# Calculating the continued fraction of $\sqrt{47}$ using a different result

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.

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I think I found the answer: Those two cont'd fractions are nearly identical because $47=6^2+11$, and thus when calculating the continued fraction the second step in calculating $\sqrt{47}$ coincides with the first step of $\alpha$, thus those two expressions are the same from that point onwards. I.e., $(\sqrt{47}-6)^{-1}=\alpha$ –  ClausW Mar 4 '12 at 13:52
Hint: $6=\sqrt{36}$ and $11=47-36$ (consider the conjugate!). (seeing your answer : yes you are right!) –  Raymond Manzoni Mar 4 '12 at 13:54

$(\sqrt{47}-6)(\sqrt{47}+6)=47-36=11$, 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 $$\sqrt{47}=6+\frac1{\left(\frac1{\sqrt{47}-6}\right)}=6+\frac1\alpha=[6,\overline{1,5,1,12}]\;.$$

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Just my comment of a minute ago, only you explained it better. Thanks. :) –  ClausW Mar 4 '12 at 13:54
@Claus: I saw your comment just after I posted; good job. –  Brian M. Scott Mar 4 '12 at 13:55
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.
$$\iff\ \sqrt{47}\: =\: 6 + \dfrac{1}{\overline{1,5,1,12}}\: =\: 6 + \dfrac{1}\alpha\ \iff\ \alpha \:=\: \dfrac{1}{\sqrt{47}-6}\: =\: \dfrac{\sqrt{47}+6}{11}$$