Problem 22 - Advanced Engineering Mathematics

Question

Find the circle and radius of convergence of the given power series. \(\sum_{k=1}^{\infty} \frac{1}{k}\left(\frac{\dot{i}}{1+i}\right)^{k} z^{k}\)

Step-by-Step Solution

Answer

The radius of convergence is \(\sqrt{2}\), and the circle of convergence is \(|z| < \sqrt{2}\).

Step 1: Understand the Power Series

The given power series is \( \sum_{k=1}^{\infty} \frac{1}{k}\left(\frac{i}{1+i}\right)^{k} z^{k} \). This series is expressed in terms of \( z^k \), with coefficients \( \frac{1}{k}\left(\frac{i}{1+i}\right)^{k} \). Our task is to find the circle and radius of convergence.

Step 2: Find the General Term

The general term of the series is \( a_k = \frac{1}{k}\left(\frac{i}{1+i}\right)^{k} z^k \). To examine the radius of convergence, we'll focus on the coefficient portion, \( \frac{1}{k}\left(\frac{i}{1+i}\right)^{k} \).

Step 3: Simplify the Coefficient

The coefficient can be further simplified. Note that \( \left(\frac{i}{1+i}\right)\) can be rewritten by multiplying numerator and denominator by the complex conjugate of the denominator. \[ \frac{i}{1+i} = \frac{i(1-i)}{(1+i)(1-i)} = \frac{i - i^2}{1 + 1} = \frac{1 + i}{2}. \]

Step 4: Rewrite the General Term

Substituting the simplified coefficient gives us:\( a_k = \frac{1}{k}\left(\frac{1+i}{2}\right)^{k} z^k \).Now the power series is \( \sum_{k=1}^{\infty} \frac{1}{k}\left(\frac{1+i}{2}\right)^{k} z^k \).

Step 5: Use the Root Test for Convergence

To determine the radius of convergence, apply the root test, which involves:\[ \lim_{k \to \infty} \sqrt[k]{\left| a_k \right|} = \lim_{k \to \infty} \sqrt[k]{\left| \frac{1}{k}\left(\frac{1+i}{2}\right)^{k} \right|} = \lim_{k \to \infty} \left| \frac{1+i}{2} \right|. \]The magnitude \( |1+i| = \sqrt{2} \), so the series has:\[ \lim_{k \to \infty} \left( \frac{\sqrt{2}}{2} \right) = \frac{1}{\sqrt{2}}. \]

Step 6: Determine the Radius and Region of Convergence

The radius of convergence \( R \) is \( \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \). Thus, the circle of convergence is \( |z| < \sqrt{2} \).

Key Concepts

Complex Numbers in Power Series

Complex numbers are an integral part of many mathematical concepts, including power series used in this context. A complex number has two components: a real part and an imaginary part, and is expressed as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part represented by \( i \), which satisfies \( i^2 = -1 \). In the given problem, the term \( \frac{i}{1+i} \) makes use of division of complex numbers. This division is often simplified using the complex conjugate. The conjugate of \( 1+i \) is \( 1-i \), and when we multiply both the numerator and denominator by this conjugate, we simplify the term to \( \frac{1+i}{2} \). This simplifies calculations and is crucial in simplifying power series' coefficients for further analysis.

Understanding and Working with Power Series

A power series is a type of infinite series that is written in the form \( \sum_{k=0}^{\infty} a_k z^k \), where \( z \) is a complex variable. It represents a function as the sum of its components, allowing for the approximation of mathematical functions. In this context, the given series \( \sum_{k=1}^{\infty} \frac{1}{k}\left(\frac{1+i}{2}\right)^{k} z^k \) is centered around the variable \( z \) and involves complex coefficients. A critical aspect is determining where the series converges, which depends on the values of \( z \) that make the series meaningful. Convergence dictates whether a series reliably approximates a function in a specific region, expressed as a circle of convergence, determined by its radius.

Applying the Root Test for Convergence

The root test is a method used to determine the convergence of a power series, especially useful in calculating the radius of convergence. It examines the absolute magnitude of terms in a series and is executed by finding the limit \( \lim_{k \to \infty} \sqrt[k]{|a_k|} \), where \( a_k \) is the term of the series. In our problem, simplifying \( a_k = \frac{1}{k}\left(\frac{1+i}{2}\right)^{k} z^k \) gives consideration to the coefficient \( \left(\frac{1+i}{2}\right)^k \), and using its magnitude, \( |1+i| = \sqrt{2} \). The root test works by calculating \( \lim_{k \to \infty} \left( \frac{\sqrt{2}}{2} \right) = \frac{1}{\sqrt{2}} \). This leads to the determination of the circle of convergence radius \( R = \sqrt{2} \), meaning the series converges for values of \( z \) such that \( |z| < \sqrt{2} \).