Question

Show that if \(d(n)\) is \(O(f(n))\) and \(e(n)\) is \(O(g(n)),\) then the product \(d(n) e(n)\) is \(O(f(n) g(n))\)

Short answer

The product \(d(n) e(n)\) is \(O(f(n) g(n))\) because \(|d(n) e(n)| \leq K |f(n) g(n)|\) for some constant \(K\) and sufficiently large \(n\).

Step-by-step solution

Understand the Definitions

Before proving the statement, recall the definition of Big-O notation. Big-O notation describes an upper bound on the time complexity or growth of a function. Specifically, a function \(d(n)\) is \(O(f(n))\) if there exist constants \(C > 0\) and \(n_0\) such that for all \(n \geq n_0\), \(|d(n)| \leq C|f(n)|\). Similarly, \(e(n)\) is \(O(g(n))\) if there exist constants \(C' > 0\) and \(n_0'\) such that for all \(n \geq n_0'\), \(|e(n)| \leq C'|g(n)|\).

Express the Given Conditions

From the definitions, we know there exist constants \(C\), \(C'\), \(n_0\), and \(n_0'\) such that: 1. \(|d(n)| \leq C |f(n)|\) for all \(n \geq n_0\)2. \(|e(n)| \leq C' |g(n)|\) for all \(n \geq n_0'\)

Combine the Inequalities

To show the product \(d(n) e(n)\) is \(O(f(n) g(n))\), consider the product of the given inequalities:\(|d(n) e(n)| = |d(n)||e(n)| \leq C |f(n)| C' |g(n)|\)Simplifying the right-hand side, we get:\(|d(n) e(n)| \leq CC' |f(n)||g(n)|\)

Establish a New Constant and Range

Define a new constant \(K = CC'\). Then the inequality becomes:\(|d(n) e(n)| \leq K |f(n) g(n)|\)To bound this for all \(n \geq \max(n_0, n_0')\), since \(n_0\) and \(n_0'\) are the starting points for the original inequalities.

Conclude the Proof

Thus, we have shown that \(|d(n) e(n)| \leq K |f(n) g(n)|\) for all \(n \geq \max(n_0, n_0')\). By definition, this means that \(d(n) e(n)\) is \(O(f(n) g(n))\).

Key concepts

Time Complexity

Time complexity is a way to describe the efficiency of an algorithm in terms of how the runtime increases as the input size grows. It focuses on the number of basic steps that the algorithm takes as a function of the input size, denoted as 'n'. Detecting time complexity allows us to compare different algorithms to determine which one is faster or more scalable.
Examples of common time complexities include:

• O(1) - Constant time
• O(n) - Linear time
• O(log n) - Logarithmic time
• O(n^2) - Quadratic time.

Growth of a Function

Understanding the growth of a function is fundamental in computing time complexities. This concept relates to how a function increases as its input grows. For example, a linear function like f(n) = n grows at a steady rate, while a quadratic function like f(n) = n^2 grows much faster as the input size increases. By visualizing these functions as graphs, we can better understand and compare their growth rates.
When we analyze an algorithm, we usually look at its worst-case scenario to ensure its efficiency under all conditions. This involves focusing on upper bounds of growth.

Inequalities in Algorithms

Inequalities play a crucial role in determining the performance bounds of algorithms. Using inequalities, we can establish relationships between different growth functions. For instance, in our example problem, the goal is to prove an inequality that shows the product of two functions’ growth rates.
Consider two functions, d(n) and e(n). If we know that d(n) is O(f(n)) and e(n) is O(g(n)), we can use the inequalities:

• |d(n)| ≤ C|f(n)| for some constant C
• |e(n)| ≤ C'|g(n)| for some constant C'.

Combining these, we aim to show that |d(n)e(n)| ≤ K|f(n)g(n)| for some constant K, proving the composite growth rate.

Constants in Big O Notation

In Big O notation, constants help set the upper bounds on the growth of functions. They are crucial in inequalities. In our earlier example, constants C and C’ bound d(n) and e(n) concerning f(n) and g(n), respectively. By combining these inequalities, we introduce a new constant, K, which acts as a product of C and C’.
The constant helps simplify comparative analysis without affecting the overall growth behavior, ensuring our Big O analysis remains focused on the dominant factors as n grows large.

Combining Big O Functions

Combining Big O functions is an essential technique for analyzing the time complexity of algorithms comprised of multiple segments. When we combine two Big O functions, we look at how their growth rates interact. For instance, in our problem, we combine two inequalities:

• |d(n)| ≤ C|f(n)|
• |e(n)| ≤ C’|g(n)|

for all n greater than a certain starting point. By combining these, we get a new function representing the bound:

• |d(n)e(n)| ≤ CC’|f(n)g(n)|

This process allows us to understand the composite behavior, thus proving that the product d(n)e(n) is O(f(n)g(n)), combining their upper bounds into a single, more comprehensive expression.