# Need to show $x_1=x_2=x_3=\ldots=x_n=\phi$ for the given sequence of $n$ positive positive numbers.

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

1. $x_1\cdot x_2\cdot x_3 \cdots x_n > 1$
2. $x_1 - x_2 = -\frac{x_2-x_3}{x_2x_3}$
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:

1. 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.

1. 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.

• Let $f(x) = 1+ {1 \over x}$ and examine how $f(f(...f(x)...))$ behaves and look for fixed points. For $x \ge 1$, $f$ is bounded and monotonic. – copper.hat Feb 3 '17 at 6:25
• @copper.hat It seems to be behaving sort of like $f^n(x)=\frac{F_n x+F_{n-1}}{F_{n-1}x+F_{n-2}}$, where $F_n$ is the $n$th Fibonacci number. – Hungry Blue Dev Feb 3 '17 at 6:43
• @copper.hat Am I correct? – Hungry Blue Dev Feb 3 '17 at 6:46
• @Astrobleme Not too surprising, since if $x_1=x_2$ then it's the root of $x^2-x-1=0$ i.e. the golden ratio. – dxiv Feb 3 '17 at 6:49
• For each $i$ we have $|x_i-x_{i+1}|=|(x_{i+1}-x_{i+2})/(x_ix_{i+1})|\leq|x_{i+1}-x_{i+2}|$, because $x_j>1$ for each $j$. – Matemáticos Chibchas Feb 3 '17 at 7:44

Using $(2)$ repeatedly you get $$\lvert x_1 - x_2 \rvert = \frac{\lvert x_2-x_3\rvert }{x_2x_3} = \frac{\lvert x_3-x_4\rvert }{x_2x_3^2x_4} = \ldots = \frac{\lvert x_{n}-x_1 \rvert }{x_2x_3^2x_4^2\cdots x_{n-1}^2x_n^2x_1} \\ = \frac{\lvert x_1-x_2 \rvert }{x_2x_3^2x_4^2\cdots x_{n-1}^2x_n^2x_1^2 x_2} = \frac{\lvert x_1-x_2 \rvert }{(x_1\cdots x_n)^2}$$ In $(1)$ you have shown that the denominator is strictly greater than one, therefore $x_1 - x_2 = 0$ must hold, and consequently, all $x_i$ are equal.