Help Infinite continued fractions

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Homework Help Overview

The discussion revolves around evaluating an infinite continued fraction defined by the recursive relationship involving the constants -2 and 1. Participants are exploring the implications of this structure and its representation through a quadratic equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to derive the sum of the continued fraction and are questioning the validity of the results obtained from the quadratic formula. There is a suggestion to write additional approximations to clarify the behavior of the sequence.

Discussion Status

Some participants have provided insights into the nature of the roots of the quadratic equation and have engaged in rewriting the continued fraction in a different form to facilitate solving for x. Multiple interpretations of the results are being explored, particularly regarding which root is relevant.

Contextual Notes

There are indications of confusion regarding the feasibility of the negative root as a solution and the implications of the an value being -2. The discussion reflects uncertainty about the assumptions underlying the continued fraction setup.

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Help! Infinite continued fractions!

Homework Statement


-2 + 1
-2 + 1
-2 + 1
-2 + 1
...

an = -2, bn = 1

Homework Equations


What is the sum of this indefinite continued fraction, putting this into a quadratic equation? x^2 + 2x - 1 = 0

The Attempt at a Solution


Using the quadratic formula, I got .4142 and -2.4142, but there is only 1 sum. The an value is -2 (so is that even possible?). Help quickly!
 

Attachments

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It's going to be the negative root, isn't it? Write down a few more approximations, like -2+1/-2 and -2+1/(-2+1/(-2)) to make it clear.
 


From the .doc image:

x = [-2, -2, -2,...] = [-2],[0] = [-3, 1, 1, 2],[0] =[tex]-1 - \sqrt{}2[/tex]
~ -2.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727
 
Last edited:


[tex]\text{Let }x=-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\dots}}}}\quad\to\quad x+2=\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\dots}}}}\quad\to\quad \frac{1}{x+2}=-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\dots}}}}[/tex]

But the right hand side of the last expression is x.

Therefore, x = 1/(x+2).

Solve that for x.
 

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