Infinite continued fraction?

x1x1x1x=1 \Large \cfrac x{1 - \cfrac x{1 - \cfrac x{1 - \cfrac x{\ddots }}}} = 1

Let fn(x)f_n(x) denote the recurrence relation, fn(x)=x1fn1(x)f_n(x) = \dfrac x{1- f_{n-1}(x)} , where n=2,3,4,n = 2,3,4,\ldots with initial term f1(x)=xf_1(x) = x . Find the value of xx satisfying limnfn(x)=1\displaystyle \lim_{n\to\infty} f_n (x) = 1.

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

Note by Aareyan Manzoor
3 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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

Let yy equal the continued fraction x1x1x1.\cfrac{x}{1-\cfrac{x}{1-\cfrac{x}{1-\ddots}}}. We know y=1.y=1. Also, we know x1x1x1=x1y.\cfrac{x}{1-\cfrac{x}{1-\cfrac{x}{1-\ddots}}}=\dfrac{x}{1-y}. However, since y=1,y=1, this has a denominator of 0.0. Thus, there is no solution.

Note that this only works because the continued fraction equals 1.1. The method Otto and Ivan talk about is better for the fraction not equalling 1.1.

Trevor B. - 3 years, 9 months ago

Log in to reply

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

Otto Bretscher - 3 years, 9 months ago

Log in to reply

view this. the topmost solution has

x12+kx222kx1x2(1k)x22+k1kx322kx2x3(1k1k)x32+(k1k1k)x422k(1k1k1.....n-1 times)xn12+(k1k1k1.....n times)xn22kxnxn1x_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 k1k1k1.....n times=1\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 - 3 years, 9 months ago

Log in to reply

No, how can you say the last term is equal to 11? All you can say is that

x12+x22++xn12+k1k1xn22k(x1x2+x2x3++xn1xn)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 n1n-1 levels. You cannot simply let nn going to infinity and assume that that last term will magically turn to 11.

Ivan Koswara - 3 years, 9 months ago

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 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 - 3 years, 9 months ago

Log in to reply

@Aareyan Manzoor The solution used in the first problem cannot be generalized to infinite number of variables.

Ivan Koswara - 3 years, 9 months ago

Log in to reply

@Ivan Koswara Why not? Can you elaborate there?

Aareyan Manzoor - 3 years, 9 months ago

Log in to reply

@Aareyan Manzoor Exactly because of this issue: there is no solution to the equation you shown, so you cannot pick kk to allow you to add up everything.

Ivan Koswara - 3 years, 9 months ago

Log in to reply

@Ivan Koswara Nice! thinking reversely! Do you think this is worth studying further?

Aareyan Manzoor - 3 years, 9 months ago

Log in to reply

@Aareyan Manzoor I did find the following interesting: for what kk does

k1k1k1\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 k14k \le \frac{1}{4} works, tending to 114k2\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 11).

Ivan Koswara - 3 years, 9 months ago

Log in to reply

@Ivan Koswara Yes, this is exactly right, Ivan! The iteration f(x)=k1xf(x)=\frac{k}{1-x} has fixed points at 1±14k2\frac{1\pm\sqrt{1-4k}}{2} for 0k0.25 0\leq k\leq 0.25, and the sequence with the seed x0=kx_0=k will increase towards the lower of those fixed point in that case... all this is quite easy to verify.

Otto Bretscher - 3 years, 9 months ago

Log in to reply

@Otto Bretscher I find that k<0k < 0 also converge. And for completeness, we also need to prove that k>14k > \frac{1}{4} don't converge.

Ivan Koswara - 3 years, 9 months ago

Log in to reply

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

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

Otto Bretscher - 3 years, 9 months ago

Log in to reply

@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 - 3 years, 9 months ago

Log in to reply

@Ivan Koswara Simple...you find the fixed points by letting f(x)=xf(x)=x ;)

Otto Bretscher - 3 years, 9 months ago

Log in to reply

@Otto Bretscher x=k1xx2x+k=0x=1±14k2x=\dfrac{k}{1-x}\\x^2-x+k=0\\x=\dfrac{1\pm\sqrt{1-4k}}{2}

Aareyan Manzoor - 3 years, 9 months ago

Log in to reply

@Aareyan Manzoor yes exactly

Otto Bretscher - 3 years, 9 months ago

Log in to reply

@Aareyan Manzoor So, are we all convinced that your problem has no solutions?

Otto Bretscher - 3 years, 9 months ago

Log in to reply

@Otto Bretscher I am convinced that there are no solution.

Aareyan Manzoor - 3 years, 9 months ago

Log in to reply

@Aareyan Manzoor Actually, we did not have to do all this work. It suffices to point out that the iteration function f(x)=k1xf(x)=\frac{k}{1-x} has a pole at 1 for any non-zero kk.

Otto Bretscher - 3 years, 9 months ago

Log in to reply

@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 - 3 years, 9 months ago

Log in to reply

@Aareyan Manzoor Fascinating indeed!

Otto Bretscher - 3 years, 9 months ago

Log in to reply

@Otto Bretscher ...oh, I misinterpreted the meaning of "fixed points" there; I thought fixed points of the limit (with respect to kk) instead of ff (with respect to xx).

Additionally, the fact that there is no fixed points thus it doesn't converge hinges on the fact that ff is continuous, so that might need a mention.

Ivan Koswara - 3 years, 9 months ago

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 !

Aareyan Manzoor - 3 years, 9 months ago

Log in to reply

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

Otto Bretscher - 3 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...