Problem 21 - 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. $$ f(z)=(z-1) e^{-2 z}, z_{0}=1 $$

Step-by-Step Solution

Answer

The Taylor series of \( f(z) \) centered at \( z_0 = 1 \) has a radius of convergence \( R = \infty \).

Step 1: Understanding the Function and Center Point

We have the function \( f(z) = (z-1) e^{-2z} \) which we need to expand in a Taylor series about the point \( z_0 = 1 \). This means we want to express \( f(z) \) as a power series around \( z = 1 \).

Step 2: Expanding \( e^{-2z} \) as a Series

First, recall that the exponential function \( e^x \) can be expanded as a Taylor series: \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}. \]Substitute \( x = -2z \) to get the series expansion of \( e^{-2z} \): \[ e^{-2z} = \sum_{n=0}^{\infty} \frac{(-2z)^n}{n!}. \]

Step 3: Substitute the Function into the Series

Now substitute \( t = z - 1 \) to focus the expansion at \( z_0 = 1 \). Rewriting \( e^{-2z} \) with \( z = t + 1 \), we have: \[ e^{-2(z+1)} = \sum_{n=0}^{\infty} \frac{(-2)^{n} (t+1)^{n}}{n!}. \]We then multiply this whole expression by \( z - 1 = t \), resulting in: \[ (z-1) e^{-2z} = t \sum_{n=0}^{\infty} \frac{(-2)^{n} (t+1)^{n}}{n!}. \]

Step 4: Simplify the Expression

Simplify the expression by expanding \( (t+1)^n \) using the binomial theorem:\[ (t+1)^n = \sum_{k=0}^{n} \binom{n}{k} t^k. \]Thus, the series becomes:\[ (z-1) e^{-2z} = \sum_{n=0}^{\infty} t \sum_{k=0}^{n} \frac{(-2)^n \binom{n}{k} t^k}{n!}. \] Further simplify to get the coefficient of \( t^m \) by combining the terms.

Step 5: Determining the Radius of Convergence

Since the function involves the exponential function \( e^{-2z} \), it will converge for all \( z \in \mathbb{C} \). Thus, the radius of convergence \( R \) for this series is infinity. Finally, substitute back \( t = z - 1 \) for the complete Taylor series expression around \( z_0 = 1 \).

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when working with power series, like the Taylor series. It helps us understand the domain within which a series converges, meaning it settles down to a finite value.

For any given power series \( \sum_{n=0}^{\infty} c_n (z-z_0)^n \), the radius of convergence \( R \) defines the interval or disk in the complex plane where this series converges. This radius can be found using the formula:

• Lloyd's Ratio Test: \( R= \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|c_n|}} \)
• Alternately, the Radius can be determined from the root test or comparison test.

When \( R = \infty \), the series converges for all values of \( z \). In this problem, because of the exponential nature of the function being expanded, our series converges for all values of \( z \) in the complex plane. Thus, the radius of convergence is infinite, meaning it doesn't restrict convergence to any particular neighborhood of \( z_0 \).

Knowing the radius of convergence helps in identifying where the function accurately represents through its series expansion.

Exponential Function

The exponential function is one of the fundamental functions in mathematics, commonly denoted as \( e^x \). It's unique because it rates of change are proportional to the value of the function itself. This property makes it incredibly useful for differential equations, modeling growth and decay processes, and much more.

In the context of series expansion, the exponential function is beautifully expressed as an infinite sum using the Taylor series:

• Base formula: \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \)

This expansion holds true for all complex numbers \( x \), demonstrating the exponential function's universality and rich behavior.

In our problem, the function \( e^{-2z} \) was expanded. By substituting \( x = -2z \) into the exponential series, it allows us to approximate \( e^{-2z} \) as a sum of infinitely many polynomials. Such transformation aids in the function's manipulation and is foundational in deriving solutions for various mathematical problems.

Series Expansion

Series expansion is a powerful tool in mathematics that allows us to represent complex functions as infinite sums of simpler terms, often polynomials. This technique is vital for approximating functions, especially within a specific interval or region.

The Taylor series is a primary example of series expansion. It represents a function \( f(x) \) in terms of its derivatives at a single point \( z_0 \). The general form is:

• \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(x-z_0)^n \)

In our exercise, this expansion is performed at the center point \( z_0 = 1 \), which simplifies the computations around that point by effectively transforming it into a polynomial expression. It provides an approximation of the function near \( z_0 \) and helps better understand the function's behavior.

When performing series expansion, attention to convergence is important, ensuring the series actually approximates the function adequately over the region of interest. This adds depth to the calculus toolbox, giving students an edge in both theoretical and applied contexts.

Binomial Theorem

The Binomial Theorem is a mathematical treasure that facilitates the expansion of expressions raised to a power. It lets us express \( (a+b)^n \) as a sum involving terms of \( a^k \) and \( b^{n-k} \) multiplied by combinatorial coefficients \( \binom{n}{k} \).

In formula, it is:

• \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k\)

This theorem is particularly useful for expanding powers of binomials, hence its application in this exercise's simplification steps.

In the original problem, the expression \((t+1)^n\) used within the Taylor series expands thanks to the Binomial Theorem. The coefficients \( \binom{n}{k} \) arise from combinatorics, counting the ways to choose \( k \) items from \( n \), adding another layer of understanding to polynomial expressions.

By understanding and applying the Binomial Theorem, one can elegantly expand and simplify polynomial-like expressions, demonstrating the deep connection between algebra and calculus.