So I attempted the 2020 paper for BMO2 (https://bmos.ukmt.org.uk/home/bmo2-2020.pdf), but I've been stuck on Q4.
A sequence $b_1,b_2,b_3,…$ of non-zero real numbers has the property that $b_{n+2}=\frac{b^2_{n+1}−1}{b_n}$ for all positive integers n. Suppose that $b_1=1$ and $b_2=k$ where $1<k<2$. Show that there is some constant $B$,depending on $k$, such that $−B≤b_n≤B$ for all $n$. Also show that, for some $1<k<2$, there is a value of $n$ such that $b_n>2020$.
The only idea that came to mind was to try and solve for an nth term (which I could not do) or to use induction to prove the existence of bounds, which doesn't seem to work out since we have the division.
Looking online I found someone's solution, in which they gave the nth term as
$$ b_n = \frac{\sin(n\gamma)}{\sin(\gamma)} $$
where $\gamma = \arccos(k/2)$
And from here it's trivial.
However they did not say how they derived this formula apart from "intuition". The only intuitive link I see between the nth term and the recurrence is that both feel "bouncy".
The only method I can think of that would be helpful here is generating functions, but as far as I'm aware they cannot help when we have products of sequences except for the special case of convolutions and convolutions multiplied by binomial coefficients.
So my questions are:
- How could you derive this nth term?
- Does this fit into a larger pattern of recurrence relations that can be solved using slightly "squished" trig function?
- How could this be solved using only elementary mathematics? (As the competition is intended so every question can, at least in theory, be done using clever tricks without higher maths (given that the question did not ask for an nth term, I presume there is a method to solve this that does not actually require finding one)
Best Answer
A while ago somebody posted a similar question (or maybe the same) but I forgot where it was.
$$b_{n+2} b_n = b_{n+1}^2 - 1, b_{n+1} b_{n-1} = b_n^2 - 1\\ \implies b_{n+1}^2-b_{n+2}b_n = b_n^2-b_{n+1}b_{n-1}\\ \implies b_{n+1}(b_{n+1}+b_{n-1})=b_n(b_n+b_{n+2})\\ \implies \frac{b_{n+1}+b_{n-1}}{b_n}=\frac{b_{n+2}+b_n}{b_{n+1}}$$
Therefore $$\frac{b_{n+1}+b_{n-1}}{b_n} \equiv \frac{b_3+b_1}{b_2}=\frac{k^2-1+1}{k}=k\\ \implies b_{n+1}=k b_n-b_{n-1}$$
This is a second order linear difference equation with characteristic function $\lambda^2-k\lambda+1=0$ which has two complex roots. So you can follow this link to get the solution which then would have the trig term.