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Chapter 9: Problem 22

Find the interval of convergence of the power series, and find a familiar function that is represented by the power series on that interval. $$ \begin{aligned} 1-(x+3)+(x+3)^{2}-(x+3)^{3} \\\\+\cdots+(-1)^{k}(x+3)^{k} & \ldots \end{aligned} $$

Short Answer

Expert verified
The interval of convergence is \(-4 < x < -2\), representing the function \(\frac{1}{4+x}\).

Step by step solution

01

Identify the Power Series

The given series is \(1 - (x+3) + (x+3)^2 - (x+3)^3 + \cdots + (-1)^k (x+3)^k + \cdots\). This can be rewritten as \(\sum_{k=0}^{\infty} (-1)^k (x+3)^k\), which is a geometric series with first term \(a = 1\) and common ratio \(r = -(x+3)\).
02

Determine Convergence Criteria

For a geometric series \(\sum_{k=0}^{\infty} ar^k\) to converge, the absolute value of the common ratio must be less than 1, i.e., \(|r| < 1\). Therefore, we have the inequality \(|-(x+3)| < 1\).
03

Simplify the Inequality

Simplify \(|-(x+3)| < 1\) to \(|x+3| < 1\). This means \(-1 < x+3 < 1\).
04

Solve the Inequality for x

Solve the inequality \(-1 < x+3 < 1\) by subtracting 3 from all parts to isolate \(x\): \(-4 < x < -2\). This is our interval of convergence.
05

Recognize the Series as a Known Function

The series \(\sum_{k=0}^{\infty} (-1)^k (x+3)^k\) inside the interval \(-4 < x < -2\) represents the function \(\frac{1}{1 - (-1)(x+3)}\), or \( \frac{1}{1 + (x+3)} \).
06

Verify the Function Representation

Simplify \(\frac{1}{1 + (x+3)}\) to \(\frac{1}{4+x}\), which confirms the series corresponds to this familiar function within the interval of convergence.

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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 fascinating concept in mathematics. It is one where each term is derived by multiplying the previous term by a constant factor, called the common ratio. When dealing with geometric series, there are two main things to focus on:
  • The first term of the series, denoted as \(a\).
  • The common ratio, denoted as \(r\).
In the series provided, which is \(1 - (x+3) + (x+3)^2 - (x+3)^3 + \cdots\), we identify it as a geometric series with the first term \(a = 1\) and the common ratio \(r = -(x+3)\). This identification is crucial for evaluating and exploring further properties, such as its convergence.
Interval of Convergence
Convergence is a crucial aspect when dealing with series, as it helps determine the values for which the series sums to a finite number. For a geometric series, the series converges if the absolute value of the common ratio is less than 1, expressed mathematically as \(|r| < 1\).
  • In our example, we have \(|-(x+3)| < 1\).
  • This inequality simplifies to \(|x+3| < 1\), indicating the limit within which our variable must fall.
Further simplifying gives us \(-4 < x < -2\). This range is known as the interval of convergence, and only within this interval does the series behave nicely, summing to a definite value.
Function Representation
One powerful tool in mathematics is representing functions using series. It allows us to approximate and understand functions in different contexts.This series, \(\sum_{k=0}^{\infty} (-1)^k (x+3)^k\), can be expressed as a known function within its interval of convergence \(-4 < x < -2\).
  • The representation is found using the formula \(\frac{1}{1 - r}\), given that the series is geometric.
  • Plugging in, we find the function to be \(\frac{1}{1 - (-1)(x+3)}\), which simplifies to \(\frac{1}{1 + (x+3)}\) or equivalently \(\frac{1}{4+x}\).
This transformation is a powerful demonstration of how series can provide insights into the functions they represent, especially within specific intervals where they converge.

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Most popular questions from this chapter

Prove: (a) If \(f\) is an even function, then all odd powers of \(x\) in its Maclaurin series have coefficient \(0 .\) (b) If \(f\) is an odd function, then all even powers of \(x\) in its Maclaurin series have coefficient \(0 .\)

Confirm that the integral test is applicable and use it to determine whether the series converges. $$ \text { (a) } \sum_{k=1}^{\infty} \frac{1}{5 k+2} \quad \text { (b) } \sum_{k=1}^{\infty} \frac{1}{1+9 k^{2}} $$

Find the radius of convergence and the interval of convergence. $$ \sum_{k=0}^{\infty} \frac{(-2)^{k} x^{k+1}}{k+1} $$

Use Maclaurin series to approximate the integral to three decimal-place accuracy. $$\int_{0}^{1} \sin \left(x^{2}\right) d x$$

Use a graphing utility to confirm that the integral test applies to the series \(\sum_{k=1}^{\infty} k^{2} e^{-k},\) and then determine whether the series converges.
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