G4.2 Suppose $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{8} x^{8}$ is a polynomial of degree 8 with real coefficients $a_{0}, a_{1}, \ldots, a_{8}$. If $P(k)=\frac{1}{k}$ when $k=1,2, \ldots, 9$, and $b=P(10)$, find the value of $b$.
$$
P(k)=\frac{1}{k} \text {, for } k=1,2, \ldots, 9 \text {. }
$$

Let $F(x)=x P(x)-1$, then $F(k)=k P(k)-1=0$, for $k=1,2, \ldots, 9$. $F(x)$ is a polynomial of degree 9 and the roots are $1,2, \ldots, 9$.
$$
F(x)=x P(x)-1=c(x-1)(x-2) \ldots(x-9)
$$
$P(x)=\frac{c(x-1)(x-2) \cdots(x-9)+1}{x}$, which is a polynomial of degree 8 .
Compare the constant term of $c(x-1)(x-2) \ldots(x-9)+1=0$ :
$$
\begin{array}{l}
-c \cdot 9!+1=0 \\
c=\frac{1}{9!} \Rightarrow P(x)=\frac{(x-1)(x-2) \cdots(x-9)+9!}{9!x} \\
P(10)=\frac{9!+9!}{9!\cdot 10}=\frac{1}{5}
\end{array}
$$