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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
8 months, 2 weeks 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 \(1.\) Trevor B. · 8 months, 1 week ago

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@Trevor B. This approach is equivalent to ours, pointing out that 1 fails to be a fixed point of the iteration. Otto Bretscher · 8 months, 1 week ago

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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 · 8 months, 2 weeks ago

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@Aareyan Manzoor 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 \(1\). Ivan Koswara · 8 months, 2 weeks ago

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@Ivan Koswara 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? Aareyan Manzoor · 8 months, 2 weeks ago

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@Aareyan Manzoor The solution used in the first problem cannot be generalized to infinite number of variables. Ivan Koswara · 8 months, 2 weeks ago

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@Ivan Koswara Why not? Can you elaborate there? Aareyan Manzoor · 8 months, 2 weeks ago

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@Aareyan Manzoor 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. Ivan Koswara · 8 months, 2 weeks ago

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@Ivan Koswara Nice! thinking reversely! Do you think this is worth studying further? Aareyan Manzoor · 8 months, 2 weeks ago

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@Aareyan Manzoor 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 \(1\)). Ivan Koswara · 8 months, 2 weeks ago

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@Ivan Koswara 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. Otto Bretscher · 8 months, 2 weeks ago

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@Otto Bretscher I find that \(k < 0\) also converge. And for completeness, we also need to prove that \(k > \frac{1}{4}\) don't converge. Ivan Koswara · 8 months, 2 weeks ago

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@Ivan Koswara 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. Otto Bretscher · 8 months, 2 weeks ago

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@Otto Bretscher 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. Ivan Koswara · 8 months, 2 weeks ago

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@Ivan Koswara Simple...you find the fixed points by letting \(f(x)=x\) ;) Otto Bretscher · 8 months, 2 weeks ago

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@Otto Bretscher ...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. Ivan Koswara · 8 months, 2 weeks ago

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@Otto Bretscher \[x=\dfrac{k}{1-x}\\x^2-x+k=0\\x=\dfrac{1\pm\sqrt{1-4k}}{2}\] Aareyan Manzoor · 8 months, 2 weeks ago

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@Aareyan Manzoor So, are we all convinced that your problem has no solutions? Otto Bretscher · 8 months, 2 weeks ago

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@Otto Bretscher I am convinced that there are no solution. Aareyan Manzoor · 8 months, 2 weeks ago

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@Aareyan Manzoor 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 \(k\). Otto Bretscher · 8 months, 2 weeks ago

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@Otto Bretscher 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! Aareyan Manzoor · 8 months, 2 weeks ago

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@Aareyan Manzoor Fascinating indeed! Otto Bretscher · 8 months, 2 weeks ago

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@Aareyan Manzoor yes exactly Otto Bretscher · 8 months, 2 weeks ago

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@Aareyan Manzoor What do you think @Otto Bretscher? sorry if this gives a way answer to your question but please have a look ! Aareyan Manzoor · 8 months, 2 weeks ago

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@Aareyan Manzoor Looks interesting... I will study your work when I have a block of free time Otto Bretscher · 8 months, 2 weeks ago

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