Chapter 3: Problem 5
For the following functions \(f\) defined in a neighborhood of the point \(a \in \mathbb{C}\) determine the TAYLOR series at \(a\) and the convergence radius: (a) \(f(z)=\exp (z), \quad a=1\), (b) \(f(z)=\frac{1}{z}\), (c) \(f(z)=\frac{1}{z^{2}-5 z+6}, \quad a=0\) (d) \(f(z)=\frac{1}{(z-1)(z-2)}, \quad a=0\)
Short Answer
Step by step solution
Understanding the Problem
Function (a) - Taylor Series for \( f(z) = \exp(z) \) at \( a = 1 \)
Function (b) - Taylor Series for \( f(z) = \frac{1}{z} \) at \( a = 0 \)
Function (c) - Taylor Series for \( f(z) = \frac{1}{z^2 - 5z + 6} \) at \( a = 0 \)
Function (d) - Taylor Series for \( f(z) = \frac{1}{(z-1)(z-2)} \) at \( a = 0 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radius of Convergence
Imagine plotting each term of a power series as a circle centered on a point \( a \) (our expansion point). Every "circle" corresponds to an area where the series converges. The radius of this circle is our radius of convergence.
- For entire functions like the exponential function \( \exp(z) \), the series converges everywhere, giving an infinite radius of convergence, \( R = \infty \).
- For rational functions or functions with poles, the radius of convergence is determined by the nearest pole to the point \( a \).
- This is because a power series cannot converge at the poles of the function it represents.
Complex Functions
In the given exercises, several types of functions are considered:
- Polynomials and rational functions reveal their behavior through their zeros and poles.
- The exponential function, for example, is an entire function, meaning it's differentiable everywhere in the complex plane.
- Rational functions like \( \frac{1}{z} \) or \( \frac{1}{z-1} \) exhibit singular points (poles), where the function takes on infinite values.
Partial Fraction Decomposition
For example, consider the function \( f(z) = \frac{1}{z^2 - 5z + 6} \). To express it in a more "workable" form, you first factor the denominator: \( (z-2)(z-3) \). Through partial fraction decomposition, this then splits into simpler fractions, \( \frac{A}{z-2} + \frac{B}{z-3} \), which can each be expanded separately.
- This method simplifies complex expressions into forms where a power series expansion can be easily applied.
- Each term can be individually expanded in their respective regions of convergence.