Example 3 Find the periodic continued fraction of $\xi_{0}=\sqrt{73}$.

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To make condition (16) hold, $\xi_{0}$ should be expressed as
$$\xi_{0}=(\sqrt{73}+0) / 1, \quad q_{0}=1, c_{0}=0, d=73$$
$a_{0}=[\sqrt{73}]=8$. Using equation (22), we obtain
$$\begin{array}{c}
c_{1}=8 \cdot 1-0=8, \quad q_{1}=\left(73-8^{2}\right) / 1=9 \\
\xi_{1}=(\sqrt{73}+8) / 9, \quad a_{1}=1
\end{array}$$

Below, we use equation (34) to find:
$$\begin{array}{ll}
c_{0}=0, & q_{0}=1, \\
\xi_{0}=(\sqrt{73}+0) / 1, & a_{0}=8 ; \\
c_{1}=8, & q_{1}=9, \\
\xi_{1}=(\sqrt{73}+8) / 9, & a_{1}=1 ; \\
c_{2}=1 \cdot 9-8=1, & q_{2}=1+(8-1) \cdot 1=8, \\
\xi_{2}=(\sqrt{73}+1) / 8, & a_{2}=1 ; \\
c_{3}=1 \cdot 8-1=7, & q_{3}=9+(1-7) \cdot 1=3, \\
\xi_{3}=(\sqrt{73}+7) / 3, & a_{3}=5 ; \\
c_{4}=5 \cdot 3-7=8, & q_{4}=8+(7-8) \cdot 5=3, \\
\xi_{4}=(\sqrt{73}+8) / 3, & a_{4}=5 ; \\
c_{5}=5 \cdot 3-8=7, & q_{5}=3+(8-7) \cdot 5=8, \\
\xi_{5}=(\sqrt{73}+7) / 8, & a_{5}=1 ; \\
c_{6}=1 \cdot 8-7=1, & q_{6}=3+(7-1) \cdot 1=9, \\
\xi_{6}=(\sqrt{73}+1) / 9, & a_{6}=1 ; \\
c_{7}=1 \cdot 9-1=8, & q_{7}=8+(1-8) \cdot 1=1, \\
\xi_{7}=(\sqrt{73}+8) / 1, & a_{7}=16 ; \\
c_{8}=16 \cdot 1-8=8, & q_{8}=9+(8-8)=9, \\
\xi_{8}=(\sqrt{73}+8) / 9, & a_{8}=1 .
\end{array}$$

This results in $\xi_{8}=\xi_{1}$, hence we have
$$\xi_{0}=\sqrt{73}=\langle 8, \overline{1,1,5,5,1,1,16}\rangle .$$

Let $d>1$ be a non-square number, and $\xi_{0}$ is given by equation (16). According to Theorem 6, $\xi_{j}$ can be expressed by equation (21), but on the other hand, there is a general relationship between $\xi_{0}$ and $\xi_{j}$ —— equation (17) in §3. Therefore, from these two relationships, we can derive the relationship between $\left\{c_{n}\right\},\left\{q_{n}\right\}$ and $\left\{h_{n}\right\},\left\{k_{n}\right\}$. The theorem below characterizes this relationship.