Going from Infinite Continued Fraction to real numbers

Question

I have been given the infinite continued fraction of $\left[12, 2, 2, 12,\dots\right]$ where the first 12 is the natural number part and the remaining numbers repeat. I am familiar with how to convert real numbers into continued fractions, but I am stuck as to how to convert a continued fraction into a real number. I know that since this is infinite, it must be a irrational number, so it must be a $\sqrt{n}$ or $a + \sqrt{n}$. What is a good way to start approaching this question? Thank you.

Answer 1

Hint: Call the value $x$. Subtract $12$, invert, subtract $2$, invert, subtract $2$, and invert. You get $x$ back again.

Write this down and solve for $x$.

$$x = \frac1{\frac1{\frac1{x-12}-2}-2} $$

Answer 2

Edit (almost 7 years later): I now know of a different way to solve such problems; so, I thought I'd provide an update! I wrote about the magic box in The Aperiodical (link) and thought it would be worthwhile to apply it for your problem:

The final convergent given is equal to the expression about which you have asked; let us denote it as $y$. Next, we can proceed by solving the quadratic equation:

$$\frac{62y + 25}{5y + 2} = y$$

You could do this by hand, but I outsourced it to WolframAlpha: $y = 6 + \sqrt{41}$

You can then ask WolframAlpha to compute the continued fraction for this value of $y$, and you will see that it returns the result of $[12, 2, 2, 12, 2, 2, 12, \ldots]$ as desired. (WA computation)

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Original response (from 2017): Another approach (I'm not sure if your first comma was intended to be a semicolon; in any event, you can modify slightly if your notation represents something different from the equations below, but the method is roughly the same):

If I am understanding your notation correctly, then you are considering the continued fraction:

Because of the repetition, this could be rewritten as:

Now, you have an equation with one variable. In trying to solve for $x$, you will come to see that this is a quadratic equation; thus, two possible solutions will arise. However, one of them will be negative and the other one will be positive. Since your constructed number is manifestly positive, you'll want the latter.