Determine the inequality relationship between a logarithmic expression involving the largest prime factor of n and log(log(n)) | Step-by-Step Solution

What We're Solving:

We need to determine whether the expression e^(γ) n log(p_k) + O(1) is less than or equal to e^(γ) n log(log(n)), where p_k appears to be the k-th prime (likely the largest prime factor of n), γ is the Euler-Mascheroni constant, and O(1) represents a bounded term.

The Approach:

Our strategy is to:

• 1. Understand what each component represents
• 2. Simplify the inequality by removing common factors
• 3. Analyze the relationship between log(p_k) and log(log(n))
• 4. Consider what happens as n grows large

The key insight is that we're really comparing how fast the logarithm of a prime grows versus log(log(n)).

Step-by-Step Solution:

Step 1: Simplify the inequality Since e^(γ) * n appears on both sides and is positive, we can divide both sides by it: log(p_k) + O(1)/(e^(γ) * n) ≤ log(log(n))

Step 2: Consider the asymptotic behavior As n → ∞, the term O(1)/(e^(γ) * n) → 0, so our inequality becomes: log(p_k) ≤ log(log(n)) (asymptotically)

Step 3: Interpret what p_k represents If p_k is the largest prime factor of n, then we need to think about how large the biggest prime factor of n can be. In the worst case, p_k could be as large as n itself (when n is prime).

Step 4: Analyze the key relationship

• If p_k ≈ n (worst case), then log(p_k) ≈ log(n)
• We're asking if log(n) ≤ log(log(n))
• But log(n) > log(log(n)) for all n > 1!

Step 5: Consider the context However, if this inequality is supposed to hold in some specific context (like average case or for most values of n), we need additional information about how p_k relates to n.

The Answer:

The inequality is generally FALSE if p_k can be the largest prime factor of n, because log(p_k) can be as large as log(n), and log(n) > log(log(n)) for n > 1.

However, if there are additional constraints on p_k (such as it being bounded by some function smaller than n), the answer might be different. The problem might be asking about a specific theorem or context where such bounds exist.

Memory Tip:

Remember that log(log(n)) grows much slower than log(n)! Think of it this way: if n = 10^6, then log(n) ≈ 14, but log(log(n)) ≈ 2.6. The "double log" function is one of the slowest-growing functions you'll encounter in mathematics - it's asking a lot for something to be bounded by it!

Great work tackling this challenging problem! Asymptotic analysis can be tricky, but you're developing the right intuition by thinking about how different functions grow relative to each other.