(41) Fibonacci & Lucas numbers

[Albert Chan]

Starting from Binet's formula, where \(φ = {1+\sqrt5 \over 2}\)

\(\sqrt5 F_n = φ^n - (1-φ)^n = φ^n - (-1/φ)^n = φ^n - i^{2n} φ^{-n}
= i^n \left(({φ \over i})^n - ({φ \over i})^{-n} \right)
\)

\(\large F_n = {2 \over \sqrt5}\; i^n \sinh(n \ln({φ \over i})) \)


Let \(z_3 = \ln({φ \over i}) = \ln(φ) - {\pi \over 2} i \)

\(\large \cosh z_3 = {{φ \over i} + {i \over φ} \over 2}
= {-i φ\; +\; i (φ-1) \over 2} = {-i\over2} \)

\(\large \sinh z_3 = ± \sqrt{ \cosh^2 z_3 - 1 } = ± \sqrt{{-5\over4}} = {-\sqrt5 i\over 2} = {\sqrt5 \over 2i}\quad \) // principle branch

\(\large F_n = \left({2i \over \sqrt5}\right) i^{n-1} \sinh(n z_3)
= i^{n-1} {\sinh(n z_3) \over \sinh(z_3)} \quad\) // matching previous post