Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-18, find the terms through \(x^{5}\) in the Maclaurin series for \(f(x)\). Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\frac{1}{1+x} \ln \left(\frac{1}{1+x}\right)=\frac{-\ln (1+x)}{1+x} $$

Step-by-Step Solution

Answer

The Maclaurin series for \( f(x) \) is: \(-x + \frac{x^2}{2} - \frac{2x^3}{3} + x^4 - x^5\).

Step 1: Identify Known Maclaurin Series

The known Maclaurin series you need are for the functions \( \ln(1+x) \) and \( \frac{1}{1+x} \). The Maclaurin series for \( \ln(1+x) \) is:\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} + \cdots \]And the Maclaurin series for \( \frac{1}{1+x} \) is:\[ \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots \]

Step 2: Express the Function

The function given is \( f(x) = \frac{-\ln(1+x)}{1+x} \). We express it as the product of two series, one being negative of the other. We can also write it as:\[ f(x) = (-1) \times \frac{1}{1+x} \ln(1+x) \]

Step 3: Multiply the Series

We need to multiply the two known series \((-\ln(1+x))\) and \(\frac{1}{1+x}\). Substitute the series expressions from Step 1 into the function:\[ f(x) = -(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}) \times (1 - x + x^2 - x^3 + x^4 - x^5) \]

Step 4: Calculate Product Terms

Multiply the terms, focusing on those that will contribute up to \( x^5 \):- Constant term: \( 0 \) (as no constant term exists in \(-\ln(1+x)\)).- Coefficient of \( x \): \( -x \).- Coefficient of \( x^2 \): \(- \frac{x^2}{2} + x^2 = \frac{x^2}{2} \).- Coefficient of \( x^3 \): \(-\frac{x^3}{3} + x^3 = \frac{2x^3}{3} \).- Coefficient of \( x^4 \): \(-\frac{x^4}{4} + x^3 - x^4 = -x^4 + \frac{x^3}{2} \).- Coefficient of \( x^5 \): \(-\frac{x^5}{5} + x^4 - x^5 \) contributing to further reduced terms.

Step 5: Combine and Simplify

Combine all the calculated terms to get the Maclaurin series of \( f(x) \) up to \( x^5 \):\[ f(x) = -x + \frac{x^2}{2} - \frac{2x^3}{3} + x^4 - x^5 \]

Key Concepts

Calculus

In calculus, one of the essential topics is the concept of series and how functions can be expressed as infinite sums of terms. This specific exercise deals with the familiar concept of the Maclaurin series. The Maclaurin series is a type of Taylor series that is centered at zero. It is used to represent functions as a series of ascending powers of the initial variable, usually denoted as \(x\). Calculus allows us to dive deep into the behavior of functions at any point by expanding them into such series, thereby simplifying complex functions into more manageable forms.

• The Maclaurin series serves as a powerful tool to study, approximate, and analyze functions.
• Using derivatives, one can derive terms of the Maclaurin series till a desired order, providing a polynomial approximation of the function.

This technique is extensively useful in both theoretical and practical applications in engineering, physics, and computer science, enhancing our understanding of function behavior and offering solutions to complex mathematical problems.

Series Expansion

Series expansion is a method used in mathematical analysis to represent functions as sums of an infinite sequence of terms. The goal is often to approximate more complicated expressions using series, which are typically easier to handle. The problem presented uses the known Maclaurin series for functions \(\ln(1+x)\) and \(\frac{1}{1+x}\) as a starting point.

• To find the series expansion, we explore each known series individually and then manipulate them algebraically to form the desired series.
• Terms are multiplied individually to produce new coefficients, focusing on obtaining terms through specific powers, like up to \(x^5\).

This method is invaluable in deriving series representations for functions where direct calculation of derivatives might be complex or cumbersome. By breaking complex functions down into lower-degree polynomials, series expansions allow us to closely approximate functions within a specified range.

Mathematical Analysis

Mathematical analysis provides the fundamental techniques necessary to describe and solve problems involving continuous change. It blends rigorous theoretical approaches with practical computational methods. This exercise utilizes mathematical analysis to derive the Maclaurin series for a compound function \(f(x) = \frac{-\ln(1+x)}{1+x}\). By leveraging known series and their properties, this kind of analysis allows us to find and confirm solutions methodically and effectively.

• Mathematical analysis emphasizes precision and the ethical use of mathematical laws and formulas.
• Through critical thinking and methodical steps, mathematical analysis reduces complex functions into simple polynomial expressions within the context of series.
• It paves the way for deeper insights into function behavior and interaction through differential and integral calculus.

The use of series expansion in mathematical analysis exemplifies how theory-driven approaches solve real-world math problems, bridging the gap between abstract concepts and practical applications.