Question

Show that \(\lceil f(n)\rceil\) is \(O(f(n)),\) if \(f(n)\) is a positive nondecreasing function that is always greater than 1

Short answer

Since \( \lceil f(n) \rceil \leq 2f(n)\), \( \lceil f(n) \rceil \) is \( O(f(n))\).

Step-by-step solution

- Understanding the Ceiling Function

The ceiling function, denoted as \(\lceil x \rceil\), maps any real number to the smallest following integer. For instance, \( \lceil 3.2 \rceil = 4\).

- Expressing \( \lceil f(n) \rceil\)

Because \( f(n) \) is a positive nondecreasing function greater than 1, we know \( f(n) \geq 1\). Therefore, \( \lceil f(n) \rceil \leq f(n) + 1 \).

- Relating \( \lceil f(n) \rceil \) to \( f(n) \)

Since \( \lceil f(n) \rceil \leq f(n) + 1\), and \( f(n) \) is a positive function that grows without decreasing, we can establish an inequality: \( \lceil f(n) \rceil \leq 2f(n)\).

- Establishing Big-O Notation

The Big-O notation is used to describe an upper bound for a function to within a constant factor. Here, \( \lceil f(n) \rceil\) is arguably bounded above by \( 2f(n)\), which demonstrates that it grows linearly with \( f(n)\).

- Concluding the Proof

To conclude, since \( \lceil f(n) \rceil\) is at most \( 2f(n)\), and the Big-O notation disregards constant coefficients, it implies that \( \lceil f(n) \rceil \) is \( O(f(n))\).

Key concepts

Ceiling Function

The ceiling function, represented by \(\lceil x \rceil\), is a mathematical function that rounds any given number up to the nearest integer, regardless of whether the number is already an integer. For instance, \(\lceil 3.2 \rceil = 4\). This function is essential when dealing with real numbers and ensuring that results are within the integer domain.
If we apply the ceiling function to a positive and nondecreasing function \(f(n)\), which is greater than 1, such as \(f(n) ≈ 2.3\), the ceiling will round it up to \(3\). This property helps us understand how \(\lceil f(n) \rceil\) relates to \(f(n)\) itself.

Nondecreasing Function

A nondecreasing function is one that never decreases as its input increases. More formally, a function \(f(n)\) is nondecreasing if, for every \(n\) and \(m\) where \(n < m\), the inequality \(f(n) \leq f(m)\) holds true. This implies that the value of \(f(n)\) will either stay the same or increase as \(n\) increases.
An example might be a function representing the total cost of items in a cart as more items are added. Here, understanding \(f(n)\) as nondecreasing allows us to confidently discuss the properties of \(\lceil f(n) \rceil\), knowing \(f(n)\) will not drop below its previous values.

Upper Bound

An upper bound in mathematics refers to a value that is greater than or equal to every element in a given set. When we talk about \( \lceil f(n) \rceil\) being \(O(f(n))\), we are essentially claiming that \(\lceil f(n) \rceil\) does not grow faster than a certain multiple of \(f(n)\).
In our context, since \(f(n)\) is always positive and doesn't decrease, and given that we're looking at the upper ceiling of \(f(n)\), we can state that \(\lceil f(n) \rceil \leq 2f(n)\). Here, 2 acts as a constant factor, making it easy to express \( \lceil f(n) \rceil\) as \(O(f(n))\). This helps in understanding that as \(n \to \infty\), the growth rate of \(\lceil f(n) \rceil\) is bounded or capped relative to the growth of \(f(n)\), thus simplifying its analysis.