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Magazine Chapter 19: Problem 2
In Problems 1 and 2 , show that \(z=0\) is a removable singularity of the given function. Supply a definition of \(f(0)\) so that \(f\) is analytic at \(z=0\). $$ f(z)=\frac{\sin 4 z-4 z}{z^{2}} $$
Short Answer
Expert verified
The singularity at \( z=0 \) is removable; define \( f(0) = 0 \) to make \( f \) analytical at \( z=0 \).
Step by step solution
01
Understand the given function
We have the function \( f(z) = \frac{\sin 4z - 4z}{z^2} \). This function will not be analytic at \( z = 0 \) due to division by zero. We aim to determine if this singularity is removable.
02
Use power series expansion
To analyze the singularity at \( z = 0 \), we'll expand \( \sin(4z) \) into its power series: \( \sin(4z) = 4z - \frac{(4z)^3}{3!} + \frac{(4z)^5}{5!} - \cdots \). Subtracting \( 4z \) from this series gives us \( \sin(4z) - 4z = -\frac{(4z)^3}{3!} + \frac{(4z)^5}{5!} - \cdots \).
03
Simplify using Taylor series
Substitute the series expansion from Step 2 back into the original function: \( f(z) = \frac{-\frac{(4z)^3}{3!} + \frac{(4z)^5}{5!} - \cdots}{z^2} \). Divide each term in the series by \( z^2 \): \( f(z) = -\frac{64z^2}{3!} + \frac{(4z)^3}{5!z^2} - \cdots = -\frac{64z}{6} + \frac{1024z^3}{120z^2} - \cdots \). Simplifying further, we get \( f(z) = -\frac{64}{6}z + \cdots \).
04
Determine if singularity is removable
Notice that the lowest power term is \( -\frac{64}{6}z \), indicating that terms with \( z^2 \) and higher vanish when \( z = 0 \). Therefore, \( z=0 \) itself doesn't cause a problem in terms of convergence if appropriately defined.
05
Define the function at z=0
To make \( f \) analytic at \( z=0 \), ensure that \( f(0) \) is equal to the limit as \( z \) approaches 0: \( f(0) = \lim_{z \to 0} \frac{\sin 4z - 4z}{z^2} = 0 \), thus making the function smooth and removing the singularity.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Removable Singularities
A singularity in complex analysis is a point where a function is not defined or not differentiable. A removable singularity is a specific type where, despite the initial undefined state at that point, the singularity can be "removed" by redefining the function appropriately. For instance, consider the function given in the problem:
- The function is originally undefined at \( z = 0 \) due to division by zero.
- Through mathematical manipulation such as power series expansion or limits, one can redefine the value at this point.
- This allows the function to become smooth and analytic, meaning differentiable everywhere, including at the singularity's location.
Analytic Functions
An analytic function in complex analysis is a function that is locally given by a convergent power series. This means the function can be expressed as a power series around any point within its domain, where the coefficients are derived from the function's derivatives at that point. Analyticity is a strong form of differentiability that implies:
- The function is differentiable not just once, but infinitely many times.
- The power series expansion matches the function exactly within a region of convergence.
Power Series Expansion
A power series is an infinite series of the form: \[S(z) = a_0 + a_1z + a_2z^2 + a_3z^3 + \cdots\]where each \( a_n \) is a coefficient and \( z \) is the variable. In complex analysis, power series are crucial since they can represent functions around points where they are analytic. The key elements include:
- The range of \( z \) where the series converges is called the radius of convergence.
- For every point within this radius, the series provides an accurate representation of the function.
- In the problem, we expanded \( \sin(4z) \) into a power series, which helped identify the removable singularity at \( z=0 \).
Taylor Series
The Taylor series is a special classification of power series and is used to approximate functions near a particular point using derivatives. It is written as:\[T(z) = f(a) + f'(a)(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \frac{f'''(a)}{3!}(z-a)^3 + \cdots\]where \( a \) is the center of the series, and \( f^n(a) \) are the derivatives of the function evaluated at \( a \). The attributes of a Taylor Series include:
- Convergence around the point \( a \) where the function is analytic.
- Exact representation of the function within its radius of convergence.
- It can also be extended further to become a Maclaurin series when \( a = 0 \).
