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

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 \(1.\)

Log in to reply

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?

Log in to reply

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 \(1\).

Log in to reply

I dont understand what you're implying here.Assuming you have seen the solution to that problem, we see that it is

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?Log in to reply

Log in to reply

Log in to reply

Log in to reply

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 \(1\)).

Log in to reply

Log in to reply

Log in to reply

If \(k<0\), then all the terms will be negative, so, we certainly don't get solutions to your problem.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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.

Log in to reply

What do you think @Otto Bretscher? sorry if this gives a way answer to your question but please have a look !

Log in to reply

Looks interesting... I will study your work when I have a block of free time

Log in to reply