The question states:
There are $n$ positive numbers $x_1,x_2,\ldots, x_n$, where $n \geq 3$, satisfy $$ x_1=1+\frac{1}{x_2},\ x_2 = 1+\frac{1}{x_3},\ \cdots, \ x_{n-1}=1+\frac{1}{x_n} $$ and also, $$ x_n=1+\frac{1}{x_1} $$ Show that
- $x_1\cdot x_2\cdot x_3 \cdots x_n > 1$
- $x_1 - x_2 = -\frac{x_2-x_3}{x_2x_3}$
- $x_1 = x_2 = \cdots = x_n$
Hence find the value of $x_1$
I could do $(1)$ and $(2)$ quite easily. Here's how:
- Plugging in $$ x_1=1+\frac{1}{x_2},\ x_2 = 1+\frac{1}{x_3},\ \cdots, \ x_{n-1}=1+\frac{1}{x_n} $$ and $$ x_n=1+\frac{1}{x_1} $$
in $x_1\cdot x_2\cdot x_3 \cdots x_n$, we get $$ \left(1+\frac{1}{x_1}\right)\cdot \left(1+\frac{1}{x_2}\right)\cdot \left(1+\frac{1}{x_3}\right)\cdots \left(1+\frac{1}{x_n}\right)\\ =1+k>1 $$ where $k$ is a positive number.
- Just by substituting the expressions for $x_1$ and $x_2$, we get $$ x_1-x_2=1+\frac{1}{x_2}-1-\frac{1}{x_3}=\frac{1}{x_2}-\frac{1}{x_3}\\ =-\frac{x_2-x_3}{x_2x_3} $$
Now I've tried for a day but I couldn't find a way to prove that they're all equal. A nudge in the proper direction would be appreciated.
Edit
Following copper.hat's advice, I decided to analyse the function $f(x)=1+\frac{1}{x}$. Here's what I did:
$$ f(x)=1+\frac{1}{x} \\ (f \circ f)(x)=f^2(x)=1+\frac{1}{1+\frac{1}{x}}\\ \vdots\\ \lim_{n \to \infty}\underbrace{f \circ f \cdots f}_{n \text{ times}}=\lim_{n \to \infty}f^n(x)=1+\frac{1}{1+\frac{1}{\ddots}}=\phi $$
Therefore, intuitively it's easy to see that all the terms are $\phi$. But how do I prove this rigorously? Or how should I write it to prove the third claim? Basically this post is about seeking guidance in mastering Mathematical lingo now.