Chapter 6: Problem 26
Use partial fractions as an aid in obtaining the Maclaurin series for the given function. Give the radius of convergence \(R\) of the series. $$ f(z)=\frac{z-7}{z^{2}-2 z-3} $$
Short Answer
Expert verified
Maclaurin series: \( \sum_{n=0}^{\infty} \left( \frac{-2}{3^{n+1}} + (-1)^{n+1} \right) z^n \). Radius of convergence \( R = 1 \).
Step by step solution
01
Partial Fraction Decomposition
First, decompose the given function \( f(z) = \frac{z-7}{z^2 - 2z - 3} \). Start by factoring the denominator: \( z^2 - 2z - 3 = (z-3)(z+1) \). Therefore, we can express \( f(z) \) as follows: \( \frac{z-7}{(z-3)(z+1)} = \frac{A}{z-3} + \frac{B}{z+1} \). To find \( A \) and \( B \), use the cover-up method or equate coefficients after multiplying through by the denominator.
02
Solve for Coefficients A and B
Equate \( z-7 = A(z+1) + B(z-3) \). Expand and collect like terms: \( A(z+1) + B(z-3) = Az + A + Bz - 3B = (A+B)z + (A-3B) \). Equate coefficients with \( z-7 \):\(A+B = 1\) and \(A-3B=-7\). Solve these equations to find \(A = 2\) and \(B = -1\).
03
Express the Function with Coefficients
Now we have: \( \frac{z-7}{(z-3)(z+1)} = \frac{2}{z-3} - \frac{1}{z+1} \). Next, express each term as a power series valid at \(z=0\).
04
Find Maclaurin Series for Each Term
For \( \frac{2}{z-3} = \frac{-2}{3-z}\), use the series \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \) with \( u = \frac{z}{3} \):\( \frac{-2}{3} \sum_{n=0}^{\infty} \left(\frac{z}{3}\right)^n = \sum_{n=0}^{\infty} \frac{-2}{3^{n+1}} z^n \). For \( \frac{-1}{z+1} \), use \( \frac{1}{1+v} = \sum_{n=0}^{\infty} (-1)^n v^n \) with \( v = -z \): \( \sum_{n=0}^{\infty} (-1)^{n+1} z^n \).
05
Combine Series and Find Result
Combine the results: \( \sum_{n=0}^{\infty} \left( \frac{-2}{3^{n+1}} + (-1)^{n+1} \right) z^n \). This is the Maclaurin series representation of \( f(z) \).
06
Determine the Radius of Convergence
The series \( \frac{-2}{3-z} \) converges for \( |z/3|<1 \), i.e., \(|z|<3\). The series \( \frac{-1}{z+1} \) converges for \(|z|<1\). The overall radius of convergence is where both series converge, thus \( R = 1 \).
07
Conclusion
The Maclaurin series for the function is \( \sum_{n=0}^{\infty} \left( \frac{-2}{3^{n+1}} + (-1)^{n+1} \right) z^n \) with the radius of convergence \( R = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Maclaurin Series
The Maclaurin Series is a powerful tool used in complex analysis to express functions as infinite series. This expands a function into terms of powers of the variable around zero. If you have a function, like our example function, you can break it down into simpler components that are easier to work with. A Maclaurin Series takes the form:
In this exercise, we used the decomposition into simpler fractions to separately compute series for each fraction. This allowed us to express the resulting function in a form that is an infinite sum, making it both easier to handle and to analyze mathematically.
- \( f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \)
In this exercise, we used the decomposition into simpler fractions to separately compute series for each fraction. This allowed us to express the resulting function in a form that is an infinite sum, making it both easier to handle and to analyze mathematically.
Partial Fraction Decomposition
Partial Fraction Decomposition is a method used in simplifying complex rational expressions. When dealing with functions like \( \frac{z-7}{z^2 - 2z - 3} \), breaking it down into simpler fractions makes it easier to use in further calculations.
- You start by factoring the denominator.
- This allows you to express the function as a sum of fractions with simpler denominators.
- \( \frac{z-7}{(z-3)(z+1)} = \frac{A}{z-3} + \frac{B}{z+1} \)
Radius of Convergence
The Radius of Convergence is a critical concept when working with series, indicating where a series representation of a function converges. It's essential when examining the validity of a series solution, showing the values for which the series holds true.
- For each component series in the partial fractions, determine its radius.
- The smallest radius ensures overall convergence.