Problem 3
Question
Express the given function as a power series in \(x\) with base point \(0 .\) Calculate the radius of convergence \(R\). \(\frac{1}{2-x}\)
Step-by-Step Solution
Verified Answer
The function is \( \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} x^n \) and the radius of convergence is 2.
1Step 1: Recall the geometric series formula
The geometric series formula is given by \( \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \) for \(|r| < 1\). We will use this formula to express the given function as a power series.
2Step 2: Rewrite the function to match the formula
The given function is \( \frac{1}{2-x} \). To use the geometric series formula, we rewrite it as \( \frac{1}{2(1-\frac{x}{2})} \). Here, \( a = \frac{1}{2} \) and \( r = \frac{x}{2} \).
3Step 3: Write the power series expression
Using the geometric series formula, substitute \( a = \frac{1}{2} \) and \( r = \frac{x}{2} \) into \( \sum_{n=0}^{\infty} ar^n \). Thus, the power series becomes \( \sum_{n=0}^{\infty} \frac{1}{2} \left( \frac{x}{2} \right)^n = \frac{1}{2} \sum_{n=0}^{\infty} \left( \frac{x}{2} \right)^n \).
4Step 4: Simplify the power series
The expression simplifies to \( \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} x^n \). This is the power series representation of the function centered at \( x = 0 \).
5Step 5: Determine the radius of convergence
The radius of convergence \( R \) is determined by the condition \(|r|<1\), which is \(|\frac{x}{2}| < 1\) or \(|x| < 2\). Therefore, the radius of convergence is \( R = 2 \).
Key Concepts
Understanding Geometric SeriesRadius of ConvergenceThe Convergence of Series
Understanding Geometric Series
A geometric series refers to a series where each term is a fixed multiple, or ratio, of the preceding one. A simple way to think about this is to consider the sequence of terms produced by multiplying a starting value repeatedly by a constant factor, known as the common ratio. Geometric series are everywhere in mathematics and appear in numerous applications, ranging from finance to physics.
Mathematically, the geometric series can be expressed as:
In practice, when you're given a function like \( \frac{1}{2-x} \), you can identify it as a geometric series by rewriting it to match the format \( \frac{a}{1-r} \), illustrating how a power series can be derived.
Mathematically, the geometric series can be expressed as:
- The sum, \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term.
- \( r \) is the common ratio, and \( n \) is the term number.
In practice, when you're given a function like \( \frac{1}{2-x} \), you can identify it as a geometric series by rewriting it to match the format \( \frac{a}{1-r} \), illustrating how a power series can be derived.
Radius of Convergence
The radius of convergence, denoted as \( R \), signifies the range of input values (or domain) within which a power series converges to a finite sum. It is a critical concept when dealing with power series as it tells us where a series representation of a function is valid.
Simply put, the radius of convergence helps identify the interval centered on a given point, usually denoted \( x = a \), over which the series converges. The interval is often expressed as \(|x - a| < R\).
In the context of the given function \( \frac{1}{2-x} \), the rearrangement to \( \frac{1}{2(1-\frac{x}{2})} \) indicates that \( r = \frac{x}{2} \). To determine \( R \), you find \(|r| < 1\) leading to \(|\frac{x}{2}| < 1\), simplifying to \(|x| < 2\). Thus, the radius of convergence is 2, meaning the series will converge only when \( x \) is within 2 units from 0.
Simply put, the radius of convergence helps identify the interval centered on a given point, usually denoted \( x = a \), over which the series converges. The interval is often expressed as \(|x - a| < R\).
In the context of the given function \( \frac{1}{2-x} \), the rearrangement to \( \frac{1}{2(1-\frac{x}{2})} \) indicates that \( r = \frac{x}{2} \). To determine \( R \), you find \(|r| < 1\) leading to \(|\frac{x}{2}| < 1\), simplifying to \(|x| < 2\). Thus, the radius of convergence is 2, meaning the series will converge only when \( x \) is within 2 units from 0.
The Convergence of Series
The convergence of a series refers to the behavior of the series as the number of terms increases towards infinity. A series that converges has terms that approach a single, finite value. When working with power series, convergence is a fundamental aspect, defining the set of points where the series precisely maps onto the function it represents.
Determining convergence involves various tests, including the ratio test, the root test, and more. However, for geometric series, the convergence is straightforward due to the condition \(|r| < 1\).
Understanding convergence helps ensure that mathematical operations on series, such as integration and differentiation, are valid within the series' radius of convergence. For the function \( \frac{1}{2-x} \), expressed as a series \( \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} x^n \), it converges for \(|x| < 2\), as determined earlier by the radius of convergence. Thus, the convergence tells us the series accurately represents the function within this interval.
Determining convergence involves various tests, including the ratio test, the root test, and more. However, for geometric series, the convergence is straightforward due to the condition \(|r| < 1\).
Understanding convergence helps ensure that mathematical operations on series, such as integration and differentiation, are valid within the series' radius of convergence. For the function \( \frac{1}{2-x} \), expressed as a series \( \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} x^n \), it converges for \(|x| < 2\), as determined earlier by the radius of convergence. Thus, the convergence tells us the series accurately represents the function within this interval.
Other exercises in this chapter
Problem 2
Evaluate \(\lim _{n \rightarrow \infty} a_{n}\) for the given sequence \(\left\\{a_{n}\right\\}\). $$ a_{n}=\frac{3 n^{2}+n+4}{2 n^{3}+1} $$
View solution Problem 2
State what conclusion, if any, may be drawn from the Divergence Test. $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{1+\sqrt{n}}} $$
View solution Problem 3
Use the Ratio Test to determine the convergence or divergence of the given series. \(\sum_{n=1}^{\infty} \frac{2^{n}}{n^{3}}\)
View solution Problem 3
The given series may be shown to converge by using the Alternating Series Test. Show that the hypotheses of the Alternating Series Test are satisfied. $$ \sum_{
View solution