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Chapter 6: Problem 66

Use partial fractions to find the power series of each function. $$ \frac{3}{(x+2)(x-1)} $$

Short Answer

Expert verified
The power series is \( \sum_{n=0}^{\infty} \left[ (-1)\left(\frac{-x}{2}\right)^n + 2x^n \right] \), valid for \(|x| < 1\).

Step by step solution

01

Partial Fraction Decomposition

First, decompose the given fraction \(\frac{3}{(x+2)(x-1)}\) into partial fractions. Assume that it can be written as \(\frac{A}{x+2} + \frac{B}{x-1}\). We need to find the constants \(A\) and \(B\).
02

Equalize and Multiply by the Denominator

Set \(\frac{3}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}\). Multiply through by \((x+2)(x-1)\) to eliminate the denominators:\[ 3 = A(x-1) + B(x+2) \].
03

Solve for Constants A and B

Expand and simplify the equation: \[ A(x - 1) + B(x + 2) = 3 \]leads to \[ (A + B)x + (-A + 2B) = 3 \].Matching coefficients, solve the system of equations:1. \(A + B = 0\)2. \(-A + 2B = 3\).Solve these equations to get \(A = -2\) and \(B = 2\).
04

Write the Decomposed Function

Re-write the original fraction using the constants \(A\) and \(B\) found in step 3:\[\frac{3}{(x+2)(x-1)} = \frac{-2}{x+2} + \frac{2}{x-1}\].
05

Find the Power Series for Each Fraction

Using the decomposed fractions, find the power series for each segment. For \(\frac{-2}{x+2}\), write it as \(-2 \cdot \frac{1}{2}\cdot\frac{1}{1 - (-x/2)}\) leading to the series:\[ -1 \sum_{n=0}^{\infty} \left(\frac{-x}{2}\right)^n \], valid for \(|x|<2\).For \(\frac{2}{x-1}\), write it as \(2\cdot\frac{1}{1 - x}\) leading to the series:\[ 2\sum_{n=0}^{\infty} x^n \], valid for \(|x|<1\).
06

Combine Power Series

Combine the two series to get the overall power series:\[ \sum_{n=0}^{\infty} \left[ (-1)\left(\frac{-x}{2}\right)^n + 2x^n \right] \]. This series is valid within the region \(|x| < 1\) as both individual series must converge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power series
A power series is a series of the form: \[ \sum_{n=0}^{\infty} c_n (x - a)^n \] where \( c_n \) are coefficients and \( a \) is the center of the series. Power series are essential tools in mathematical analysis for representing functions. They allow functions to be expressed as sums of infinite terms, each powered by a variable. This representation is invaluable for solving problems, especially when dealing with functions that are difficult to handle directly.
  • They can converge or diverge, meaning they may or may not represent the function everywhere.
  • Often used for approximation in problems involving calculus and analysis.
Breaking complex functions into power series can be especially useful for integration and differentiation, as power series terms are simple polynomials.
Partial fraction decomposition
Partial fraction decomposition is a technique to simplify complex rational expressions, helping us manipulate them into simpler forms. When using partial fractions, we break down a complex fraction into a sum of simpler fractions, each with linear or quadratic denominators. This simplification is particularly useful in calculus when integrating rational functions or finding series expansion.
  • Assumes original function can be split into a sum of simpler fractions.
  • Involves determining constants to match the original expression.
  • Helpful for managing complex expressions efficiently, such as in power series.
For the function \( \frac{3}{(x+2)(x-1)} \), we express it as \( \frac{A}{x+2} + \frac{B}{x-1} \), solving for constants \( A \) and \( B \).
Convergence of series
Understanding the convergence of series is crucial when working with power series. A series converges if the sum of its infinite terms approaches a finite value. In the context of power series, we must determine the range of \( x \) for which the series accurately represents the function.
  • The interval of convergence is the set of \( x \) values where the series converges.
  • Radius of convergence, \( R \), defines the distance from the center where convergence occurs.
  • Divergence outside this interval means the power series does not represent the function accurately.
In the given exercise, the separate power series have different validity ranges (\(|x| < 2\) and \(|x| < 1\)). The overall power series converges within \(|x| < 1\), ensuring both parts overlap.
Mathematical proofs
Mathematical proofs provide a rigorous foundation for mathematical statements, ensuring validity and reliability. For a concept like partial fractions and power series, proofs enable us to understand why certain steps and transformations are valid.
  • Usually involve logical reasoning and systematic derivation of results.
  • Require validating each step of manipulation, ensuring algebraic transformations maintain equality.
  • In the original solution, proving the correctness of the partial fraction decomposition involves solving equalities.
By solving the system of equations for the constants \( A \) and \( B \), as shown in the exercise, the correct decomposition is confirmed. This solid foundation ensures further manipulations, like finding power series, are based on accurate mathematical expressions.

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Most popular questions from this chapter

Suppose that \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges to a function \(y\) such that \(y^{\prime \prime}-y^{\prime}+y=0\) where \(y(0)=0\) and \(y^{\prime}(0)=1 .\) Find a formula that relates \(a_{n+2}, a_{n+1},\) and \(a_{n}\) and compute \(a_{1}, \ldots, a_{5}\)

Suppose that a set of standardized test scores is normally distributed with mean \(\mu=100\) and standard deviation \(\sigma=10\). Set up an integral that represents the probability that a test score will be between 90 and 110 and use the integral of the degree 10 Maclaurin polynomial of \(\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}\) to estimate this probability.

Find the Maclaurin series of \(F(x)=\int_{0}^{x} f(t) d t\) by integrating the Maclaurin series of \(f\) term by term. If \(f\) is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero. $$ F(x)=\int_{0}^{x} \frac{\sin t}{t} d t ; \quad f(t)=\frac{\sin t}{t}=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{2 n}}{(2 n+1) !} $$

Use the substitution \((b+x)^{r}=(b+a)^{r}\left(1+\frac{x-a}{b+a}\right)^{r} \quad\) in the binomial expansion to find the Taylor series of each function with the given center. $$ \sqrt{x} \text { at } x=9 $$

Find the Maclaurin series of \(F(x)=\int_{0}^{x} f(t) d t\) by integrating the Maclaurin series of \(f\) term by term. If \(f\) is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero. $$ F(x)=\int_{0}^{x} \frac{1-\cos t}{t^{2}} d t ; \quad f(t)=\frac{1-\cos t}{t^{2}}=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{2 n}}{(2 n+2) !} $$
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