attempt5

formal source

 112    2φ - 1 ≈ 2.236
 113    φ^2.236 ≈ 2.963
 114    Divided by φ: 1.83 (too small) -/
 115noncomputable def attempt5 : ℝ := phi^(phi + 1/phi) / phi
 116
 117/-! ## Continued Fraction Connection -/
 118
 119/-- e has a beautiful continued fraction:
 120
 121    e = 2 + 1/(1 + 1/(2 + 1/(1 + 1/(1 + 1/(4 + 1/(1 + 1/(1 + ...)))))))
 122
 123    Pattern: [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, ...]
 124
 125    φ has: [1; 1, 1, 1, 1, ...] (all 1s)
 126
 127    Both are "simple" continued fractions in some sense. -/
 128def eContinuedFraction : List ℕ := [2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8]
 129
 130def phiContinuedFraction : List ℕ := [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
 131
 132/-! ## The J-Cost Connection -/
 133
 134/-- In RS, e appears in probability distributions:
 135
 136    Boltzmann: P ∝ exp(-E/kT)
 137    J-cost: P ∝ exp(-J/J₀)
 138
 139    The exponential (base e) is fundamental for probability normalization.
 140
 141    Why e specifically? Because:
 142    d/dx e^x = e^x
 143
 144    Only exponential maintains shape under differentiation. -/
 145theorem e_from_normalization :