Example 1 Given the function $f_{1}(x)=\frac{2 x-1}{x+1}$, for positive integer $n$, define $f_{n+1}(x)=f_{1}\left[f_{n}(x)\right]$. Find the analytical expression for $f_{1234}(x)$.
Analysis If we start from the given conditions and gradually derive the analytical expression of $f_{1234}(x)$, the workload would be enormous, which is clearly impractical. Therefore, we can generalize the original problem and directly derive the analytical expression of $f_{n}(x)$.

Solution According to the given conditions, we have
$$
\begin{array}{c}
f_{1}(x)=\frac{2 x-1}{x+1}, \\
f_{2}(x)=f_{1}\left[f_{1}(x)\right]=f_{1}\left(\frac{2 x-1}{x+1}\right)=\frac{x-1}{x}, \\
f_{3}(x)=f_{1}\left[f_{2}(x)\right]=f_{1}\left(\frac{x-1}{x}\right)=\frac{x-2}{2 x-1}, \\
f_{1}(x)=f_{1}\left[f_{3}(x)\right]=f_{1}\left(\frac{x-2}{2 x-1}\right)=\frac{1}{1-x}, \\
f_{5}(x)=f_{1}\left[f_{1}(x)\right]=f_{1}\left(\frac{1}{1-x}\right)=\frac{x+1}{2-x}, \\
f_{6}(x)=f_{1}\left[f_{5}(x)\right]=f_{1}\left(\frac{x+1}{2-x}\right)=x, \\
f_{7}(x)=f_{1}\left[f_{6}(x)\right]=f_{1}(x) .
\end{array}
$$

This indicates that $\left\{f_{n}(x)\right\}$ is a periodic function sequence with the smallest positive period of 6. Therefore, the analytical expression of $f_{n}(x)$ is
$$
f_{n}(x)=\left\{\begin{array}{l}
\frac{2 x-1}{x+1}(n=6 k+1), \\
\frac{x-1}{x}(n=6 k+2), \\
\frac{x-2}{2 x-1}(n=6 k+3), \\
\frac{1}{1-x}(n=6 k+4), \\
\frac{x+1}{2-x}(n=6 k+5), \\
x(n=6 k+6),
\end{array}\right.
$$

where $k$ is a non-negative integer.
By substituting $n=1234$ into the expression of $f_{n}(x)$, we get
$$
f_{1231}(x)=f_{6 \times 211+1}(x)=\frac{1}{1-x} .
$$

Explanation The solution process above demonstrates two fundamental steps: one is the generalization process, where to find the analytical expression of $f_{1234}(x)$, we first generalize from the specific to the general, directly deriving the expression of $f_{n}(x)$; the other is the specialization process, where in seeking the expression of $f_{n}(x)$, we return from the general to the specific, examining the cases when $n=1,2, \cdots, 7$ based on the initial conditions and recursive relations, to identify the underlying patterns. Then, we combine these two processes organically, thereby solving the original problem. This shows that in practical problem-solving, generalization and specialization can often be used together to complement and support each other, enhancing problem-solving efficiency.

In college entrance exams and mathematics competitions, problems related to the periodicity of functions and function sequences frequently appear. Correctly recognizing and grasping the periodic characteristics of such periodic functions (sequences) is crucial for solving problems.