STEP paper 1 1999 question - is it possible to justify this solution?

Question

I am currently a year 13 student preparing for STEP exams for university, and I came across the following problem. I solved it correct, but also solved it a second time using an alternate, much quicker method to get the same results for part a) and part c). This alternate solution, however, contained a step which I do not know how to justify.

Problem:

Let $(x_n)_{n=1}^i,\,i\in\mathbb{N}$ be a sequence satisfying $$x_1=1+\frac1{x_2},\,x_2=1+\frac1{x_3},\,\dots,\,x_i=1+\frac1{x_{1}},\ldots \quad x_n>0\,\,\forall n$$ Prove that

a) $x_n>1$ for all $n$.

b) $x_1-x_2=-\displaystyle \frac{x_2-x_3}{x_2x_3}$

c) $x_1=x_2=\dots=x_i$

My solution:

Write $$x_1=1+\cfrac1{x_2}=1+\cfrac1{1+\cfrac1{x_3}}=1+\cfrac1{1+\cfrac1{\ddots+\cfrac1{x_i}}}$$

Due to the recurrence relation being cyclical we can continue this process to get the infinite fraction

$$x_1=1+\cfrac1{1+\cfrac1{1+\ddots}}$$

which can be solved to find $x_1=\frac{1\pm\sqrt{5}}2$, giving $x_1=\frac{1+\sqrt5}2$ since $x_n>0$. Since there was nothing special about picking $x_1$ at the start, this argument also applies to $x_n$ for all $n$. This proves both a) and c).

My question:

Is it possible to justify the highlighted step in my solution? I don't think this method is justifiable as it seems too shady to me due to the fact that I am repeating a recurrence relation infinitely to reduce $x_1$ to a constant. If this step is justifiable, why?

Answer 1

Due to the recurrence relation being cyclical we can continue this process to get the infinite fraction [...] which can be solved to find $x_1=\frac{1\pm\sqrt{5}}2$, giving $x_1=\frac{1+\sqrt5}2$ since $x_n>0$

Presumably you solved it by writing $t=1+\dfrac{1}{t}$ and solving for the limit $t\,$. Problem with this "shortcut" solution is that you have to first prove that the infinite fraction converges, otherwise you cannot pass what is essentially the recurrence $\displaystyle t_{n+1}=1+\frac{1}{t_n}$ to the limit.

Answer 2

You could consider the sequence $a_0=1$ satisfying the following induction relation $$ a_{n+1}=1+\frac{1}{a_n} $$ Hence $$ a_{n}=1+\displaystyle \frac{1}{a_{n-1}}=1+\frac{1}{1+\displaystyle \frac{1}{a_{n-2}}}=1+\frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{a_{n-3}}}} $$ If you prove that $\left(a_n\right)_{n \in \mathbb{N}}$ converges ; hence the limit satisfies $$ \ell=1+\frac{1}{\ell} \Leftrightarrow \ell^2=\ell+1 $$ You'll find that with $a_n>0$ that $\displaystyle \ell=\frac{1+\sqrt{5}}{2}$

The number $\ell$ that is the limit of this sequence satisfies the expression you mention, in fact it is a continued fraction that it is written $\left[\overline{1}\right]$.

However, the question you have might not expect you find explicitly $x_1$.

How did you find $$ x_1=\frac{1+\sqrt{5}}{2} $$ ?

Answer 3

Define $f(t)=1+\dfrac{1}{t}$, before taking the infinite fraction you have shown that $g(x_1)=f^{(i)}(x_1)=x_1$. So, $x_1$ is a fixed point of $g$.

Repeating the process to obtain the infinite fraction, will not let you find $x_1$, $x_1$ is not a fixed point of $f$ but a fixed point of $g$. The infinite fraction is well-defined as long as $x_1$ is a fixed point of $g$, so doing this process will not give you anything more, as for any fixed point of $g$ the infinite fraction is well defined.

Of course a fixed point of $f$ is a fixed point of $g$, however $g$ may have different fixed points. If you prove that $g$ has a unique fixed point then you can find the value of $x_1$. The point is what you have derived reduces the problem in studying the fixed points of $g$.