I have calculated the continued fraction of $\alpha=\frac{6+\sqrt{47}}{11}$ which equals $\overline{[1,5,1,12]}$. Now I am asked to calculated the cont. fraction of $\sqrt{47}$ using this result. I am not sure whether there is a simple formula to calculate the continued fraction of $\sqrt{47}=11\alpha-6$.
I know the answer to be $\sqrt{47}=[6,\overline{1,5,1,12}]$ (checked by Mathematica) but it's not clear how to arrive at this result using our previous answer.
On
I know the answer to be $\sqrt{47}=[6,\overline{1,5,1,12}]$ (checked by Mathematica) but it's not clear how to arrive at this result using our previous answer.
No ingenuity is needed. The above observation makes the proof mechanical. The above is true
$$\iff\ \sqrt{47}\: =\: 6 + \dfrac{1}{\overline{1,5,1,12}}\: =\: 6 + \dfrac{1}\alpha\ \iff\ \alpha \:=\: \dfrac{1}{\sqrt{47}-6}\: =\: \dfrac{\sqrt{47}+6}{11}$$
$(\sqrt{47}-6)(\sqrt{47}+6)=47-36=11$, so $$(\sqrt{47}-6)\alpha=(\sqrt{47}-6)\left(\frac{\sqrt{47}+6}{11}\right)=1\;,$$ and $$\sqrt{47}-6=\frac1{\alpha}\;.$$
Clearly $\lfloor\sqrt{47}\rfloor=6$, so you know that $$\sqrt{47}=6+\frac1{\left(\frac1{\sqrt{47}-6}\right)}=6+\frac1\alpha=[6,\overline{1,5,1,12}]\;.$$