Problem 11 - 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, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\frac{1}{1+x+x^{2}} $$

Step-by-Step Solution

Answer

The Maclaurin series for \( f(x) = \frac{1}{1+x+x^2} \) up to \( x^5 \) is \( 1-x+x^3-2x^4-3x^5 \).

Step 1: Recall the Maclaurin Series Definition

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \). Our goal is to find the terms up to \( x^5 \).

Step 2: Use Geometric Series Expansion

Notice that \( \frac{1}{1+x+x^2} \) is similar to the geometric series \( \frac{1}{1-u} = 1 + u + u^2 + \cdots \), provided \( |u| < 1 \). Here, consider \( u = x + x^2 \), substituting, we have: \( \frac{1}{1+x+x^2} = 1 - (x + x^2) + (x + x^2)^2 - (x + x^2)^3 + \cdots \).

Step 3: Calculate \((x + x^2)^n\) Up to \(x^5\)

Compute the powers of \( x + x^2 \) and collect terms up to \( x^5 \): - \((x + x^2)^2 = x^2 + 2x^3 + x^4\), - \((x + x^2)^3 = x^3 + 3x^4 + 3x^5\).We only need up to \( x^5 \).

Step 4: Combine Terms

Based on the expansion in Step 3, substitute back into the series expansion:- Expansion Terms: 1st term: \(1\) 2nd term: \(- (x + x^2) = -x - x^2\) 3rd term: \((x + x^2)^2 = x^2 + 2x^3 + x^4\) 4th term: \(- (x + x^2)^3 = -x^3 - 3x^4 - 3x^5\)Combine these terms and collect like terms to form the series.

Step 5: Simplify and Write the Series

Collect all terms, simplifying like terms:\[ 1 - x - x^2 + x^2 + 2x^3 + x^4 - x^3 - 3x^4 - 3x^5 \]Simplify the expression by combining similar powers of \( x \):\[ 1 - x + x^3 - 2x^4 - 3x^5 \]This gives the series expansion up to \( x^5 \).

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In mathematics, it is often represented as:

• First term (when the common ratio is 1): \( a \)
• General form: \( a + ar + ar^2 + ar^3 + \cdots \)

The concept of a geometric series becomes very useful when dealing with functions like \( \frac{1}{1-u} \), where \( |u|<1 \). This formula expands into an infinite series: \( 1 + u + u^2 + u^3 + \cdots \).

This approach simplifies exploring functions like \( \frac{1}{1+x+x^2} \), which can be rewritten using the series expansion technique with \( u = x + x^2 \).
This allows us to substitute directly into known series forms, making it simpler to calculate higher powers, like finding terms up to \( x^5 \).
Understanding geometric series is crucial since this infinite expansion is the backbone of many other series expansions used in calculus and analysis.

Power Series Expansion

A power series is an infinite series of the form \( a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots \), where \( a_n \) are coefficients and \( x \) is a variable. Power series are vital in approximating complex functions, especially when direct computation is challenging.

For example, the Maclaurin series is a special type of power series that provides a way to express functions as an infinite sum of terms calculated from the derivatives of a function at a single point (usually \( x=0 \)).
This approach is incredibly useful as it serves as a bridge in transforming functions into sums, permitting easier manipulation and evaluation.

In our exercise, transforming functions to a power series allows us to manipulate \( \frac{1}{1+x+x^2} \) efficiently by combining it with the geometric series expansion.

• A key advantage is obtaining polynomial approximations of functions, crucial in both theoretical and practical applications.
• Power series offer insights into function behavior and approximations over specific intervals.

By comparing terms systematically using algebraic operations on known series, we can match the necessary terms up to \( x^5 \).

Calculus

Calculus is a field of mathematics that focuses on limits, functions, derivatives, integrals, and infinite series. Its fundamental idea is to study how things change and to provide a framework for modeling dynamic systems.

The Maclaurin series, a core topic in calculus, is a specific type of Taylor series centered at \( x=0 \). The series approximates a function using polynomials and is beneficial for modeling real-world situations where exact solutions are challenging.

Calculating Maclaurin series involves essential calculus concepts:

• Derivatives: To find coefficients for the series expansion, we need the function's derivatives at zero.
• Symmetry and substitutions: These simplify the application of the series to different functions, adjusting to different input variables like \( x, x^2 \), etc.

By applying calculus techniques, like derivatives and limits, we achieve a more comprehensive understanding of series expansion and make problem-solving more intuitive.
Using calculus to evaluate derivatives helps systematically find the terms in our Maclaurin series up to \( x^5 \), revealing how the function behaves around zero.