\[ \Large \cfrac x{1 - \cfrac x{1 - \cfrac x{1 - \cfrac x{\ddots }}}} = 1 \]

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.

## Comments

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TopNewestThere 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 \(1.\) – Trevor B. · 10 months, 3 weeks ago

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– Otto Bretscher · 10 months, 3 weeks ago

This approach is equivalent to ours, pointing out that 1 fails to be a fixed point of the iteration.Log in to reply

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? – Aareyan Manzoor · 11 months ago

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

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 \(1\). – Ivan Koswara · 11 months ago

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Neededfor 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? – Aareyan Manzoor · 11 months agoLog in to reply

– Ivan Koswara · 11 months ago

The solution used in the first problem cannot be generalized to infinite number of variables.Log in to reply

– Aareyan Manzoor · 11 months ago

Why not? Can you elaborate there?Log in to reply

– Ivan Koswara · 11 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.Log in to reply

– Aareyan Manzoor · 11 months ago

Nice! thinking reversely! Do you think this is worth studying further?Log in to reply

\[\displaystyle \Large{ \frac{k}{1 - \frac{k}{1 - \frac{k}{1 - \ldots}}} }\]

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 \(1\)). – Ivan Koswara · 11 months ago

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– Otto Bretscher · 11 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.Log in to reply

– Ivan Koswara · 11 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.Log in to reply

If \(k<0\), then all the terms will be negative, so, we certainly don't get solutions to your problem. – Otto Bretscher · 11 months ago

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– Ivan Koswara · 11 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.Log in to reply

– Otto Bretscher · 11 months ago

Simple...you find the fixed points by letting \(f(x)=x\) ;)Log in to reply

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. – Ivan Koswara · 11 months ago

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– Aareyan Manzoor · 11 months ago

\[x=\dfrac{k}{1-x}\\x^2-x+k=0\\x=\dfrac{1\pm\sqrt{1-4k}}{2}\]Log in to reply

– Otto Bretscher · 11 months ago

So, are we all convinced that your problem has no solutions?Log in to reply

– Aareyan Manzoor · 11 months ago

I am convinced that there are no solution.Log in to reply

– Otto Bretscher · 11 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 \(k\).Log in to reply

– Aareyan Manzoor · 11 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!Log in to reply

– Otto Bretscher · 11 months ago

Fascinating indeed!Log in to reply

– Otto Bretscher · 11 months ago

yes exactlyLog in to reply

@Otto Bretscher? sorry if this gives a way answer to your question but please have a look ! – Aareyan Manzoor · 11 months ago

What do you thinkLog in to reply

– Otto Bretscher · 11 months ago

Looks interesting... I will study your work when I have a block of free timeLog in to reply