Problem 15 - Advanced Engineering Mathematics

Question

In Problems 13-22, expand the given function in a Taylor series centered at the indicated point. Give the radius of convergence of each series. $$ f(z)=\frac{1}{3-z}, z_{0}=2 i $$

Step-by-Step Solution

Answer

The Taylor series expansion of \( \frac{1}{3-z} \) around \( z_0 = 2i \) has a radius of convergence \( \sqrt{13} \).

Step 1: Define the Taylor Series

To expand a function in a Taylor series centered at a point \( z_0 \), the general form is given by: \[f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n\] Our function is \( f(z) = \frac{1}{3-z} \) and we need to expand it around \( z_0 = 2i \).

Step 2: Manipulate the Function Form

We rewrite \( f(z) = \frac{1}{3-z} \) to make it easier to identify the series:\[f(z) = \frac{1}{1 - (z - 3)}\]This resembles the form of the geometric series with \( a = 1 \) and \( x = z - 3 \).

Step 3: Substitute and Expand the Geometric Series

Using the geometric series formula \( \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \) for \(|x| < 1\), substitute \( x = (z - 3) \) into the geometric series:\[f(z) = \sum_{n=0}^{\infty} (z - 3)^n = \sum_{n=0}^{\infty} ((z - 2i) + (2i - 3))^n\]

Step 4: Express in Terms of the Center Point

Replace \( z - 3 \) with \( (z - 2i) + (2i - 3) \) and expand using the binomial theorem:\[f(z) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{n}{k} (z - 2i)^k (2i - 3)^{n-k}\]

Step 5: Determine the Radius of Convergence

The function \( \frac{1}{3-z} \) is analytic where \( z eq 3 \). The distance from the center \( z_0 = 2i \) to \( z = 3 \) is the radius of convergence. Calculate the distance using \( |z_0 - 3| \):\[R = |2i - 3| = \sqrt{(-3)^2 + (2)^2} = \sqrt{13}\] Thus, the radius of convergence is \( \sqrt{13} \).

Key Concepts

Radius of Convergence

When you work with Taylor series, one of the key concepts to understand is the radius of convergence. This is crucial because it tells you the interval within which the series will converge to the original function. For a Taylor series centered at a point \( z_0 \), the radius of convergence \( R \) is the distance from \( z_0 \) to the nearest point where the function is not analytic.
The radius of convergence is calculated using the formula \( |z_0 - z| \), where \( z \) is the point where the function ceases to be analytic.

• In many cases, like with our function \( f(z) = \frac{1}{3-z} \), this happens at the point where the function is undefined or has a singularity.
• For this function, the problematic point is \( z = 3 \), and it's important to note that the function is not analytic there. Hence, you measure the distance from the center to this problematic point.

In the original exercise, we centered the series at \( z_0 = 2i \). Using the distance formula for complex numbers, you can calculate \( R = \sqrt{(-3)^2 + (2)^2} = \sqrt{13} \).
This means that the Taylor series will converge for values of \( z \) within a distance of \( \sqrt{13} \) from \( 2i \). Understanding this gives you a boundary within which you can rely on the Taylor expansion to accurately approximate \( f(z) \).

Geometric Series

The geometric series is a fundamental concept within the study of sequences and series, serving as a basic building block for many other series, including Taylor series. A geometric series has the form \( \sum_{n=0}^{\infty} x^n \), which converges to \( \frac{1}{1-x} \) when \(|x| < 1\).

• It is simple because each term is a fixed multiple of the previous term, given by the common ratio \( x \).
• Understanding this pattern allows you to transform many functions into a series using geometric progression.

In the exercise, the function \( f(z)=\frac{1}{3-z} \) was manipulated into a form \( \frac{1}{1-(z-3)} \), which closely resembles the format of a geometric series. This allows you to use the geometric series formula to expand the function around another point.
By setting \( x = z-3 \), you replace a more complex function with an infinite sum that is easier to handle, especially when seeking derivatives or integrals.

Complex Analysis

Complex analysis revolves around studying functions that are based on complex numbers. Unlike real-valued functions, complex functions can behave differently, having properties and applications unique to the complex plane. One of the essential aspects of complex analysis is the function's analyticity, where a complex function is differentiable at every point in its domain.

• Understanding analyticity helps in determining the convergence and behavior of functions, as encountered in Taylor series expansions.
• Functions that are not analytic at certain points indicate potential singularities, which are crucial for deciding the radius of convergence.

For the function \( f(z) = \frac{1}{3-z} \), complex analysis provides the tools to analyze where the function is analytic and determine how the series behaves at and around the center point \( z_0 = 2i \).
The series expansion techniques like Taylor series allow us to work with complex functions effectively, aiding in solving real-world problems where complex numbers play a role. Understanding these core principles enriches your knowledge of both mathematics and its applications in engineering and physics.