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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.

Bill Dubuque
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Blue Cat Blues
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    Firstly, I'd try to solve it for $n=3$, to get the hang of it. Secondly, you may try and solve the system of equations directly. Try writing $x_{n-1}$ as a function of $x_1$, and think about how to continue from there. – Rei Henigman Feb 25 '23 at 20:02
  • You say: and also $x_n=1+\frac{1}{x_1}$ - May be you mean $x_{n+1}$ instead of $x_1$? – NoChance Feb 25 '23 at 20:18
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    @NoChance No, I am fairly certain $x_1$ was intended. This cyclicity is what makes it an interesting problem – FShrike Feb 25 '23 at 20:33
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    I can't do better than suggest OP take Rei's advice: try it out for $n=3$ and $n=4.$ I think OP should be able to have a go at this question and I don't want to post an answer until I have seen some effort from OP.... – Adam Rubinson Feb 25 '23 at 20:38
  • Thanks for the riddle, and +1 for your different solution. I'll add this question to the would-never-be-published riddle book I'm writing – Rei Henigman Feb 26 '23 at 06:36
  • Follows immediately from the formula for the $n$'th composition of a Mobius function here in the linked dupe (this is what is done implicitly in Jagy's answer below – Bill Dubuque Feb 26 '23 at 18:27
  • See https://math.stackexchange.com/a/3530501/42969 or https://math.stackexchange.com/q/2609631/42969 – Martin R Feb 26 '23 at 18:27

2 Answers2

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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 $$ x = \frac{(B+C) x +B }{B x + C }\; \; ? \; \; $$

Will Jagy
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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.

FShrike
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  • $x =1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}$ then $x = \frac1{x-1}$ so $x^2-x-1 = 0$. Thus, the other solution is $-1/\phi$. You can add this as an edit maybe – D S Feb 26 '23 at 05:23
  • @DS It doesn’t work like that. If $x$ equals that fraction, it must be the Golden Ratio, since that fraction converges to one and only one thing. I don’t think I can use continued fraction theory in the instance that $x<0$ – FShrike Feb 26 '23 at 10:08
  • but didn't you say, "There is another solution, the rational conjugate of $\phi$ ,by inspection. I am not sure how to prove this is the only negative solution." I just thought I'll add it in the comments – D S Feb 27 '23 at 05:32
  • @DS I did say that, but I don’t think your proof works – FShrike Feb 27 '23 at 07:15