Chapter 8: Problem 9
Express the given function as a power series in \(x\) with base point \(0 .\) Calculate the radius of convergence \(R\). \(\frac{x^{3}}{1+x^{4}}\)
Short Answer
Expert verified
The power series is \( x^3 - x^7 + x^{11} - x^{15} + \cdots \) with radius of convergence \( R = 1 \).
Step by step solution
01
Understand the Function
The function given is \( \frac{x^3}{1 + x^4} \). We want to express this as a power series about the base point 0. This means we'll use a series expansion centered at \( x = 0 \).
02
Identify the Geometric Series Structure
The expression \( \frac{1}{1 - u} \) can be expanded into a geometric series \( 1 + u + u^2 + u^3 + \cdots \), provided that \( |u| < 1 \). We rewrite the denominator as \( 1 - (-x^4) \) to meet this format.
03
Rewrite the Function Using Geometric Series
Using the geometric series formula, rewrite the denominator: \( \frac{1}{1 + x^4} = 1 - x^4 + x^8 - x^{12} + \cdots \).
04
Apply the Power of x
Multiply the series by \( x^3 \) to get the final power series representation: \( x^3 (1 - x^4 + x^8 - x^{12} + \cdots) = x^3 - x^7 + x^{11} - x^{15} + \cdots \).
05
Find the Radius of Convergence
The geometric series \( 1 + u + u^2 + \cdots \) converges when \( |u| < 1 \). We have \( u = x^4 \), so \( |x^4| < 1 \) which simplifies to \( |x| < 1 \). This means the radius of convergence \( R \) is 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
A geometric series is a sum of terms where each term is a constant multiple, known as the common ratio, of the previous term. The general form of a geometric series is:
- \( a + ar + ar^2 + ar^3 + \cdots \)
Radius of Convergence
The radius of convergence of a power series is a crucial concept for determining the interval in which the series converges. It is dependent on both the function and the series form. For the function \[ y = \frac{x^3}{1 + x^4} \]we expressed it using a geometric series. After rewriting it, the series becomes:\[ x^3 (1 - x^4 + x^8 - x^{12} + \cdots) \]This series converges when \(|x^4| < 1\), meaning \(|x| < 1\). Here, the radius of convergence \( R \) is 1. This value determines how far from the center point, \( x = 0 \), the series will accurately represent the function.
Series Expansion
Series expansion is a technique used to approximate functions using a series of simpler terms, often in the form of power series, which are infinite sums involving powers of a variable. By understanding how to expand the function\[ \frac{x^3}{1 + x^4} \] we express it in a form where it becomes easier to analyze and calculate values. Using the geometric series structure, we start with:\[ \frac{1}{1 + x^4} = 1 - x^4 + x^8 - x^{12} + \cdots \]Multiplying throughout by \(x^3\), every term is adjusted accordingly, resulting in:\[ x^3 - x^7 + x^{11} - x^{15} + \cdots \]This step transforms the original function into a power series that can be used for further analysis, particularly over its radius of convergence.
Convergence Interval
The convergence interval is the set of all values for which a power series converges to a function. For any power series, the endpoints of this interval are determined by the radius of convergence and can sometimes include or exclude the endpoints themselves. For the expanded series:\[ x^3 - x^7 + x^{11} - x^{15} + \cdots \]The condition for convergence derived from the geometric series form is \(|x| < 1\), leading to a convergence interval of \[ (-1, 1) \]This interval provides a domain where the power series accurately replicates the original function \( \frac{x^3}{1 + x^4} \) and where calculations using the power series remain valid.