Chapter 1: Q28p (page 32)

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

Short Answer

Expert verified

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

Step by step solution

Given information

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

Definition of Maclaurin series

The definition of the Maclaurin series is defined,

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

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

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

State the Maclaurin series.

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

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

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

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

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

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

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

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

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

Short Answer

Step by step solution

Given information

Definition of Maclaurin series

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

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

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sin[ln(1+x)]

Short Answer

Step by step solution

Given information

Definition of Maclaurin series

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

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Nuclear Physics

Magnetism

Magnetism and Electromagnetic Induction

Space Physics

Electric Charge Field and Potential

Study anywhere. Anytime. Across all devices.

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