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I saw a funky looking formula for the mth Fibonacci number on twitter

\begin{equation}\label{eqn:fibonacci_sinh:20}

F_m = \frac{2}{\sqrt{5} i^m} \sinh\lr{ m \ln\lr{i\phi} },

\end{equation}

where

\begin{equation}\label{eqn:fibonacci_sinh:60}

\phi = \frac{ 1 + \sqrt{5} }{2},

\end{equation}

is the golden ratio.

This certainly doesn’t look like it’s a representation of the sequence

\begin{equation}\label{eqn:fibonacci_sinh:40}

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \cdots

\end{equation}

We can verify that it works in Mathematica, as seen in fig. 1.

fig. 1. Verification of hyperbolic sine representation of mth Fibonacci numbe

Recall that we previously found this formula for the mth Fibonacci number

\begin{equation}\label{eqn:fibonacci_sinh:80}

F_m = \inv{\sqrt{5}} \lr{ \phi^m – { \bar{\phi}}^m },

\end{equation}

where \( \bar{\phi} \) is the conjugate of the golden ratio

\begin{equation}\label{eqn:fibonacci_sinh:100}

\bar{\phi} = \frac{ 1 – \sqrt{5} }{2}.

\end{equation}

Let’s see how these are equivalent. First observe that the golden conjugate is easily related to the inverse of the golden ratio

\begin{equation}\label{eqn:fibonacci_sinh:120}

\begin{aligned}

\inv{\phi}

&=

\frac{2}{1 + \sqrt{5}} \\

&=

\frac{2\lr{ 1 – \sqrt{5}} }{1^2 – \lr{\sqrt{5}}^2 } \\

&=

-\frac{1 – \sqrt{5} }{2} \\

&=

-\bar{\phi}.

\end{aligned}

\end{equation}

Substitution gives

\begin{equation}\label{eqn:fibonacci_sinh:140}

F_m = \inv{\sqrt{5}} \lr{ \phi^m – \lr{\frac{-1}{\phi}}^m }.

\end{equation}

Multiplying by \( i^m \), we have

\begin{equation}\label{eqn:fibonacci_sinh:160}

\begin{aligned}

i^m F_m

&= \inv{\sqrt{5}} \lr{ i^m \phi^m – \inv{(-i)^m} \lr{\frac{-1}{\phi}}^m } \\

&= \inv{\sqrt{5}} \lr{ \lr{ i \phi} ^m – \lr{i \phi}^{-m} } \\

\end{aligned}

\end{equation}

We can write any exponent in terms of \( e \)

\begin{equation}\label{eqn:fibonacci_sinh:180}

a^m = e^{\ln a^m} = e^{m \ln a},

\end{equation}

so

\begin{equation}\label{eqn:fibonacci_sinh:200}

\begin{aligned}

i^m F_m

&= \inv{\sqrt{5}} \lr{ e^{m \ln \lr{ i \phi}} – e^{-m \ln\lr{i \phi} } } \\

&= \inv{\sqrt{5}} 2 \sinh\lr{ m \ln \lr{ i \phi } },

\end{aligned}

\end{equation}

as we wanted to show. It’s a bit strange looking, but we see why it works.

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