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Chapter 1: Problem 198

Let \(\mathbb{Z} / n \mathbb{Z}\) be the set \(\\{0,1, \ldots, n-1\\}\) with addition modulo \(n\). Consider subsets \(S_n\) of \(\mathbb{Z} / n \mathbb{Z}\) such that \(\left(S_n+k\right) \cap S_n\) is nonempty for every \(k\) in \(\mathbb{Z} / n \mathbb{Z}\). Let \(f(n)\) denote the minimal number of elements in such a subset. Find $$ \lim _{n \rightarrow \infty} \frac{\ln f(n)}{\ln n} $$ or show that this limit does not exist.

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Understanding the Problem

Given a set \(\backslashmathbb\backslash{Z\backslash}/ n \backslashmathbb\backslash{Z\backslash}\) where \(\backslashmathbb\backslash{Z\backslash}/ n \backslashmathbb\backslash{Z\backslash} = \{0,1, \backslashldots, n-1\}\), subsets \(S_n\) need to be identified such that \((S_n + k) \backslash cap S_n\) is nonempty for every \(k \in \backslashmathbb\backslash{Z\backslash}/ n \backslashmathbb\backslash{Z\backslash}\), and find the limit of \(\backslashln f(n)\) divided by \(\backslashln n\) as \(\backslash n \backslash to \backslash infty\), where \(\backslashf(n)\) denotes the minimal number of elements in such a subset.
02

Analyzing Conditions for \(S_n\)

To meet the condition that \(S_n + k\) intersects with \(S_n\) for every \(k\), \(S_n\) must have an element for each possible residue class modulo \(n\). This means covering every element in \( \mathbb{Z} / n \mathbb{Z}\).
03

Determining Lower Bound for \(S_n\)

Since there are \(n\) possible residue classes, the minimal \(S_n\) must have at least \( \backslashlceil \backslashfrac{n}{2} \backslashrceil\) elements. This ensures that \(S_n + k\) will overlap with \(S_n\) regardless of \(k\).
04

Calculating \(f(n)\)

Given \(S_n\) must have at least \( \backslashlceil \backslashfrac{n}{2} \backslashrceil\) elements, \(f(n) = \backslashlceil \backslashfrac{n}{2} \backslashrceil\). As \( \backslashn\) approaches infinity, \(f(n) \backslashapprox \backslashfrac{n}{2}\).
05

Finding the Limit

Consider \( \backslashlim_ {n \rightarrow \backslashinfty} \backslashfrac { \backslashln f(n) } { \backslashln n } \). Since \( f(n) \backslashapprox \backslashfrac{n}{2} \), we have \(\ln f(n) \backslashapprox \ln( \backslashfrac{n}{2}),\). This simplifies to \ln n - \backslashln 2.\
06

Final Calculation

The limit becomes \( \backslashlim_{n \rightarrow \backslashinfty} \backslashfrac{ \backslashln n - \backslashln 2 } { \backslashln n } = \backslashlim_{n \backslashrightarrow \backslashinfty} (1 - \backslashfrac { \backslax\backln 2} { \backslashln n }) \). As \( \backslashn \rightarrow \backslashinfty \, \backslashfrac { \backslashln 2 } { \backslashln n } \rightarrow 0 \), yielding the limit as 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Subset Intersection
Subset intersection is a concept where we look at common elements present in two or more sets. In modular arithmetic, we must consider the intersection of a subset with shifted versions of itself. For the problem at hand, this means finding a subset \( S_n \subseteq \mathbb{Z} / n \mathbb{Z} \) such that \( (S_n + k) \cap S_n eq \emptyset \) for every \( k \in \mathbb{Z} / n \mathbb{Z} \).
This demands each shift by \( k \) of the subset needs to intersect with the original subset at least once. This can be understood better by visualizing a set and its translations.
Modular Addition
Modular addition is the operation of adding numbers within a modular system. For example, in \( \mathbb{Z} / n \mathbb{Z} \), adding two elements \( a \) and \( b \) results in \( (a + b) \mod n \).
To solve our problem, we particularly deal with subset translations using modular addition. If \( S_n \) is a subset of \( \mathbb{Z}/n \mathbb{Z} \), then \( S_n + k \) refers to the set obtained by adding \( k \) to each element of \( S_n \) and then taking the result modulo \(n\).
Ensuring that \( (S_n + k) \cap S_neq \emptyset \) for every \(k\) means through modular addition, each shifted set intersects the original one.
Asymptotic Analysis
Asymptotic analysis helps us understand the behavior of functions as the input grows towards infinity. Specifically, we look at the rate of growth or shrinkage.
In our problem, we're interested in the limit of the ratio \( \frac{\ln f(n)}{\ln n} \) as \( n \) approaches infinity. Through asymptotic analysis, we determine that since \( f(n) \approx \frac{n}{2} \), then \( \ln f(n) \approx \ln \frac{n}{2} = \ln n - \ln 2 \).
Thus, evaluating \( \lim_{n \to \infty} \frac{\ln n - \ln 2}{\ln n} = \lim_{n \to \infty} (1 - \frac{\ln 2}{\ln n}) = 1 \).
Limit Calculation
Calculating limits involves finding the value a function approaches as the input grows larger or smaller. In our context, we deal with the limit \( \lim_{n \to \infty} \frac{\ln f(n)}{\ln n} \).
First, we understand that \( f(n) = \lceil \frac{n}{2} \rceil \approx \frac{n}{2} \) and thus \( \ln f(n) \approx \ln \frac{n}{2} \).
This reduces to \( \lim_{n \to \infty} \frac{\ln (\frac{n}{2})}{\ln n} \) equals \( \lim_{n \to \infty} \frac{\ln n - \ln 2}{\ln n} \).
Solving gives us \( \lim_{n \to \infty} (1 - \frac{\ln 2}{\ln n}) = 1 \) since \( \frac{\ln 2}{\ln n} \) tends towards 0 as \( n \to \infty \).
Therefore, the limit is 1.

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