Example 9 (2005 Shanghai College Entrance Examination Question) In the rectangular coordinate plane, it is known that points $P_{1}(1,2), P_{2}\left(2,2^{2}\right)$, $P_{3}\left(3,2^{3}\right), \cdots, P_{n}\left(n, 2^{n}\right)$, where $n$ is a positive integer. For any point $A_{0}$ in the plane, let $A_{1}$ be the point symmetric to $A_{0}$ with respect to $P_{1}$, $A_{2}$ be the point symmetric to $A_{1}$ with respect to $P_{2}$, ..., and $A_{n}$ be the point symmetric to $A_{n-1}$ with respect to $P_{n}$.
(1) Find the coordinates of the vector $\overrightarrow{A_{0} A_{2}}$.
(2) When point $A_{0}$ moves on the curve $C$, the trajectory of point $A_{2}$ is the graph of the function $y=f(x)$, where $f(x)$ is a periodic function with a period of 3, and when $x \in(0,3]$, $f(x)=\lg x$. Find the analytical expression of the function whose graph is the curve $C$ in the interval (1,4$]$.
(1) Let point $A_{1}(x, y)$, the coordinates of the point $A_{1}$ symmetric to point $P_{1}$ are $A_{1}(2-x, 4-y)$, and the coordinates of the point $A_{1}$ symmetric to point $P_{2}$ are $\Lambda_{2}(2+x, 4+y)$. Therefore, $\overrightarrow{A_{1} A_{2}}=(2,4)$.
(2) Since $\overrightarrow{A_{11} A_{2}}=(2,4)$, the graph of $f(x)$ is obtained by translating the curve $C$ 2 units to the right and 4 units up. Therefore, the curve $C$ is the graph of the function $y=g(x)$. Here, $g(x)$ is a periodic function with a period of 3, and when $x \in(-2,1]$, $g(x)=\lg (x+2)-4$. Thus, when $x \in(1,4]$, $g(x)=\lg (x-1)-4$.
(3) $\overrightarrow{A_{0} A_{n}}=\overrightarrow{A_{0} A_{2}}+\overrightarrow{A_{2} A_{4}}+\cdots+\overrightarrow{A_{n-2} A_{n}}$, since $\overrightarrow{A_{2 k-2} A_{2 k}}=2 \overrightarrow{P_{2 k-1} P_{2 k}}$, we get
$$
\begin{aligned}
\overrightarrow{A_{0} A_{n}} & =2\left(\overrightarrow{P_{1} P_{2}}+\overrightarrow{P_{3} P_{4}}+\cdots+\overrightarrow{P_{n-1} P_{n}}\right) \\
& =2\left[(1,2)+\left(1,2^{3}\right)+\cdots+\left(1,2^{n-1}\right)\right] \\
& =2\left(\frac{n}{2}, \frac{2\left(2^{n}-1\right)}{3}\right)=\left(n, \frac{4\left(2^{n}-1\right)}{3}\right) .
\end{aligned}
$$