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Allaun Silverfox 22b0159924 math(worksheet): G3_WORKSHEET.md — pure arithmetic, no English in formulas

2026-06-23 05:29:41 -05:00

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WORKSHEET G3 — Eigensolid Fixed Point

Verifiable with any calculator. No English inside formulas.

PART A: The Crossing Operator C (applied numerically)

INPUT: p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05) Check: 0.3+0.1+0.15+0.05+0.2+0.08+0.07+0.05 = 1.0 ✓

FORMULA:

C(p)₁ = C(p)₂ = (p₁ + p₂) / 2
C(p)₃ = C(p)₄ = (p₃ + p₄) / 2
C(p)₅ = C(p)₆ = (p₅ + p₆) / 2
C(p)₇ = C(p)₈ = (p₇ + p₈) / 2

WORK:

C(p)₁ = C(p)₂ = (0.3 + 0.1) / 2 = 0.4 / 2 = 0.2
C(p)₃ = C(p)₄ = (0.15 + 0.05) / 2 = 0.2 / 2 = 0.1
C(p)₅ = C(p)₆ = (0.2 + 0.08) / 2 = 0.28 / 2 = 0.14
C(p)₇ = C(p)₈ = (0.07 + 0.05) / 2 = 0.12 / 2 = 0.06

OUTPUT: C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

Check: 0.2+0.2+0.1+0.1+0.14+0.14+0.06+0.06 = 1.0 ✓

PART B: Idempotence C∘C = C (verified numerically)

INPUT: C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

WORK:

C(C(p))₁ = C(C(p))₂ = (0.2 + 0.2) / 2 = 0.4 / 2 = 0.2
C(C(p))₃ = C(C(p))₄ = (0.1 + 0.1) / 2 = 0.2 / 2 = 0.1
C(C(p))₅ = C(C(p))₆ = (0.14 + 0.14) / 2 = 0.28 / 2 = 0.14
C(C(p))₇ = C(C(p))₈ = (0.06 + 0.06) / 2 = 0.12 / 2 = 0.06

OUTPUT: C(C(p)) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

VERIFICATION: C(C(p)) = C(p) ✓

RESULT: C∘C = C. One application reaches the fixed point.

PART C: Image M ≅ Δ₃ (verified numerically)

Image M: All vectors with p₁=p₂, p₃=p₄, p₅=p₆, p₇=p₈.

Map from Δ₃ to M:

(q₁, q₂, q₃, q₄) ↦ (q₁/2, q₁/2, q₂/2, q₂/2, q₃/2, q₃/2, q₄/2, q₄/2)

TEST: (q₁, q₂, q₃, q₄) = (0.4, 0.2, 0.28, 0.12) Check: 0.4 + 0.2 + 0.28 + 0.12 = 1.0 ✓

WORK:

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)
     = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

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)

VERIFICATION: This equals C(p) from Part A. ✓

Inverse map:

ψ(p₁, p₂, ..., p₈) = (2p₁, 2p₃, 2p₅, 2p₇)

TEST: ψ(0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

= (2×0.2, 2×0.1, 2×0.14, 2×0.06)
= (0.4, 0.2, 0.28, 0.12)

VERIFICATION: ψ(φ(q)) = q ✓ and φ(ψ(C(p))) = C(p) ✓

PART D: Contraction (verified numerically)

INPUT: p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05) INPUT: q = (0.2, 0.2, 0.1, 0.1, 0.15, 0.1, 0.1, 0.05)

From Part A: C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

Compute C(q):

C(q)₁ = C(q)₂ = (0.2 + 0.2) / 2 = 0.2
C(q)₃ = C(q)₄ = (0.1 + 0.1) / 2 = 0.1
C(q)₅ = C(q)₆ = (0.15 + 0.1) / 2 = 0.125
C(q)₇ = C(q)₈ = (0.1 + 0.05) / 2 = 0.075

OUTPUT: C(q) = (0.2, 0.2, 0.1, 0.1, 0.125, 0.125, 0.075, 0.075) Check: 0.2+0.2+0.1+0.1+0.125+0.125+0.075+0.075 = 1.0 ✓

STEP 1: Compute d_F(p, q)

√(p₁q₁) = √(0.3×0.2) = √0.06   = 0.24495
√(p₂q₂) = √(0.1×0.2) = √0.02   = 0.14142
√(p₃q₃) = √(0.15×0.1) = √0.015 = 0.12247
√(p₄q₄) = √(0.05×0.1) = √0.005 = 0.07071
√(p₅q₅) = √(0.2×0.15) = √0.03  = 0.17321
√(p₆q₆) = √(0.08×0.1) = √0.008 = 0.08944
√(p₇q₇) = √(0.07×0.1) = √0.007 = 0.08367
√(p₈q₈) = √(0.05×0.05) = √0.0025= 0.05000

Sum:

S_pq = 0.24495 + 0.14142 + 0.12247 + 0.07071
     + 0.17321 + 0.08944 + 0.08367 + 0.05000
     = 0.97587

d_F(p,q) = 2·arccos(0.97587)

Type into calculator: 2 * arccos(0.97587)

STEP 2: Compute d_F(C(p), C(q))

√(C(p)₁·C(q)₁) = √(0.2×0.2)  = 0.2
√(C(p)₂·C(q)₂) = √(0.2×0.2)  = 0.2
√(C(p)₃·C(q)₃) = √(0.1×0.1)  = 0.1
√(C(p)₄·C(q)₄) = √(0.1×0.1)  = 0.1
√(C(p)₅·C(q)₅) = √(0.14×0.125) = √0.0175 = 0.13229
√(C(p)₆·C(q)₆) = √(0.14×0.125) = 0.13229
√(C(p)₇·C(q)₇) = √(0.06×0.075) = √0.0045 = 0.06708
√(C(p)₈·C(q)₈) = √(0.06×0.075) = 0.06708

Sum:

S_CpCq = 0.2 + 0.2 + 0.1 + 0.1 + 0.13229 + 0.13229 + 0.06708 + 0.06708
       = 0.99874

d_F(C(p), C(q)) = 2·arccos(0.99874)

Type into calculator: 2 * arccos(0.99874)

STEP 3: Compare

Value	Calculator Input	Expected
arccos(0.97587)	`arccos(0.97587)`	~0.2197
arccos(0.99874)	`arccos(0.99874)`	~0.0502
d_F(p,q)	`2*arccos(0.97587)`	~0.4394
d_F(C(p),C(q))	`2*arccos(0.99874)`	~0.1004

VERIFICATION: d_F(C(p), C(q)) ≈ 0.1004 < d_F(p,q) ≈ 0.4394

CONTRACTION CONFIRMED: C shrinks Fisher distance.

Contraction ratio: 0.1004 / 0.4394 ≈ 0.228

Type: 0.1004 / 0.4394 → ~0.23

PART E: Strict Inequality (when p, q differ within a pair)

INPUT: p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05) INPUT: r = (0.1, 0.3, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)

p and r differ only within pair 1: (0.3, 0.1) vs (0.1, 0.3).

C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06) C(r) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

C(p) = C(r) because C averages each pair.

d_F(C(p), C(r)) = 0

d_F(p, r):

√(0.3×0.1) = √0.03 = 0.17321
√(0.1×0.3) = √0.03 = 0.17321
√(0.15×0.15) = 0.15
√(0.05×0.05) = 0.05
√(0.2×0.2) = 0.2
√(0.08×0.08) = 0.08
√(0.07×0.07) = 0.07
√(0.05×0.05) = 0.05

Sum = 0.17321 + 0.17321 + 0.15 + 0.05 + 0.2 + 0.08 + 0.07 + 0.05
    = 0.94642

d_F(p,r) = 2·arccos(0.94642)

Type into calculator: 2 * arccos(0.94642)

Expected: ~0.66 (significantly > 0)

VERIFICATION: d_F(C(p), C(r)) = 0 < 0.66 = d_F(p,r)

STRICT INEQUALITY CONFIRMED: When p, r differ within a pair, C collapses them completely.

PART F: Information Loss (computed numerically)

FORMULA:

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)) ]

where sₖ = p_{2k-1} + p_{2k}.

INPUT: p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)

Pair 1: s₁ = 0.3 + 0.1 = 0.4

a = 0.3/0.4 = 0.75, b = 0.1/0.4 = 0.25
KL = 0.75·ln(0.75/0.5) + 0.25·ln(0.25/0.5)
   = 0.75·ln(1.5) + 0.25·ln(0.5)
   = 0.75·0.40547 + 0.25·(-0.69315)
   = 0.30410 - 0.17329
   = 0.13081

Term 1: s₁ × KL = 0.4 × 0.13081 = 0.05232

Pair 2: s₂ = 0.15 + 0.05 = 0.2

a = 0.15/0.2 = 0.75, b = 0.05/0.2 = 0.25
KL = 0.75·ln(1.5) + 0.25·ln(0.5)
   = 0.13081 (same as above)

Term 2: s₂ × KL = 0.2 × 0.13081 = 0.02616

Pair 3: s₃ = 0.2 + 0.08 = 0.28

a = 0.2/0.28 = 0.71429, b = 0.08/0.28 = 0.28571
KL = 0.71429·ln(0.71429/0.5) + 0.28571·ln(0.28571/0.5)
   = 0.71429·ln(1.42858) + 0.28571·ln(0.57142)
   = 0.71429·0.35668 + 0.28571·(-0.55962)
   = 0.25477 - 0.15989
   = 0.09488

Term 3: s₃ × KL = 0.28 × 0.09488 = 0.02657

Pair 4: s₄ = 0.07 + 0.05 = 0.12

a = 0.07/0.12 = 0.58333, b = 0.05/0.12 = 0.41667
KL = 0.58333·ln(0.58333/0.5) + 0.41667·ln(0.41667/0.5)
   = 0.58333·ln(1.16667) + 0.41667·ln(0.83333)
   = 0.58333·0.15415 + 0.41667·(-0.18232)
   = 0.08992 - 0.07597
   = 0.01395

Term 4: s₄ × KL = 0.12 × 0.01395 = 0.00167

TOTAL:

I_loss(p) = 0.05232 + 0.02616 + 0.02657 + 0.00167
          = 0.10672 nats

In bits: 0.10672 / ln(2) = 0.10672 / 0.69315 = 0.15396 bits

VERIFY on calculator:

0.05232 + 0.02616 + 0.02657 + 0.00167

Result: ~0.1067

SUMMARY (all verified numerically)

Claim	Verification Method	Result
C(p) computed	8 additions, 4 divisions	✓ sums to 1.0
C∘C = C	Apply C twice, compare	✓ identical
Image M ≅ Δ₃	Map (0.4,0.2,0.28,0.12) → C(p)	✓ inverse works
Contraction d_F(C(p),C(q)) < d_F(p,q)	Calculator: arccos comparison	✓ 0.1004 < 0.4394
Strict inequality	p, r differ in pair 1 only	✓ 0 < 0.66
Information loss	4 KL divergences, weighted	✓ 0.1067 nats

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WORKSHEET G3 — Eigensolid Fixed Point

Verifiable with any calculator. No English inside formulas.

PART A: The Crossing Operator C (applied numerically)

PART B: Idempotence C∘C = C (verified numerically)

PART C: Image M ≅ Δ₃ (verified numerically)

PART D: Contraction (verified numerically)

PART E: Strict Inequality (when p, q differ within a pair)

PART F: Information Loss (computed numerically)

SUMMARY (all verified numerically)

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