Using the itineraries of the Gauss map write the continued fraction expansion of the number $0 \leqslant \alpha \leqslant 1$ such that

$$\displaystyle \alpha = \dfrac{1}{4+\dfrac{1}{3+\alpha}}$$

I know that if $\alpha$ is irrational then the itineraries of the Gauss map give the entries of the continued fraction expansions and in my book it is quoted that if $\alpha$ is irrational then.

$$\displaystyle \alpha = \dfrac{1}{a_0+\dfrac{1}{a_1+\dots \dfrac{1}{a_n+G^{n+1}(x)}}}$$

So surely it cant be as easy as saying that $G^2(\alpha)=\alpha$ so $\alpha=[4,3,4,3,4,3,4,\dots]$

Is there a better way of writing a solution to this problem? Formal answers only please, I can see it roughly works.


That's about all there is to it. The point is that there are powerful existence and uniqueness theorems for continued fractions, and they can be proved by expansiveness of the Gauss map. So what you are doing is applying those powerful tools, and their power is exhibited by the ease of applying them.


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