) ≈ 0.1004 < d_F(p,q) ≈ 0.4394
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**CONTRACTION CONFIRMED:** C shrinks Fisher distance.
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**Contraction ratio:** 0.1004 / 0.4394 ≈ 0.228
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Type: `0.1004 / 0.4394` → ~0.23
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---
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## PART E: Strict Inequality (when p, q differ within a pair)
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**INPUT:** p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
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**INPUT:** r = (0.1, 0.3, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
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p and r differ only within pair 1: (0.3, 0.1) vs (0.1, 0.3).
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**C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)**
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**C(r) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)**
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C(p) = C(r) because C averages each pair.
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**d_F(C(p), C(r)) = 0**
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**d_F(p, r):**
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```
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√(0.3×0.1) = √0.03 = 0.17321
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√(0.1×0.3) = √0.03 = 0.17321
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√(0.15×0.15) = 0.15
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√(0.05×0.05) = 0.05
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√(0.2×0.2) = 0.2
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√(0.08×0.08) = 0.08
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√(0.07×0.07) = 0.07
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√(0.05×0.05) = 0.05
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Sum = 0.17321 + 0.17321 + 0.15 + 0.05 + 0.2 + 0.08 + 0.07 + 0.05
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= 0.94642
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```
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**d_F(p,r) = 2·arccos(0.94642)**
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Type into calculator: `2 * arccos(0.94642)`
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Expected: ~0.66 (significantly > 0)
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**VERIFICATION:** d_F(C(p), C(r)) = 0 < 0.66 = d_F(p,r)
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**STRICT INEQUALITY CONFIRMED:** When p, r differ within a pair, C collapses them completely.
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---
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## PART F: Information Loss (computed numerically)
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**FORMULA:**
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```
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I_loss(p) = Σₖ₌₁⁴ sₖ · [ (p_{2k-1}/sₖ)·ln((p_{2k-1}/sₖ)/(1/2)) + (p_{2k}/sₖ)·ln((p_{2k}/sₖ)/(1/2)) ]
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```
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where sₖ = p_{2k-1} + p_{2k}.
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**INPUT:** p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
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**Pair 1:** s₁ = 0.3 + 0.1 = 0.4
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```
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a = 0.3/0.4 = 0.75, b = 0.1/0.4 = 0.25
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KL = 0.75·ln(0.75/0.5) + 0.25·ln(0.25/0.5)
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= 0.75·ln(1.5) + 0.25·ln(0.5)
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= 0.75·0.40547 + 0.25·(-0.69315)
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= 0.30410 - 0.17329
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= 0.13081
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```
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Term 1: s₁ × KL = 0.4 × 0.13081 = 0.05232
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**Pair 2:** s₂ = 0.15 + 0.05 = 0.2
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```
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a = 0.15/0.2 = 0.75, b = 0.05/0.2 = 0.25
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KL = 0.75·ln(1.5) + 0.25·ln(0.5)
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= 0.13081 (same as above)
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```
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Term 2: s₂ × KL = 0.2 × 0.13081 = 0.02616
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**Pair 3:** s₃ = 0.2 + 0.08 = 0.28
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```
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a = 0.2/0.28 = 0.71429, b = 0.08/0.28 = 0.28571
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KL = 0.71429·ln(0.71429/0.5) + 0.28571·ln(0.28571/0.5)
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= 0.71429·ln(1.42858) + 0.28571·ln(0.57142)
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= 0.71429·0.35668 + 0.28571·(-0.55962)
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= 0.25477 - 0.15989
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= 0.09488
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```
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Term 3: s₃ × KL = 0.28 × 0.09488 = 0.02657
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**Pair 4:** s₄ = 0.07 + 0.05 = 0.12
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```
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a = 0.07/0.12 = 0.58333, b = 0.05/0.12 = 0.41667
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KL = 0.58333·ln(0.58333/0.5) + 0.41667·ln(0.41667/0.5)
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= 0.58333·ln(1.16667) + 0.41667·ln(0.83333)
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= 0.58333·0.15415 + 0.41667·(-0.18232)
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= 0.08992 - 0.07597
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= 0.01395
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```
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Term 4: s₄ × KL = 0.12 × 0.01395 = 0.00167
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---
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**TOTAL:**
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```
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I_loss(p) = 0.05232 + 0.02616 + 0.02657 + 0.00167
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= 0.10672 nats
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```
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**In bits:** 0.10672 / ln(2) = 0.10672 / 0.69315 = 0.15396 bits
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**VERIFY on calculator:**
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```
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0.05232 + 0.02616 + 0.02657 + 0.00167
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```
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Result: ~0.1067
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---
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## SUMMARY (all verified numerically)
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| Claim | Verification Method | Result |
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|-------|---------------------|--------|
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| C(p) computed | 8 additions, 4 divisions | ✓ sums to 1.0 |
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| C∘C = C | Apply C twice, compare | ✓ identical |
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| Image M ≅ Δ₃ | Map (0.4,0.2,0.28,0.12) → C(p) | ✓ inverse works |
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| Contraction d_F(C(p),C(q)) < d_F(p,q) | Calculator: arccos comparison | ✓ 0.1004 < 0.4394 |
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| Strict inequality | p, r differ in pair 1 only | ✓ 0 < 0.66 |
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| Information loss | 4 KL divergences, weighted | ✓ 0.1067 nats |
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