Question

In Problems 11-16 use an appropriate series in (2) to find the Maclaurin series of the given function. Write your answer in summation notation.

$\frac{\mathbf{x}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}$

Short answer

$\sum _{n=0}^{\infty }{(-1)}^n{x}^{2n+1}$

Step-by-step solution

Definition

Maclaurin seriesare a type of series expansion in which all terms are nonnegative integer powers of the variable.

Find Maclaurin series

Consider the function$\frac{x}{1+x^2} ---\left(1\right)$

We have Maclaurin series of:

$$\begin{gathered} \frac{1}{1-x}=1+x+x^2+x^3+\dots \\ =\sum _{n=0}^{\infty }{x}^n \end{gathered}$$

We can rewrite equation (1) in the following form & replace $x$as $-x^2$in (2):

$$\begin{gathered} \frac{x}{1+x^2}=x·\frac{1}{1+x^2} \\ =x·\frac{1}{1-\left(-x^2\right)} \\ =x·\left(1+\left(-x^2\right)+{\left(-x^2\right)}^2+{\left(-x^2\right)}^3+\dots \right)\text{From (2)} \\ =x·1+x·\left(-x^2\right)+x·{\left(-x^2\right)}^2+x·{\left(-x^2\right)}^3+\dots \\ =x-x^3+x^5-x^7+\dots \\ =\sum _{n=0}^{\infty }{(-1)}^n{x}^{2n+1} \end{gathered}$$