Example 1 Find the value of $\xi_{0}=\langle-1,1,4, \overline{3,1,1,1,3,7}\rangle$.

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Solve for $\xi_{0}$, the maximal pure cyclic part is $\xi_{3}=\langle\overline{3,1,1,1,3,7}\rangle . \xi_{3}$ satisfies
$$\xi_{3}=\left\langle 3,1,1,1,3,7, \xi_{3}\right\rangle$$

First, calculate $\langle 3,1,1,1,3\rangle$ and $\langle 3,1,1,1,3,7\rangle$. We have
$$\begin{aligned}
\langle 3,1,1,1,3\rangle & =\langle 3,1,1,4 / 3\rangle=\langle 3,1,7 / 4\rangle \\
& =\langle 3,11 / 7\rangle=40 / 11 \\
\langle 3,1,1,1\rangle= & \langle 3,1,2\rangle=\langle 3,3 / 2\rangle=11 / 3 .
\end{aligned}$$

Using formula (17) from § 3, we get
$$\langle 3,1,1,1,3,7\rangle=(7 \cdot 40+11) /(7 \cdot 11+3)=291 / 80$$

and
$$\xi_{3}=\left\langle 3,1,1,1,3,7, \xi_{3}\right\rangle=\left(291 \xi_{3}+40\right) /\left(80 \xi_{3}+11\right),$$

which leads to $2 \xi_{3}^{2}-7 \xi_{3}-1=0$. From this and $\xi_{3}>1$, we get
$$\xi_{3}=(7+\sqrt{57}) / 4$$

Therefore,
$$\begin{aligned}
\xi_{0} & =\langle-1,1,4,(7+\sqrt{57}) / 4\rangle=\langle-1,1,(\sqrt{57}+1) / 2\rangle \\
& =\langle-1,(\sqrt{57}+3) /(\sqrt{57}+1)\rangle=(3-\sqrt{57}) / 24
\end{aligned}$$