<s>[INST] Closed form for this continued fraction Is there a closed form for this continued fraction?
$$x+\frac{1}{x+\frac{1}{x+\frac{1}{...}}}$$
 [/INST] Here's a cute, handwavy way to do it:
$$ f(x) = x + \cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}}$$
Notice then that
$$ \frac{1}{f(x)} = \cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}} $$
Thus
\begin{align}
 x+\frac{1}{f(x)} &= x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}} \\
&=f(x)
\end{align}
So, we have a functional relationship
$$ x + \frac{1}{f(x)} = f(x) $$
or equivalently
$$ xf(x) + 1 = f(x)^{2} $$
This quadratic equation may be solved quite simply, we have candidate solutions:
$$ f_{\pm}(x) = \frac{x\pm\sqrt{x^{2}+4}}{2}$$
We should note that $f(1)$ as a continued fraction is precisely the golden ratio. Hence we deduce
$$ f(x) = \frac{x+\sqrt{x^{2}+4}}{2}. $$
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