Chapter 1: Q5P (page 32)

$x^{2} l n (1 - x)$

Short Answer

Expert verified

The series is $x^{2} \ln (1 - x) = - \sum_{n = 1}^{\infty} \frac{x^{n + 2}}{n}$

The sum is given below.

$S_{1} = - x^{3}$

$S_{2} = - x^{3} - \frac{- x^{4}}{2}$

$S_{3} = - x^{3} - \frac{- x^{4}}{2} - \frac{- x^{5}}{3}$

The graphs are shown below.

Step by step solution

Given Information

The Maclaurin series.

Definition of the Maclaurin Series.

A Maclaurin series is a function with an expansion series that gives the sum of the function's derivatives.

Find the series and sum.

The Maclaurin series is given below.

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

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

$x^{2} \ln (1 - x) = - x^{3} - \frac{x^{4}}{2} - \frac{x^{5}}{3} - \dots$

The general series is given below $x^{2} \ln (1 - x) = - x^{3} - \frac{x^{4}}{2} - \frac{x^{5}}{3} - \dots - \frac{x^{n + 2}}{n}$

$x^{2} \ln (1 - x) = - \sum_{n = 1}^{\infty} \frac{x^{n + 2}}{n}$

The sum is given below.

$S_{1} = - x^{3}$

$S_{2} = - x^{3} - \frac{- x^{4}}{2}$

$S_{3} = - x^{3} - \frac{- x^{4}}{2} - \frac{- x^{5}}{3}$

Plot the graphs.

The graph of the function with $S_{1}, S_{2}, S_{3}$ is given below.

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$x^{2} l n (1 - x)$

Short Answer

Step by step solution

Given Information

Definition of the Maclaurin Series.

Find the series and sum.

Plot the graphs.

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x2ln(1-x)

Short Answer

Step by step solution

Given Information

Definition of the Maclaurin Series.

Find the series and sum.

Plot the graphs.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

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Turning Points in Physics

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Study anywhere. Anytime. Across all devices.

$x^{2} l n (1 - x)$