Problem 32 - Advanced Engineering Mathematics

Question

(a) Consider the function \(f(z)=\operatorname{Ln}(1+z)\). What is the radius of the largest circle centered at the origin within which \(f\) is analytic? (b) Expand \(f\) in a Maclaurin series. What is the radius of convergence of this series? (c) Use the result in part (b) to find a Maclaurin series for \(\operatorname{Ln}(1-z)\) (d) Find a Maclaurin series for \(\operatorname{Ln}\left(\frac{1+z}{1-z}\right)\).

Step-by-Step Solution

Answer

The radius of analyticity is 1. The Maclaurin series for \( \operatorname{Ln}(1+z) \) converges with radius 1, and \( \operatorname{Ln}\left(\frac{1+z}{1-z}\right) = 2\sum_{n=0}^{\infty} \frac{z^{2n+1}}{2n+1} \).

Step 1: Understanding Analyticity

The function is analytic where the expression inside the logarithm is not equal to zero. Since the function is \( f(z) = \operatorname{Ln}(1 + z) \), it is analytic where \( 1 + z eq 0 \). This implies that \( z eq -1 \). Thus, the largest circle centered at the origin where the function is analytic is determined by the distance to \( z = -1 \), which is 1.

Step 2: Calculate Radius of Analyticity

Since \( f(z) = \operatorname{Ln}(1 + z) \) is analytic for \( |z| < 1 \), the radius of the largest circle centered at the origin within which \( f \) is analytic is \( R = 1 \).

Step 3: Expand into Maclaurin Series

The Maclaurin series expansion of \( f(z) = \operatorname{Ln}(1 + z) \) is derived from integrating the series \( \sum_{n=0}^{\infty} (-1)^n z^n \), resulting in: \[ \operatorname{Ln}(1+z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} z^n \].

Step 4: Determine Radius of Convergence

The series \( \operatorname{Ln}(1+z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} z^n \) converges for \( |z| < 1 \). Hence, the radius of convergence of the Maclaurin series is \( R = 1 \).

Step 5: Maclaurin Series for \( \operatorname{Ln}(1-z) \)

To find the Maclaurin series for \( \operatorname{Ln}(1-z) \), substitute \( z \) with \(-z\) in the series for \( \operatorname{Ln}(1+z) \): \[ \operatorname{Ln}(1-z) = -\sum_{n=1}^{\infty} \frac{1}{n} z^n \]. This series converges for \( |z| < 1 \).

Step 6: Maclaurin Series for \( \operatorname{Ln}\left(\frac{1+z}{1-z}\right) \)

Use the property \( \operatorname{Ln}\left(\frac{1+z}{1-z}\right) = \operatorname{Ln}(1+z) - \operatorname{Ln}(1-z) \) and substitute the series obtained in Steps 3 and 5 to find: \[ \operatorname{Ln}\left(\frac{1+z}{1-z}\right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} z^n + \sum_{n=1}^{\infty} \frac{1}{n} z^n = 2\sum_{n=0}^{\infty} \frac{z^{2n+1}}{2n+1} \]. This series converges for \( |z| < 1 \).

Key Concepts

Maclaurin series

A Maclaurin series is a special type of Taylor series where the function is expanded about zero. It's an infinite series of terms calculated from the derivatives of a function at a single point, in this case, everything is centered at zero.
The general formula for the Maclaurin series of a function \( f(z) \) is given by:

• \( f(z) = f(0) + \frac{f'(0)}{1!} z + \frac{f''(0)}{2!} z^2 + \frac{f'''(0)}{3!} z^3 + \ldots \)

In our exercise, we expand the logarithmic function \( \operatorname{Ln}(1+z) \) into a Maclaurin series. Through integration and manipulation, we finally reach the series:

• \( \operatorname{Ln}(1 + z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} z^n \)

This series allows us to approximate the natural logarithm near \( z=0 \), and is powerful in calculating various logarithmic approximations.

Analytic functions

Analytic functions are functions that are smooth and differentiable over their domain. For a function to be analytic at a point, it must be expressible as a power series within some radius of that point.
In our exercise, the function \( f(z) = \operatorname{Ln}(1+z) \) is analytic in the domain where \( 1+z eq 0 \). This means, as long as \( z eq -1 \), there are no breaks or sharp turns in the function's plot, and it can be examined and expanded using series.
The "largest circle centered at the origin" within which the function is analytic is a visual way of representing the limits of such expansions, and it shows the boundaries up to which the function remains smooth.

Radius of convergence

The radius of convergence of a power series is the distance from the center of the circle within which the series converges. It indicates how close or far from the center you can go before the series is no longer valid.
In the context of our exercise, the Maclaurin series \( \operatorname{Ln}(1+z) \) is valid for \( |z| < 1 \). This means the radius of convergence for this series is 1.
The radius of convergence is crucial when working with series, as it dictates the domain over which our series representation of the function is applicable. You can calculate it using the formula:

• \( R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}} \)

where \( a_n \) are the coefficients of the series. In our case, with the conditions satisfied for \(|z| < 1\), the series provides accurate results inside this radius.