Chapter 1: Q32p (page 32)

$\ln (1 + x e^{x})$

Short Answer

Expert verified

$\ln (1 + x e^{x}) = x + \frac{x^{2}}{2} - \frac{x^{3}}{6} - \frac{x^{4}}{12} + \dots$

Step by step solution

Given information

The Maclaurin expansion of the function $\ln (1 + x e^{x})$ is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

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

$e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots$

$x e^{x} = x + x^{2} + \frac{x^{3}}{2} + \frac{x^{4}}{6} + \dots$

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

State the Maclaurin series.

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

$e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots$

$x e^{x} = x + x^{2} + \frac{x^{3}}{2} + \frac{x^{4}}{6} + \dots$

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

Replace $x$ with $[x + x^{2} + \frac{x^{3}}{2} + \frac{x^{4}}{6} + \dots]$ in the first equation.

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

$\ln (1 + x e^{x}) = \frac{24 x + 12 x^{2} - 4 x^{3} - 2 x^{4}}{24} + \dots$

$\ln (1 + x e^{x}) = x + \frac{x^{2}}{2} - \frac{x^{3}}{6} - \frac{x^{4}}{12} + \dots$

Thus, the final answer is $\ln (1 + x e^{x}) = x + \frac{x^{2}}{2} - \frac{x^{3}}{6} - \frac{x^{4}}{12} + \dots$

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$\ln (1 + x e^{x})$

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 Maclaurin series to evaluate

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ln1+xex

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 Maclaurin series to evaluate

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Work Energy and Power

Electric Charge Field and Potential

Physics of Motion

Thermodynamics

Modern Physics

Study anywhere. Anytime. Across all devices.

$\ln (1 + x e^{x})$