Chapter 5: Problem 3
Suppose that \(\sum_{n=0}^{\infty} a_{n} x^{n}\) has an interval of convergence of \((-1,1)\). Find the interval of convergence of \(\sum_{n=0}^{\infty} a_{n}\left(\frac{x}{2}\right)^{n}\).
Short Answer
Expert verified
The interval of convergence is
\((-2, 2)\).
Step by step solution
01
Understanding the series transformation
The original power series is given by \( \sum_{n=0}^{\infty} a_{n} x^{n} \) with an interval of convergence \((-1,1)\). The new series is \( \sum_{n=0}^{\infty} a_{n}\left(\frac{x}{2}\right)^{n} \). We need to find the interval of convergence for this new series.
02
Substituting into the new series
Substitute \( y = \frac{x}{2} \) into the original series. This transforms \( \sum_{n=0}^{\infty} a_{n} y^{n} \), which has the same form as the original series but with an input \( y \) instead of \( x \).
03
Finding the interval of convergence for the substitute
The interval of convergence for the series \( \sum a_n x^n \) is \( -1 < x < 1 \). When substituting \( y = \frac{x}{2} \), we need to find the interval for \( y \), so it becomes \(-1 < \frac{x}{2} < 1\).
04
Solving for x
Solve the inequality \(-1 < \frac{x}{2} < 1\) to find the range of \( x \). Multiply the entire inequality by 2, yielding \(-2 < x < 2\).
05
Determining the interval of convergence
The inequality solved in the previous step tells us that the interval of convergence for the series \( \sum_{n=0}^{\infty} a_{n}\left(\frac{x}{2}\right)^{n} \) is \(-2 < x < 2\). This is the result for convergence of the new series.
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 Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series is a way of expressing a function as an infinite sum of terms. Each term in the series is based on a constant coefficient and is multiplied by a power of the variable. The general form of a power series can be expressed as:
- \( \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) are constant coefficients, and \( x \) is the variable.
Series Transformation
Series transformation involves altering a given power series into a new series. This could involve changing the variable or applying some operation to the coefficients or terms. In our example, we deal with transforming the original series:
- From \( \sum_{n=0}^{\infty} a_{n} x^{n} \)
- To \( \sum_{n=0}^{\infty} a_{n}\left(\frac{x}{2}\right)^{n} \)
Inequality Solving
Inequality solving is a technique used to determine the values or range of values that a certain expression can take. In context with power series, inequalities help us explore the interval of convergence.In our transformed series example, we face the inequality:
- \(-1 < \frac{x}{2} < 1\)
- Results in \(-2 < x < 2\)
Substitution Method
The substitution method is an approach used to simplify and solve complex mathematical expressions or equations. This technique entails replacing a variable with another expression or variable to make calculations more manageable.Consider our example, where substitution was crucial. The original power series was in terms of \( x \), and by substituting \( y = \frac{x}{2} \), the series is transformed into a familiar form:
- \( \sum_{n=0}^{\infty} a_{n} y^{n} \)