Problem 26 - Advanced Engineering Mathematics

Question

In Problems 25 and 26, use partial fractions as an aid in obtaining the Maclaurin series for the given function. Give the radius of convergence of the series. $$ f(z)=\frac{z-7}{z^{2}-2 z-3} $$

Step-by-Step Solution

Answer

The Maclaurin series of the function is found using partial fractions, with a radius of convergence of 1.

Step 1: Factor the Denominator

First, factor the denominator \( z^2 - 2z - 3 \). This can be factored as \( (z - 3)(z + 1) \).

Step 2: Set Up Partial Fractions

Express \( \frac{z-7}{(z-3)(z+1)} \) as a sum of partial fractions: \( \frac{z-7}{(z-3)(z+1)} = \frac{A}{z-3} + \frac{B}{z+1} \).

Step 3: Solve for Coefficients

Multiply both sides by \((z-3)(z+1)\) to clear the denominators: \( z-7 = A(z+1) + B(z-3) \). Expand and collect like terms to get \( z-7 = (A+B)z + (A - 3B) \). Equate coefficients to get two equations: \[A + B = 1 \] and \[A - 3B = -7\].

Step 4: Solve the System of Equations

Solve the system: From \(A + B = 1\), we have \(A = 1 - B\). Substitute into \(A - 3B = -7\): \(1 - B - 3B = -7\) gives \(-4B = -8\), so \(B = 2\). Substituting back, \(A = 1 - 2 = -1\).

Step 5: Write Partial Fraction Decomposition

Plug values back into the partial fraction decomposition to get: \( \frac{z-7}{(z-3)(z+1)} = \frac{-1}{z-3} + \frac{2}{z+1} \).

Step 6: Find Maclaurin Series for Each Term

Express \( \frac{-1}{z-3} \) and \( \frac{2}{z+1} \) as power series centered at \( z = 0 \): \[ \frac{-1}{z-3} = -\frac{1}{3} \cdot \frac{1}{1-\frac{z}{3}} = -\frac{1}{3} \sum_{n=0}^{\infty} \left(\frac{z}{3}\right)^n \] and \[ \frac{2}{z+1} = \frac{2}{1-(-z)} = 2 \sum_{n=0}^{\infty} (-z)^n \].

Step 7: Combine Series and Find Radius of Convergence

Combine series: \[ f(z) = -\frac{1}{3} \sum_{n=0}^{\infty} \left(\frac{z}{3}\right)^n + 2 \sum_{n=0}^{\infty} (-z)^n \]. The series \( \frac{1}{1-\frac{z}{3}} \) converges for \(|z/3| < 1\), giving a radius of convergence \(|z| < 3\) and \(2 \sum (-z)^n\) converges for \(|z|<1\), with a radius of 1. The smaller radius of convergence dominates, so \(R = 1\).

Key Concepts

Partial Fractions

Partial fractions are a technique used to break down complex rational expressions into simpler fractions, which can be easier to work with, especially when integrating or differentiating. In the given exercise, we started from the function \( f(z) = \frac{z-7}{z^2 - 2z - 3} \). This expression has a quadratic denominator, and by using partial fraction decomposition, it can be split into two simpler fractions.

• First, factor the denominator \( z^2 - 2z - 3 \) to get \((z - 3)(z + 1)\).
• Express the original function as the sum of two fractions: \( \frac{A}{z-3} + \frac{B}{z+1} \).
• Find the coefficients \(A\) and \(B\) by solving a system of linear equations.

Once you have \(A\) and \(B\), the function becomes easier to analyze. This transformation is especially helpful in finding series expansions, such as the Maclaurin series, where simpler expressions enhance computational efficiency.

Radius of Convergence

The radius of convergence is an essential concept when working with power series, such as a Maclaurin series. It tells us the range of values for which the series will converge to the function it's meant to represent. In the exercise, after obtaining the series for each partial fraction, we determined their respective radii of convergence.

• The term \(-\frac{1}{3}\sum_{n=0}^{\infty} \left(\frac{z}{3}\right)^n\) converges for \( \left| z/3 \right| < 1\), implying the radius here is 3.
• For the series \(2\sum_{n=0}^{\infty} (-z)^n\), convergence occurs for \( \left| z \right| < 1 \), giving a radius of 1.

The overall series will use the smallest radius of these, since it's the more restrictive. Therefore, the series for the given function converges for \(|z| < 1\). Understanding the radius of convergence ensures accurate representation of functions within specified limits and facilitates error management during approximations.

Power Series

Power series play a crucial role in approximating functions. They are infinite polynomial expressions centered around a specific point. The Maclaurin series, a type of power series, is especially popular because it's centered at zero.The exercise illustrates transforming the function \( f(z) = \frac{z-7}{(z-3)(z+1)} \) into a power series using partial fractions.

• By converting \(\frac{-1}{z-3}\) and \(\frac{2}{z+1}\) into series, the function becomes a sum of simpler series.
• For each fraction, we express it as a geometric series: \(-\frac{1}{3} \sum_{n=0}^{\infty} \left(\frac{z}{3}\right)^n\) and \(2 \sum_{n=0}^{\infty} (-z)^n\).

Each term is pivotal in developing an understanding of both the behavior of the function and its approximate values at different points. Power series allow us to handle complex problems by breaking them into manageable pieces while maintaining accuracy for practical calculations.