Problem 45 - Discrete Mathematics with Applications

Question

Let \(a, b, k \in \mathbb{N}, b \geq 2,\) and \(n=b^{k} .\) Consider the function \(f\) defined by \(f(n)=a f(n / b)+g(n) .\) Show that \(f(n)=a^{k} f(1)+\sum_{i=0}^{k-1} a^{i} g\left(n / b^{i}\right)\)

Step-by-Step Solution

Answer

To prove that the given function \(f(n)\) satisfies the given equation, we followed these steps: 1. Wrote down the recursive function for different values of \(k\) to observe the pattern. 2. Noticed the pattern for \(f(n)\) as \(f(n) = a^k f\left(\frac{n}{b^{k-1}}\right) + \sum_{i=0}^{k-1} a^i g\left(\frac{n}{b^i}\right)\) for different values of \(k\). 3. Proved the formula using induction, showing it holds true for the base case (\(k=1\)) and for \(k+1\) when assuming it holds for \(n = b^k\). Thus, by induction, the formula holds for all positive integers \(k\).

Step 1: Writing down the recursive function for different values of k

Let's consider a few cases for different values of \(k\) to get insight into the problem. For \(k=1\), \(n=b^1\) \[ f(n) = af\left(\frac{n}{b}\right) + g(n) \] For \(k=2\), \(n=b^2\) \[ f(n) = af\left(\frac{n}{b}\right) + g(n) = af\left(\frac{b^2}{b}\right) + g(b^2) \] Substitute \(f\left(\frac{b^2}{b}\right)\) with the \(k=1\) expression above \[ f(n) = a\left[af\left(\frac{b}{b}\right) + g(b)\right] + g(b^2) \] For \(k=3\), \(n=b^3\) \[ f(n) = af\left(\frac{n}{b}\right) + g(n) = af\left(\frac{b^3}{b}\right) + g(b^3) \] Substitute \(f\left(\frac{b^3}{b}\right)\) with the \(k=2\) expression above \[ f(n) = a\left[a\left[af\left(\frac{b^2}{b}\right) + g(b^2)\right] + g(b)\right] + g(b^3) \]

Step 2: Observing the pattern

We can observe the following pattern emerge in the recursive function: For \(k=1\), \[ f(n) = a^1f\left(\frac{n}{b^0}\right) + a^0g\left(\frac{n}{b^1}\right) \] For \(k=2\), \[ f(n) = a^2f\left(\frac{n}{b^1}\right) + a^1g\left(\frac{n}{b^1}\right) + a^0g\left(\frac{n}{b^2}\right) \] For \(k=3\), \[ f(n) = a^3f\left(\frac{n}{b^2}\right) + a^2g\left(\frac{n}{b^1}\right) + a^1g\left(\frac{n}{b^2}\right) + a^0g\left(\frac{n}{b^3}\right) \] In each case, we can rewrite the expressions as: \[ f(n) = a^k f\left(\frac{n}{b^{k-1}}\right) + \sum_{i=0}^{k-1} a^i g\left(\frac{n}{b^i}\right) \]

Step 3: Proving the formula using induction

We will now prove the formula using induction. Base case, \(k=1\): \[ f(n) = a^1f\left(\frac{n}{b^0}\right) + a^0g\left(\frac{n}{b^1}\right) \] The base case holds true. Inductive hypothesis, assume the formula holds for \(n = b^k\): \[ f(n) = a^k f\left(\frac{n}{b^{k-1}}\right) + \sum_{i=0}^{k-1} a^i g\left(\frac{n}{b^i}\right) \] We want to prove it holds for \(n = b^{k+1}\), \[ f(n) = af\left(\frac{n}{b}\right) + g(n) \] Substitute the inductive hypothesis: \[ f(n) = a\left[a^k f\left(\frac{n}{b^k}\right) + \sum_{i=0}^{k-1} a^i g\left(\frac{n}{b^{i+1}}\right) \right] + g(n) \] Simplify the expression: \[ f(n) = a^{k+1} f\left(\frac{n}{b^{k}}\right) + \sum_{i=0}^{k-1} a^{i+1} g\left(\frac{n}{b^{i+1}}\right) + g(n) \] We can rewrite this as: \[ f(n) = a^{k+1} f\left(\frac{n}{b^{k}}\right) + \sum_{i=0}^{k} a^{i} g\left(\frac{n}{b^{i+1}}\right) \] Thus, by induction, the formula holds for all positive integers \(k\). The exercise is proved to be true.

Key Concepts

Recursive Functions

Recursive functions are fundamental in the world of mathematics and computer science. They are essentially functions that refer to themselves during their execution. To understand recursive functions, consider a simple example: the factorial of a number, denoted as \(n!\), can be defined recursively by stating that \(1! = 1\) and for all integers \(n > 1\), \(n! = n \times (n-1)!\). The function calls itself with a simpler version of the original problem, which eventually reaches the base case, \(1!\), ensuring the recursion ends.

In the context of the exercise, a recursive function is given by \(f(n) = a f(\frac{n}{b}) + g(n)\), where the function \(f\) calls itself with a smaller argument \(\frac{n}{b}\) and adds a term \(g(n)\). Recognizing and analyzing these recursive calls are crucial since they reveal patterns that enable us to solve or prove properties about the function, as seen in the step-by-step solution of the textbook exercise.

Recursive functions are not only theoretical constructs; they are also widely used in programming to solve problems by breaking them down into subproblems of the same type.

Proof by Induction

Proof by induction is a powerful mathematical technique used to prove statements about integers. It works on the principle of domino effect; if the first domino falls (the base case) and knocking down one domino causes the next one to fall (the inductive step), then all the dominos will fall. In mathematical terms, we start by proving that a statement is true for a basic case—usually when \(n = 1\). This serves as our first domino. Then, we assume the statement is true for an arbitrary number \(n = k\) and prove that the statement must then be true for the next number \(n = k + 1\).

In our exercise, the base case is verified at \(k = 1\), and the inductive step involves demonstrating that if the function's equation holds true for a certain \(k\), it must also hold for \(k + 1\), thereby confirming the formula for all positive integers. This methodical approach helps in simplifying complex problems into comprehensible pieces and is a testament to the elegance of mathematical reasoning. Additionally, it is essential for students to grasp induction as it is not only used in discrete mathematics but also an array of mathematical disciplines.

Discrete Mathematics

Discrete mathematics is a branch of mathematics dealing with discrete elements that use algebra and arithmetic. It is the cornerstone of computer science and involves topics such as logic, set theory, combinatorics, graph theory, and algorithms. Discrete mathematics does not involve continuous variables; instead, it is concerned with discrete entities.

The textbook exercise presented incorporates concepts from discrete mathematics, specifically recursive functions, which are sequences of values defined by repetitive processes. Each term is derived from the preceding terms through a fixed repetitive rule, and discrete mathematics provides the tools needed to analyze these sequences.

Understanding discrete mathematics is crucial for tackling problems in algorithm analysis and design, decision-making, software engineering, and cryptography, among others. Its principles are foundational for comprehending not only the theoretical aspects of computing but also the practical concerns involved in writing algorithms and developing computational models. Therefore, learning discrete mathematics forms part of the fundamental literacy in the digital age for students not just in mathematics, but also in computer science and related fields.