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For even index 2k, L(2k) = 2·cosh(2k·ln φ) and F(2k) = (2/√5)·sinh(2k·ln φ). Corollary: L(2k)² − 5·F(2k)² = 4. Direct from Binet formulas and cosh²−sinh²=1. For odd indices the roles swap.
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Title CCFU Proof 6 — Lucas = 2cosh, Fibonacci = (2/√5)sinh (even index)
For even index 2k, L(2k) = 2·cosh(2k·ln φ) and F(2k) = (2/√5)·sinh(2k·ln φ). Corollary: L(2k)² − 5·F(2k)² = 4. Direct from Binet formulas and cosh²−sinh²=1. For odd indices the roles swap.
Work type Research papers, Thesis, Lecture notes
Tags lucas, golden ratio, cosh, proof, ccfu, hyperbolic, fibonacci, sinh, mathematics
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Identifier 2605155667382
Entry date May 15, 2026, 5:32 PM UTC
License All rights reserved
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Author 100.00 %. Holder Captain Cookie Face Universe. Date May 15, 2026.
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