Example. Expand the function

$$
f(x)=\frac{3}{2-x-x^{2}}
$$

into a Taylor series in powers of $x\left(x_{0}=0\right)$.
## SOLUTION.

1. To use tabular expansions, we decompose the given function into elementary fractions:

$$
\frac{3}{2-x-x^{2}}=\frac{1}{1-x}+\frac{1}{x+2} \text {. }
$$

2. Using the tabular expansion

$$
\frac{1}{1-t}=1+t+t^{2}+\ldots+t^{n}+\ldots=\sum_{n=0}^{\infty} t^{n}, \quad t \in(-1,1)
$$

$$
\begin{gathered}
\text { we get } \frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}, \quad x \in(-1,1) \\
\frac{1}{2+x}=\frac{1}{2} \cdot \frac{1}{1+x / 2}=\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{n}}{2^{n}}=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{n}}{2^{n+1}}, \quad x \in(-2,2) .
\end{gathered}
$$

Thus,

$$
\frac{3}{2-x-x^{2}}=\sum_{n=0}^{\infty}\left(1+\frac{(-1)^{n}}{2^{n+1}}\right) x^{n}
$$

3. The domain of convergence of the obtained series is the intersection of the above-mentioned domains of convergence, i.e., $(-1,1) \cap(-2,2)=(-1,1)$.

Answer. $\frac{3}{2-x-x^{2}}=\sum_{n=0}^{\infty}\left(1+\frac{(-1)^{n}}{2^{n+1}}\right) x^{n} \quad \forall x \in(-1,1)$.

Conditions of the Problems. Expand the functions into a Taylor series $\left(x_{0}=0\right)$.

1. $\frac{5}{6-x-x^{2}}$.
2. $\ln \left(1-x-20 x^{2}\right)$.
3. $\sin \left(x+\frac{\pi}{6}\right)$.
4. $\frac{x}{\sqrt{9+x^{2}}}$.
5. $\frac{x}{3+2 x}$.
6. $\sqrt[4]{16+x}$.
7. $x e^{2 x^{2}+1}$.
8. $\frac{1}{\sqrt[3]{27-x}}$.
9. $\quad \ln \left(12 x^{2}+7 x+1\right)$.
10. $\cos ^{2} 3 x$.

## Answers.

1. $\sum_{n=0}^{\infty}\left(\frac{1}{3^{n+1}}+\frac{(-1)^{n}}{2^{n+1}}\right) x^{n}, \quad x \in(-2,2)$.
2. $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} 4^{n}-5^{n}}{n} x^{n}, \quad x \in\left[-\frac{1}{5}, \frac{1}{5}\right)$.
3. $\frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} x^{2 n+1}+\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n)!} x^{2 n}, \quad x \in(-\infty,+\infty)$.
4. $\quad \frac{x}{3}+\sum_{n=1}^{\infty} \frac{(-1)^{n}(2 n-1)!!}{n!2^{n} 3^{2 n+1}} x^{2 n+1}, \quad x \in(-3,3)$.
5. $\sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{n}}{3^{n+1}} x^{n+1}, \quad x \in\left[-\frac{3}{2}, \frac{3}{2}\right)$.
6. $2+\frac{x}{32}+\sum_{n=2}^{\infty}(-1)^{n-1} \frac{3 \cdot 7 \cdot 11 \ldots(4 n-5)}{2^{6 n-1} n!} x^{n}, \quad x \in[-16,16]$.
7. $e \sum_{n=0}^{\infty} \frac{2^{n}}{n!} x^{2 n+1}, \quad x \in(-\infty,+\infty)$.
8. $\frac{1}{3}+\sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \ldots(3 n-2)}{n!3^{4 n+1}} x^{n}, \quad x \in(-27,27)$.
9. $\sum_{n=1}^{\infty}(-1)^{n-1} \frac{4^{n}+3^{n}}{n} x^{n}, \quad x \in\left(-\frac{1}{4}, \frac{1}{4}\right]$.
10. $1+\sum_{n=1}^{\infty} \frac{(-1)^{n} 6^{2 n}}{2(2 n)!} x^{2 n}, \quad x \in(-\infty,+\infty)$.

### 10.14. Approximate Calculations Using Power Series

PROBLEM STATEMENT. Calculate the integral

$$
\int_{0}^{b} f(x) d x
$$

with accuracy $\alpha$, where $f(x)$ can be expanded into a power series with radius of convergence $R>b$.

PLAN OF SOLUTION. Practical calculations usually boil down to summing a certain numerical series. In this case, such a series is obtained by expanding the integrand in a power series and integrating it term by term.

1. Expand the integrand in a power series in powers of $x$

$$
f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}
$$

and determine its domain of convergence.

2. A power series can be integrated term by term over any interval belonging to the interval of convergence. Therefore, integrating the obtained series term by term and using the Newton-Leibniz formula, we get

$$
\int_{0}^{b} f(x) d x=\int_{0}^{b} \sum_{n=0}^{\infty} c_{n} x^{n} d x=\sum_{n=0}^{\infty} c_{n} \frac{b^{n+1}}{n+1}
$$

3. Calculate the sum of the numerical series with the given accuracy (by estimating the remainder of the series).

NOTE. If the integrand is expanded in a series not in powers of $x$, but in powers of $x-b / 2$, the series will converge faster, i.e., fewer terms may be required to achieve the desired accuracy, but the final calculations will be more complex.