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Find the value of $x$, if:

$$\large 1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}$$

My attempt:

Noting that: $$\large x=1+\frac{1}{2+\frac{1}{x}}$$

$$x=\frac{1+\sqrt{3}}{2}$$

question: Is my solution correct?

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Yes, it is correct. Good job. =D –  Darth Geek Jul 22 at 10:44
1  
Well, you've left out a lot of algebra between the 2nd and 3rd equations. –  Gerry Myerson Jul 22 at 10:48

2 Answers 2

up vote 3 down vote accepted

Your observation is valid if you know that the continued fraction converges. This is a "simple" continued fraction (technical term), so it's not hard to show it converges. Check a few of the convergents and it will be apparent that these form a couple of sequences, depending on whether the last "denominator" is 1 or 2. However there is a theory that shows the eventually "periodic" continued fractions (like this) are precisely the quadratic irrationals. So your deduction can be made rigorous.

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If others would like to know how he arrived at that:

Like you stated:

$$\large x=1+\frac{1}{2+\frac{1}{x}}$$

$$\large x= 1 + \frac{1}{\frac{2x+1}{x}}$$ $$\large x = 1 + \frac{x}{2x+1}$$ $$\large 2x^2 + x = 2x + 1 + x$$ $$\large 2x^2 - 2x - 1 =0$$ Quadratic Formula: $$x = \frac{2 \pm \sqrt{4 + 8}}{4}$$ $$x = \frac{2 \pm 2\sqrt{3}}{4}$$ $$x = \frac{1 \pm \sqrt{3}}{2}$$

Because $x = \frac{1 - \sqrt{3}}{2}$ is extraneous, our final solution is $$x = \frac{1 + \sqrt{3}}{2}$$

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