CCFU Proof 13 — Parallelism Identity Π(ln φ) = arctan(2)

2605165676497

CCFU Proof 13 — Parallelism Identity Π(ln φ) = arctan(2)

05/16/2026

Given.
Let φ = (1+√5)/2, so φ² = φ + 1 and φ > 1.

Definition.
The Lobachevsky parallelism function in H² (K = −1):
    Π(d) = 2·arctan(e^{−d})
Equivalently: Π(d) = arccos(tanh d)

Claim.
    Π(ln φ) = arctan(2)

Proof via double angle formula.

    Π(ln φ) = 2·arctan(e^{−ln φ})
            = 2·arctan(1/φ)

Using 2·arctan(x) = arctan(2x/(1−x²)) when x² < 1.

    x = 1/φ
    x² = 1/φ² < 1  ✓

Since x > 0 and x² < 1, the result lies in (0, π/2),
so no branch correction is needed.

    2x/(1−x²) = (2/φ) / (1 − 1/φ²)
              = (2/φ) / ((φ²−1)/φ²)
              = 2φ/(φ²−1)
              = 2φ/φ          [since φ²−1 = φ]
              = 2

Therefore Π(ln φ) = 2·arctan(1/φ) = arctan(2).  ∎

Verification via arccos form.

    tanh(ln φ) = (φ − 1/φ)/(φ + 1/φ) = (φ²−1)/(φ²+1) = φ/(φ+2)

Since 2φ − 1 = √5:
    √5·φ = (2φ − 1)·φ = 2φ² − φ = 2(φ+1) − φ = φ + 2
Therefore:
    φ/(φ+2) = φ/(√5·φ) = 1/√5

    arccos(1/√5) = arctan(2)  ✓

Both definitions give the same result.  ∎