Proving the Validity of n 100n ÷ log n on Using Asymptotic Notations

When dealing with the question of whether the mathematical expression n 100n ÷ log n on holds true, it is essential to use the formal definitions of asymptotic notations, specifically the Big O notation. Understanding the underlying concepts and applying the definitions accurately can help clarify the validity of such claims.

Formal Definition of Big O Notation

Big O notation () is a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity. We say that fn o(gn) if for every positive constant c, there exists a constant n0, such that for all n n0, fn c cdot gn .

Validating the Expression

To determine if n 100n ÷ log n on is true or false, we should substitute n 100n ÷ log n for fn and n for gn. Thus, we have to prove whether

fn o(gn) is true by showing that for every c 0, there exists an n0 such that for all n n0, n 100n ÷ log n c cdot n.

Step-by-Step Analysis

Let's consider the expression more closely:

1. Substitute n 100n ÷ log n for fn and n for gn in the definition above.

2. We need to show that n 100n ÷ log n c cdot n for all n n0.

3. Simplify the inequality:

n 100n ÷ log n c cdot n

100n ÷ log n (c - 1) cdot n

100 ÷ log n c - 1

log n 100 ÷ (c - 1)

From the previous steps, we see that for any positive constant c, we can choose a constant n0 such that log n 100 ÷ (c - 1) for all n n0. Since the logarithmic function grows very slowly compared to n, the inequality holds for large values of n.

Conclusion

Based on the above analysis, we can conclude that n 100n ÷ log n on is true. The logarithmic term 100n ÷ log n grows slower than n, eventually becoming negligible as n increases, thus satisfying the Big O condition.

Additional Insight

If we further refine the expression to n 100n ÷ log n O(n), we can show that there exist constants N and C such that for n ge; N, n 100n ÷ log n is less than or equal to C cdot n. If we let N ≥ 2, then as long as C ≥ 100 ÷ log 2, the inequality holds. This implies that n 100n ÷ log n O(n).

Thus, the statement n 100n ÷ log n on is indeed true, and it follows that n 100n ÷ log n O(n).

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