Problem 45 - Discrete Mathematics with Applications

Question

Let \(f_{1}(n)=O(g(n))\) and \(f_{2}(n)=k f_{1}(n),\) where \(k\) is a positive constant. Show that \(f_{2}(n)=O(g(n))\)

Step-by-Step Solution

Answer

Given that \(f_1(n)=O(g(n))\), we have constants \(c_1\) and \(N_1\) such that \(0 \leq f_1(n) \leq c_1g(n)\) for all \(n \geq N_1\). Since \(f_2(n) = kf_1(n)\), we can write \(f_2(n) \leq k(c_1g(n))\) for all \(n \geq N_1\). Thus, by setting \(c_2 = kc_1\) and \(N_2 = N_1\), we have \(f_2(n) \leq c_2g(n)\) for all \(n \geq N_2\), showing that \(f_{2}(n) = O(g(n))\).

Step 1: First, we know that \(f_1(n)=O(g(n))\), which means there exist positive constants \(c_1\) and \(N_1\) such that \(0 \leq f_1(n) \leq c_1g(n)\) for all \(n \geq N_1\). We are also given that \(f_2(n) = kf_1(n)\), where \(k\) is a positive constant. #Step 2: Write an inequality for f_2(n)#

Now, let's express \(f_2(n)\) using the given relationship with \(f_1(n)\) and the inequality defined for \(g(n)\): \(f_2(n) = kf_1(n) \leq k(c_1g(n))\), given that \(n \geq N_1\). #Step 3: Determine the conditions for f_2(n) = O(g(n))#

Step 2: In order to show that \(f_2(n) = O(g(n))\), we need to find a constant \(c_2\) and a constant \(N_2\) such that \(f_2(n) \leq c_2g(n)\) for all \(n \geq N_2\). From our inequality in Step 2, we can substitute \(c_2 = kc_1\) and \(N_2 = N_1\) to have: \(f_2(n) \leq c_2g(n)\) for all \(n \geq N_2\). #Step 4: Conclude the proof#

Thus, we have shown that \(f_2(n) = O(g(n))\) as there exist positive constants \(c_2\) and \(N_2\) such that \(f_{2}(n) \leq c_2 * g(n)\) for all \(n \geq N_2\).

Key Concepts

Big O Notation

Big O Notation is a way to express the upper-bound behavior of a function as its input grows. It evaluates how quickly function outputs rise based on input increase. Usually, it's used to describe the worst-case complexity or performance of algorithms, making it a critical concept in computer science.

In simpler terms, when we say a function \(f(n)\) is \(O(g(n))\), it means no matter what happens, \(f(n)\)'s growth rate won't exceed \(g(n)\)'s growth rate multiplied by any constant. This means:

• There exist constant values for \(c\) and a starting point \(N\)
• For all \(n\) starting from \(N\), \(f(n)\) is bounded above by \(c\) times \(g(n)\)

Thus, Big O Notation gives us a way to standardize the function's growth pattern using simpler standard functions.

Inequality

An inequality helps compare the scales of two functions. In the context of Big O, inequalities allow us to articulate how one function bounds another. We describe a function \(f(n)\) as smaller than or equal to another function, adjusted by a constant factor.

For instance, if we know \(f_1(n) \leq c_1g(n)\) for all sufficiently large \(n\), it shows how function \(f_1\) is growth-controlled by function \(g\) with adjustment factor \(c_1\). Such inequalities help us set conditions under which one function can majorly dominate another one.

When they involve constants and large inputs, inequalities become an efficient way to demonstrate Big O behavior by showing function control over predictable growth boundaries.

Constants

Constants in asymptotic notation, like Big O, play a crucial role. These constants control the bounding effect of the reference function. They help us precisely define how much larger one function can be compared to another.

When we conclude that \(f(n)\) is \(O(g(n))\), we're essentially describing the existence of a positive constant \(c\) that allows as many repetitions of \(g(n)\) as needed, so \(f(n)\) remains underneath. Therefore, the function's growth is predictable within a scaled factor.

• These constants are independent of input size \(n\)
• Show relationship strength in function bounding

This is why constants are vital for clearly showing a function is bounded by another and determining precise limits of big O relationships.