Example 1 Is there a mapping $f: \mathbf{R}^{+} \rightarrow \mathbf{R}$, satisfying
$$
f^{(1989)}(x)=\frac{x}{x+1}
$$
( $\mathbf{R}^{+}$ denotes the set of positive real numbers)?
The mapping $f(x)=\frac{1}{\frac{1}{x}+\frac{1}{1989}}$ satisfies the requirements. In fact, $\mathrm{E} , \frac{1}{f(x)}$
$$
\begin{array}{l}
=\frac{1}{x}+\frac{1}{1989}, \frac{1}{f^{(2)}(x)}=\frac{1}{f(x)}+\frac{1}{1989}=\frac{1}{x}+\frac{2}{1989}, \cdots, \\
\frac{1}{f^{(k)}(x)}=\frac{1}{x}+\frac{k}{1989}, \cdots, \frac{1}{f^{(1989)}(x)}=\frac{1}{x}+1 .
\end{array}
$$