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Chapter 1: Q18MP (page 1)

Find the Maclaurin series of the following functions.
$arc t a n x = {鈭�}_{0}^{x} \frac{d u}{1 + u^{2}}$

Short Answer

Expert verified

Maclaurin series of $\arctan x is x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + 鈥� . .$

Step by step solution

01

The Maclaurin series and the given function.

Function is $\arctan x = {鈭�}_{0}^{x} \frac{d u}{1 + u^{2}}$

The Maclaurin series of ${鈭�}_{0}^{x} \frac{d u}{1 + u^{2}}$ is given as follows:

${鈭�}_{0}^{x} \frac{d u}{1 + u^{2}} = {鈭�}_{0}^{x} (1 - u^{2} + u^{4} - u^{6} + 鈥� . .) d u {鈭�}_{0}^{x} \frac{d u}{1 + u^{2}} = {(u - \frac{u^{3}}{3} + \frac{u^{5}}{5} - \frac{u^{7}}{7} + 鈥� . .)}_{0}^{x} {鈭�}_{0}^{x} \frac{d u}{1 + u^{2}} = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + 鈥�$

Now, we have:

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

02

Calculation by the Maclaurin series.

Substitute $u^{2}$ for $x$ in expansion of $\frac{1}{1 + x}$ as follows:

$\begin{array}{rcl} \frac{1}{1 + u^{2}} & = & 1 - u^{2} + {(u^{2})}^{2} - {(u^{2})}^{3} + 鈥� . . \\ \frac{1}{1 + u^{2}} & = & 1 - u^{2} + u^{4} - u^{6} + 鈥� . . \end{array}$

Apply integration for $\lim$ $0$ to $x$ in the above equation as follows:

${鈭�}_{0}^{x} \frac{d u}{1 + u^{2}} = {鈭�}_{0}^{x} (1 - u^{2} + u^{4} - u^{6} + 鈥� . .) d u {鈭�}_{0}^{x} \frac{d u}{1 + u^{2}} = {(u - \frac{u^{3}}{3} + \frac{u^{5}}{5} - \frac{u^{7}}{7} + 鈥� . .)}_{0}^{x} {鈭�}_{0}^{x} \frac{d u}{1 + u^{2}} = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + 鈥� . .$

Thus, the Maclaurin series of $\arctan x is x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + 鈥� . .$

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Most popular questions from this chapter

$x \sqrt{1 + x}$

$\ln \sqrt{\frac{1 + x}{1 - x}} = {鈭�}_{0}^{x} \frac{d t}{1 - t^{2}}$

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations.

$\lim_{x 鈫� 0} \frac{\sin x - x}{x^{3}}$ .

In $(x + \sqrt{1 + x^{2}}) = {鈭�}_{0}^{x} \frac{dt}{\sqrt{1 + t^{2}}}$

Write the Maclaurin series for $\frac{1}{\sqrt{1 + x}}$ in $鈭�$ form using the binomial coefficient notation. Then find a formula for the binomial coefficients in terms ofn as we did in Example $2$ above

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