Problem 6 - Arbeitsbuch Mathematik für Ingenieure

Question

Gibt es eine auf \(\mathbb{C}\) reguläre Funktion \(f\) mit den Funktionswerten \(f\left(\frac{1}{n}\right)=\) a) \(1,0,1,0,1,0, \ldots\) b) \(\frac{1}{2}, 0, \frac{1}{4}, 0, \frac{1}{8}, 0, \ldots\) c) \(\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8} \ldots\) für \(n=1,2,3,4,5,6, \ldots\) ?

Step-by-Step Solution

Answer

A and B are not possible; C is possible.

Step 1 - Understand the Function Values

Examine the given function values for each part. We need a function defined on \(\frac{1}{n}\) for \(n = 1, 2, 3, \ldots\) that satisfies the given sequences.

Step 2 - Check Regularity in \(\mathbb{C}\)

A function is regular (holomorphic) in \(\mathbb{C}\) if it is differentiable everywhere in \(\mathbb{C}\). This implies continuity and differentiability are required, even at the accumulation point \(0\).

Step 3a - Analyze Sequence for Part (a)

The sequence given is \(1, 0, 1, 0, 1, 0, \ldots\). Consider \(f(1) = 1\), \(f(1/2) = 0\), \(f(1/3) = 1\), etc. The values alternate between 1 and 0. If \(f\) were continuous at \(0\), then by letting \ approach 0, it would imply \(f(1/n)\rightarrow f(0)\), which is a contradiction because the sequence does not converge as \(n\rightarrow \infty\). Thus, such a function does not exist.

Step 3b - Analyze Sequence for Part (b)

The sequence given is \(\frac{1}{2}, 0, \frac{1}{4}, 0, \frac{1}{8}, 0, \ldots\). Consider \(f(1) = \frac{1}{2}\), \(f(1/2) = 0\), \(f(1/3) = \frac{1}{4}\), etc. Here, similar to part (a), the values do not converge as \(n\rightarrow \infty\) since the subsequence of function values involving fractional powers decreases towards 0 and the others are exactly 0. Therefore, no regular function on \(\mathbb{C}\) could match this sequence.

Step 3c - Analyze Sequence for Part (c)

The sequence given is \(\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots\). Here, \(f(1) = \frac{2}{3}\), \(f(1/2) = \frac{3}{4}\), \(f(1/3) = \frac{4}{5}\), etc. As \(n\rightarrow \infty\), these values converge to 1. If there were such a function and it was regular, it should converge to \((\frac{n+1}{n})\rightarrow 1\) implying \(f(\frac{1}{n}) \rightarrow 1 \rightarrow f(0)\). This is consistent with the requirement since the function could be manipulated to fit the criteria.

Key Concepts

complex analysis

Complex analysis is the study of functions that operate on complex numbers. These functions, known as holomorphic or analytic functions, are differentiable at every point in their domain. In simpler terms, a function is holomorphic if you can compute its derivative at any complex number in its domain and it will still be a complex number. An essential aspect of complex analysis is the implication that holomorphic functions are infinitely differentiable, and they obey certain convergence properties that make them incredibly useful in both theoretical and applied mathematics. To consider if a function is holomorphic, we typically look into its continuity and differentiability properties.

function continuity

Function continuity is a core concept in mathematics where a function is continuous if, roughly speaking, you can draw its graph without lifting your pen. For a function in complex analysis to be continuous at a point, it means the function's value changes smoothly around that point. In mathematical terms, if a function \( f \) is continuous at a point \( z_0 \), then for any given small number \( \epsilon > 0 \), there exists a small number \( \delta > 0 \) such that whenever \( |z - z_0| < \delta \), it follows that \( |f(z) - f(z_0)| < \epsilon \). This concept is crucial in determining if holomorphic functions are possible under given conditions. For example, in parts (a) and (b) of the exercise, the function values lack the required smoothness near the accumulation point 0.

sequence convergence

Sequence convergence, especially in the context of complex numbers, is about whether the values of a sequence approach a particular point as the sequence progresses. Mathematically, a sequence \( \{a_n\} \) converges to a point \( L \) if for every \( \epsilon > 0 \), there is a positive integer \( N \) such that for all \( n > N \), \( |a_n - L| < \epsilon \). This concept helps analyze the behavior of function values as \( n \) grows. In the exercise parts (a) and (b), the sequences do not converge to a single value, which violates the requirements for having a holomorphic function matching those function values. Meanwhile, in part (c), the sequence converges to 1, making it possible for a holomorphic function to exist that fits this sequence.

accumulation point

An accumulation point (or limit point) of a sequence is a point where elements of the sequence get arbitrarily close to as the sequence progresses. In the complex plane, if a sequence \( \{z_n\} \) accumulates at a point \( L \), then for any neighborhood around \( L \), there are infinitely many points of the sequence within that neighborhood. For example, the sequence \( \frac{1}{n} \) has 0 as an accumulation point because as \( n \) approaches infinity, \( \frac{1}{n} \) gets closer to 0. This concept is vital to understand why a holomorphic function can't fit the alternating sequence in part (a) of the exercise without being discontinuous, which contradicts the requirement for holomorphic functions to be continuous.