$e_{n+1} = K e_n e_{n-1} $ is $|e_{n+1}| = C|e_n|^{\varphi}$?

Question

if $ e_{n+1} = K e_n e_{n-1} $ ($K$ is a constant, and $e_n$ is a serise),
then, $ | e_{n+1} | = C|e_n|^\varphi$($C$ is constant) and $\varphi$ is golden ratio.

Is this true? If yes, How can I show this?

Answer 1

Take the logarithm of both sides! Letting $k_n=\log e_n$ the recurrence becomes $k_{n+1}=\log K+k_n+k_{n-1}$, and then letting $m_n=k_n+\log K$ we get $m_{n+1} = k_{n+1}+\log K$ $=\log K+k_n+k_{n-1} + \log K$ = $m_n+m_{n-1}$, so $m_n$ satisfies the Fibonacci recurrence; this means in particular that $m_n = C_1\phi^n + C_2\psi^n$ where as typical $\psi = \bar{\phi} = -\phi^{-1}$ is the conjugate of the golden ratio). If your initial conditions are such that $C_2$ is zero then your formula will hold exactly (as you can see by unwinding the transformations from $e_n$ to $m_n$), but otherwise the relation $e_{n+1} = Ce_n^\phi$ will only hold asymptotically.