Prove that (log2 n)^3 is smaller than n when n goes very large.
without using limit

To prove that \((\log_2 n)^3\) is smaller than \(n\) when \(n\) is very large without using limits, we can compare the growth rates of both functions through direct analysis and properties of logarithms and polynomials.

### Step 1: Properties of Logarithms and Polynomials
We need to show that there exists a threshold \(n_0\) such that for all \(n > n_0\), \((\log_2 n)^3 < n\).

### Step 2: Analyzing the Functions
Let's denote \(f(n) = (\log_2 n)^3\) and \(g(n) = n\).

For large \(n\), the function \(g(n) = n\) grows exponentially faster than the function \(f(n) = (\log_2 n)^3\), which only grows polynomially with respect to \(\log_2 n\).

### Step 3: Inequality Setup
Consider the inequality:

\[
(\log_2 n)^3 < n
\]

Taking the base-2 logarithm of both sides, we get:

\[
\log_2((\log_2 n)^3) < \log_2(n)
\]

Using the power rule for logarithms, \(\log_2(a^b) = b \log_2(a)\), we can simplify the left-hand side:

\[
3 \log_2(\log_2 n) < \log_2 n
\]

### Step 4: Isolate the Logarithm Term
Now, we divide both sides by \(\log_2 n\):

\[
3 \frac{\log_2(\log_2 n)}{\log_2 n} < 1
\]

This inequality can be further simplified to:

\[
\frac{\log_2(\log_2 n)}{\log_2 n} < \frac{1}{3}
\]

### Step 5: Bounding the Ratio
We need to show that for sufficiently large \(n\), the term \(\frac{\log_2(\log_2 n)}{\log_2 n}\) becomes very small. Observe that as \(n\) grows, \(\log_2 n\) grows much faster than \(\log_2(\log_2 n)\).

### Step 6: Direct Comparison for Large \(n\)
To get a concrete understanding, consider the behavior for large \(n\):

- For \(n = 2^{10} = 1024\), \(\log_2 n = 10\), and \(\log_2(\log_2 n) = \log_2(10)\).
- For \(n = 2^{20}\), \(\log_2 n = 20\), and \(\log_2(\log_2 n) = \log_2(20)\).

As \(n\) increases, the ratio \(\frac{\log_2(\log_2 n)}{\log_2 n}\) decreases:

\[
\frac{\log_2(20)}{20} \approx \frac{4.32}{20} \approx 0.216
\]

For larger \(n\), this ratio continues to decrease.

### Step 7: Conclusion
Thus, for sufficiently large \(n\), \(\frac{\log_2(\log_2 n)}{\log_2 n} < \frac{1}{3}\), which implies:

\[
3 \frac{\log_2(\log_2 n)}{\log_2 n} < 1
\]

Therefore,

\[
(\log_2 n)^3 < n \quad \text{for sufficiently large} \; n
\]

This completes the proof that \((\log_2 n)^3\) is smaller than \(n\) for very large \(n\).

Would you like more details or have any questions? Here are five additional questions you might find interesting:

1. Can you explain the differences between polynomial and exponential growth in more detail?
2. How does the change of base formula for logarithms work in general?
3. What are other examples of functions with different growth rates?
4. How can we apply similar techniques to other logarithmic inequalities?
5. What are the implications of understanding logarithmic versus polynomial growth in computer science?

Tip: When dealing with inequalities involving logarithms and polynomials, understanding their fundamental growth behaviors can often simplify proofs and comparisons.

Learn how to prove that (log base 2 of n)^3 is smaller than n for very large n using properties of logarithms, polynomials, and inequalities. Understand the growth rates of functions without relying on limits. Suitable for advanced high school students and beyond.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation