Help Infinite continued fractions

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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!
 

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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] =-1 - \sqrt{}2
~ -2.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727
 
Last edited:


\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}}}}

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

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

Solve that for x.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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