Problem 1
Question
In each of Exercises \(1-10\), express the given function as a power series in \(x\) with base point \(0 .\) Calculate the radius of convergence \(R\). \(\frac{1}{1-2 x}\)
Step-by-Step Solution
Verified Answer
The power series is \( \sum_{n=0}^{\infty} 2^n x^n \) with radius of convergence \( R = \frac{1}{2} \).
1Step 1: Identify the Function Form
Notice that the function \( \frac{1}{1-2x} \) is already in the form \( \frac{1}{1-ax} \), where \( a = 2 \). This resembles the geometric series form \( \sum_{n=0}^{\infty} (ax)^n \).
2Step 2: Express as a Power Series
Using the geometric series formula, \( \frac{1}{1-ax} = \sum_{n=0}^{\infty} (ax)^n \), substitute \( a = 2 \) to get the power series representation: \( \frac{1}{1-2x} = \sum_{n=0}^{\infty} (2x)^n = \sum_{n=0}^{\infty} 2^n x^n \).
3Step 3: Check for Convergence
A geometric series \( \sum_{n=0}^{\infty} (ax)^n \) converges when \( |ax| < 1 \). For our series, this condition becomes \( |2x| < 1 \).
4Step 4: Solve for Radius of Convergence
The inequality \( |2x| < 1 \) simplifies to \( |x| < \frac{1}{2} \). This tells us that the power series converges when \( |x| < \frac{1}{2} \), meaning the radius of convergence \( R \) is \( \frac{1}{2} \).
Key Concepts
Radius of ConvergenceGeometric SeriesFunction Representation
Radius of Convergence
The radius of convergence is a key concept when working with power series. It determines the interval around the base point within which the series converges and maintains validity. To calculate this, we examine the condition for convergence in the series: for the series \( \sum_{n=0}^{\infty} (ax)^n \), convergence is achieved when \( |ax| < 1 \).
In our example, the series \( \sum_{n=0}^{\infty} (2x)^n \) converges when \(|2x| < 1\). By solving this inequality, we attain \( |x| < \frac{1}{2} \). Thus, the radius of convergence \( R \) is \( \frac{1}{2} \). This informs us on how far from the base point 0 we can consider the series to still represent the function accurately within a particular range.
In our example, the series \( \sum_{n=0}^{\infty} (2x)^n \) converges when \(|2x| < 1\). By solving this inequality, we attain \( |x| < \frac{1}{2} \). Thus, the radius of convergence \( R \) is \( \frac{1}{2} \). This informs us on how far from the base point 0 we can consider the series to still represent the function accurately within a particular range.
Geometric Series
A geometric series is a special kind of series with a constant ratio between successive terms. It forms a crucial foundation in understanding power series and their applications. The general form of a geometric series is \( \sum_{n=0}^{\infty} r^n \), where \( r \) is the common ratio.
For convergence, the absolute value of the common ratio must be less than 1, i.e., \( |r| < 1 \). In the context of power series in the form \( \frac{1}{1-ax} \), the geometric series formula \( \sum_{n=0}^{\infty} (ax)^n \) is derived. This formula helps us to transform functions like \( \frac{1}{1-2x} \) into their series representation, making it easier to work with or understand the approximation of complex functions.
For convergence, the absolute value of the common ratio must be less than 1, i.e., \( |r| < 1 \). In the context of power series in the form \( \frac{1}{1-ax} \), the geometric series formula \( \sum_{n=0}^{\infty} (ax)^n \) is derived. This formula helps us to transform functions like \( \frac{1}{1-2x} \) into their series representation, making it easier to work with or understand the approximation of complex functions.
Function Representation
Representing functions as power series is an effective method in mathematical analysis and applications, especially when dealing with complex functions or finding approximations. A power series is an infinite series of the form \( \sum_{n=0}^{\infty} c_n x^n \), where \( c_n \) are coefficients. This representation is highly versatile, allowing for functions to be expressed in a way that can be readily analyzed or manipulated.
In the exercise example, we use the geometric series representation \( \frac{1}{1-2x} = \sum_{n=0}^{\infty} 2^n x^n \). This breakdown into a series makes it easier to approximate the function for values within the radius of convergence \( R = \frac{1}{2} \). Power series also prove useful in solving differential equations, integrating functions, and performing operations like differentiation term by term, all of which leverage the polynomial form that we derive from the series representation.
In the exercise example, we use the geometric series representation \( \frac{1}{1-2x} = \sum_{n=0}^{\infty} 2^n x^n \). This breakdown into a series makes it easier to approximate the function for values within the radius of convergence \( R = \frac{1}{2} \). Power series also prove useful in solving differential equations, integrating functions, and performing operations like differentiation term by term, all of which leverage the polynomial form that we derive from the series representation.
Other exercises in this chapter
Problem 1
In each of Exercises \(1-12,\) use the Ratio Test to determine the convergence or divergence of the given series. \(\sum_{n=1}^{\infty} \frac{n}{e^{n}}\)
View solution Problem 1
Use Theorem 2 and, where necessary, limit formula (8.5.1) to calculate the radius of convergence \(R\). Determine the interval of convergence \(I\) by checking
View solution Problem 1
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