Example 1. Expand the function $f(x)=\frac{1}{1-x}$ into a Taylor series in powers of: a) $x-1$, b) $x$, c) $x+1$.
Solution. a) Expansion in a series of powers of $(x-1)$ is impossible, since $f(x)$ is not defined at the point $x=1$. In other words, a series of powers of $(x-1)$ for this function does not exist.

Cases b) and c) will be considered together. Expansion in powers of $x$ and $(x+1)$ is possible, because $f(x)$ is infinitely differentiable in the neighborhoods of the points $x=0$ and $x=-1$.

The coefficients of the expansions are given in Theorem 18. First, we compute the general derivatives of all orders.

$$
\begin{gathered}
f^{\prime}(x)=\frac{1}{(1-x)^{2}}, f^{\prime \prime}(x)=\frac{2}{(1-x)^{3}}, f^{\prime \prime \prime}(x)=\frac{2 \cdot 3}{(1-x)^{3}} \\
f^{I V}(x)=\frac{2 \cdot 3 \cdot 4}{(1-x)^{4}}, \ldots, f^{(n)}(x)=\frac{1 \cdot 2 \cdot \ldots \cdot n}{(1-x)^{n}}=\frac{n!}{(1-x)^{n}}, \ldots
\end{gathered}
$$

It remains to substitute $x=0$ or $x=-1$ and divide the results by the corresponding factorial.

b) $c_{0}=1, c_{1}=1, c_{2}=\frac{2}{2}=1, c_{3}=\frac{3!}{3!}=1, \ldots, c_{n}=\frac{n!}{n!}=1, \ldots$, and

$$
\frac{1}{1-x}=1+x+x^{2}+x^{3}+\ldots+x^{n}+\ldots \quad(|x|<1)
$$

The convergence condition $|x|<1$ is related to the fact that the radius of convergence for the obtained series is 1 (verify this!).

$$
\begin{aligned}
& \text { c) } c_{0}=\frac{1}{2}, c_{1}=\frac{1}{2^{2}}, c_{2}=\frac{1}{2^{3}}, \ldots, c_{n}=\frac{1}{2^{n+1}}, \ldots, \text { and } \\
& \frac{1}{1-x}=\frac{1}{2}+\frac{1}{2^{2}}(x+1)+\frac{1}{2^{3}}(x+1)^{2}+\ldots+\frac{1}{2^{n+1}}(x+1)^{n}+\ldots \\
& (|x+1|<2)
\end{aligned}
$$

The radius of convergence for this series is 2 (obtain this!).

Note. Clearly, the form of the series depends not only on the function itself but also on the center of the series.