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Chapter 10: Problem 50

Use a geometric series to represent each of the given functions as a power series about \(x=0,\) and find their intervals of convergence. $$ \text { a. }f(x)=\frac{5}{3-x} \quad \text { b. } g(x)=\frac{3}{x-2} $$

Short Answer

Expert verified
a. \((-3, 3)\) with series \( \frac{5}{3} \sum_{n=0}^{\infty} \left(\frac{x}{3}\right)^n \); b. \((-2, 2)\) with series \(-\frac{3}{2} \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n\).

Step by step solution

01

Identify the Function Format

The function in part (a) is given as \( f(x) = \frac{5}{3-x} \). We want to express it as a geometric series, which generally has the form \( \frac{1}{1-u} \), where \( |u| < 1 \). We need to rewrite our function to match this form.
02

Rewrite the Function for Series Expansion (a)

Rewrite \( \frac{5}{3-x} \) as a series by factoring the 5: \( 5 \cdot \frac{1}{3-x} \). Now further manipulate so that the denominator matches \( 1-u \), yielding \( 5 \cdot \frac{1}{3(1 - \frac{x}{3})} \). Thus, \( u = \frac{x}{3} \).
03

Express Function as a Geometric Series (a)

Using the geometric series expansion \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \), substitute \( u = \frac{x}{3} \). Then \( \frac{1}{3-x} = \frac{1}{3} \sum_{n=0}^{\infty} \left( \frac{x}{3} \right)^n \). So, \( f(x) = 5 \cdot \frac{1}{3} \sum_{n=0}^{\infty} \left( \frac{x}{3} \right)^n = \frac{5}{3} \sum_{n=0}^{\infty} \left( \frac{x}{3} \right)^n \).
04

Find Interval of Convergence for Part (a)

Since \( u = \frac{x}{3} \) must satisfy \( \left| \frac{x}{3} \right| < 1 \), solving gives \( |x| < 3 \). Thus, the interval of convergence for \( f(x) \) is \( (-3, 3) \).
05

Identify the Function Format for Part (b)

The function \( g(x) = \frac{3}{x-2} \) does not directly fit the needed geometric series form \( \frac{1}{1-u} \). We need to rearrange so that a similar pattern emerges.
06

Rewrite the Function for Series Expansion (b)

Consider \( g(x) = \frac{3}{x-2} = -\frac{3}{2-x} \). Factoring and rewriting gives \( -3 \cdot \frac{1}{2-x} = -3 \cdot \frac{1}{2(1-\frac{x}{2})}\). Thus, \( u = \frac{x}{2} \).
07

Express Function as a Geometric Series (b)

Substitute into the geometric series \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \) with \( u = \frac{x}{2} \). This yields \( \frac{1}{2-x} = \frac{1}{2} \sum_{n=0}^{\infty} \left( \frac{x}{2} \right)^n \). Therefore, \( g(x) = -3 \cdot \frac{1}{2} \sum_{n=0}^{\infty} \left( \frac{x}{2} \right)^n = -\frac{3}{2} \sum_{n=0}^{\infty} \left( \frac{x}{2} \right)^n \).
08

Find Interval of Convergence for Part (b)

The condition \( |u| < 1 \) where \( u = \frac{x}{2} \) gives \( |x| < 2 \). Therefore, the interval of convergence for \( g(x) \) is \( (-2, 2) \).

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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 an infinite sum of terms in the form of \(a_n (x - c)^n\), where \(a_n\) are coefficients, \(x\) is a variable, and \(c\) is the center of the series expansion. In simple terms, it allows us to express complex functions as a combination of simpler polynomial terms. Power series representations are particularly useful for approximations and solving problems that involve calculus.
  • Centered at a Point: The power series is typically centered at a specific point, often denoted as \(c\). In the given exercise, the power series is about \(x = 0\), which means the series is expanded around zero.
  • Infinite Series: Unlike finite polynomials, power series can continue indefinitely. Higher powers of \(x-c\) provide more detailed behavior near the center of the series.
  • Convergence Radius: Not all power series converge everywhere; they often have a radius of convergence, which determines where the series representation is valid.
Understanding the concept of power series aids in simplifying complex functions into manageable pieces. This becomes evident when solving for series expansion and checking convergence.
Interval of Convergence
The interval of convergence refers to the set of \(x\)-values where a power series converges to a definite value. A power series converges if the sum of its terms approaches a finitelimit as the number of terms grows. Determining this range is essential as it defines where the series reliably represents the function.
  • Absolute Value Conditions: For a power series \(\sum_{n=0}^{\infty} a_n (x-c)^n\), convergence depends on the inequality \(|x-c| < R\), where \(R\) represents the radius of convergence.
  • Open Intervals: The result is often an interval of \((c-R, c+R)\), where the series converges in this range. For example, in the exercise provided, the interval of convergence for the function \(f(x)\) is \((-3, 3)\) for which the series will hold true.
  • Boundary Points: Sometimes, the endpoints or boundaries \(c-R\) and \(c+R\) might also be included鈥攖his needs to be verified separately.
Recognizing the interval of convergence is critical for trusted numerical calculations and function analysis using power series.
Geometric Series Expansion
A geometric series is a special kind of series in mathematics characterized by each term being a constant multiple of the previous term. The form is typically \(\frac{a}{1-r}\), which simplifies to a summation using the formula \(\sum_{n=0}^{\infty} ar^n\).For functions to be expanded as a geometric series, they first need to be structured to fit the \(\frac{1}{1-u}\) mold, where \(|u| < 1\).
  • Matching Forms: In the solved exercise, functions \(f(x)\) and \(g(x)\) are rewritten to fit the required form by manipulation of constants and expressions. This allows the application of the formula \(\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n\).
  • Applications: Geometric series make the calculation of repeating decimal patterns and interest calculations more feasible. They are used extensively in calculus for function approximations near the expansion center.
  • Convergence Check: A valid expansion must also see that \(|u| < 1\) ensures mathematical correctness over the computed interval.
Geometric series provide a powerful tool for handling functions by breaking them into simpler series components while keeping track of limitations via convergence intervals.

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Most popular questions from this chapter

The sequence \(\\{n /(n+1)\\}\) has a least upper bound of 1 Show that if \(M\) is a number less than \(1,\) then the terms of 1 \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\} .\)

Which of the sequences converge, and which diverge? Give reasons for your answers. $$ a_{n}=\frac{2^{n}-1}{3^{n}} $$

Pythagorean triples A triple of positive integers \(a, b,\) and \(c\) is called a Pythagorean triple if \(a^{2}+b^{2}=c^{2} .\) Let \(a\) be an odd positive integer and let $$ b=\left\lfloor\frac{a^{2}}{2}\right\rfloor \text { and } c=\left\lceil\frac{a^{2}}{2}\right\rceil $$ be, respectively, the integer floor and ceiling for \(a^{2} / 2\) a. Show that \(a^{2}+b^{2}=c^{2} .\) (Hint: Let \(a=2 n+1\) and express \(b\) and \(c\) in terms of \(n .\) b. By direct calculation, or by appealing to the accompanying figure, find $$ \lim _{a \rightarrow \infty} \frac{\left\lfloor\frac{a^{2}}{2}\right\rfloor}{ | \frac{a^{2}}{2} \rceil} $$

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{2}{1+e^{n}} $$

Establish the equations in Exercise 68 by combining the formal Taylor series for \(e^{i \theta}\) and \(e^{-i \theta} .\)
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