Problem 45 - A First Course in Complex Analysis with Applications

Question

Suppose that a complex function \(f\) is analytic in a domain \(D\) that contains \(z_{0}=0\) and \(f\) satisfies \(f^{\prime}(z)=4 z+f^{2}(z)\). Suppose further that \(f(0)=1\). (a) Compute \(f^{\prime}(0), f^{\prime \prime}(0), f^{\prime \prime \prime}(0), f^{(4)}(0)\), and \(f^{(5)}(0)\). (b) Find the first six terms of the Maclaurin expansion of \(f\).

Step-by-Step Solution

Answer

The derivatives are \( f'(0)=1 \), \( f''(0)=6 \), \( f'''(0)=14 \), \( f^{(4)}(0)=64 \), \( f^{(5)}(0)=328 \); the series is \( 1 + z + 3z^2 + \frac{7}{3}z^3 + \frac{8}{3}z^4 + \frac{41}{30}z^5 \).

Step 1: Understand the Problem

We need to compute derivatives of the function \( f \) at \( z_0 = 0 \) and find the first six terms of the Maclaurin expansion. Since the derivative of \( f \) is given by \( f'(z) = 4z + f^2(z) \), we'll use this to find the derivatives recursively.

Step 2: Calculate \( f'(0) \)

Using the derivative formula \( f'(z) = 4z + f^2(z) \) and knowing that \( f(0) = 1 \), substitute \( z = 0 \):\[f'(0) = 4(0) + (1)^2 = 1\]So, \( f'(0) = 1 \).

Step 3: Calculate \( f''(0) \)

Differentiate the given equation \( f'(z) = 4z + f^2(z) \) with respect to \( z \):\[f''(z) = 4 + 2f(z)f'(z)\]Substitute \( f(0) = 1 \) and \( f'(0) = 1 \):\[f''(0) = 4 + 2(1)(1) = 6\]Thus, \( f''(0) = 6 \).

Step 4: Calculate \( f'''(0) \)

Differentiate \( f''(z) = 4 + 2f(z)f'(z) \) with respect to \( z \):\[f'''(z) = 2f'(z)^2 + 2f(z)f''(z)\]Substitute \( f(0) = 1 \), \( f'(0) = 1 \), and \( f''(0) = 6 \):\[f'''(0) = 2(1)^2 + 2(1)(6) = 2 + 12 = 14\]Thus, \( f'''(0) = 14 \).

Step 5: Calculate \( f^{(4)}(0) \)

Further differentiate \( f'''(z) = 2f'(z)^2 + 2f(z)f''(z) \) to find:\[f^{(4)}(z) = 6f'(z)f''(z) + 2f(z)f'''(z)\]Substitute \( f'(0) = 1 \), \( f''(0) = 6 \), and \( f'''(0) = 14 \):\[f^{(4)}(0) = 6(1)(6) + 2(1)(14) = 36 + 28 = 64\]Thus, \( f^{(4)}(0) = 64 \).

Step 6: Calculate \( f^{(5)}(0) \)

Differentiate \( f^{(4)}(z) = 6f'(z)f''(z) + 2f(z)f'''(z) \):\[f^{(5)}(z) = 6(f''(z)^2 + f'(z)f'''(z)) + 2f'(z)f'''(z)\]Substitute \( f'(0) = 1 \), \( f''(0) = 6 \), and \( f'''(0) = 14 \):\[f^{(5)}(0) = 6(6^2 + 1 \cdot 14) + 2 \cdot 1 \cdot 14 = 6(36 + 14) + 28 = 300 + 28 = 328\]Thus, \( f^{(5)}(0) = 328 \).

Step 7: Construct the Maclaurin Series

The Maclaurin series for \( f(z) \) is given by:\[f(z) = f(0) + \frac{f'(0)}{1!}z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \frac{f^{(4)}(0)}{4!}z^4 + \frac{f^{(5)}(0)}{5!}z^5 + \cdots\]Substitute the computed derivatives:\[f(z) = 1 + z + 3z^2 + \frac{7}{3}z^3 + \frac{8}{3}z^4 + \frac{41}{30}z^5 + \cdots\]This gives the first six terms of the Maclaurin expansion.

Key Concepts

Analytic Function

In the realm of complex analysis, an analytic function is one that is indefinitely differentiable over a certain domain. This means it has derivatives of all orders, and these derivatives also behave well in the domain. Analytic functions are crucial because they are expressed as power series within their radius of convergence. This imbues them with incredibly useful properties.

For a function to be analytic at a point, the limiting process of defining the derivative must work at that point and in some neighborhood around it. This local power series expansion is an expression of the function in terms of its values and derivatives at a particular point, such as zero. Complex functions, like the one in our exercise, often have much richer behaviors and traits than their real counterparts, due to the additional dimension provided by complex numbers.

• They satisfy the Cauchy-Riemann equations, a set of conditions for a function to be analytic in the field of complex numbers.
• An important property of analytic functions is their local conformance, meaning they preserve angles locally.

Understanding the role of analytic functions helps simplify complex expressions, as you can compute derivatives recursively as seen in the exercise step-by-step solution.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series, which represents a function as an infinite sum of terms calculated from the derivatives of a function evaluated at a single point, here at zero. It's incredibly useful for approximating functions and understanding their behavior near that point.

In our exercise, the Maclaurin series allows us to express the analytic function in terms of its derivatives at 0. The series is structured as follows:

\[ f(z) = f(0) + \frac{f'(0)}{1!}z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \frac{f^{(4)}(0)}{4!}z^4 + \frac{f^{(5)}(0)}{5!}z^5 + \cdots\]

• The Maclaurin series helps in simplifying complex functions into polynomials, which are often easier to work with.
• It also facilitates solving differential equations, as it allows functions to be approximated by finite sums instead of dealing with continuous expressions.

In the problem, the first six terms of the Maclaurin expansion were found using recursive derivative calculations, showcasing its practical application in complex analysis.

Derivative Calculations

Derivative calculations for complex functions follow many of the same rules as in calculus with real numbers. However, they require careful consideration of complex components. In our exercise, the function's derivative is known: \( f'(z) = 4z + f^2(z) \). By substituting known values and differentiating recursively, various derivatives are determined.

Understanding the process of taking derivatives in this context involves:

• Differentiating iteratively, building off previous calculations.
• Applying chain and power rules of differentiation adjusted for complex numbers.
• Using initial conditions, like \( f(0) = 1 \) here, to simplify and solve for constants.

Each derivative provides more insight into the behavior of the function around a point, allowing us to express the function efficiently as a series. This task showcases that even with complex numbers, derivative procedures strengthen the understanding of function behavior, enabling the construction of useful approximations like the Maclaurin series.

By practicing recursive differentiation in complex settings, students can gain competence in this essential method.