Big O Notation Analysis of Expression Pairs

This article analyzes the relationships between pairs of expressions using Big O notation. It determines if one expression is of the same order as, higher order than, or lower order than the other.

Big O Notation Big O notation is a mathematical notation used to describe the limiting behavior of a function when the argument tends towards a particular value or infinity. In computer science, it is commonly used to classify the efficiency of algorithms.

Analysis of Expression Pairs

(a) A = n^3 - 100n, B = n^2 + 50n: To determine the relationship between A and B, we can simplify the expressions. For A, we have A = n^3 - 100n, and for B, we have B = n^2 + 50n. As n approaches infinity, the dominant term in both A and B is the highest power of n. In this case, it is n^3 for A and n^2 for B. Therefore, we can say that A is of order n^3 and B is of order n^2. Since n^3 is higher than n^2, we can conclude that A is of higher order than B, and thus A = O(B).

(b) A = log2(n^2), B = log2.7(n^4): For A, we have A = log2(n^2), and for B, we have B = log2.7(n^4). Simplifying further, we have A = 2log2(n) and B = 4log2(n). As n approaches infinity, the logarithmic term dominates both A and B. Therefore, we can say that A is of order log(n) and B is of order log(n). Since log(n) and log(n) are of the same order, we can conclude that A is of the same order as B, and thus A = O(B).

(c) A = 1010000, B = 0: For A, we have A = 1010000, and for B, we have B = 0. As A is a constant and B is zero, we can say that A is of higher order than B, and thus A = O(B).

(d) A = 2nlogn, B = n^10 + 8n^2: For A, we have A = 2nlog(n), and for B, we have B = n^10 + 8n^2. Simplifying further, we have A = nlog(n) and B = n^10. As n approaches infinity, the logarithmic term in A dominates the polynomial term in B. Therefore, we can say that A is of higher order than B, and thus A = O(B).

(e) A = 2n, B = 2n + log(n): For A, we have A = 2n, and for B, we have B = 2n + log(n). As n approaches infinity, the logarithmic term in B becomes negligible compared to the linear term. Therefore, we can say that A is of higher order than B, and thus A = O(B).

(f) A = 33n,B = 32n: For A, we have A = 33n, and for B, we have B = 32n. As n approaches infinity, the linear term dominates both A and B. Therefore, we can say that A is of higher order than B, and thus A = Ω(B).

(g) A = (V2)logn, B = Vlog n: For A, we have A = (V2)log(n), and for B, we have B = Vlog(n). As n approaches infinity, the logarithmic term in A dominates the square root term in B. Therefore, we can say that A is of higher order than B, and thus A = Ω(B).