Example 2 Find the periodic continued fraction of $\xi_{0}=(\sqrt{14}+1) / 2$.

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We solve according to the method of Theorem 6, which requires finding the smallest $k>h \geqslant 0$ such that $\xi_{h} = \xi_{k}$, i.e., equation (24) holds. To satisfy condition (16), $\xi_{0}$ should be expressed as
$$\begin{array}{c}
\xi_{0}=(\sqrt{56}+2) / 4, \quad d=56 \\
c_{0}=2, \quad q_{0}=4, \quad a_{0}=\left[\xi_{0}\right]=2
\end{array}$$

Now, we use the recursive formulas (22) and (23) to find $c_{j}, q_{j}, \xi_{j}, a_{j}$:
$$\begin{array}{ll}
c_{0}=2, & q_{0}=4, \\
\xi_{0}=(\sqrt{56}+2) / 4, & a_{0}=2 ; \\
c_{1}=2 \cdot 4-2=6, & q_{1}=\left(56-6^{2}\right) / 4=5, \\
\xi_{1}=(\sqrt{56}+6) / 5, & a_{1}=2 ; \\
c_{2}=2 \cdot 5-6=4, & q_{2}=\left(56-4^{2}\right) / 5=8, \\
\xi_{2}=(\sqrt{56}+4) / 8, & a_{2}=1 ; \\
c_{3}=1 \cdot 8-4=4, & q_{3}=\left(56-4^{2}\right) / 8=5, \\
\xi_{3}=(\sqrt{56}+4) / 5, & a_{3}=2 ; \\
c_{4}=2 \cdot 5-4=6, & q_{4}=\left(56-6^{2}\right) / 5=4, \\
\xi_{4}=(\sqrt{56}+6) / 4, & a_{4}=3 ; \\
c_{5}=3 \cdot 4-6=6, & q_{5}=\left(56-6^{2}\right) / 4=5, \\
\xi_{5}=(\sqrt{56}+6) / 5, & a_{5}=2 .
\end{array}$$

This gives the smallest values $h=1, k=5$ such that $\xi_{k}=\xi_{h}$, thus we obtain
$$\xi_{0}=(\sqrt{14}+1) / 2=\langle 2, \overline{2,1,2,3}\rangle .$$