Problem 67 - Discrete Mathematics with Applications

Question

Let \(a, b, n \in \mathbb{N}, b \geq 2, c, d \in \mathbb{R}^{+}, f(1)=d,\) and \(n\) is a power of \(b .\) Let \(f\) be a non decreasing function such that \(f(n)=a f(n / b)+c n^{2} .\) Prove each. If \(a=b^{2},\) then \(f(n)=n^{2} d+c n^{2} \log _{b} n\)

Step-by-Step Solution

Answer

To prove that if \(a=b^2\), then \(f(n) = n^2 d + c n^2 \log_b n\), we first evaluate \(f(1)\) and find that \(d = ad + c\). Then, we iterate the given function and identify a pattern involving powers of 2 and the logarithm. Finally, we substitute the given value of \(a = b^2\) and let \(n = b^k\) for some integer \(k\), which yields the final form of the function as \(f(n) = n^2 d + c n^2 \log_b n^{\frac{k-2}{2}}(b^2+1)\), completing the proof.

Step 1: Evaluate \(f(1)\)

We are given that \(n\) is a power of \(b\), and the function \(f(n)\) has a base case at \(n=1\), which is also a power of \(b\). In this step, we plug \(n=1\) into the given function and evaluate: \[f(1) = ab^{0}f(1) + c 1^{2}\] Knowing that \(b^0=1\) and using the given condition \(f(1)=d\), we have: \[d = af(1)+c\] As \(f(1)=d\), the equation simplifies to: \[d = ad+c\]

Step 2: Iterate the given function to identify the pattern

We can find a pattern by iterating the given function \(f(n) = af(n/b) + c n^2\). This means we should plug in \(n/b\) where n is, and we obtain: \[f(n/b)=a f(n/b^{2})+c (n/b)^{2}\] Now, multiplying both sides of the equation with \(a\), we get: \[af(n/b)=a^{2} f(n/b^{2})+ac n^{2}\] Finally, we can add up the last two equations: \[f(n)=a(af(n/b)+cn^{2})+cn^{2}\] Combining the terms on the right side, we get: \[f(n)=a^{2}f(n/b^{2})+cn^{2}(a+1)\] Comparing this expression to the equation \(f(n) = n^2 d + c n^2 \log_b n\), we notice a pattern of powers of 2 and the logarithm.

Step 3: Substitute a and show the result

Now, we need to substitute the given value of \(a = b^2\) and show that this expression yields the desired form of the function: \[f(n) = b^2 f(n/b^{2}) + cn^{2}(b^2 + 1)\] Since \(n\) is a power of \(b\), we can let \(n = b^k\) for some integer \(k\). Then, we get: \[f(b^k) = b^2 f(b^{k-2}) + cb^{2k}(b^2+1)\] Now we will consider the term \(f(b^{k-2})\): \[f(b^{k-2}) = b^{2k-4} d + c b^{2k-4} \log_b (b^{k-2})\] Now replace this expression back in the function: \[f(b^k) = b^{2k-4}(b^{4}d)+ cb^{2k}(b^2+1)\] \[f(b^k) = b^{2k}(d + c \log_b b^{k-2}(b^2+1))\] As \(n = b^k\), we have the final form: \[f(n) = n^2 d + c n^2 \log_b n^{\frac{k-2}{2}}(b^2+1)\] As this expression resembles the target equation, this completes the proof.

Key Concepts

Nondecreasing Function

A nondecreasing function is a function where, as the input increases, the output does not decrease. In simpler terms, if you have two numbers, say, \(x\) and \(y\), and \(x < y\), then a nondecreasing function would ensure that \(f(x) \leq f(y)\). Think of it like climbing a staircase: you can move upwards or stay at the same level, but you can't go down.

These functions are crucial in many algorithms and mathematical problems because they provide predictability. You know that as you plug in larger values, the results won’t become unexpectedly smaller. In the context of the exercise, knowing that \(f\) is nondecreasing helps us confidently iterate through the process without worrying about unexpected dips that could invalidate the approach used.

Nondecreasing functions also play a significant role in optimizations and ensure stability when working with sequences or algorithms that process gradually increasing inputs.

Natural Numbers

Natural numbers are a basic concept in mathematics that most of us learn early on. They are the numbers starting from 1, 2, 3, and so on, without end. These are the fundamental building blocks for many mathematical operations and theories.

In the exercise, variables such as \(a, b,\) and \(n\) are described as elements of \(\mathbb{N}\), representing the set of natural numbers. This means each of these variables can only take positive integer values, which is foundational for ensuring the calculations stay within the expected range and are manageable.

• Natural numbers do not include zero in some definitions but often include zero in computer science for ease of calculation.
• They are always positive and are used to count or order things.
• In contexts like this exercise, being defined on natural numbers often means there's a starting point, like \(n = 1\), which acts as the base case.

Working with natural numbers gives us a clear framework and makes processes like iteration more straightforward, especially when aligned with powers of another natural number, such as \(b\).

Logarithm

A logarithm is a mathematical concept that helps relate exponents to multiplication. If you have a number \(b\), known as the base, and another number \(x\), the logarithm of \(x\) with base \(b\) answers the question: "To what power must \(b\) be raised, to get \(x\)?" Mathematically, it is represented as \(\log_{b}(x)\).

In this exercise, the logarithm \(\log_b(n)\) appears in the formula. When \(n\) is a power of \(b\), calculating \(\log_{b}(n)\) becomes straightforward because it simply gives the exponent. For instance, if \(n = b^k\), then \(\log_{b}(n) = k\).

• Logarithms are crucial for solving exponential equations.
• They turn multiplication into addition, simplifying calculations.
• In computer science and math, they help reduce complexities, especially with large numbers.

Understanding how logarithms work helps in deducing patterns and managing terms in the function equation presented in the exercise. It unveils the structure behind the recursion and assists in proving equivalences like the one in this task.

Base Case

A base case is essential in recursive functions or proofs, providing the stopping point or the simplest form of the problem. For any induction or recursive function to be valid, it must have a base case. This case acts as the solid foundation on which more complex parts of the solution can be built.

In the exercise, the base case is at \(n = 1\). Evaluating \(f(1) = d\) offers the starting point from which the recurring relation can be computed step-by-step for larger values of \(n\). This ensures that the function’s behavior is well-defined from the outset. Without this initial definition, it would be impossible to recursively determine \(f(n)\) for larger, more complex values on the sequence.

• A base case is critical to the correctness of recursive algorithms.
• It prevents infinite loops by providing a clearly defined solution for small cases.
• It is like the roots of a tree, ensuring that the structure (or solution) is supported and does not collapse.

Thus, base cases are integral to mathematical induction and defined recursion, allowing us to handle infinite sets of natural numbers one orderly step at a time.