/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 \(3-10\) Find a power series rep... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(3-10\) Find a power series representation for the function and determine the interval of convergence. $$ f(x)=\frac{1+x}{1-x} $$

Short Answer

Expert verified
The power series representation is \( 1 + 2 \sum_{n=0}^{\infty} x^{n+1} \) with an interval of convergence \( -1 < x < 1 \).

Step by step solution

01

Identify the Base Function

The given function is \( f(x) = \frac{1+x}{1-x} \). Notice that \( \frac{1}{1-x} \) resembles the geometric series \( \sum_{n=0}^{\infty} x^n \), valid for \( |x| < 1 \).
02

Decompose the Function

Rewrite \( f(x) = \frac{1}{1-x} + \frac{x}{1-x} \). This separates the constant term from the additional \( x \) in the numerator.
03

Find the Power Series Representation

First, substitute the geometric series for \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \). Since \( \frac{x}{1-x} \) can also be expressed by multiplying each term of the series by \( x \), it becomes \( \sum_{n=0}^{\infty} x^{n+1} = \sum_{n=1}^{\infty} x^n \). Therefore, \( f(x) = \sum_{n=0}^{\infty} x^n + \sum_{n=1}^{\infty} x^n \).
04

Simplify the Power Series

Combine the series from Step 3: \( \sum_{n=0}^{\infty} x^n + \sum_{n=1}^{\infty} x^n = 1 + 2x + 2x^2 + 2x^3 + \cdots = 1 + 2 \sum_{n=1}^{\infty} x^n \). This simplifies further to give the series \( 1 + 2 \sum_{n=0}^{\infty} x^{n+1} \).
05

Determine the Interval of Convergence

For geometric series, the convergence is within \( |x| < 1 \). Thus, the interval of convergence for our series is \( -1 < x < 1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Geometric Series
The geometric series is a fundamental concept in calculus and sequences. It can be defined as the sum of a sequence where each term after the first is the preceding term multiplied by a constant, called the common ratio. For example, a simple geometric series can look like this:
  • 1, x, x^2, x^3, ...
In mathematical terms, it is expressed as:\[\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \ldots\]where \( a \) is the first term, and \( r \) is the common ratio. For the series to converge, the absolute value of the common ratio \( |r| \) must be less than 1.
Understanding the geometric series concept is crucial when dealing with power series representations of functions. In this particular problem, the function \( \frac{1}{1-x} \) can be represented as a geometric series \( \sum_{n=0}^{\infty} x^n \), as long as \( |x| < 1 \).
This is a powerful way to express functions, giving us tools to easily handle complex fractions by breaking them into infinite series.
Interval of Convergence
The interval of convergence is a range of values for which a power series converges. Understanding this interval is important because it defines where the expression for our power series accurately represents the original function.
For a geometric series, the interval of convergence is determined by the condition \( |x| < 1 \). This criterion ensures that the sum of the infinite series is finite and meaningful. For the given function \( \frac{1+x}{1-x} \), the power series remains valid within the interval below:
  • \(-1 < x < 1\)
By ensuring that \( |x| \) is less than 1, we know that each term in the series becomes smaller, allowing the series to sum up to a finite value. If \( |x| \geq 1 \), the terms do not converge to zero fast enough, and the series diverges.
Understanding intervals of convergence helps in analyzing where functions can be expanded into series solutions and the limits of these expansions.
Function Decomposition
Function decomposition involves breaking a complex function into simpler parts to make calculations easier. In the context of our exercise, decomposing the function helps us derive its power series representation.
Given the function \( \frac{1+x}{1-x} \), we decompose it as follows:
  • \( \frac{1}{1-x} \) — representing the standard geometric series.
  • \( \frac{x}{1-x} \) — a shifted series by multiplying \( x \), which shifts the indices of the sum.
By separating these parts, each can be expressed independently as a geometric series, allowing simple computations:
  • \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \)
  • \( \frac{x}{1-x} = \sum_{n=1}^{\infty} x^n \)
Combining these results leads to a unified series that neatly represents the original function:\[ 1 + 2 \sum_{n=1}^{\infty} x^n \]This decomposition allows difficult problems to be solved by leveraging the properties of simpler series representations, greatly simplifying the handling of complex functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty} \frac{n^{2}+1}{5^{n}}$$

Find the sum of the series. \(1-\ln 2+\frac{(\ln 2)^{2}}{2 !}-\frac{(\ln 2)^{3}}{3 !}+\cdots\)

Suppose that the power series \(\Sigma c_{a}(x-a)^{n}\) satisfies \(c_{n} \neq 0\) for all \(n\) . Show that if \(\lim _{n \rightarrow \infty}\left|c_{n} / c_{n+1}\right|\) exists, then it is equal to the radius of convergence of the power series.

(a) Show that the function defined by $$f(x)=\left\\{\begin{array}{ll}{e^{-1 / x^{2}}} & {\text { if } x \neq 0} \\\ {0} & {\text { if } x=0}\end{array}\right.$$ is not equal to its Maclaurin series. (b) Graph the function in part (a) and comment on its behavior near the origin.

In Section 4.8 we considered Newton's method for approximating a root \(r\) of the equation \(f(x)=0,\) and from an initial approximation \(x_{1}\) we obtained successive approximations \(X_{2}, X_{3}, \ldots,\) where $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ Use Taylor's Inequality with \(n=1, a=x_{n},\) and \(x=r\) to show that if \(f^{\prime \prime}(x)\) exists on an interval I containing \(r, x_{n}\) and \(x_{n+1},\) and \(\left|f^{\prime \prime}(x)\right| \leqslant M,\left|f^{\prime}(x)\right| \geqslant K\) for all \(x \in I,\) then $$\left|x_{n+1}-r\right| \leqslant \frac{M}{2 K}\left|x_{n}-r\right|^{2}$$ [This means that if \(x_{n}\) is accurate to \(d\) decimal places, then \(x_{n+1}\) is accurate to about 2\(d\) decimal places. More precisely, if the error at stage \(n\) is at most \(10^{-m}\) , then the error at stage \(n+1\) is at most \((M / 2 K) 10^{-2 m} \cdot ]\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.