Example 12 Let the sequence of positive real numbers $x_{0}, x_{1}, \cdots, x_{1995}$ satisfy the following two conditions:
(1) $x_{0}=x_{1995}$; (2) $x_{i-1}+\frac{2}{x_{i-1}}=2 x_{i}+\frac{1}{x_{i}}, i=1,2, \cdots, 1995$. Find the maximum value of $x_{0}$ in all sequences that satisfy the conditions.
Solution: From (2), we get $x_{i-1}+\frac{2}{x_{i-1}}=2 x_{i}+\frac{1}{x_{i}}$, then $x_{i}=\frac{x_{i}-1}{2}$ or $x_{i}=\frac{1}{x_{i}-1}$. Let $f(x)=\frac{x}{2}, g(x)=\frac{1}{x}$, thus $x_{0}=x_{1995}=$ (1995th composition of $f$ and $g$ about $x_{0}$). Suppose there are $t$ times $g$, and $1995-t$ times $f$, then $x_{0}=x_{0}^{(-1)^{t}} \cdot 2^{s}$, where $|s| \leqslant 1995$, and $s \equiv 1995-t(\bmod 2)$.
If $t$ is odd, then $s$ is even, we have $x_{0}=\frac{1}{x_{0}} \cdot 2^{3}$, which gives $x_{0} \leqslant 2^{997}$; if $t$ is even, then $s$ is odd, we have $x_{0}=x_{0} \cdot 2^{s}$, which gives $S=0$, a contradiction, hence $x_{0} \leqslant 2^{997}$.
The sequence $x_{0}=2^{997}, x_{1}=2^{996}, \cdots, x_{1994}=2^{-997}, x_{1995}=2^{997}=x_{0}$ satisfies conditions (1) and (2), thus the maximum value of $x_{0}$ is $2^{997}$.