Problem 5 - Advanced Engineering Mathematics

Question

In Problems 1-12, expand the given function in a Maclaurin series. Give the radius of convergence of each series. $$ f(z)=e^{-2 z} $$

Step-by-Step Solution

Answer

Radius of convergence = ∞

Step 1: Understand the Maclaurin Series Formula

The Maclaurin series of a function \( f(z) \) is given by the infinite sum \( \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \), where \( f^{(n)}(0) \) is the \( n \)-th derivative of \( f(z) \) evaluated at \( z=0 \). For \( e^{-2z} \), this series will need the derivatives of \( f(z)=e^{-2z} \).

Key Concepts

Radius of Convergence

The radius of convergence is a fundamental concept when working with power series, including Maclaurin series. When we express a function as an infinite sum of terms, we need to know for which values of the variable these terms actually converge to the function. This is defined by the radius of convergence, denoted usually by "R." In simple terms, it tells us how far from the central point (usually the origin) we can go while ensuring the series still gives a valid approximation of the function.
To find the radius of convergence of the Maclaurin series for a function like \( e^{-2z} \), we can use the formula associated with power series:

• For a power series \( \sum_{n=0}^{\infty} c_n z^n \), the radius of convergence \( R \) is given by \( \frac{1}{\limsup_{n \to \infty} |c_n|^{1/n}} \)

This formula can help us determine the set of values for \( z \) where the series will remain valid. For exponential functions, like \( e^{-2z} \), the entire complex plane is covered, meaning \( R \) is infinite. This means the series converges for any value of \( z \), which is a significant property of such functions.

Infinite Series Expansion

An infinite series expansion is a way to express a function as the sum of an infinite number of terms. It is a powerful tool in mathematics, allowing complex functions to be approximated by simpler terms. The Maclaurin series is a specific type of infinite series expansion centered around \( z = 0 \).
For the function \( f(z) = e^{-2z} \), we can expand it as a Maclaurin series using its derivatives taken at \( z = 0 \). This involves calculating each derivative of the function, evaluating them at the origin, and plugging these values into the general formula:

• \( f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \)

Given the exponential nature of \( e^{-2z} \), this particular series will follow a predictable pattern with terms that involve powers of \( z \) scaled by factorials. This makes the infinite series an especially convenient form for calculation and analysis in many practical applications.

Derivatives of Exponential Functions

When dealing with exponential functions like \( f(z) = e^{-2z} \), understanding derivatives is crucial for forming series expansions like the Maclaurin series. Exponential functions have distinctive properties that simplify the process of taking derivatives. In general:

• The derivative of \( e^{az} \) with respect to \( z \) is \( ae^{az} \).

For our function, each derivative will multiply the exponent by \(-2\), leading to an elegant property where the form of the function remains unchanged, except for a multiplying constant. The successive derivatives of the function \( e^{-2z} \) at the origin demand applying this rule recursively:

• \( f'(z) = -2e^{-2z} \)
• \( f''(z) = 4e^{-2z} \)

These expressions are then evaluated at \( z = 0 \) to contribute to the series expansion coefficients, making the exponential functions wonderfully suited for such mathematical operations, given their regular patterns.