Find the value of $x_1$

Question

The positive numbers $x_1,x_2,\cdots,x_n$ $n\ge3$ 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 $$x_n=1+\frac{1}{x_1}$$ Find the value of $x_1.$

My first thought is that this question has an unknown number of variables$:$ $x_1,\cdots, x_n.$ That makes it seem rather complicated. I might, if necessary, try to understand the result by choosing an easy value for $n$ (maybe $n = 3$). If I manage to prove some of the results in this special case, I will certainly go back to the general case$:$ doing the special case might help me tackle the general case.

Also, I don't think that I can apply any inequality here. I can see that each $x_i>1.$ But now I'm stuck. No idea is striking to my mind.

Any help is greatly appreciated.

Answer 1

the framework for this is worth knowing. You have a Mobius transformation; these are functions $f(x) = \frac{ax+b}{cx+d} $ where $a,b,c,d$ are constants and $ad-bc \neq 0.$ The good bit is that such a function has an easy inverse, $g(x) = \frac{dx - b}{-cx + a}$

You have $f(x) = \frac{x+1}{x}.$ We find compositions (iterating) $f^{(2)}x = \frac{2x+1}{x+1},$ then $f^{(3)}x = \frac{3x+2}{2x+1},$ $f^{(4)}x = \frac{5x+3}{3x+2}.$

If you know how to multiply matrices, we are just finding the coefficient matrix $$ M_n = \left( \begin{array}{cc} 1&1 \\ 1&0 \\ \end{array} \right)^n $$
or $$ M_n = \left( \begin{array}{cc} F_{n+1}&F_n \\ F_n&F_{n-1} \\ \end{array} \right) $$ with Fibonacci numbers indexed so $F_5 = 5$

Note that $ \det \left( \begin{array}{cc} 1&1 \\ 1&0 \\ \end{array} \right) = -1 \; , \;$ so that $ \det M_n = (-1)^n \;$ and ....

What does it say if $$ x_1 = \frac{F_{n+1} x_1 +F_n }{F_n x_1 +F_{n-1} }\; \; ? \; \; $$ Use the quadratic formula

P. S. All you really need is that $F_{n+1} = F_n + F_{n-1}.$ If we had $$ x = \frac{(B+C) x +B }{B x + C }\; \; ? \; \; $$

Answer 2

Let $\alpha:=x_1$.

You're told: $$\alpha=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots+\cfrac{1}{1+\cfrac{1}{\alpha}}}}}$$

We could replace this $n$-tiered continued fraction by a $2n$-tiered continued fraction by substituting $\alpha=1+\cdots$:

$$\alpha=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots+\cfrac{1}{1+\cfrac{1}{\alpha}}}}}}}}}}$$

Again:

$$\alpha=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots+\cfrac{1}{1+\cfrac{1}{\alpha}}}}}}}}}}}}}}}$$

We would like to say "etc." and pass to a limit.

Since everything is positive, according to one of the conditions in the analysis in my question, if $\alpha$ is positive then it must equal the infinite continued fraction: $$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}$$Which famously converges to the Golden Ratio, $\phi$.

There is another solution, the rational conjugate of $\phi$, by inspection. I am not sure how to prove this is the only negative solution. However, you asked about positive numbers, so all is well.

There is another treatment of this problem via real analytic technique by Michael Penn, here. This is the same idea as Will Jagy's.