Problem 43 - A First Course in Complex Analysis with Applications

Question

The function \(f(z)=\sec z\) is analytic at \(z=0\) and hence possesses a Maclaurin series representation. We could, of course, use (7), but there are several alternative ways of obtain the coefficients of the series $$ \sec z=a_{0}+a_{1} z+a_{2} z^{2}+a_{3} z^{3}+\cdots $$ One way is to equate coefficients on both sides of the identity \(1=(\sec z) \cos z\) or $$ 1=\left(a_{0}+a_{1} z+a_{2} z^{2}+a_{3} z^{3}+\cdots\right)\left(1-\frac{z^{2}}{2 !}+\frac{z^{4}}{4 !}-\frac{z^{6}}{6 !}+\cdots\right) $$ Find the first three nonzero terms of the Maclaurin series of \(f .\) What is the radius of convergence \(R\) of the series?

Step-by-Step Solution

Answer

The first three nonzero terms are: \(1 + \frac{z^2}{2} - \frac{5z^4}{24}\). The radius of convergence \(R\) is \(\frac{\pi}{2}\).

Step 1: Maclaurin Series of \(\cos z\)

First, we recall the Maclaurin series expansion for the cosine function:\[\cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots\]We will use this series to expand \(1 = (\sec z)(\cos z)\).

Step 2: Maclaurin Series of \(\sec z\) as Power Series

Assume the series expansion for \(\sec z\) is given by:\[\sec z = a_0 + a_1 z + a_2 z^2 + a_3 z^3 + \cdots\]This series is multiplied by the cosine series we established in Step 1.

Step 3: Equate Series to Obtain Constant Term

For the constant term in the identity \[1 = (a_0 + a_1 z + a_2 z^2 + \cdots)(1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots)\]We equate the coefficients:\[a_0 = 1\]

Step 4: Equate Coefficients of \(z^2\)

Focus on the \(z^2\) term in the expansion:\[a_0 \left(-\frac{z^2}{2!}\right) + a_2 \cdot 1 = 0\]Substitute \(a_0 = 1\) to solve for \(a_2\):\[-\frac{1}{2} + a_2 = 0\]\[a_2 = \frac{1}{2}\]

Step 5: Equate Coefficients of \(z^4\)

Focus on the \(z^4\) term in the expansion:\[a_0 \left(\frac{z^4}{4!}\right) + a_2 \left(-\frac{z^2}{2!}\right) + a_4 \cdot 1 = 0\]Substitute \(a_0 = 1\) and \(a_2 = \frac{1}{2}\) to solve for \(a_4\):\[\frac{1}{4!} - \frac{1}{2} \cdot \frac{1}{2} + a_4 = 0\]\[a_4 = \frac{1}{24} - \frac{1}{4} = \frac{1}{24} - \frac{6}{24} = -\frac{5}{24}\]

Step 6: Determining the Radius of Convergence

The radius of convergence \(R\) for the Maclaurin series expansion of \(\sec z\) can be determined by examining the nearest singularity of \(\sec z\), which occurs where \(\cos z = 0\). This first happens at \(z = \frac{\pi}{2}\), so:\[R = \frac{\pi}{2}\]

Key Concepts

Analytic Functions

An analytic function is a type of function that is smooth and well-behaved in the complex plane. These functions can be represented by a power series around any point within a disk of convergence. For example, the secant function, \( f(z) = \sec z \),is analytic around \( z = 0 \). This means it can be expressed by a Maclaurin series, which is a special form of Taylor series centered at zero. When a function is analytic, it implies that the function behaves nicely without any abrupt changes, making it easier to analyze and work with mathematically.
Presence of such a series also allows mathematicians to perform various operations like differentiation or integration term by term.

Radius of Convergence

The radius of convergence, denoted here as \( R \), is a measure of the interval in the complex plane within which a power series converges absolutely. In simpler terms, it shows how far from the center (in this case, \( z = 0 \)) we can move and still have the series converging to the function.
Knowing the radius helps us understand where our series truncation remains a good approximation of the actual function.

• For the \( \sec z \) series, the nearest singularity occurs where the cosine function equates to zero.
• This is because \( \sec z \) becomes undefined at those points.
• The first occurrence is at \( z = \frac{\pi}{2} \).

The corresponding radius of convergence for this series is thus \( R = \frac{\pi}{2} \).

Series Expansion

A series expansion, like the Maclaurin series, is an infinite sum of terms calculated from the values of a function's derivatives at a single point.
It's a powerful tool that transforms complex operations into simpler algebraic ones. For the function \( f(z)=\sec z \),
we express it as:\[ \sec z = a_{0} + a_{1} z + a_{2} z^{2} + a_{3} z^{3} + \cdots \]

• Each coefficient \( a_n \) contains valuable information about the function's behavior.
• In the example, we found \( a_0 = 1 \), \( a_2 = \frac{1}{2} \), and\( a_4 = -\frac{5}{24} \).

This method of expansion is essential for approximating functions and solving differential equations, among other tasks.

Complex Functions

Complex functions involve variables that have a real and an imaginary part, typically written as \( z = x + yi \), where \( x \) and \( y \) are real numbers, and \( i \) is the imaginary unit. Functions like \( \sec z \) take these complex inputs and output complex numbers as well.

• These functions require special methods, such as series expansion, to better understand their behavior.
• Complex analysis reveals much about physical phenomena, fluid dynamics, and electrical engineering.
• With \( \sec z \), understanding its Maclaurin series provides insights into its behavior near the origin of the complex plane.

This means that exploration of \( \sec z \) in the context of complex variables takes advantage of the rich structure provided by complex numbers, allowing for greater flexibility and depth in analysis.