Problem 2 - Complex Analysis

Question

Decide, whether there are analytic functions \(f_{j}: \mathbb{E} \rightarrow \mathbb{C}, 1 \leq j \leq 4\), with (a) \(f_{1}\left(\frac{1}{2 n}\right)=f_{1}\left(\frac{1}{2 n-1}\right)=\frac{1}{n}, \quad n \geq 1\) (b) \(f_{2}\left(\frac{1}{n}\right)=f_{2}\left(-\frac{1}{n}\right)=\frac{1}{n^{2}}, \quad n \geq 1\) (c) \(f_{3}^{(n)}(0)=(n !)^{2}, \quad n \geq 0\) (d) \(f_{4}^{(n)}(0)=\frac{n !}{n^{2}}, \quad n \geq 0\).

Step-by-Step Solution

Answer

Only \( f_4 \) can exist as an analytic function on \( \mathbb{E} \).

Step 1: Analyze f_1 Condition

We need to construct an analytic function \( f_1(z) \) where \( f_1(\frac{1}{2n}) = f_1(\frac{1}{2n-1}) = \frac{1}{n} \) for \( n \geq 1 \). An analytic function in \( \mathbb{E} \) (unit disk) cannot have multiple equal values for multiple inputs unless it is constant due to the identity theorem for analytic functions. Thus, such an \( f_1 \) cannot exist except as a constant, which it cannot be since it would contradict the initial conditions.

Step 2: Analyze f_2 Condition

We need \( f_2(\frac{1}{n}) = f_2(-\frac{1}{n}) = \frac{1}{n^2} \). If \( f_2 \) were entire (analytic across the complex plane), then symmetry with respect to positive and negative inputs would imply it is even, and such values mean \( f_2 \) must be constant everywhere, contradicting the given pairs. Thus no such \( f_2 \) can exist.

Step 3: Analyze f_3 Condition

The function \( f_3 \) with condition \( f_3^{(n)}(0) = (n!)^2 \) implies that its Taylor series at \( z = 0 \) is \( f_3(z) = \sum_{n=0}^{\infty} \frac{(n!)^2}{n!} z^n = \sum_{n=0}^{\infty} n! z^n \). However, \( n! \) grows too rapidly for convergence in any neighborhood of 0 (as factorials have super-exponential growth), hence such an \( f_3 \) cannot be entire.

Step 4: Analyze f_4 Condition

The function \( f_4 \) with condition \( f_4^{(n)}(0) = \frac{n!}{n^2} \) implies that its Taylor series at \( z = 0 \) is \( f_4(z) = \sum_{n=0}^{\infty} \frac{1}{n^2} z^n \). This series converges for \( |z| < 1 \), and hence such an \( f_4 \) does exist as an analytic function within the unit disk \( \mathbb{E} \).

Key Concepts

Identity Theorem

The identity theorem is an essential concept in complex analysis, particularly when dealing with analytic functions. An analytic function is simply a function that is differentiable at every point in its domain. The identity theorem states that if two analytic functions agree on a set that has an accumulation point, they must be identical across their entire domain.
In simpler terms, if you and your friend write down numbers to fill in a long list and you both have the same numbers in a part of the list with many entries, the entire list must be identical if you both were writing from some analytic function.
Let's consider the function in the step-by-step solution denoted as \( f_1 \). The specific condition \( f_1(\frac{1}{2n}) = f_1(\frac{1}{2n-1}) = \frac{1}{n} \) for \( n \geq 1 \), would require \( f_1 \) to have the same output for different inputs repeatedly, across infinitely many points. According to the identity theorem, this situation leads to the conclusion that \( f_1 \) would have to be a constant function, but a constant function cannot satisfy the given conditions where outputs vary. Hence, such a function cannot exist apart from being constant, which contradicts the initial conditions.

Taylor Series

The Taylor series is a powerful tool used for representing functions as infinite sums of their derivatives at a single point. It's like a recipe where you know all ingredients: the function can be "reconstructed" with these derivatives around that point.
Consider a Taylor series expansion in the original exercise. For example, the third condition is elaborated as \( f_3^{(n)}(0) = (n!)^2 \). If this series were expanded, it would look like \( f_3(z) = \sum_{n=0}^{\infty} \frac{(n!)^2}{n!} z^n = \sum_{n=0}^{\infty} n! z^n \).
Here's the catch—factorials grow very quickly—imagine multiplying a series of numbers rapidly getting bigger! For the Taylor series to effectively "build" the function accurately, each term has to be meaningful in context and not shoot to infinity too rapidly. Sadly, a factorial grows too fast for convergence near zero here, and thus, \( f_3 \) cannot represent an entire function through its Taylor series. Instead, there is no circle around 0 where this series can converge.

Entire Functions

When we talk about entire functions, we're referring to functions that are analytic across the entire complex plane. This means they’re smooth and differentiable everywhere in the complex world—quite literally everywhere.
Looking at the exercise, let's zoom onto condition \( f_4 \) with \( f_4^{(n)}(0) = \frac{n!}{n^2} \). Here, we see a Taylor series represented as \( f_4(z) = \sum_{n=0}^{\infty} \frac{1}{n^2} z^n \).
This specific series, unlike factorials, decreases quickly enough (i.e., the term \( \frac{1}{n^2} \) shrinks fast enough) that it will converge whenever \( |z| < 1 \). This means that \( f_4 \) is not just analytic at one point but throughout the entire unit disk \( \mathbb{E} \) without any edge cases falling apart. Thus, \( f_4 \) can indeed exist as an analytic function in this bounded space, but it becomes entire only within \( \mathbb{E} \) and not beyond.