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# Infinite continued fraction?

$\Large \cfrac x{1 - \cfrac x{1 - \cfrac x{1 - \cfrac x{\ddots }}}} = 1$

Let $$f_n(x)$$ denote the recurrence relation, $$f_n(x) = \dfrac x{1- f_{n-1}(x)}$$, where $$n = 2,3,4,\ldots$$ with initial term $$f_1(x) = x$$. Find the value of $$x$$ satisfying $$\displaystyle \lim_{n\to\infty} f_n (x) = 1$$.

In other words, find the value of $$x$$ satisfying the equation above with the expression on the left hand side representing an infinitely nested fraction.

Note by Aareyan Manzoor
1 year, 4 months ago

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There is no solution to the above, per the following:

Let $$y$$ equal the continued fraction $$\cfrac{x}{1-\cfrac{x}{1-\cfrac{x}{1-\ddots}}}.$$ We know $$y=1.$$ Also, we know $$\cfrac{x}{1-\cfrac{x}{1-\cfrac{x}{1-\ddots}}}=\dfrac{x}{1-y}.$$ However, since $$y=1,$$ this has a denominator of $$0.$$ Thus, there is no solution.

Note that this only works because the continued fraction equals $$1.$$ The method Otto and Ivan talk about is better for the fraction not equalling $$1.$$ · 1 year, 4 months ago

This approach is equivalent to ours, pointing out that 1 fails to be a fixed point of the iteration. · 1 year, 4 months ago

view this. the topmost solution has

$x_1^2+kx_2^2≥2\sqrt{k}x_1x_2\\ (1-k)x_2^2+\dfrac{k}{1-k}x_3^2≥2\sqrt{k}x_2x_3\\ \left(1-\dfrac{k}{1-k}\right)x_3^2+\left(\dfrac{k}{1-\dfrac{k}{1-k}}\right)x_4^2≥2\sqrt{k}\\ \cdots\cdots \\ \left(\underbrace{1-\dfrac{k}{1-\dfrac{k}{1-.....}}}_{\text{n-1 times}}\right)x_{n-1}^2+\left(\underbrace{\dfrac{k}{1-\dfrac{k}{1-\dfrac{k}{1-.....}}}}_{\text{n times}}\right)x_{n}^2≥2\sqrt{k}x_nx_{n-1}$ we had the last term $\underbrace{\dfrac{k}{1-\dfrac{k}{1-\dfrac{k}{1-.....}}}}_{\text{n times}}=1$ but when n aproaches $$\infty$$? we know that the max exists from here, then there must exist a k, right? · 1 year, 4 months ago

No, how can you say the last term is equal to $$1$$? All you can say is that

$x_1^2 + x_2^2 + \ldots + x_{n-1}^2 + \frac{k}{1-\frac{k}{1-\ldots}} x_n^2 \ge 2\sqrt{k} \cdot (x_1x_2 + x_2x_3 + \ldots + x_{n-1}x_n)$

where the tower has $$n-1$$ levels. You cannot simply let $$n$$ going to infinity and assume that that last term will magically turn to $$1$$. · 1 year, 4 months ago

I dont understand what you're implying here.Assuming you have seen the solution to that problem, we see that it is Needed for that term to be one.From the second problem we see that the maximum value exists! why cant we conclude that there exists a x? · 1 year, 4 months ago

The solution used in the first problem cannot be generalized to infinite number of variables. · 1 year, 4 months ago

Why not? Can you elaborate there? · 1 year, 4 months ago

Exactly because of this issue: there is no solution to the equation you shown, so you cannot pick $$k$$ to allow you to add up everything. · 1 year, 4 months ago

Nice! thinking reversely! Do you think this is worth studying further? · 1 year, 4 months ago

I did find the following interesting: for what $$k$$ does

$\displaystyle \Large{ \frac{k}{1 - \frac{k}{1 - \frac{k}{1 - \ldots}}} }$

converge, and to what limit? From testing a few things, I suspect all $$k \le \frac{1}{4}$$ works, tending to $$\frac{1 - \sqrt{1 - 4k}}{2}$$. If this is proven, this instantly tells that there is no solution to your equation (because the limit is never $$1$$). · 1 year, 4 months ago

Yes, this is exactly right, Ivan! The iteration $$f(x)=\frac{k}{1-x}$$ has fixed points at $$\frac{1\pm\sqrt{1-4k}}{2}$$ for $$0\leq k\leq 0.25$$, and the sequence with the seed $$x_0=k$$ will increase towards the lower of those fixed point in that case... all this is quite easy to verify. · 1 year, 4 months ago

I find that $$k < 0$$ also converge. And for completeness, we also need to prove that $$k > \frac{1}{4}$$ don't converge. · 1 year, 4 months ago

About $$k>0.25$$... if there are no fixed points, then it cannot converge.

If $$k<0$$, then all the terms will be negative, so, we certainly don't get solutions to your problem. · 1 year, 4 months ago

Well, I mean, how do you prove there are no fixed points? And I generalized the problem, although it's not necessary for the problem given in the note. · 1 year, 4 months ago

Simple...you find the fixed points by letting $$f(x)=x$$ ;) · 1 year, 4 months ago

...oh, I misinterpreted the meaning of "fixed points" there; I thought fixed points of the limit (with respect to $$k$$) instead of $$f$$ (with respect to $$x$$).

Additionally, the fact that there is no fixed points thus it doesn't converge hinges on the fact that $$f$$ is continuous, so that might need a mention. · 1 year, 4 months ago

$x=\dfrac{k}{1-x}\\x^2-x+k=0\\x=\dfrac{1\pm\sqrt{1-4k}}{2}$ · 1 year, 4 months ago

So, are we all convinced that your problem has no solutions? · 1 year, 4 months ago

I am convinced that there are no solution. · 1 year, 4 months ago

Actually, we did not have to do all this work. It suffices to point out that the iteration function $$f(x)=\frac{k}{1-x}$$ has a pole at 1 for any non-zero $$k$$. · 1 year, 4 months ago

That was the motivation of the problem... It was undefined at the point x=1. I was interested to see how the combination of your two "irrelevant to this note" problems actually produced something which was fascinating though disproven! · 1 year, 4 months ago

Fascinating indeed! · 1 year, 4 months ago

yes exactly · 1 year, 4 months ago

What do you think @Otto Bretscher? sorry if this gives a way answer to your question but please have a look ! · 1 year, 4 months ago