How to solve $a_{n} = \frac{2a_{n-1}^2}{a_{n-2}}$, where $a_0 = 1$ and $a_1 = 2$

Question

I'm trying to solve this reccurence relation as a part of the "linear recurrence relations with constant coefficients" chapter, with the hint that the next step is using logarithms.

The problem I'm facing is what to do after the equation is logarithmized (since it's not linear anymore? and the coefficients aren't constant?)

The usual way of solving $a_{n+3} - 2a_{n+2} -4a_{n+1} + 8a_n = 0$ for example, would be to find the characteristic polynomial, which would be $x^3 - 2x^2 - 4x + 8 = 0$ and using it's solutions to predict the form of $a_n$ and solving it using the starting conditions $a_0, a_1, a_2$. I just don't see how a similair way of solving would be relevant here.

Answer 1

Hint :

1. Show inductively that $a_n > 0$ for every $n \geq 0$.

2. Define $b_n = \ln(a_n)$. Show that $(b_n)$ satisfies $$b_n = \ln(2)+ 2b_{n-1}-b_{n-2}$$

3. Solve the recursion for $(b_n)$ (it is now linear, with constant coefficients)

4. Deduce the result for $a_n$.