Problem 23
Question
Find the circle and radius of convergence of the given power series. $$ \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k 2^{k}}(z-1-i)^{k} $$
Step-by-Step Solution
Verified Answer
The circle of convergence is \( |z - (1+i)| < 2 \), and the radius is 2.
1Step 1: Identify the Power Series Structure
The given power series is \( \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k 2^{k}}(z-1-i)^{k} \). It is centered at \( z_0 = 1+i \) with terms of the form \( a_k (z - z_0)^k \) where \( a_k = \frac{(-1)^k}{k2^k} \).
2Step 2: Apply the Radius of Convergence Formula
Use the formula for the radius of convergence: \( R = \frac{1}{\limsup_{k \to \infty} \sqrt[k]{|a_k|}} \). The terms are \( a_k = \frac{1}{k2^k} \), so \( |a_k| = \frac{1}{k2^k} \).
3Step 3: Calculate \( \limsup \)
Find \( \limsup_{k \to \infty} \sqrt[k]{|a_k|} = \limsup_{k \to \infty} \sqrt[k]{\frac{1}{k2^k}} = \limsup_{k \to \infty} \left( \frac{1}{k^{1/k}2} \right) = \frac{1}{2} \) since \( k^{1/k} \to 1 \) as \( k \to \infty \).
4Step 4: Solve for the Radius of Convergence
Substitute the \( \limsup \) result into the radius formula: \( R = \frac{1}{\frac{1}{2}} = 2 \).
5Step 5: Determine the Circle of Convergence
Since the power series is centered at \( z_0 = 1+i \) and has a radius \( R = 2 \), the circle of convergence is \( |z - (1+i)| < 2 \).
Key Concepts
Radius of ConvergencePower SeriesCircle of Convergence
Radius of Convergence
In complex analysis, the concept of the "Radius of Convergence" is essential for understanding the behavior of power series. When we have a power series like \( \sum_{k=1}^{\infty} a_k (z - z_0)^k \), it converges within a certain distance from the center \( z_0 \). This distance is precisely the radius of convergence.
To find this radius mathematically, we often use the formula:
\[ R = \frac{1}{\limsup_{k \to \infty} \sqrt[k]{|a_k|}}\]
This expression requires calculating the 'limsup', which gives us a kind of limiting behavior of the sequence \(|a_k|\). The use of the root \( k \) in \( \sqrt[k]{|a_k|} \) helps in evaluating series where terms shrink rapidly with higher powers.
The formula tells us how far \( z \) can move away from \( z_0 \) such that the series remains convergent. Beyond this radius, the series diverges.
To find this radius mathematically, we often use the formula:
\[ R = \frac{1}{\limsup_{k \to \infty} \sqrt[k]{|a_k|}}\]
This expression requires calculating the 'limsup', which gives us a kind of limiting behavior of the sequence \(|a_k|\). The use of the root \( k \) in \( \sqrt[k]{|a_k|} \) helps in evaluating series where terms shrink rapidly with higher powers.
The formula tells us how far \( z \) can move away from \( z_0 \) such that the series remains convergent. Beyond this radius, the series diverges.
- For example, if the result of \( R \) is 2, entire values satisfying \(|z - z_0| < 2\) make the series converge.
- This approach simplifies the analysis of infinite series and is a cornerstone of power series analytics in complex numbers.
Power Series
A "Power Series" is a series of the form:
\[ \sum_{k=0}^{\infty} a_k (z - z_0)^k\]
It's an infinite sum consisting of powers of \( z - z_0 \), where \( a_k \) represents the coefficient of each term. These play a crucial role in many areas of mathematics because they can express functions as infinite polynomials.
Power series are utilized to approximate functions, which is particularly useful in applied mathematics and physics. They allow us to work with simplicity, as we represent complex functions in the form of approachable polynomials. The center, \( z_0 \), is the point around which these approximations are made.
\[ \sum_{k=0}^{\infty} a_k (z - z_0)^k\]
It's an infinite sum consisting of powers of \( z - z_0 \), where \( a_k \) represents the coefficient of each term. These play a crucial role in many areas of mathematics because they can express functions as infinite polynomials.
Power series are utilized to approximate functions, which is particularly useful in applied mathematics and physics. They allow us to work with simplicity, as we represent complex functions in the form of approachable polynomials. The center, \( z_0 \), is the point around which these approximations are made.
- In this form, the series is highly flexible—different center points lead to different series behaviors.
- Calculating the convergence properties is important, as it affects how the series portrays a function across a range of values.
Circle of Convergence
The "Circle of Convergence" relates directly to the radius of convergence and refers to the geographic scope where the power series converges on the complex plane. Once we know the radius \( R \) of a series centered at \( z_0 \), this circle can be defined by the inequality:
\[ |z - z_0| < R\]
Visualize this as an actual circle on the complex plane centered around \( z_0 \). Within this circle, all points \( z \) make the series converge.
It's crucial to realize that the circle boundary itself—where \( |z - z_0| = R \)—doesn't ensure convergence or divergence, and it must be tested separately. Typically, the interior of the circle guarantees convergence, but the boundary may require further analysis depending on the nature of \( a_k \).
\[ |z - z_0| < R\]
Visualize this as an actual circle on the complex plane centered around \( z_0 \). Within this circle, all points \( z \) make the series converge.
It's crucial to realize that the circle boundary itself—where \( |z - z_0| = R \)—doesn't ensure convergence or divergence, and it must be tested separately. Typically, the interior of the circle guarantees convergence, but the boundary may require further analysis depending on the nature of \( a_k \).
- The idea of this circle helps us to understand where a series is reliable and can be used to model functions or compute values.
- Practically, knowing the circle can also guide what inputs would lead to meaningful and accurate series expansions in applications.
Other exercises in this chapter
Problem 23
Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C} \frac{z}{z^{4}-1} d z, C:|z|=2 $$
View solution Problem 23
In Problems 23 and 24, expand \(f(z)=\frac{1}{(z-2)(z-1)^{3}}\) in a Laurent series valid for the given annular domain. $$ 0
View solution Problem 24
Evaluate the Cauchy principal value of the given improper integral. $$ \int_{0}^{\infty} \frac{1}{x^{6}+1} d x $$
View solution Problem 24
Determine the order of the poles for the given function. $$ f(z)=\frac{e^{z}-1}{z^{2}} $$
View solution