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Chapter 1: Q28p (page 32)

$s i n [l n (1 + x)]$

Short Answer

Expert verified

$\sin [\ln (1 + x)] = x - \frac{x^{2}}{2} + \frac{x^{3}}{6} + 鈥�$

Step by step solution

01

Given information

The Maclaurin expansion of the function $\sin [\ln (1 + x)]$ is to be evaluated.

02

Definition of Maclaurin series

The definition of the Maclaurin series is defined,

$s i n (x) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + 鈥�$

$l n (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + 鈥�$

03

Part-A Step 3: Begin by stating the Maclaurin series

State the Maclaurin series.

$\sin (x) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + 鈥�$

$\ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + 鈥�$

04

Part-B Step 3: Use the first Maclaurin series to evaluate

Substitute $x$ with $[x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + 鈥�]$ in the first series and evaluate.

$\sin [\ln (1 + x)] = [x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + 鈥�] - \frac{{[x - \frac{x^{2}}{2} + 鈥�]}^{3}}{6} + 鈥�$

$\sin [\ln (1 + x)] = [x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + 鈥�] - \frac{x^{3} - \frac{3 x^{4}}{2} + \frac{3 x^{5}}{4} - \frac{x^{6}}{8} + 鈥�}{6} + 鈥�$

$\sin [\ln (1 + x)] = x - \frac{x^{2}}{2} + \frac{x^{3}}{6} + 鈥�$

Thus, the final answer is $\sin [\ln (1 + x)] = x - \frac{x^{2}}{2} + \frac{x^{3}}{6} + 鈥�$

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Most popular questions from this chapter

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.

$\lim_{x 鈫� 0} \frac{1 - \cos x}{x^{2}}$ .

Estimate the Error if ${鈭�}_{n = 1}^{鈭�} \frac{x^{n}}{n^{3}}$ is approximated by the sum of its first three terms for $| x | < \frac{1}{2}$ .

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges.鈥�

${鈭�}_{n = 0}^{鈭�} \frac{(- 1)^{n} (x)^{\frac{1}{2}}}{nlnn}$

Use the special comparison test to find whether the following series converge or diverge.

${鈭�}_{n = 9}^{鈭�}$ $\frac{(2 n + 1) (3 n - 5)}{\sqrt{n^{2} - 73}}$

Show that the interval of convergence of the series ${鈭�}_{n = 1}^{鈭�} \frac{x^{n}}{n^{2} + n}$ is $| x | 鈮� 1 (For x = 1$ . (For , this is the series of Problem 9.) Using theorem $(14.4)$ , show that for $x = \frac{1}{2}$ , four terms will give two decimal place accuracy.

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