Chapter 9: Problem 11
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, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\frac{1}{1+x+x^{2}} $$
Short Answer
Expert verified
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 by step solution
01
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 \( x^5 \).
02
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: \( \frac{1}{1+x+x^2} = 1 - (x + x^2) + (x + x^2)^2 - (x + x^2)^3 + \cdots \).
03
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 \( x^5 \).
04
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.
05
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 \( x^5 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
- First term (when the common ratio is 1): \( a \)
- General form: \( a + ar + ar^2 + ar^3 + \cdots \)
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.
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.
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:
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.
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.
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.